Samacheer Kalvi Class 12 Chemistry Solutions Chapter 4 Transition and Inner Transition Elements

Part - I - Text Book Evaluation

I. Choose the Correct Answer

 

Question 1. Sc (Z = 21) is a transition element but Zinc (Z = 30) is not because
(a) \( \mathrm{Zn}^{2+} \) ions are colourless and form white compounds
(b) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled
(c) last electron as assumed to be added to 4s level in case of zinc
(d) both Sc and Zn do not exhibit variable oxidation states
Answer: (b) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled
In simple words: Scandium has some empty space in its d-orbitals which allows it to be a transition element. Zinc has all its d-orbitals completely full, so it does not count as a transition element.

🎯 Exam Tip: Remember that for an element to be a transition metal, it must have a partially filled d-orbital in its elemental state or in any of its common oxidation states.

 

Question 2. Which of the following d block element has half-filled penultimate d sub shell as well as half-filled valence sub shell?
(a) Cr
(b) Pd
(c) Pt
(d) none of these
Answer: (a) Cr
In simple words: Chromium is special because it has electrons spread out to make its d-subshell half-full, and its outermost s-subshell is also half-full. This makes it very stable.

🎯 Exam Tip: Chromium and Copper show unusual electronic configurations to achieve extra stability through half-filled or fully-filled subshells.

 

Question 3. Among the transition metals of 3d series, the one that has highest negative \( \mathrm{E}^{\circ}_{\mathrm{M}^{2+}/\mathrm{M}} \) standard electrode potential is
(a) Ti
(b) Cu
(c) Mn
(d) Zn
Answer: (a) Ti
In simple words: Among the 3d transition metals, Titanium is the most likely to lose electrons and become a \( \mathrm{Ti}^{2+} \) ion, which means it has the most negative standard electrode potential. This means it is easily oxidized.

🎯 Exam Tip: A high negative standard electrode potential indicates a strong tendency for the metal to be oxidized, meaning it is a good reducing agent.

 

Question 4. Which one of the following ions has the same number of unpaired electrons as present in \( \mathrm{V}^{3+} \)?
(a) \( \mathrm{Ti}^{3+} \)
(b) \( \mathrm{Cr}^{2+} \)
(c) \( \mathrm{Ni}^{2+} \)
(d) \( \mathrm{Cr}^{3+} \)
Answer: (c) \( \mathrm{Ni}^{2+} \)
In simple words: \( \mathrm{V}^{3+} \) has 2 unpaired electrons. Out of the given options, \( \mathrm{Ni}^{2+} \) also has 2 unpaired electrons, meaning their magnetic properties would be similar.

🎯 Exam Tip: To determine the number of unpaired electrons, first write the electronic configuration of the neutral atom, then remove electrons from the outermost shell (usually s, then d) based on the ion's charge, and then count the unpaired d-electrons.

 

Question 5. The magnetic moment of \( \mathrm{Mn}^{2+} \) ion is
(a) 5.92BM
(b) 2.80BM
(c) 8.95BM
(d) 3.90BM
Answer: (a) 5.92BM
In simple words: The \( \mathrm{Mn}^{2+} \) ion has 5 unpaired electrons, which gives it a calculated magnetic moment of about 5.92 Bohr Magnetons.

🎯 Exam Tip: The spin-only magnetic moment can be calculated using the formula \( \mu = \sqrt{n(n+2)} \) BM, where n is the number of unpaired electrons. For \( \mathrm{Mn}^{2+} \), n = 5, so \( \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \).

 

Question 6. The catalytic behaviour of transition metals and their compounds is ascribed mainly due to
(a) their magnetic behaviour
(b) their unfilled d orbitals
(c) their ability to adopt variable oxidation states
(d) their chemical reactivity
Answer: (c) their ability to adopt variable oxidation states
In simple words: Transition metals are good catalysts because they can easily change their oxidation state. This helps them to form temporary bonds with reactants and speed up chemical reactions.

🎯 Exam Tip: Variable oxidation states allow transition metals to participate in redox reactions by readily accepting and donating electrons, which is key to catalysis.

 

Question 7. The correct order of increasing oxidizing power in the series
(a) \( \mathrm{VO}_{2}^{+} < \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} < \mathrm{MnO}_{4}^{-} \)
(b) \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} < \mathrm{VO}_{2}^{+} < \mathrm{MnO}_{4}^{-} \)
(c) \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} < \mathrm{MnO}_{4}^{-} < \mathrm{VO}_{2}^{+} \)
(d) \( \mathrm{MnO}_{4}^{-} < \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} < \mathrm{VO}_{2}^{+} \)
Answer: (a) \( \mathrm{VO}_{2}^{+} < \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} < \mathrm{MnO}_{4}^{-} \)
In simple words: The power of these ions to oxidize other substances increases from \( \mathrm{VO}_{2}^{+} \) to \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \) and then to \( \mathrm{MnO}_{4}^{-} \). This means \( \mathrm{MnO}_{4}^{-} \) is the strongest oxidizer among them.

🎯 Exam Tip: Oxidizing power generally increases with the increasing positive oxidation state of the central metal atom within similar compounds. Permanganate (Mn in +7 state) is a very strong oxidizing agent.

 

Question 8. In acid medium, potassium permanganate oxidizes oxalic acid to
(a) oxalate
(b) Carbon dioxide
(c) acetate
(d) acetic acid
Answer: (b) Carbon dioxide
In simple words: When potassium permanganate reacts with oxalic acid in an acidic solution, it breaks down the oxalic acid and changes it into carbon dioxide gas.

🎯 Exam Tip: Remember the common oxidation products of organic compounds. Oxalic acid is a dicarboxylic acid that gets fully oxidized to carbon dioxide.

 

Question 9. Which of the following statements is not true?
(a) on passing \( \mathrm{H}_{2}\mathrm{S} \), through acidified \( \mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} \) solution, a milky colour is observed
(b) \( \mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} \) is preferred over \( \mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} \) in volumetric analysis
(c) \( \mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} \) solution in acidic medium is orange in colour
(d) \( \mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} \) solution becomes yellow on increasing the pH beyond 7
Answer: (b) \( \mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} \) is preferred over \( \mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} \) in volumetric analysis
In simple words: The statement that sodium dichromate is better for volumetric analysis than potassium dichromate is false. Potassium dichromate is usually preferred because it is more pure and easier to weigh accurately.

🎯 Exam Tip: In volumetric analysis, a primary standard needs to be pure, stable, and have a high molar mass, which are characteristics of potassium dichromate rather than sodium dichromate.

 

Question 10. Permanganate ion changes to ______ in acidic medium
(a) \( \mathrm{MnO}_{4}^{2-} \)
(b) \( \mathrm{Mn}^{2+} \)
(c) \( \mathrm{Mn}^{3+} \)
(d) \( \mathrm{MnO}_{2} \)
Answer: (b) \( \mathrm{Mn}^{2+} \)
In simple words: When a permanganate ion is in an acidic environment, it changes its form and becomes a \( \mathrm{Mn}^{2+} \) ion. This change often shows as a colour fading from purple.

🎯 Exam Tip: Remember the common reduction products of permanganate ions: \( \mathrm{Mn}^{2+} \) in acidic medium, \( \mathrm{MnO}_{2} \) in neutral/weakly alkaline medium, and \( \mathrm{MnO}_{4}^{2-} \) in strongly alkaline medium.

 

Question 11. How many moles of \( \mathrm{I}_{2} \) are liberated when 1 mole of potassium dichromate react with potassium iodide?
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3
In simple words: For every 1 mole of potassium dichromate that reacts with potassium iodide, 3 moles of iodine gas are produced.

🎯 Exam Tip: To solve this, balance the redox reaction: \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 6\mathrm{I}^{-} + 14\mathrm{H}^{+} \implies 2\mathrm{Cr}^{3+} + 3\mathrm{I}_{2} + 7\mathrm{H}_{2}\mathrm{O} \). The stoichiometry directly shows that 1 mole of dichromate liberates 3 moles of iodine.

 

Question 12. The number of moles of acidified \( \mathrm{KMnO}_{4} \) required to oxidize 1 mole of ferrous oxalate \( (\mathrm{FeC}_{2}\mathrm{O}_{4}) \) is
(a) 5
(b) 3
(c) 0.6
(d) 1.5
Answer: (c) 0.6
In simple words: To fully change 1 mole of ferrous oxalate, you need 0.6 moles of acidified potassium permanganate. This specific amount is needed for the chemical reaction to happen correctly.

🎯 Exam Tip: In this reaction, both \( \mathrm{Fe}^{2+} \) (to \( \mathrm{Fe}^{3+} \)) and \( \mathrm{C}_{2}\mathrm{O}_{4}^{2-} \) (to \( \mathrm{CO}_{2} \)) are oxidized by \( \mathrm{KMnO}_{4} \). Calculate the total change in oxidation state for ferrous oxalate (3 for iron, 6 for oxalate, so total 9 electrons lost by one mole of \( \mathrm{FeC}_{2}\mathrm{O}_{4} \)), and 5 electrons gained by \( \mathrm{MnO}_{4}^{-} \) (Mn+7 to Mn+2). Then apply stoichiometry.

 

Question 13. Which one of the following statements related to lanthanons is incorrect?
(a) Europium shows +2 oxidation state
(b) The basicity decreases as the ionic radius decreases from Pr to Lu.
(c) All the lanthanons are much more reactive than aluminium.
(d) \( \mathrm{Ce}^{4+} \) solutions are widely used as oxidising agents in volumetric analysis.
Answer: (c) All the lanthanons are much more reactive than aluminium.
In simple words: It is not true that all lanthanons are much more reactive than aluminum. Their reactivity varies, and some are less reactive.

🎯 Exam Tip: Lanthanons generally exhibit similar chemical properties due to the shielding effect of 4f electrons, but their reactivity is not universally greater than a common metal like aluminum.

 

Question 14. Which of the following lanthanoid ions is diamagnetic?
(a) \( \mathrm{Eu}^{2+} \)
(b) \( \mathrm{Yb}^{2+} \)
(c) \( \mathrm{Ce}^{2+} \)
(d) \( \mathrm{Sm}^{2+} \)
Answer: (b) \( \mathrm{Yb}^{2+} \)
In simple words: The \( \mathrm{Yb}^{2+} \) ion is diamagnetic because all of its electrons are paired up. This means it is not attracted to a magnetic field.

🎯 Exam Tip: Diamagnetic substances have all electrons paired, while paramagnetic substances have unpaired electrons and are attracted to magnetic fields. To determine this, write out the electron configuration of each ion.

 

Question 15. Which of the following oxidation states is most common among the lanthanoids?
(a) 4
(b) 2
(c) 3
(d) 3
Answer: (d) 3
In simple words: The most common and stable oxidation state for lanthanoid elements is +3. They usually prefer to form ions with a +3 charge.

🎯 Exam Tip: The +3 oxidation state is characteristic of lanthanoids because the ionization energies to remove three electrons (two 6s and one 4f or 5d) are relatively low and lead to stable configurations.

 

Question. Assertion : \( \mathrm{Ce}^{4+} \) is used as an oxidizing agent in volumetric analysis Reason : \( \mathrm{Ce}^{4+} \) has the tendency of attaining +3 oxidation state.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false
(d) Both assertion and reason are false.
Answer: (a) Both assertion and reason are true and reason is the correct explanation of assertion.
In simple words: Both parts are correct: \( \mathrm{Ce}^{4+} \) helps oxidize other things, and this is because it likes to become \( \mathrm{Ce}^{3+} \) by taking an electron. The reason explains why it's a good oxidizing agent.

🎯 Exam Tip: For assertion-reason questions, first check if both statements are true individually. If they are, then check if the reason correctly explains the assertion.

 

Question 17. The most common oxidation state of actinoids is
(a) +2
(b) +3
(c) +4
(d) +6
Answer: (b) +3
In simple words: Like lanthanoids, the most common oxidation state for actinoid elements is +3. However, actinoids can show a wider range of oxidation states compared to lanthanoids.

🎯 Exam Tip: While +3 is common for actinoids, be aware that many actinoids also exhibit higher oxidation states (like +4, +5, +6, +7) due to the small energy difference between 5f, 6d, and 7s orbitals.

 

Question 18. The actinoid elements which show the highest oxidation state of +7 are
(a) Np, Pu, Am
(b) Li, Fm, Th
(c) U, Th, Md
Answer: (a) Np, Pu, Am
In simple words: Neptunium (Np), Plutonium (Pu), and Americium (Am) are the actinoid elements that can reach the highest oxidation state of +7.

🎯 Exam Tip: It's important to remember that while +3 is common, some actinoids can achieve very high oxidation states, which are distinct characteristics of their chemistry.

 

Question 19. Which one of the following is not correct?
(a) \( \mathrm{La(OH)}_3 \) is less basic than \( \mathrm{Lu(OH)}_3 \)
(b) In lanthanoid series ionic radius of \( \mathrm{Ln}^{3+} \) ions decreases
(c) La is actually an element of transition metal series rather than lanthanide series
(d) Atomic radii of \( \mathrm{Zr} \) and \( \mathrm{Hf} \) are same because of lanthanide contraction,
Answer: (a) \( \mathrm{La(OH)}_3 \) is less basic than \( \mathrm{Lu(OH)}_3 \)
In simple words: The statement that \( \mathrm{La(OH)}_3 \) is less basic than \( \mathrm{Lu(OH)}_3 \) is wrong. Actually, \( \mathrm{La(OH)}_3 \) is more basic because its ion is larger, which makes the bond to oxygen weaker.

🎯 Exam Tip: Basicity of lanthanide hydroxides decreases across the series due to lanthanide contraction, which increases the covalent character of the \( \mathrm{M-OH} \) bond. Therefore, \( \mathrm{La(OH)}_3 \) is the most basic, and \( \mathrm{Lu(OH)}_3 \) is the least basic.

II. Answer the following questions

 

Question 1. What are transition metals? Give four examples
Answer: A transition metal is an element whose atom has a partially filled d subshell in its elemental state or in any of its common oxidation states. These elements are located in the central part of the periodic table, found between the s-block and p-block elements. Their chemical properties bridge the highly reactive s-block metals and the non-metallic p-block elements.
Examples of transition metals include Iron (Fe), Copper (Cu), Tungsten (W), and Titanium (Ti). The d-orbitals in these metals allow them to participate in many chemical reactions.
In simple words: Transition metals are elements that have d-orbitals that are not completely full. They are found in the middle of the periodic table. Examples are Iron, Copper, Tungsten, and Titanium.

🎯 Exam Tip: When defining transition metals, always mention the partially filled d-orbital condition (in elemental or ionic state) and provide a few common examples.

 

Question 2. Explain the oxidation states of 4d series elements.
Answer: The 4d series elements show a range of oxidation states, typically varying from +3 for Yttrium (Y) up to +8 for Ruthenium (Ru). The highest oxidation states in the 4d series are observed when these elements form compounds with highly electronegative elements such as Oxygen (O), Fluorine (F), and Chlorine (Cl). For example, Ruthenium can exhibit a remarkable +8 oxidation state in compounds like \( \mathrm{RuO}_{4} \). This variability in oxidation states is a key characteristic of transition metals, allowing them to form diverse compounds.
In simple words: Elements in the 4d series can have many different oxidation states, from +3 to +8. They show their highest oxidation states when they combine with strong electron-attracting elements like oxygen or fluorine.

🎯 Exam Tip: Highlight that the variability in oxidation states is due to the involvement of both ns and (n-1)d electrons, and that higher oxidation states are usually stabilized by highly electronegative elements.

 

Question 3. What are inner transition elements?
Answer: Inner transition elements are a special group of elements that have partially filled f-orbitals in their elemental or ionic forms. They are generally categorized into two main series:

• **Lanthanoids:** This series includes 14 elements, starting from Cerium (\( _{58}\mathrm{Ce} \)) to Lutetium (\( _{71}\mathrm{Lu} \)). These elements follow Lanthanum (\( _{57}\mathrm{La} \)) and are defined by the filling of the 4f orbitals.
• **Actinoids:** This series also consists of 14 elements, ranging from Thorium (\( _{90}\mathrm{Th} \)) to Lawrencium (\( _{103}\mathrm{Lr} \)). They follow Actinium (\( _{89}\mathrm{Ac} \)) and are characterized by the filling of the 5f orbitals.

Since their f-orbitals are located "inner" to the penultimate shell, these elements are known as f-block elements. They exhibit unique chemical behaviors due to the nature of these inner f-electrons.
In simple words: Inner transition elements are called f-block elements. They have f-orbitals that are not completely full. They are divided into two groups: lanthanoids (4f series) and actinoids (5f series).

🎯 Exam Tip: Clearly distinguish between lanthanoids (4f) and actinoids (5f) and state that their characteristic feature is the partial filling of f-orbitals, which are located deep within the atom.

 

Question 4. Justify the position of lanthanides and actinides in the periodic table.
Answer: The position of lanthanides and actinides at the bottom of the periodic table is justified by their unique electronic configurations and chemical properties.
**Lanthanides:**

• Their actual position is in group 3 and period 6 of the periodic table.
• In this sixth period, electrons preferentially fill the inner 4f-subshell.
• The fourteen elements following lanthanum (from Cerium to Lutetium) exhibit very similar chemical properties due to the poor shielding effect of the 4f electrons. This similarity means they cannot be placed in 14 separate groups.
• Their position is justified by: (1) General electronic configuration: \( [\mathrm{Xe}] 4\mathrm{f}^{1-14} 5\mathrm{d}^{0,1} 6\mathrm{s}^2 \); (2) A common oxidation state of +3; (3) Similar physical and chemical properties.

**Actinides:**

• Their actual position is in group 3 and period 7 of the periodic table.
• In this seventh period, electrons preferentially fill the inner 5f-subshell.
• The fourteen elements following Actinium (from Thorium to Lawrencium) also show similar chemical properties, largely due to the shielding effect of the 5f electrons, leading to their placement together.
• Their position is justified by: (1) General electronic configuration: \( [\mathrm{Rn}] 5\mathrm{f}^{2-14} 6\mathrm{d}^{0-2} 7\mathrm{s}^2 \); (2) A common oxidation state of +3; (3) Similar physical and chemical properties.

Placing them separately at the bottom keeps the main body of the periodic table organized.
In simple words: Lanthanides and actinides are placed at the bottom because they fill inner f-orbitals. This makes them all very similar to each other, so they are grouped together instead of taking up many spots in the main table.

🎯 Exam Tip: Emphasize "lanthanide contraction" for lanthanides and similar "actinide contraction" for actinides as the key reason for their similar properties and thus their unique placement.

 

Question 5. What are actinides? Give three examples.
Answer: Actinides are a series of 14 elements found in the f-block of the periodic table, from Thorium (Th) to Lawrencium (Lr). These elements follow Actinium (Ac) and are characterized by the filling of the 5f orbital. All actinides are radioactive, and most of the heavier ones are extremely unstable with very short half-lives. They are often produced artificially in laboratories rather than occurring naturally. They are similar to lanthanides but generally show a wider range of oxidation states.
Examples of actinides include Thorium, Uranium, and Neptunium.
In simple words: Actinides are a group of 14 radioactive elements that fill the 5f-orbitals. Many are made in labs and do not last long. Thorium, Uranium, and Neptunium are examples.

🎯 Exam Tip: Remember that all actinides are radioactive. Give common examples and briefly mention their unique characteristics like radioactivity and artificial synthesis for heavier members.

 

Question 6. Describe the preparation of potassium dichromate.
Answer: The process of preparing potassium dichromate involves several steps, starting from chromite ore (\( \mathrm{FeCr}_{2}\mathrm{O}_{4} \)).
**Step 1: Oxidation of Chromite Ore:** Chromite ore is finely ground and then fused with sodium carbonate (\( \mathrm{Na}_{2}\mathrm{CO}_{3} \)) in the presence of air. This process oxidizes chromium from the +3 to the +6 oxidation state, forming yellow sodium chromate (\( \mathrm{Na}_{2}\mathrm{CrO}_{4} \)).
\( 4\mathrm{FeCr}_{2}\mathrm{O}_{4} + 8\mathrm{Na}_{2}\mathrm{CO}_{3} + 7\mathrm{O}_{2} \implies 8\mathrm{Na}_{2}\mathrm{CrO}_{4} + 2\mathrm{Fe}_{2}\mathrm{O}_{3} + 8\mathrm{CO}_{2} \)
**Step 2: Conversion to Sodium Dichromate:** The yellow sodium chromate solution is then acidified with sulfuric acid (\( \mathrm{H}_{2}\mathrm{SO}_{4} \)). This converts the chromate ion into the orange dichromate ion.
\( 2\mathrm{Na}_{2}\mathrm{CrO}_{4} + 2\mathrm{H}^{+} \implies \mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} + 2\mathrm{Na}^{+} + \mathrm{H}_{2}\mathrm{O} \)
**Step 3: Precipitation of Potassium Dichromate:** Finally, potassium chloride (\( \mathrm{KCl} \)) is added to the orange sodium dichromate solution. Since potassium dichromate is less soluble than sodium dichromate, it crystallizes out as orange crystals.
\( \mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} + 2\mathrm{KCl} \implies \mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} + 2\mathrm{NaCl} \)
In simple words: Potassium dichromate is made from chromite ore. First, the ore is heated with sodium carbonate and air to make yellow sodium chromate. Then, acid is added to change it into orange sodium dichromate. Finally, potassium chloride is added to get solid potassium dichromate.

🎯 Exam Tip: Remember the color changes (yellow to orange) and the key reagents (sodium carbonate, air, sulfuric acid, potassium chloride) for each step of the process.

 

Question 7. What is lanthanide contraction and what are the effects of lanthanide contraction?
Answer: **Lanthanide contraction** is the steady decrease in the atomic and ionic radii of the lanthanoid elements as you move across the 4f series (from Cerium to Lutetium). This happens because the 4f electrons provide very poor shielding from the increasing nuclear charge. As a result, the outer electrons are pulled closer to the nucleus, making the atoms and ions smaller. This effect is very significant for the elements that come after the lanthanides in the periodic table.
**Effects of Lanthanide Contraction:**

• **Basicity Differences:** The basic character of lanthanoid hydroxides \( (\mathrm{Ln(OH)}_3) \) decreases from \( \mathrm{Ce}^{3+} \) to \( \mathrm{Lu}^{3+} \). As the ionic size decreases, the \( \mathrm{Ln-OH} \) bond becomes more covalent and less ionic, reducing basicity.
• **Similar Atomic Radii:** Elements in the second (4d) and third (5d) transition series that are in the same group (e.g., Zirconium and Hafnium) often have very similar atomic radii. This is because the lanthanide contraction almost perfectly cancels out the expected increase in size from adding an extra shell.
• **Separation Difficulties:** The very similar ionic radii of lanthanides make it challenging to separate them from each other using chemical methods.

In simple words: Lanthanide contraction means lanthanide atoms get smaller and smaller as you go along the series. This is because their inner electrons do not block the nuclear pull well. This makes elements after them in the periodic table have similar sizes and makes it harder to separate lanthanides from each other.

🎯 Exam Tip: Clearly define lanthanide contraction as the decrease in atomic/ionic radii. For effects, focus on the consequences for chemical properties (basicity) and the similarity in sizes of 4d and 5d transition elements.

 

Question 8. Complete the following.
Answer:
(a) \( 3\mathrm{MnO}_{4}^{2-} + 4\mathrm{H}^{+} \implies 2\mathrm{MnO}_{4}^{-} + \mathrm{MnO}_{2} + 2\mathrm{H}_{2}\mathrm{O} \)
(b) \( \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CH}_{3} \xrightarrow{\text{acidified } \mathrm{KMnO}_{4}} \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COOH} \)
(c) \( 2\mathrm{MnO}_{4}^{-} + 10\mathrm{Fe}^{2+} + 16\mathrm{H}^{+} \implies 2\mathrm{Mn}^{2+} + 10\mathrm{Fe}^{3+} + 8\mathrm{H}_{2}\mathrm{O} \)
(d) \( 2\mathrm{KMnO}_{4} \xrightarrow{\text{Red hot}} \mathrm{K}_{2}\mathrm{MnO}_{4} + \mathrm{MnO}_{2} + \mathrm{O}_{2} \)
(e) \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 6\mathrm{I}^{-} + 14\mathrm{H}^{+} \implies 2\mathrm{Cr}^{3+} + 3\mathrm{I}_{2} + 7\mathrm{H}_{2}\mathrm{O} \)
(f) \( \mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} + 2\mathrm{KCl} \implies \mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} + 2\mathrm{NaCl} \)
In simple words: These are balanced chemical equations showing how different compounds react. They show what goes in and what comes out in specific chemical changes.

🎯 Exam Tip: For completing reactions, identify the reactants and conditions (like acidic medium or heating), predict the products based on known redox behaviors or precipitation rules, and then balance the equation for atoms and charge.

 

Question 9. What are interstitial compounds?
Answer: Interstitial compounds are formed when very small atoms, such as hydrogen (H), boron (B), carbon (C), or nitrogen (N), get trapped in the tiny empty spaces (interstitial holes) within the crystal lattice of a transition metal. These small atoms do not form chemical bonds in the usual way but rather fit into the existing metallic structure. This creates new compounds with distinct properties. They are often non-stoichiometric, meaning their elemental ratios are not fixed whole numbers, like Titanium carbide (\( \mathrm{TiC} \)) or Zirconium hydride (\( \mathrm{ZrH}_{1.92} \)).
In simple words: Interstitial compounds are made when small atoms like hydrogen or carbon get stuck inside the gaps of a metal's structure. These compounds are not perfectly structured and have special properties.

🎯 Exam Tip: Key features to remember are: formation with small non-metal atoms, trapping in metal lattice interstitial sites, and resulting non-stoichiometry. Give examples like metal carbides or hydrides.

 

Question 10. Calculate the number of unpaired electrons in \( \mathrm{Ti}^{3+} \), \( \mathrm{Mn}^{2+} \), and calculate the spin-only magnetic moment.
Answer:
**For \( \mathrm{Ti}^{3+} \):**
Electronic configuration of \( \mathrm{Ti}^{3+} \) is \( [\mathrm{Ar}]3\mathrm{d}^1 \).
It has only one unpaired electron, so \( \mathrm{n} = 1 \).
Spin-only magnetic moment \( \mu = \sqrt{\mathrm{n}(\mathrm{n}+2)} = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 \text{ BM} \).

**For \( \mathrm{Mn}^{2+} \):**
Electronic configuration of \( \mathrm{Mn}^{2+} \) is \( [\mathrm{Ar}]3\mathrm{d}^5 \).
It has five unpaired electrons, so \( \mathrm{n} = 5 \).
Spin-only magnetic moment \( \mu = \sqrt{\mathrm{n}(\mathrm{n}+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM} \).
In simple words: For \( \mathrm{Ti}^{3+} \), there's 1 unpaired electron, so its magnetic strength is about 1.732. For \( \mathrm{Mn}^{2+} \), there are 5 unpaired electrons, making its magnetic strength about 5.92.

🎯 Exam Tip: Always start by writing the electronic configuration of the neutral atom, then remove electrons from the outermost s-orbital first, then from the d-orbital, to get the ion's configuration. Then apply the spin-only magnetic moment formula.

 

Question 11. Write the electronic configuration of \( \mathrm{Ce}^{4+} \) and \( \mathrm{Co}^{2+} \).
Answer:
**Electronic configuration of \( \mathrm{Ce}^{4+} \):**
The atomic number of Cerium (\( \mathrm{Ce} \)) is 58. Its neutral configuration is \( [\mathrm{Xe}] 4\mathrm{f}^1 5\mathrm{d}^1 6\mathrm{s}^2 \). When it forms a \( \mathrm{Ce}^{4+} \) ion, it loses four electrons (two from 6s, one from 5d, and one from 4f).
Therefore, the electronic configuration of \( \mathrm{Ce}^{4+} \) is \( [\mathrm{Xe}] 4\mathrm{f}^0 5\mathrm{d}^0 6\mathrm{s}^0 \). This empty f-subshell contributes to its stability.

**Electronic configuration of \( \mathrm{Co}^{2+} \):**
The atomic number of Cobalt (\( \mathrm{Co} \)) is 27. Its neutral configuration is \( [\mathrm{Ar}] 3\mathrm{d}^7 4\mathrm{s}^2 \). When it forms a \( \mathrm{Co}^{2+} \) ion, it loses two electrons from the outermost 4s orbital.
Therefore, the electronic configuration of \( \mathrm{Co}^{2+} \) is \( [\mathrm{Ar}] 3\mathrm{d}^7 \).
In simple words: For \( \mathrm{Ce}^{4+} \), all outer electrons are lost, leaving it with an empty 4f shell. For \( \mathrm{Co}^{2+} \), it loses its two 4s electrons, leaving 7 electrons in its 3d shell.

🎯 Exam Tip: For transition metals and f-block elements, remember to remove electrons from the outermost s-orbital first when forming cations, then from the d- or f-orbitals as needed.

 

Question 12. Explain briefly +2 states become more and more stable in the first half of the first-row transition elements with increasing atomic number.
Answer: In the first half of the first-row transition elements, as the atomic number increases, the +2 oxidation state becomes more and more stable. This occurs because, after the removal of the two 4s valence electrons to form a \( \mathrm{M}^{2+} \) ion, the 3d orbitals gradually fill up with electrons. The increasing nuclear charge, combined with the way the 3d orbitals are filled, leads to a smaller size and a more effective nuclear charge on the d-electrons. This pulls the remaining d-electrons closer to the nucleus, making the \( \mathrm{M}^{2+} \) cation more stable. This trend is observed from Scandium (Sc) to Manganese (Mn), where \( \mathrm{Mn}^{2+} \) (with its stable half-filled \( 3\mathrm{d}^5 \) configuration) is particularly stable.
In simple words: In the first half of transition metals, the +2 charged state gets more stable as the atomic number grows. This happens because the 3d electrons are pulled closer to the center, making the ion more secure.

🎯 Exam Tip: Connect the increasing stability of the +2 oxidation state to the progressive filling of 3d orbitals and the effect of increasing nuclear charge. Highlight the specific stability of \( \mathrm{Mn}^{2+} \) due to its half-filled d-orbital.

 

Question 13. Which is more stable? \( \mathrm{Fe}^{3+} \) or \( \mathrm{Fe}^{2+} \) – explain.
Answer: \( \mathrm{Fe}^{3+} \) is more stable than \( \mathrm{Fe}^{2+} \). Here's why:

• **Electronic configuration of \( \mathrm{Fe}^{3+} \):** \( [\mathrm{Ar}] 3\mathrm{d}^5 \). This configuration has five unpaired electrons, which is a half-filled d-subshell. Half-filled subshells are known for their extra stability.
• **Electronic configuration of \( \mathrm{Fe}^{2+} \):** \( [\mathrm{Ar}] 3\mathrm{d}^6 \). This configuration has four unpaired electrons and a partially filled d-subshell, which is less stable compared to the half-filled or fully-filled configurations.

Therefore, \( \mathrm{Fe}^{3+} \) with its stable half-filled \( 3\mathrm{d}^5 \) configuration is more stable than \( \mathrm{Fe}^{2+} \). For a similar reason, \( \mathrm{Mn}^{2+} \) is very stable due to its \( 3\mathrm{d}^5 \) configuration.
In simple words: \( \mathrm{Fe}^{3+} \) is more stable than \( \mathrm{Fe}^{2+} \) because \( \mathrm{Fe}^{3+} \) has a half-filled d-orbital, which is a very stable arrangement of electrons.

🎯 Exam Tip: When comparing stability, always refer to the electronic configuration and identify if it achieves a half-filled or fully-filled d-orbital, as these are strong indicators of stability.

 

Question 14. Explain the variation in \( \mathrm{E}^{\circ}_{\mathrm{M}^{3+}/\mathrm{M}^{2+}} \) 3d series.
Answer: The standard reduction potential for the \( \mathrm{M}^{3+}/\mathrm{M}^{2+} \) half-cell indicates the relative stability between the +3 and +2 oxidation states of transition metals. Transition metals in their higher oxidation states often tend to be oxidizing agents. Here's a look at the variation in \( \mathrm{E}^{\circ}_{\mathrm{M}^{3+}/\mathrm{M}^{2+}} \) values for the 3d series:

The negative values for Titanium, Vanadium, and Chromium indicate that their higher oxidation states (\( \mathrm{Ti}^{3+} \), \( \mathrm{V}^{3+} \), \( \mathrm{Cr}^{3+} \)) are preferred, meaning they are relatively stable and less easily reduced. The high positive reduction potential of \( \mathrm{M}^{3+}/\mathrm{M}^{2+} \) for Manganese (\( \mathrm{Mn}^{3+}/\mathrm{Mn}^{2+} \)) indicates that \( \mathrm{Mn}^{2+} \) is much more stable than \( \mathrm{Mn}^{3+} \). This is due to the very stable half-filled \( 3\mathrm{d}^5 \) electronic configuration of \( \mathrm{Mn}^{2+} \). Similarly, a positive value for Cobalt shows that \( \mathrm{Co}^{2+} \) is more stable than \( \mathrm{Co}^{3+} \).
In simple words: The standard reduction potential helps us compare how stable the +3 and +2 forms of a metal are. Negative values mean the +3 state is stable, while large positive values, like for Manganese, mean the +2 state is much more stable than the +3 state.

🎯 Exam Tip: Explain that a negative \( \mathrm{E}^{\circ} \) value for \( \mathrm{M}^{3+}/\mathrm{M}^{2+} \) means \( \mathrm{M}^{3+} \) is stable and hard to reduce, while a positive value means \( \mathrm{M}^{3+} \) is easily reduced to \( \mathrm{M}^{2+} \), making \( \mathrm{M}^{2+} \) more stable. Link this to electronic configurations (e.g., \( \mathrm{Mn}^{2+} \) has \( 3\mathrm{d}^5 \)).

 

Question 15. Compare lanthanides and actinides.
Answer: Here is a comparison between lanthanides and actinides:

In simple words: Lanthanides fill 4f orbitals, have higher binding energy, and usually don't form complexes or oxocations. Actinides fill 5f orbitals, have lower binding energy, often form complexes and oxocations, and show a wider range of colours and oxidation states.

🎯 Exam Tip: Focus on the key distinguishing features: the f-orbital being filled (4f vs. 5f), binding energy, complex formation tendency, typical colouration, and the maximum oxidation states exhibited.

 

Question 16. Explain why Cr\(^{2+}\) is strongly reducing while Mn\(^{3+}\) is strongly oxidizing.
Answer: The standard reduction potential (E°) for Cr\(^{3+}\)/Cr\(^{2+}\) is -0.41 V, which is a negative value. This shows that Cr\(^{2+}\) can easily lose electrons and change into Cr\(^{3+}\), making it a strong reducing agent. On the other hand, the E° value for Mn\(^{3+}\)/Mn\(^{2+}\) is +1.57 V, which is a positive value. This means Mn\(^{3+}\) can easily gain electrons and change into Mn\(^{2+}\), making it a strong oxidizing agent. The ability to switch between oxidation states makes these elements behave in unique ways.
In simple words: Cr\(^{2+}\) likes to give away electrons, so it is a good reducer. Mn\(^{3+}\) likes to take electrons, so it is a good oxidizer. This is because of how easily they change their electron count.

🎯 Exam Tip: Remember that a negative E° value indicates a strong reducing agent, while a positive E° value indicates a strong oxidizing agent. Relate these values directly to the ease of electron loss or gain.

 

Question 17. Compare the ionization enthalpies of first series of the transition elements.
Answer: The ionization energy of transition elements is typically between the ionization energies of s-block and p-block elements. As we move from left to right across a transition metal series, the ionization enthalpy generally increases. This increase is mainly due to the increase in nuclear charge, which happens as d-electrons are filled into orbitals, pulling them closer to the nucleus. The following table shows the general trends in ionization enthalpy for the first series of transition elements:

In simple words: The energy needed to remove an electron from transition metals changes across the series. It generally goes up because the nucleus pulls electrons more strongly as you add more protons. Some parts show small changes or no change due to electron arrangements.

🎯 Exam Tip: When discussing ionization enthalpy trends, always mention the effect of increasing nuclear charge and any specific electronic configurations (like half-filled or fully-filled orbitals) that lead to deviations from the general trend.

 

Question 18. Actinoid contraction is greater from element to element than the lanthanoid contraction why?
Answer: Actinoid contraction is more pronounced, or greater, from one element to the next compared to lanthanoid contraction. This is because the 5f electrons in actinoids are not very good at shielding the outer electrons from the nuclear charge. The 4f electrons in lanthanoids offer slightly better shielding. Because of this poor shielding in actinoids, the effective nuclear charge felt by the outer electrons increases more significantly, leading to a greater reduction in atomic and ionic radii with each increasing atomic number. This effect is a key reason for the differences in properties between the two series.
In simple words: Actinoids shrink more than lanthanoids as atomic number increases. This happens because 5f electrons in actinoids are not good at blocking the nucleus's pull, so the outer electrons feel a stronger squeeze.

🎯 Exam Tip: Focus on the shielding effectiveness of f-electrons. Remember that 5f electrons have poorer shielding than 4f electrons, leading to a stronger contraction effect in the actinoid series.

 

Question 19. Out of Lu(OH)\(_{3}\) and La(OH)\(_{3}\) which is more basic and why?
Answer: La(OH)\(_{3}\) is more basic than Lu(OH)\(_{3}\). This is because the basic character of hydroxides in the lanthanoid series decreases as the ionic radius decreases from Lanthanum (La) to Lutetium (Lu). Due to lanthanoid contraction, the size of the M\(^{3+}\) ions decreases across the series. A larger ionic size for La\(^{3+}\) means the La-OH bond is more ionic and easier to break, releasing more OH\(^{-}\) ions, making it more basic. Conversely, the smaller Lu\(^{3+}\) ion leads to a more covalent Lu-OH bond, reducing its basicity.

In simple words: La(OH)\(_{3}\) is more basic. This is because lanthanum ions are bigger than lutetium ions. Bigger ions make the bond with hydroxide more like salt, so it breaks easily to release hydroxide, making it a stronger base.

🎯 Exam Tip: Remember that basicity of lanthanoid hydroxides is inversely related to the ionic radius. A larger cation means a more ionic bond with hydroxide and thus higher basicity.

 

Question 20. Why europium (II) is more stable than cerium (II)
Answer: Europium (II) is more stable than cerium (II) due to its electronic configuration. Europium (II) has a half-filled 4f\(^{7}\) orbital, which is a very stable configuration. Cerium (II), on the other hand, does not have such a stable configuration in its 4f shell. The special stability of half-filled (f\(^{7}\)) and completely filled (f\(^{14}\)) f-orbitals often explains why certain oxidation states are preferred. This is clearly seen in the comparison below:

In simple words: Europium (II) is more stable because it has a special electron setup (half-filled f-orbitals) which is very strong and preferred. Cerium (II) does not have this special setup, so it is less stable.

🎯 Exam Tip: When explaining stability differences, always refer back to the electron configuration, specifically mentioning the stability associated with half-filled (f\(^{7}\)) or completely filled (f\(^{14}\)) orbitals.

 

Question 21. Why do zirconium and Hafnium exhibit similar properties?
Answer: Zirconium (Zr) and Hafnium (Hf) exhibit very similar properties because they have almost identical atomic sizes. This is a direct consequence of the lanthanoid contraction. The elements in the 4f series, which come before Hf, experience a significant decrease in atomic radius due to the poor shielding of the 4f electrons. This contraction effectively cancels out the expected increase in size when moving down a group in the periodic table. As a result, Hf ends up being nearly the same size as Zr, leading to very similar chemical behaviors and properties.
In simple words: Zirconium and Hafnium are very similar because they are almost the same size. This size similarity happens because of a 'shrinkage' effect called lanthanoid contraction, which makes Hafnium much smaller than it would normally be.

🎯 Exam Tip: The key phrase here is "lanthanoid contraction." Always mention it when explaining the similar properties of elements in the same group but different periods, especially for those after the lanthanoids.

 

Question 22. Which is stronger reducing agent Cr\(^{2+}\) or Fe\(^{2+}\)?
Answer: Cr\(^{2+}\) is a stronger reducing agent than Fe\(^{2+}\). We can tell this by looking at their standard electrode potentials (E°) for the reduction reactions. The E° value for Cr\(^{3+}\)/Cr\(^{2+}\) is negative (-0.41 V), while the E° value for Fe\(^{3+}\)/Fe\(^{2+}\) is positive (+0.77 V). A more negative reduction potential means that Cr\(^{2+}\) has a greater tendency to get oxidized (lose electrons) to form Cr\(^{3+}\). Therefore, Cr\(^{2+}\) readily gives up electrons, acting as a strong reducing agent. Fe\(^{2+}\), with its positive E° value, is less likely to lose an electron and become Fe\(^{3+}\).
In simple words: Cr\(^{2+}\) is a better electron donor than Fe\(^{2+}\). Its negative reduction potential means it prefers to lose electrons, making it a stronger reducing agent.

🎯 Exam Tip: Remember: a more negative standard reduction potential indicates a stronger reducing agent, meaning it is more easily oxidized (loses electrons). A more positive value indicates a stronger oxidizing agent.

 

Question 23. The E°M\(^{2+}\)/M value for copper is positive. Suggest a possible reason for this.
Answer: The standard electrode potential, E°\(_{M^{2+}/M}\), for copper is positive. This is unusual because for most transition metals, it is negative. The value of E° for any metal depends on the total energy changes involved in three main steps:

\( \implies \) First, atomisation: M\(_{(s)}\) + \( \Delta H_{(a)} \rightarrow M_{(g)} \) (This is the enthalpy of atomisation, which requires energy).
\( \implies \) Second, ionization: M\(_{(g)} + \Delta H_{(i)} \rightarrow M^{2+}_{(g)} \) (This is the ionization enthalpy, which also requires energy to remove electrons).
\( \implies \) Third, hydration: M\(^{2+}_{(g)} + \Delta H_{(Hyd)} \rightarrow M^{2+}_{(aq)} \) (This is the hydration enthalpy, which releases energy when ions are surrounded by water molecules).
For copper, the sum of its high enthalpy of atomisation (energy needed to turn solid copper into gas) and high ionization enthalpy (energy needed to remove electrons) is not fully compensated by its hydration enthalpy. This means that more energy is needed overall to form the aqueous ions than is released by their hydration, resulting in a positive E° value. This makes copper less likely to release electrons compared to other metals.
In simple words: Copper needs a lot of energy to turn into a gas and then to lose electrons. The energy it gets back when surrounded by water is not enough to cover these high energy costs. So, it has a positive E° value.

🎯 Exam Tip: When discussing electrode potentials, always consider the balance between atomisation enthalpy, ionization enthalpy, and hydration enthalpy. A positive E° for copper signifies that the energy input for atomisation and ionization outweighs the energy released during hydration.

 

Question 24. Describe the variable oxidation on state of 3d series elements.
Answer: The 3d series elements, also known as first-row transition elements, generally show variable oxidation states. This ability comes from the fact that the energy difference between their (n-1)d orbitals and ns orbitals is very small. Because of this small energy gap, electrons from both these orbitals can participate in chemical bonding. For example, the first element, Scandium (Sc), typically only shows a +3 oxidation state. However, other middle elements in the series, like Manganese (Mn), can show many different oxidation states, ranging from +2 to +7. The last element, Copper (Cu), usually shows +1 and +2. This range of oxidation states decreases towards the end of the series. The highest oxidation states are often seen in compounds with highly electronegative elements like oxygen and fluorine. The stability of these oxidation states is influenced by factors such as half-filled and fully-filled electronic configurations.
In simple words: Transition elements can have many different oxidation states because the energy of their d and s electrons is very similar, so both types of electrons can be used in reactions. Scandium shows only +3, while manganese can show many different ones.

🎯 Exam Tip: For variable oxidation states, highlight the small energy difference between (n-1)d and ns orbitals. Mention specific examples like Sc (+3 only) and Mn (multiple states) to illustrate the range.

 

Question 25. Which metal in the 3d series exhibits +1 oxidation state most frequently and why?
Answer: Copper (Cu) is the metal in the 3d series that most frequently exhibits the +1 oxidation state. This is because copper has an electronic configuration of \( 3d^{10}4s^{1} \). It can easily lose its single \( 4s^{1} \) electron to achieve a very stable, completely filled \( 3d^{10} \) electronic configuration. While copper can also show a +2 oxidation state, its ability to easily form the stable \( 3d^{10} \) configuration makes the +1 oxidation state quite common and stable for this element. This preference for a fully filled d-subshell drives its chemical behavior.
In simple words: Copper often shows a +1 oxidation state. It does this because it can easily lose one electron from its outer shell to get a very stable electron arrangement, which is when its d-shell is completely full.

🎯 Exam Tip: When explaining preferred oxidation states, always link it back to achieving stable electronic configurations, especially half-filled or fully-filled d-orbitals.

 

Question 26. Why first ionization enthalpy of chromium is lower than that of Zinc.
Answer: The first ionization enthalpy of chromium is lower than that of zinc because of their electronic configurations and the stability associated with them. Chromium has an electronic configuration of \( 3d^{5}4s^{1} \). It readily loses one electron from its 4s orbital to achieve a stable, half-filled \( 3d^{5} \) configuration. This makes it easier to remove the first electron. Zinc, on the other hand, has an electronic configuration of \( 3d^{10}4s^{2} \). It has a completely filled 3d orbital and a fully filled 4s orbital. Removing an electron from such a stable, completely filled configuration requires significantly more energy, hence its higher first ionization enthalpy. This shows how electron arrangement impacts chemical properties.

In simple words: Chromium's first electron is easier to remove because losing it makes the atom more stable (half-filled d-shell). Zinc's electrons are harder to remove because its outer shells are already very stable (completely full).

🎯 Exam Tip: Always compare electronic configurations and stability (half-filled or fully-filled orbitals) when explaining differences in ionization enthalpies. These configurations often explain exceptions to general trends.

 

Question 27. Transition metals show high melting points, why?
Answer: Transition metals generally have high melting points because of the strong metallic bonding present within their structures. This strong bonding is due to the presence of a large number of unpaired d-electrons. These unpaired d-electrons are available for delocalization and participate extensively in metallic bonding, forming very strong cohesive forces between the metal atoms. The more unpaired electrons available, the stronger the metallic bond, and thus the more energy (higher temperature) is required to break these bonds and melt the metal. This makes them very solid and hard materials.
In simple words: Transition metals melt at very high temperatures. This is because they have many d-electrons that are not paired up, which helps create very strong bonds between the metal atoms, making them hard to break apart.

🎯 Exam Tip: The key reason for high melting points in transition metals is the presence of unpaired d-electrons leading to strong metallic bonding. Mentioning "delocalized electrons" is also helpful.

 

III. Evaluate Yourself

 

Question 1. Compare the stability of Ni\(^{4+}\) and Pt\(^{4+}\) from their ions enthalpy value.
Answer: The stability of Ni\(^{4+}\) and Pt\(^{4+}\) can be compared by looking at their ionization enthalpy values, specifically the sum of the third and fourth ionization enthalpies (\( \text{IE}_{3} + \text{IE}_{4} \)). A lower sum indicates that less energy is required to form the M\(^{4+}\) ion, meaning it is more stable. From the given data:

• For Ni\(^{4+}\): \( \text{IE}_{3} + \text{IE}_{4} = 3395 + 5297 = 8692 \, \text{KJmol}^{-1} \)
• For Pt\(^{4+}\): \( \text{IE}_{3} + \text{IE}_{4} = 2800 + 4150 = 6950 \, \text{KJmol}^{-1} \)

Since Pt\(^{4+}\) requires less total energy to form (6950 KJmol\(^{-1}\)) compared to Ni\(^{4+}\) (8692 KJmol\(^{-1}\)), Pt\(^{4+}\) is more stable than Ni\(^{4+}\). This is a general trend in transition metals, where higher oxidation states become more stable for heavier elements in the same group, partly due to relativistic effects on electron orbitals. The following table illustrates the individual ionization energies:

In simple words: To make a \(^{4+}\) ion, it takes less energy for platinum than for nickel. This means platinum's \(^{4+}\) ion is more stable. Heavier elements in the same group often have more stable higher oxidation states.

🎯 Exam Tip: For stability comparisons based on ionization enthalpies, calculate the total energy required to reach the specific oxidation state. The ion requiring less energy for its formation is considered more stable.

 

Question 2. Why ion is more stable in +3 oxidation state than in +2 and the reverse is true for manganese?
Answer: Let's look at Iron (Fe) and Manganese (Mn):
**For Iron (Fe):**
(i) Atomic number of Fe = 26.
(ii) Electronic configuration of Fe = \( \text{[Ar]} \, 3d^{6}4s^{2} \).
(iii) After removing two electrons, \( \text{Fe}^{2+} \) has the configuration \( \text{[Ar]} \, 3d^{6} \).
(iv) Removing one more electron gives \( \text{Fe}^{3+} \) with a \( 3d^{5} \) configuration, which is a very stable half-filled d-subshell. Therefore, \( \text{Fe}^{3+} \) is more stable than \( \text{Fe}^{2+} \).
**For Manganese (Mn):**
(v) Atomic number of Mn = 25.
(vi) Electronic configuration of Mn = \( \text{[Ar]} \, 3d^{5}4s^{2} \).
(vii) After removing two outer electrons, \( \text{Mn}^{2+} \) has the configuration \( \text{[Ar]} \, 3d^{5} \), which is a very stable half-filled d-subshell.
(viii) If one more electron were removed, \( \text{Mn}^{3+} \) would have a \( 3d^{4} \) configuration, which is incompletely filled and less stable than \( \text{Mn}^{2+} \)'s \( 3d^{5} \).
Therefore, for iron, the +3 state is more stable, while for manganese, the +2 state is more stable. This highlights the importance of electron configuration in determining ion stability.
In simple words: Iron likes to be +3 because it gets a stable half-filled electron shell then. Manganese likes to be +2 because it already has a stable half-filled electron shell in that state, and losing another electron makes it less stable.

🎯 Exam Tip: Always relate the stability of oxidation states to the electronic configuration, particularly the attainment of half-filled (d\(^{5}\)) or fully-filled (d\(^{10}\)) subshells, as these configurations are highly stable.

 

12th Chemistry Guide Transition and Inner Transition Elements Additional Questions and Answers

 

Part - II - Additional Questions

 

I. Choose the correct answer

 

Question 1. Transition elements are good conductors because
(a) They are metals
(b) They are all solids
(c) They have free electrons in outer energy orbits
(d) All of the options
Answer: (c) They have free electrons in outer energy orbits
In simple words: Transition elements conduct electricity well because their outer electrons are free to move. These moving electrons carry the electric current through the material.

🎯 Exam Tip: For conductivity, the key concept is always the presence of "free" or "delocalized" electrons. While they are metals and solids, the underlying reason for conductivity is the mobility of electrons.

 

Question 2. Transition elements are
(a) All metals
(b) All non metals
(c) Metals and non metals
(d) Gases
Answer: (a) All metals
In simple words: All transition elements are considered metals. They have properties like being shiny, conducting heat and electricity, and being able to be shaped.

🎯 Exam Tip: Remember that "transition elements" is a sub-category within metals. All elements classified as transition elements will exhibit metallic properties.

 

Question 3. Transition elements form complexes very readily because of
(a) Small cation size
(b) Vacant - d - orbitals
(c) Large ionic charge
(d) All of the options
Answer: (d) All of the options
In simple words: Transition elements easily form complex compounds because their ions are small, they have empty d-orbitals to accept electrons, and they carry a significant positive charge. All these factors together help them bond with other molecules.

🎯 Exam Tip: When explaining complex formation, remember the three main reasons: small ionic size, high ionic charge, and the availability of vacant d-orbitals to accept electron pairs from ligands.

 

Question 4. The transition metal present in vitamin B12 is
(a) Fe
(b) Co
(c) Ni
(d) Na
Answer: (b) Co
In simple words: Vitamin B12, which is important for our body's nerve and blood cells, contains the transition metal Cobalt (Co).

🎯 Exam Tip: For biologically important molecules, it's crucial to memorize the key metal ions involved. Cobalt is unique to Vitamin B12 (cobalamin).

 

Question 5. Transition elements are frequently used as catalysts, because of
(a) Large ionic charge
(b) Large surface area for the reactions to be absorbed
(c) Unpaired d electrons
(d) Both (b) and (c) are correct
Answer: (d) Both (b) and (c) are correct
In simple words: Transition elements are good catalysts because they have many d-electrons that are not paired up, allowing them to easily form temporary bonds. They also provide a large surface area where other molecules can stick and react.

🎯 Exam Tip: The catalytic activity of transition metals is attributed to two main factors: their ability to provide a large surface area for adsorption of reactants and the presence of vacant d-orbitals/unpaired electrons for intermediate complex formation.

 

Question 6. d – Block elements are arranged in
(a) Three series
(b) Six series
(c) Two series
(d) Four series
Answer: (d) Four series
In simple words: The d-block elements are organized into four main rows or series on the periodic table: 3d, 4d, 5d, and 6d. Each series corresponds to the filling of a different d-subshell.

🎯 Exam Tip: Remember the four distinct d-block series (3d, 4d, 5d, 6d), corresponding to the period number minus one (e.g., 3d for Period 4).

 

Question 7. d – Block element generally for
(a) Covalent hydrides
(c) Interstitial hydrides
(d) Salt like hydrides
Answer: (c) Interstitial hydrides
In simple words: D-block elements are known for forming interstitial hydrides. These are compounds where small hydrogen atoms get stuck in the tiny gaps (interstices) within the metal's structure.

🎯 Exam Tip: Recognize interstitial hydrides as a characteristic compound type for d-block elements, formed when small hydrogen atoms occupy voids in the metal lattice.

 

Question 8. Metals which are hard and lustrous substances with high melting points form highly coloured compounds are known as
(a) Alkali metals
(b) Alkaline earth metals
(c) Transition metals
(d) None of the options
Answer: (c) Transition metals
In simple words: Transition metals are the elements known for being hard, shiny, having high melting points, and often forming compounds that are brightly colored.

🎯 Exam Tip: These descriptive properties (hardness, luster, high melting point, colored compounds) are classic indicators of transition metals, distinguishing them from alkali and alkaline earth metals.

 

Question 9. In the first transition series, the incoming electrons enters
(a) 5 d orbitals
(b) 4 d orbitals
(c) 3 d orbitals
(d) 2 d orbitals
Answer: (c) 3 d orbitals
In simple words: For the first row of transition elements, new electrons fill up the 3d orbitals. This is why it's called the 3d series.

🎯 Exam Tip: The "first transition series" is synonymous with the "3d series" because the differentiating electrons enter the 3d orbitals.

 

Question 10. Transition elements form alloys easily because they have
(a) Same atomic number
(b) Same electronic configuration
(c) Nearly same atomic size
(d) None of the options
Answer: (c) Nearly same atomic size
In simple words: Transition elements easily mix to form alloys because their atoms are very similar in size. This allows one metal's atoms to easily replace another's in a crystal structure.

🎯 Exam Tip: The primary reason for easy alloy formation among transition metals is their comparable atomic sizes, which allows for effective substitution in crystal lattices without much distortion.

 

Question 11. Transition elements that show anomalous electronic configuration in first series are.
(a) Cr and Ni
(b) Cu and Co
(c) Fe and Ni
(d) Cr and Cu
Answer: (d) Cr and Cu
In simple words: In the first series of transition elements, Chromium and Copper have unusual electron arrangements. Chromium gets a half-filled d-shell, and Copper gets a full d-shell, making them more stable.

🎯 Exam Tip: Memorize the two main exceptions in the 3d series: Chromium ([Ar]\(3d^5 4s^1\)) and Copper ([Ar]\(3d^{10} 4s^1\)). These configurations arise to achieve extra stability from half-filled or fully-filled d-orbitals.

 

Question 12. Lightest transition element is
(a) Fe
(b) Sc
(c) Os
(d) None of the options
Answer: (b) Sc
In simple words: Scandium (Sc) is the lightest element among the transition metals. It has the smallest atomic mass in the 3d series, which is the first transition series.

🎯 Exam Tip: Identify Scandium (Sc) as the first element in the 3d transition series, making it the lightest transition metal. Osmium (Os) is one of the densest, often confused for its extremeness.

 

Question 13. Densest transition element is
(a) Fe
(b) Sc
(c) Os
(d) Mn
Answer: (c) Os
In simple words: Osmium (Os) is the densest among all known elements, including transition metals. This means a small amount of it would weigh a lot.

🎯 Exam Tip: Osmium (Os) and Iridium (Ir) are known for being the two densest elements, with Osmium typically cited as the highest. This is a common factual recall question.

 

Question 14. Variable valencies of transition elements is due to
(a) Different energies of (n - 1) d electrons
(b) Different energies of ns electrons
(c) Similar energies of (n - 1)d electrons
(d) Similar energies of (n - 1)d and ns electrons
Answer: (d) Similar energies of (n - 1)d and ns electrons
In simple words: Transition elements show different valencies because the d and s electrons in their outer shells have very similar energy levels. This means both types of electrons can be involved in chemical bonds, leading to multiple oxidation states.

🎯 Exam Tip: The main reason for variable valencies in transition elements is the small energy difference between the (n-1)d and ns orbitals, allowing both to participate in bonding.

 

Question 15. Which of the following ions has minimum ionic radius
(a) Ni\(^{2+}\)
(b) Co\(^{2+}\)
(c) Cr\(^{2+}\)
(d) V\(^{2+}\)
Answer: (a) Ni\(^{2+}\)
In simple words: Among the given ions, Ni\(^{2+}\) has the smallest size. As we move across the 3d series from left to right, the ionic radii generally decrease because the nuclear charge increases, pulling the electrons closer.

🎯 Exam Tip: Remember the general trend for ionic radii in a transition series: radii decrease across the period due to increasing effective nuclear charge, despite the addition of d-electrons, which offer poor shielding.

 

Question 16. Ionic radii of zirconium and hofminium become almost identical because
(a) They are d block elements
(b) They belong to the same group
(c) Of increased nuclear charge
(d) Of lanthanoids contraction
Answer: (d) Of lanthanoids contraction
In simple words: Zirconium and Hafnium have almost the same size because of something called lanthanoid contraction. This shrinkage effect makes Hafnium much smaller than it would otherwise be, matching Zirconium's size.

🎯 Exam Tip: Lanthanoid contraction is the specific reason for the nearly identical atomic and ionic radii of elements in the second and third transition series within the same group (e.g., Zr and Hf, Nb and Ta).

 

Question 17. The colour of Fe ions is
(a) Blue
(b) Light green
(c) Very dark green
(d) Yellow
Answer: (b) Light green
In simple words: Iron ions, especially Fe\(^{2+}\) ions, are typically light green in color when in solution. This color comes from the way their d-electrons interact with light.

🎯 Exam Tip: Many transition metal ions are colored due to d-d transitions. It's useful to remember common colors for frequently encountered ions, like Fe\(^{2+}\) (light green) and Fe\(^{3+}\) (yellow/brown).

 

Question 18. Magnetic property of transition metal is due to
(a) Spin of electron
(c) Both
Answer: (a) Spin of electron
In simple words: The magnetic behavior of transition metals mainly comes from the spin of their electrons. Each electron acts like a tiny magnet, and if there are unpaired electrons, the substance shows magnetic properties.

🎯 Exam Tip: Magnetic properties (paramagnetism, ferromagnetism) in transition metals are predominantly due to unpaired electrons and their spin motion. Orbital motion also contributes but is often quenched in complexes.

 

Question 19. All ferromagnetic substances
(a) Can be magnetised
(b) Can be electrolysed
(c) Have completely filled d - orbitals
(d) None of the options
Answer: (a) Can be magnetised
In simple words: All materials that are ferromagnetic can be made into magnets. This means they can strongly attract other magnets or magnetic materials.

🎯 Exam Tip: Ferromagnetic substances are those that can be strongly magnetized, and this property is not related to electrolysis or having completely filled d-orbitals (in fact, it's often due to unpaired electrons).

 

Question 20. Maxmium paramagnetism in 3d series shown by
(a) Mn
(b) Co
(c) Ni
(d) Fe
Answer: (a) Mn
In simple words: Manganese shows the strongest paramagnetism in the 3d series. This is because manganese atoms have the highest number of unpaired electrons (five) in their d-orbitals, which makes them strongly attracted to magnetic fields.

🎯 Exam Tip: Maximum paramagnetism is typically seen in elements with the highest number of unpaired electrons. In the 3d series, this is Manganese (Mn) with five unpaired electrons in its d-orbitals (e.g., in Mn\(^{2+}\), it is \(3d^5\)).

 

Question 21. Which of the following has the minimum magnetic moment
(a) Mn\(^{2+}\)
(b) Fe\(^{2+}\)
(c) Cr\(^{2+}\)
(d) V\(^{3+}\)
Answer: (d) V\(^{3+}\)
In simple words: Vanadium in its +3 state (V\(^{3+}\)) has the smallest magnetic moment among these options. This is because it has the fewest unpaired electrons, which are responsible for magnetic properties.

🎯 Exam Tip: To determine the minimum magnetic moment, find the ion with the fewest unpaired electrons. Calculate the number of unpaired electrons for each option (Mn\(^{2+}\) \(3d^5\), Fe\(^{2+}\) \(3d^6\), Cr\(^{2+}\) \(3d^4\), V\(^{3+}\) \(3d^2\)) and then identify the lowest count.

 

Question 22. Paramagnetism in the substance increases as
(b) The number of unpaired electrons increases
Answer: (b) The number of unpaired electrons increases
In simple words: A substance becomes more paramagnetic when it has more electrons that are not paired up. Each unpaired electron acts like a tiny magnet, so more of them means a stronger overall magnetic attraction.

🎯 Exam Tip: Paramagnetism is directly proportional to the number of unpaired electrons. More unpaired electrons lead to a stronger magnetic moment and thus greater paramagnetism.

 

Question 23. The number of unpaired electrons present in Cr\(^{3+}\) ion is
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3
In simple words: A Chromium ion in its +3 state (Cr\(^{3+}\)) has three electrons that are not paired up. Its electron configuration is \(3d^3\).

🎯 Exam Tip: To find the number of unpaired electrons, first write the electron configuration of the neutral atom, then remove electrons based on the charge, starting from the outermost s-orbital, then d-orbitals. For Cr\(^{3+}\), the configuration is \(3d^3\), meaning three unpaired electrons.

 

Question 24. Which of the following has more unpaired d electrons
(a) Zn\(^{2+}\)
(b) Fe\(^{2+}\)
(c) N\(^{3-}\)
Answer: (b) Fe\(^{2+}\)
In simple words: Among the choices, the iron ion in its +2 state (Fe\(^{2+}\)) has the most unpaired d-electrons. Its electron configuration (\(3d^6\)) means it has four unpaired electrons.

🎯 Exam Tip: To identify the ion with the most unpaired d-electrons, write the electron configuration for each ion and count the unpaired electrons in the d-subshell. Zn\(^{2+}\) (\(3d^{10}\)) has zero, Fe\(^{2+}\) (\(3d^6\)) has four, and N\(^{3-}\) has no d-electrons (it's a p-block anion).

 

Question 25. The lanthanoids contraction is responsible for the fact that
(a) Zr and Y have the same radius
(b) Zr and Nb have similar oxidation state
(c) Zr and Hf have almost the same radius
(d) Zr and Zn have same oxidation state
Answer: (c) Zr and Hf have almost the same radius
In simple words: The lanthanoid contraction causes Zirconium (Zr) and Hafnium (Hf) to have almost the same atomic size. This unusual similarity in size is a direct effect of the poor shielding by 4f electrons in the elements preceding Hafnium.

🎯 Exam Tip: Lanthanoid contraction specifically explains the nearly identical sizes of elements in the same group but different periods, particularly for the second and third transition series (e.g., Zr/Hf, Nb/Ta), impacting their chemical properties.

 

Question 26. Which of the following is coloured
(a) Cu+
(b) Cu2+
(c) Ti4+
(d) V5+
Answer: (a) Cu+
In simple words: Many transition metal ions are coloured because their electrons can absorb light and jump to higher energy levels. But ions like Cu+, with completely full electron shells, usually do not show color.

🎯 Exam Tip: Remember that color in transition metal ions is generally due to unpaired d-electrons and d-d transitions.

 

Question 27. In the four successive transition elements (Cr, Mn, Fe & Co) the stability of +2 oxidation state will be are in which of the following order? (PTA - 5)
(a) Fe>Mn>Co>Cr
(b) Co>Mn>Fe>Cr
(c) Cr>Co>Mn>Fe
(d) Mn>Fe>Cr>Co
Answer: (d) Mn>Fe>Cr>Co
In simple words: The stability of the +2 state for these metals depends on how stable their electron configuration is after losing two electrons. Manganese is the most stable, followed by iron, chromium, and then cobalt. This is due to how their d-orbitals get filled.

🎯 Exam Tip: Stability of oxidation states often relates to half-filled or completely filled d-orbitals. Consider the electronic configuration for each element in its +2 state.

 

Question 28. Chromyl chloride when dissolved in NaOH solution gives yellow solution. The yellow solution contains (PTA - 3)
(a) Cr2O72-
(b) CrO42-
(c) CrO5
(d) Cr2O3
Answer: (b) CrO42-
In simple words: When chromyl chloride mixes with a sodium hydroxide solution, it forms a yellow liquid. This yellow color comes from the chromate ion, which is \( \text{CrO}_4^{2-} \). Chromate ions are known for their bright yellow color.

🎯 Exam Tip: The chromyl chloride test is a classic qualitative test for chloride ions, which produces red vapours of chromyl chloride \( \text{(CrO}_2\text{Cl}_2) \) that turn yellow in \( \text{NaOH} \) due to chromate formation.

 

Question 29. In the dichromate anion \( \text{(Cr}_2\text{O}_7)^{2-} \) (PTA - 1)
(a) 4Cr - O bonds are equivalent
(b) 6 Cr - O bonds are equivalent
(c) All Cr-O bonds are equivalent
(d) All Cr-O bonds are non-equivalent
Answer: (b) 6 Cr - O bonds are equivalent
In simple words: In the dichromate ion, there are six bonds between chromium and oxygen atoms that are all the same in length and strength. The other two Cr-O bonds (bridging ones) are different. The structure is two chromium tetrahedra linked by a shared oxygen.

🎯 Exam Tip: Visualize the structure of dichromate ion. It has two \( \text{CrO}_4 \) tetrahedra sharing one oxygen atom, leading to two distinct types of Cr-O bonds.

 

Question 30. Which of the following forms colourless compound?
(a) Sc3+
(b) V3+
(c) Ti3+
(d) Cr3+
Answer: (a) Sc3+
In simple words: Scandium ion \( \text{Sc}^{3+} \) is colourless because it has no electrons in its d-orbitals \( \text{(3d}^0) \). Without d-electrons, it cannot absorb light through d-d transitions to show color. Other options have unpaired d-electrons which lead to color.

🎯 Exam Tip: Remember that transition metal ions are typically colourless if they have either a completely empty \( \text{(d}^0) \) or a completely filled \( \text{(d}^{10}) \) d-subshell.

 

Question 31. The basic character of the transition metal monoxides follows the order (Atomic number, Ti = 22; V = 23; Cr = 24)
(a) VO > CrO > TiO > FeO
(b) Cro > VO > FeO > TiO
(c) TiO > FeO > VO > CrO
(d) TiO > VO > CrO > FeO
Answer: (d) TiO > VO > CrO > FeO
In simple words: As you move across the transition series, the basic nature of metal oxides generally decreases. So, titanium oxide is the most basic, followed by vanadium oxide, chromium oxide, and then iron oxide. This trend is linked to increasing nuclear charge and decreasing metallic character.

🎯 Exam Tip: The basic character of oxides of transition metals generally decreases as the atomic number increases in a period, and it is also higher for lower oxidation states of the same element.

 

Question 32. The correct order of ionic radii of \( \text{Y}^{3+} \), \( \text{La}^{3+} \), \( \text{Eu}^{3+} \), and \( \text{Lu}^{3+} \) is (Atomic number Y = 39; La = 57; Eu = 60; Lu = 71)
(a) \( \text{Y}^{3+} < \text{La}^{3+} < \text{Eu}^{3+} < \text{Lu}^{3+} \)
(b) \( \text{Y}^{3+} < \text{Lu}^{3+} < \text{Eu}^{3+} < \text{La}^{3+} \)
(c) \( \text{Lu}^{3+} < \text{Eu}^{3+} < \text{La}^{3+} < \text{Y}^{3+} \)
(d) \( \text{La}^{3+} < \text{Eu}^{3+} < \text{Lu}^{3+} < \text{Y}^{3+} \)
Answer: (b) \( \text{Y}^{3+} < \text{Lu}^{3+} < \text{Eu}^{3+} < \text{La}^{3+} \)
In simple words: Yttrium ion (\( \text{Y}^{3+} \)) is smaller than all lanthanide ions. Among the lanthanides, ionic size generally decreases from lanthanum (\( \text{La}^{3+} \)) to lutetium (\( \text{Lu}^{3+} \)) due to lanthanide contraction. So, lanthanum is the largest, then europium, and then lutetium.

🎯 Exam Tip: Remember the lanthanide contraction, which causes a steady decrease in ionic radii across the lanthanide series as atomic number increases, due to poor shielding by f-electrons.

 

Question 34. The ore which contains copper and iron both
(a) Malachite
(b) Chalcopyrite
(c) Chalocite
(d) Azurite
Answer: (b) Chalcopyrite
In simple words: Chalcopyrite is a common mineral that contains both copper and iron, along with sulfur. Its chemical formula is \( \text{CuFeS}_2 \), making it a major source for extracting copper.

🎯 Exam Tip: Knowing the chemical formulas of common ores helps identify their main metal components. Chalcopyrite is a sulfide mineral of copper and iron.

 

Question 35. According to Ellingham diagram, the oxidation reaction of carbon to carbon monoxide may be used to reduce which one of the following oxides at lowest temperature?
(a) Al2O3
(b) Cu2O
(c) MgO
(d) ZnO
Answer: (a) Al2O3
In simple words: Ellingham diagrams help us see which metals can be extracted from their oxides using a reducing agent like carbon. For carbon to reduce a metal oxide, the Gibbs free energy for the reduction must be negative. This means that carbon's oxidation line on the Ellingham diagram should be lower than the metal oxide's formation line at that temperature. Aluminium oxide is a very stable oxide, requiring high temperatures for reduction.

🎯 Exam Tip: In Ellingham diagrams, a metal oxide can be reduced by carbon if the standard free energy of formation for carbon monoxide is lower than that of the metal oxide at the chosen temperature.

 

Question 36. In electro chemical process (Electrolysis of fused salt) is used to extract
(a) Iron
(b) Lead
(c) Sodium
(d) Silver
Answer: (c) Sodium
In simple words: Electrolysis of melted salts is a powerful method used to get highly reactive metals. Sodium, which is very reactive, is usually taken out of its melted salts, like sodium chloride, using electricity. This process separates the metal ions.

🎯 Exam Tip: Electrolytic reduction is suitable for extracting highly electropositive metals like alkali and alkaline earth metals from their fused halides, as chemical reduction methods are often ineffective or too reactive.

 

Question 37. In metallurgy, flux is a substance used to convert
(a) Mineral into silicate
(b) Fusible impurities to soluble impurities
(c) Infusible impurities to soluble impurities
(d) none of these
Answer: (c) Infusible impurities to soluble impurities
In simple words: In metal processing, a flux is added to turn unwanted rocky materials, called infusible impurities or gangue, into a melted, easily removable slag. This helps to clean the metal. Fluxes combine chemically with impurities to form a new compound.

🎯 Exam Tip: Fluxes react with non-fusible impurities (gangue) to form fusible slag, which can be easily separated from the molten metal, simplifying the purification process.

 

Question 38. Heating of Iron pyrites in air to remove sulphur is called
(a) Fusion
(b) Calcination
(c) Smelting
(d) Roasting
Answer: (d) Roasting
In simple words: Heating a metal ore in the presence of air to remove impurities like sulfur is called roasting. This process helps to convert sulfide ores into oxides and release volatile impurities.

🎯 Exam Tip: Roasting is specifically used for sulfide ores, while calcination is used for carbonate and hydroxide ores, involving heating in the absence or limited supply of air.

 

Question 39. Which of the following metal is leached by cyanide process?
(a) Silver
(b) Sodium
(c) Aluminium
(d) Copper
Answer: (a) Silver
In simple words: The cyanide process is a special way to get valuable metals like gold and silver from their ores. It uses a cyanide solution to dissolve the metal, separating it from other materials. This chemical process is very effective for extracting precious metals.

🎯 Exam Tip: The cyanide leaching process is commonly used for the extraction of noble metals (gold and silver) due to their ability to form soluble cyanide complexes.

 

Question 40. The elements in which extra electrons enter into (n - 2) orbitals are called
(a) f - block elements
(b) d - block elements
(c) s - block elements
(d) p – block elements
Answer: (a) f - block elements
In simple words: Elements where new electrons fill up the f-orbitals, which are two shells inward from the outermost shell (n-2), are known as f-block elements. These include the lanthanides and actinides, found at the bottom of the periodic table.

🎯 Exam Tip: F-block elements are characterized by the filling of (n-2)f orbitals, d-block elements by (n-1)d orbitals, and p-block elements by np orbitals.

 

Question 41. The most common oxidation state of lanthanoides
(a) +4
(b) +2
(c) +6
(d) +3
Answer: (d) +3
In simple words: Most lanthanoid elements typically show a +3 oxidation state. This is their most stable state, although some can also show +2 or +4 states. This stability comes from their electron arrangements.

🎯 Exam Tip: The stability of the +3 oxidation state in lanthanides is primarily due to the tendency to achieve a stable f0, f7, or f14 electron configuration.

 

Question 42. ............ form complexes.
(a) lanthanoides
(b) actinides
(c) Thorium
(d) Cerium
Answer: (b) actinides
In simple words: Actinides are special elements that can easily create complex compounds. They do this more readily than lanthanides because their 5f orbitals are more exposed and can interact better. They act as good electron acceptors.

🎯 Exam Tip: Actinides form complexes more easily than lanthanides due to the lower shielding effect of 5f electrons and their larger size and lower charge-to-radius ratio, making their valence electrons more available for bonding.

 

Question 43. ............ form oxocations.
(a) Actnides
(b) lanthanoides
(c) s - block
(d) p – block
Answer: (a) Actnides
In simple words: Actinide elements are known for forming oxocations, which are positively charged ions containing oxygen atoms. For example, uranium can form the uranyl ion \( \text{UO}_2^{2+} \). This ability is a unique chemical property of actinides.

🎯 Exam Tip: Oxocations, like \( \text{UO}_2^{2+} \), are commonly observed in actinide chemistry due to their ability to form stable bonds with oxygen, which is less common in lanthanides.

 

Question 44. The correct electronic configuration of \( \text{Gd}^{3+} \) is
(a) \( \text{[Xe]4f}^{14} \)
(b) \( \text{[Xe]4f}^{7} \)
(c) \( \text{[Xe]4f}^{0} \)
(d) \( \text{[Xe]5d}^{1} \)
Answer: (b) \( \text{[Xe]4f}^{7} \)
In simple words: Gadolinium (\( \text{Gd} \)) in its neutral atom state is \( \text{[Xe]4f}^{7}\text{5d}^{1}\text{6s}^{2} \). When it forms a \( \text{Gd}^{3+} \) ion, it loses the two 6s electrons and one 5d electron, resulting in a stable half-filled \( \text{4f}^7 \) configuration. This configuration has seven electrons in its f-orbital.

🎯 Exam Tip: When forming ions, lanthanides typically lose their 6s electrons first, and sometimes a 5d electron, to achieve stable f0, f7, or f14 configurations, which are more energetically favorable.

 

Question 45. The general electronic configuration of lanthanoides
(a) \( \text{[Xe]4f}^{0-14} \)
(b) \( \text{[Xe]5d}^{0-1} \)
(c) \( \text{[Xe]4f}^{2-14}\text{5d}^{0-1}\text{6s}^{2} \)
(d) \( \text{[Xe]4f}^{0-14}\text{5d}^{1-10}\text{6s}^{2} \)
Answer: (c) \( \text{[Xe]4f}^{2-14}\text{5d}^{0-1}\text{6s}^{2} \)
In simple words: The general way to write the electron setup for lanthanides is to start with Xenon's configuration, then add between 2 and 14 electrons in the 4f shell, either 0 or 1 electron in the 5d shell, and always 2 electrons in the 6s shell. This shows how their electrons are arranged, with variability in the 4f and 5d orbitals.

🎯 Exam Tip: The 4f electrons are filled after 6s, and the 5d orbital can have zero or one electron, contributing to the variability in their electronic structure, which impacts their chemical properties.

 

Question 46. As we move from lanthanum to Lutetium, basic character of hydroxides.
(a) decreases
(b) increases
(c) decrease and then increases
(d) none of these
Answer: (a) decreases
In simple words: When we go from lanthanum to lutetium across the lanthanide series, the size of the lanthanide ions gets smaller. Because the ions are smaller, their hydroxides become less basic. This change is due to the lanthanide contraction, which makes the metal-oxygen bond more covalent.

🎯 Exam Tip: The lanthanide contraction leads to an increase in covalent character of the M-OH bond, thereby decreasing the basicity of the hydroxides across the series.

 

Question 47. The colour of \( \text{U}^{3+} \) is
(a) red
(b) green
(c) yellow
(d) pink
Answer: (a) red
In simple words: The \( \text{U}^{3+} \) ion has a distinct red color. Many actinide ions show specific colors due to the way their f-electrons interact with light. These f-f transitions absorb certain wavelengths of light.

🎯 Exam Tip: Actinide ions often display characteristic colors in solutions, which can be used for their identification and characterization, similar to transition metal ions but involving f-orbital electrons.

 

II. Assertion and Reason

 

Question 1. Assertion : For chromium, the ground state electronic configuration is \( \text{3d}^5\text{4s}^1 \) rather than \( \text{3d}^4\text{4s}^2 \). Reason : The energy of the Cr atom is lower, when the six valence electrons are in different atomic orbitals with parallel spins
(a) Both Assertion and Reason are true and Reason is the correct explanation of A.
(b) Both Assertion and Reason are true but Reason not is the correct explanation of Assertion.
(c) Assertion is true but Reason is false
(d) Assertion is false but Reason is true
Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation of A.
In simple words: The statement that chromium has a \( \text{3d}^5\text{4s}^1 \) electron setup is true. The reason is also true: this configuration is more stable because electrons are spread out and have parallel spins. This stability makes the atom's energy lower, which is why this configuration is preferred.

🎯 Exam Tip: Remember that half-filled \( \text{(d}^5) \) and completely filled \( \text{(d}^{10}) \) subshells provide extra stability to an atom's electronic configuration, leading to exceptions like chromium and copper.

 

Question 2. Assertion : The energies of the 5s and 4d orbitals are very close. Reason: The relative energies of the 4d and 5s orbitals vary with the nuclear charge and the electronic distribution.
(a) Both Assertion and Reason are true and Reason is the correct explanation of A.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false
(d) Assertion is false but Reason is true
Answer: (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
In simple words: It's true that the 5s and 4d electron energy levels are very close to each other. It's also true that how these energy levels are arranged changes based on the atom's core charge and how its electrons are spread out. However, the reason given explains *what affects* their energies, not directly *why* their energies are already close.

🎯 Exam Tip: The proximity of energy levels between (n-1)d and ns orbitals is a characteristic feature of transition elements, enabling variable oxidation states and complex chemistry.

 

Question 3. Assertion : The melting and boiling points of the transition elements are generally very high Reason: Zn, Hg and Cd have low melting and boiling points as the d - block is complete.
(a) Both Assertion and Reason are true and Reason is the correct explanation of A.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false
(d) Assertion is false but Reason is true
Answer: (b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
In simple words: Transition elements usually have very high melting and boiling points because of strong metallic bonds formed by d-electrons. It is true that zinc, mercury, and cadmium have low melting and boiling points because their d-orbitals are full, making their metallic bonds weaker. However, this reason explains why *some* d-block elements have low melting points, not why *transition elements generally* have high melting points.

🎯 Exam Tip: High melting and boiling points in most transition metals are attributed to strong metallic bonding due to the delocalization of both ns and (n-1)d electrons.

 

Question 4. Assertion: The transition metals more similar to one another than are representative metals of group 1 and group 2 Reason: Inner d orbitals are being filled
(a) Both Assertion and Reason are true and Reason is the correct explanation of A.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false
(d) Assertion is false but Reason is true
Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation of A.
In simple words: Transition metals are quite similar to each other. This is true because their inner d-orbitals are being filled, which gives them shared properties. The inner d-orbital filling causes similar characteristics across the series.

🎯 Exam Tip: The filling of inner d-orbitals means that the outermost electron shell remains largely similar, leading to comparable chemical and physical properties across the transition series.

 

Question 5. Assertion : All the transition elements are metals and good conductors of heat and electricity. Reason: The penultimate shell of electrons of all these elements is expanding, expected to have physical and chemical properties in common.
(a) Both Assertion and Reason are true and Reason is the correct explanation of A.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false
(d) Assertion is false but Reason is true
Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation of A.
In simple words: It is true that all transition elements are metals that conduct heat and electricity well. This happens because their second-to-last electron shell is growing, which means they share many physical and chemical traits. This expansion in the penultimate shell makes them act similarly and contribute to metallic bonding.

🎯 Exam Tip: The presence of delocalized electrons in the d-orbitals explains the metallic character and good conductivity of transition elements, and their similar configurations lead to common properties.

 

III. Match the following

 

Question 1. Match the catalysts given in column I with the processes given in column II

Answer:In simple words: This match connects specific catalysts with the chemical processes they help to speed up. For example, nickel with hydrogen helps turn vegetable oil into ghee, and vanadium pentoxide is used in the contact process. Copper(II) chloride is key for the Sandmeyer reaction.

🎯 Exam Tip: Memorize important catalysts and the industrial processes they are used in, as these are frequently tested questions that require specific knowledge.

 

Question 2. Match the properties given in column I with the metals in colum II

Answer:In simple words: This table matches features of metals with the metals themselves. For example, Osmium is known for its high +8 oxidation state, and manganese can reach a +7 oxidation state. Chromium is known for having a high melting point, while metals in carbonyl compounds often have a zero oxidation state.

🎯 Exam Tip: Pay attention to the maximum oxidation states, especially for group 8 (Osmium) and group 7 (Manganese) elements. Chromium's high melting point relates to strong metallic bonding due to its electron configuration.

 

Question 3. Match the statements given in column I with oxidation states given in column II

Answer:In simple words: This table pairs up properties with their correct oxidation states. For manganese dioxide (\( \text{MnO}_2 \)), manganese has an oxidation state of +4. The most common and stable oxidation state for manganese is +2. In complex manganese oxides, it can reach +7 (as in permanganates). The usual oxidation state for lanthanoids is +3.

🎯 Exam Tip: To determine oxidation states, remember the common oxidation states of oxygen (-2) and apply the rule that the sum of oxidation states in a neutral compound is zero, and in an ion, it equals the charge of the ion.

 

Question 4. Match the following column I with column II

Answer:In simple words: This table matches different chemical acids or compounds with their correct chemical formulas. Permanganic acid is \( \text{HMnO}_4 \). Chromic acid can refer to \( \text{CrO}_3 \) (chromium trioxide). Dichromic acid is usually \( \text{H}_2\text{Cr}_2\text{O}_7 \), but here it is linked to \( \text{H}_2\text{CrO}_4 \) (Chromic acid). Another form of chromic acid is matched with \( \text{Cr}_2\text{O}_3 \) (chromium(III) oxide).

🎯 Exam Tip: Familiarize yourself with the common names and chemical formulas for different acids and oxides of transition metals, especially those of chromium and manganese, as their interconversion is important.

 

Question 5. Match the following

Answer:In simple words: This table pairs specific features of lanthanoids with the correct element. Cerium (\( \text{Ce} \)) often shows a +4 oxidation state. Europium (\( \text{Eu} \)) can exhibit a +2 oxidation state. Promethium (\( \text{Pm} \)) is the only radioactive lanthanide. Gadolinium (\( \text{Gd} \)) has a stable half-filled \( \text{4f}^7 \) electron configuration.

🎯 Exam Tip: Some lanthanides show +2 or +4 oxidation states in addition to the common +3 state due to the stability associated with empty, half-filled, or completely filled f-orbitals.

 

IV. Choose the correct statements

 

Question 1. Consider the following statements which is/are correct.
I. Sc ion has no unpaired electrons and the expected magnetic moemnt value is zero B.M
II. The oxidation state of Cr in \( \text{Cr}_2\text{O}_7^{2-} \) is +4
III. The spin only magnetic moment is given by \( \sqrt{\text{n}(\text{n}+2)} \) B M
(a) I, III
(b) II, I
(c) III, II
(d) II
Answer: (a) I, III
In simple words: Statement I is correct because scandium ion \( \text{Sc}^{3+} \) has no unpaired electrons, so its magnetic moment is zero. Statement III is also correct, as the formula \( \sqrt{\text{n}(\text{n}+2)} \) is used to calculate the spin-only magnetic moment based on the number of unpaired electrons. Statement II is incorrect; the oxidation state of Cr in \( \text{Cr}_2\text{O}_7^{2-} \) is +6, not +4.

🎯 Exam Tip: To find the magnetic moment, identify the number of unpaired electrons (n) and use the spin-only formula. For oxidation states, balance the charges in the compound or ion by assuming known oxidation states for other elements (like oxygen).

 

Question 2. Consider the following statements which is / are correct
I. All Lanthanoids displaces hydrogen from acidified water.
II. The colour of tripositive Lanthanides becomes darker as one goes from \( \text{Cr}^{3+} \) to \( \text{Lu}^{3+} \)
III. The ionic size of tripositive Lanthanoid ions increases with atomic number
(a) I
(b) II
(c) II, III
(d) I, III
Answer: (a) I
In simple words: Statement I is correct because lanthanoids are reactive metals and can remove hydrogen from acidic water. Statement II is incorrect; the color of tripositive lanthanides does not follow a simple darkening trend, and the series starts from \( \text{La}^{3+} \) to \( \text{Lu}^{3+} \), not \( \text{Cr}^{3+} \). Statement III is incorrect because the ionic size of tripositive lanthanoid ions *decreases* with atomic number due to lanthanide contraction.

🎯 Exam Tip: Lanthanides are electropositive, making them good reducing agents. Remember the lanthanide contraction causes a decrease in ionic radius across the series, which impacts properties like basicity and complex formation.

 

Question 3. Consider the following statement which is / are correct.
I. The maximum oxidation state of Osmium is +8
II. The highest oxidation state of a transition metal is given by outermost 's' plus 'd' electrons.
III. The maximum magnetic moment is shown by the ion having d orbital electronic configuration is \( \text{3d}^5 \)
(a) I
(b) II
(c) I, III
(d) I, II, III
Answer: (d) I, II, III
In simple words: All three statements are correct. Osmium can indeed show a very high oxidation state of +8, as seen in \( \text{OsO}_4 \). The highest oxidation state for a transition metal typically involves all its outer s and d electrons because these are available for bonding. Also, ions with a half-filled \( \text{d}^5 \) electron configuration, like \( \text{Mn}^{2+} \), have the most unpaired electrons, leading to the largest magnetic moment.

🎯 Exam Tip: Osmium tetroxide (\( \text{OsO}_4 \)) is a key example of Osmium's +8 oxidation state. For magnetic moment, remember that more unpaired electrons lead to higher paramagnetism and a larger magnetic moment value.

 

Question 4. Consider the following statements which is / are correct
I. Compared to \( \text{Cu}^{2+} \) salts \( \text{Cu}^{+} \) salts are less stable.
II. Iron does not form interstitial compounds
III. \( \text{Ag}^{+} \) is isoelectronic with \( \text{Cd}^{2+} \)
(a) I, III
(b) II, III
(c) III
(d) II
Answer: (a) I, III
In simple words: Statement I says \( \text{Cu}^{+} \) salts are less stable than \( \text{Cu}^{2+} \) salts, which is true. Statement III says \( \text{Ag}^{+} \) and \( \text{Cd}^{2+} \) have the same number of electrons, which is also true.

🎯 Exam Tip: Remember that \( \text{Ag}^{+} \) and \( \text{Cd}^{2+} \) both have a \( \text{d}^{10} \) configuration, making them isoelectronic.

V. Choose The Incorrect Statements

 

Question 1. Consider the following statements which is / are incorrect.
I. The common oxidation state of lanthanides is +3
II. All lanthanides form oxocations
III. All lanthanides are non - radioactive
(a) I
(b) II
(c) II, III
(d) II, III
Answer: (d) II, III
In simple words: The common oxidation state for lanthanides is indeed +3. However, not all lanthanides form oxocations, and not all of them are non-radioactive; some are radioactive.

🎯 Exam Tip: Note that promethium is the only radioactive lanthanide, which makes statement III incorrect. Also, oxocations are more common for actinides.

 

Question 2. Consider the following statements which is / are incorrect.
I. The common oxidation state of Lanthanum is +4
II. \( \text{UO}_{2}^{2+} \) is colourless.
III. Actinides form oxocations.
(a) I, II
(b) II, III
(c) III, I
(d) I, II, III
Answer: (a) I, II
In simple words: Lanthanum's common oxidation state is +3, not +4. Also, the ion \( \text{UO}_{2}^{2+} \) is yellow, not colourless. Actinides do form oxocations, so statement III is correct.

🎯 Exam Tip: Remember that lanthanum (La) typically shows a +3 oxidation state, and many uranium ions are known for their distinct colours.

 

Question 3. Consider the following statements which is incorrect regarding potassium dichromate.
(a) It oxidises ferric salt to ferrous salts.
(b) It oxidises \( \text{KI} \) to \( \text{I}_{2} \)
(c) It oxidises \( \text{H}_{2}\text{S} \) to \( \text{S} \)
(d) None of the options
Answer: (a) It oxidises ferric salt to ferrous salts.
In simple words: Potassium dichromate is a strong oxidising agent. This means it oxidises other substances. Ferric salts are already in a higher oxidation state (\( \text{Fe}^{3+} \)) and cannot be oxidised further to ferrous salts (\( \text{Fe}^{2+} \)). Instead, it would oxidise ferrous salts to ferric salts.

🎯 Exam Tip: Always remember that oxidising agents increase the oxidation state of other substances and get reduced themselves.

 

Question 4. Consider the following statements which is / are correct.
I. Potassium permanganate is extracted from chromite iron ore.
II. Potassium dichromate is extracted from Pyrolusite ore.
III. Potassium permanganate is mainly used in Tanning of leather.
(a) I
(b) II
(c) I, II
(d) I, III
Answer: (c) I, II
In simple words: Both potassium permanganate and potassium dichromate are extracted from specific ores, chromite and pyrolusite, respectively. However, potassium permanganate is not mainly used in tanning leather; instead, potassium dichromate is used for chrome tanning.

🎯 Exam Tip: It is crucial to distinguish between the extraction and uses of potassium permanganate and potassium dichromate, as they are often confused.

VI. Fill In The Blanks

 

Question 1. The electronic configuration of scandium is ……………
Answer: \( \text{[Ar] 3d}^{1}\text{4s}^{2} \)
In simple words: Scandium, which is element number 21, has an electron setup where its last electrons go into the 3d and 4s shells. It finishes with one electron in 3d and two in 4s after the Argon core.

🎯 Exam Tip: Remember that for transition elements, the \( \text{4s} \) electrons are lost before the \( \text{3d} \) electrons when forming ions, but in the ground state, the \( \text{4s} \) subshell is filled first.

 

Question 2. The metal with highest electrical conductivity at room temperature is ……………
Answer: Silver
In simple words: Silver is the best metal for conducting electricity when it is at normal room temperature. This is because its electrons move very freely.

🎯 Exam Tip: High electrical conductivity is due to the presence of free electrons that can move easily through the metal lattice.

 

Question 3. Ionisation energy values are used to predict the thermodynamic stability of the compounds.
Answer: Ionisation energy
In simple words: We use the numbers for how much energy it takes to remove electrons (ionisation energy) to guess how stable chemical compounds will be. These values show how easily an atom can become an ion.

🎯 Exam Tip: Higher ionisation energy generally indicates greater stability of a species because it is harder to remove its electrons.

 

Question 4. The common oxidation state of scandium is ……………. .(MARCH 2020)
Answer: +3
In simple words: Scandium usually shows a +3 charge when it forms compounds. This is its most common way of reacting chemically.

🎯 Exam Tip: Scandium, being the first element in the 3d series, readily loses its two 4s electrons and one 3d electron to achieve a stable noble gas configuration.

 

Question 5. \( \text{Mn}^{2+} \) is …………… than \( \text{Mn}^{4+} \)
Answer: More stable than
In simple words: The \( \text{Mn}^{2+} \) ion is more stable compared to the \( \text{Mn}^{4+} \) ion. This happens because \( \text{Mn}^{2+} \) has a very stable half-filled d-orbital electron arrangement.

🎯 Exam Tip: Half-filled (\( \text{d}^{5} \)) and fully filled (\( \text{d}^{10} \)) d-orbitals are exceptionally stable configurations for transition metal ions.

 

Question 6. The oxidation state of \( \text{W} \) in \( \text{WCl}_{6} \) is ……………
Answer: +6
In simple words: In the compound \( \text{WCl}_{6} \), tungsten has a charge of +6. Each chlorine atom has a -1 charge, and since there are six of them, the tungsten atom must balance this with a +6 charge.

🎯 Exam Tip: To find the oxidation state of an element in a neutral compound, sum the known oxidation states of other elements and set the total to zero.

 

Question 7. …………… metal is unique in 3d series
Answer: Copper
In simple words: Copper is special among the 3d series metals because it is the only one that can show a +1 oxidation state in addition to its more common +2 state. It also has a unique electron configuration.

🎯 Exam Tip: Copper's unique electronic configuration (\( \text{[Ar]} \text{3d}^{10}\text{4s}^{1} \)) allows it to exhibit both +1 and +2 oxidation states, unlike many other 3d series elements that primarily show +2 and higher states.

 

Question 8. The metal shows maximum ferromagnetic character is ……………
Answer: Iron
In simple words: Iron is the metal that shows the strongest magnetic properties, known as ferromagnetism. This means it can be strongly magnetised.

🎯 Exam Tip: Ferromagnetism is a strong form of magnetism that only occurs in certain materials, like iron, cobalt, and nickel, at room temperature.

 

Question 9. Paramagnetism is the property of ……………….. .
Answer: Unpaired electrons
In simple words: Paramagnetism is a type of magnetism that happens when a substance has electrons that are not paired up. These unpaired electrons make the substance slightly attracted to a magnetic field.

🎯 Exam Tip: To determine if a substance is paramagnetic, check its electronic configuration for the presence of unpaired electrons; more unpaired electrons usually mean stronger paramagnetism.

 

Question 10. Paramagnetism is common in ……………
Answer: Transition elements
In simple words: Many transition elements are paramagnetic because they often have unpaired electrons in their d-orbitals. This allows them to be weakly attracted to magnetic fields.

🎯 Exam Tip: Transition metals frequently exhibit paramagnetism due to their partially filled d-subshells, which contain unpaired electrons.

 

Question 11. Hydroformylation catalyst is ……………
Answer: \( \text{Co}_{2}\text{(CO)}_{8} \)
In simple words: The compound \( \text{Co}_{2}\text{(CO)}_{8} \), which contains cobalt and carbon monoxide, is used as a catalyst in a reaction called hydroformylation. This reaction adds hydrogen and a formyl group to alkenes.

🎯 Exam Tip: Metal carbonyls like dicobalt octacarbonyl are common catalysts in industrial organic reactions due to their ability to activate small molecules.

 

Question 12. Potassium dichromate is in colour ……………
Answer: Orange red
In simple words: Potassium dichromate is a chemical compound that appears as an orange-red solid. Its colour comes from the charge transfer between oxygen and chromium ions.

🎯 Exam Tip: The distinct orange-red colour of dichromate ions and yellow colour of chromate ions are important visual indicators in redox reactions involving chromium.

 

Question 13. The oxidation state of chromium in potassium dichromate is ……………
Answer: +6
In simple words: In potassium dichromate \( (\text{K}_{2}\text{Cr}_{2}\text{O}_{7}) \), chromium has an oxidation state of +6. This is one of its highest and most stable oxidation states.

🎯 Exam Tip: When calculating oxidation states, remember that oxygen is usually -2 and potassium is +1, then balance the charges in the compound.

 

Question 14. The colour of potassium permanganate is ……………
Answer: dark purple colour
In simple words: Potassium permanganate is a compound that has a very strong dark purple colour. This deep colour is due to how its electrons absorb light.

🎯 Exam Tip: Potassium permanganate (\( \text{KMnO}_{4} \)) is a powerful oxidising agent often used in titrations, and its intense purple colour acts as a self-indicator.

 

Question 15. The oxidation state uranium in \( \text{UP}_{6} \) is ……………
Answer: +6
In simple words: In the compound \( \text{UP}_{6} \), the oxidation state of uranium is +6. This is determined by the -1 charge of each fluorine-like atom, so six of them make the uranium +6.

🎯 Exam Tip: Uranium can exhibit several oxidation states, with +6 being one of the most common and stable, especially in compounds like uranium hexafluoride (\( \text{UF}_{6} \)).

VII. Choose The Correct Pair

 

Question 1.
(a) Lead – Lead pencil
(b) Cerussite – Zinc
(c) Zinc sulphate – antiseptic
(d) Lead pipes – softwater
Answer: (a) Lead – Lead pencil
In simple words: This pair is correct because lead pencils actually contain graphite, which is a form of carbon, not metallic lead. The name is historical.

🎯 Exam Tip: Be aware that common names can sometimes be misleading about chemical composition, as seen with "lead pencils" containing graphite.

 

Question 2.
(a) \( \text{Ag} \) + hot \( \text{NaOH} \) – Products
(b) \( \text{Zn} \) + hot \( \text{NaOH} \) – Products
(c) \( \text{Au} \) + hot \( \text{NaOH} \) – Products
(d) \( \text{Cr} \) + hot \( \text{NaOH} \) – Products
Answer: (b) \( \text{Zn} \) + hot \( \text{NaOH} \) – Products
In simple words: Zinc reacts with hot sodium hydroxide (NaOH) to form products because zinc is an amphoteric metal. This means it can react with both acids and strong bases.

🎯 Exam Tip: Recognise amphoteric metals (like Zn, Al, Pb, Sn) that can react with both acids and strong bases, forming salts and hydrogen gas.

 

Question 3.
(a) \( \text{Cu}^{2+} \) – Diamagnetic
(b) \( \text{Cu}^{2+} \) – Colourless
(c) \( \text{Cu}^{2+} \) – Zero magnetic moment
(d) \( \text{Cu}^{2+} \) – One unpaired electron
Answer: (d) \( \text{Cu}^{2+} \) – One unpaired electron
In simple words: The \( \text{Cu}^{2+} \) ion has one electron that is not paired up. This single unpaired electron makes it paramagnetic and gives it a colour, which is usually blue.

🎯 Exam Tip: Ions with unpaired electrons are generally coloured and paramagnetic, while those with all paired electrons are colourless and diamagnetic.

 

Question 4.
(a) \( \text{TiCl}_{4} \) – Polymerisation catalyst
(b) \( \text{Ni} \) – Hydrogenation catalyst
(c) \( \text{Fe} \) – Haber's process
(d) All of the options
Answer: (d) All of the options
In simple words: All the given pairs are correct. \( \text{TiCl}_{4} \) is used as a polymerisation catalyst (Ziegler-Natta), nickel is used as a catalyst in hydrogenation reactions, and iron is a catalyst in the Haber-Bosch process to make ammonia.

🎯 Exam Tip: It is important to know the specific catalysts used in key industrial chemical processes, as they are frequently tested.

VIII. Choose The Incorrect Pair

 

Question 1.
(a) Sc – 3d series
(b) Zn – 3d series
(c) Cr – 3d series
(d) Cu – 4d series
Answer: (d) Cu – 4d series
In simple words: The first three options correctly place elements in the 3d series. However, copper is also a part of the 3d series, not the 4d series. This makes the last option the incorrect pair.

🎯 Exam Tip: Ensure you can correctly identify which transition series (3d, 4d, 5d) each element belongs to based on its atomic number and electron configuration.

 

Question 2.
(a) \( \text{Mn}^{4+} \) – \( \text{3d}^{3} \)
(b) \( \text{Mn}^{3+} \) – \( \text{3d}^{4} \)
(c) \( \text{Mn}^{5+} \) – \( \text{3d}^{5} \)
(d) \( \text{Mn}^{5+} \) – \( \text{3d}^{2} \)
Answer: (c) \( \text{Mn}^{5+} \) – \( \text{3d}^{5} \)
In simple words: Manganese's electron configuration when it has a +5 charge should be \( \text{3d}^{2} \), not \( \text{3d}^{5} \). The other pairs correctly show the d-electron count for their respective manganese ions.

🎯 Exam Tip: To find the d-electron count for an ion, first write the ground state electronic configuration of the neutral atom, then remove electrons from the outermost s-orbital first, followed by the d-orbital for transition metals.

 

Question 3.
(a) \( \text{Cu}^{+}, \text{Zn}^{2+} \) – Diamagnetic
(b) \( \text{Sc}^{3+}, \text{Ti}^{4+}, \text{V}^{5+} \) – Paramagnetic
(c) \( \text{Co}^{5+}, \text{Fe}^{2+} \) – Paramagnetic
(d) \( \text{Cu}^{2+} \) – Paramagnetic
Answer: (b) \( \text{Sc}^{3+}, \text{Ti}^{4+}, \text{V}^{5+} \) – Paramagnetic
In simple words: The ions \( \text{Sc}^{3+} \), \( \text{Ti}^{4+} \), and \( \text{V}^{5+} \) all have empty d-orbitals, meaning they have no unpaired electrons. So, they are diamagnetic, not paramagnetic. This makes option (b) the incorrect pair.

🎯 Exam Tip: Remember that ions with a \( \text{d}^{0} \) or \( \text{d}^{10} \) configuration (no unpaired electrons) are diamagnetic, while those with partially filled d-orbitals (unpaired electrons) are paramagnetic.

 

Question 4.
(a) Potassium dichromate – +6
(b) Potassium dichromate – oxidising
(c) Potassium dichromate – Chrome tanning
(d) Potassium dichromate – Bayer's reagent
Answer: (d) Potassium dichromate – Bayer's reagent
In simple words: Potassium dichromate has chromium in the +6 oxidation state, acts as an oxidising agent, and is used in chrome tanning. However, Bayer's reagent is a different chemical, specifically cold, dilute, alkaline potassium permanganate. This makes the last option the incorrect pair.

🎯 Exam Tip: It is important to distinguish between potassium dichromate and potassium permanganate, as they are both strong oxidising agents but have different uses and properties, like being part of Bayer's reagent.

IX. Choose The Odd Man Out

 

Question 1. Scandium, Titanium, Vanadium, Cadmium
Answer: Cadmium
Cadmium belongs to the 4d series of transition metals. All the other elements listed (Scandium, Titanium, Vanadium) belong to the 3d series. This difference in series makes cadmium the odd one out. Cadmium's higher atomic mass and electron shell structure place it in a different period in the periodic table compared to the other elements.
In simple words: Cadmium is the odd one out because it's from the 4d metal group, while Scandium, Titanium, and Vanadium are all from the 3d metal group.

🎯 Exam Tip: Knowing the periodic table positions and electron configurations helps identify which transition series elements belong to.

 

Question 2. Zirconium, Niobium, Technetium, Ruthenium
Answer: Technetium
Technetium is a radioactive element, meaning it naturally decays over time. Zirconium, Niobium, and Ruthenium are all stable, non-radioactive elements. This distinct nuclear property makes technetium the odd one out among the given elements. Its radioactivity makes it distinct for many applications and handling procedures.
In simple words: Technetium is different because it's radioactive, but Zirconium, Niobium, and Ruthenium are not radioactive.

🎯 Exam Tip: Familiarise yourself with key properties of elements, such as radioactivity, which can be a distinguishing factor in such questions.

 

Question 3. \( \text{U}^{3+}, \text{UO}_{2}^{2+}, \text{U}^{4+}, \text{Ce}^{3+} \)
Answer: \( \text{Ce}^{4+} \)
The ion \( \text{Ce}^{4+} \) is colourless, while \( \text{U}^{3+} \), \( \text{UO}_{2}^{2+} \), and \( \text{U}^{4+} \) are all coloured ions. This difference in visual appearance makes \( \text{Ce}^{4+} \) the odd one out. The colour of an ion often depends on its electronic configuration and the possibility of d-d or f-f transitions. \( \text{Ce}^{4+} \) has a \( \text{4f}^{0} \) configuration, which does not allow for f-f transitions, making it colourless.
In simple words: The odd one out is \( \text{Ce}^{4+} \) because it is colourless. All the uranium ions listed (\( \text{U}^{3+} \), \( \text{UO}_{2}^{2+} \), \( \text{U}^{4+} \)) have colours.

🎯 Exam Tip: Ions with empty or completely filled f-orbitals (like \( \text{Ce}^{4+} \) with \( \text{4f}^{0} \)) are typically colourless, as they lack the f-f electronic transitions responsible for colour.

X. Additional Questions – 2 Mark

 

Question 1. Write the classification of transition elements.
Answer: Transition elements are classified into different series based on the d-orbital being filled. These elements form a bridge between the s-block and p-block elements.

• The 3d series consists of 10 elements from Scandium (\( \text{Sc} \)) to Zinc (\( \text{Zn} \)), found in the 4th period.
• The 4d series includes 10 elements from Yttrium (\( \text{Y} \)) to Cadmium (\( \text{Cd} \)), found in the 5th period.
• The 5d series comprises 10 elements from Lanthanum (\( \text{La} \)) to Mercury (\( \text{Hg} \)) (excluding the lanthanides), located in the 6th period.
• The 6d series encompasses 10 elements from Actinium (\( \text{Ac} \)) to Copernicium (\( \text{Cn} \)) (excluding the actinides), found in the 7th period.

In simple words: Transition elements are grouped into four main series: 3d (Scandium to Zinc), 4d (Yttrium to Cadmium), 5d (Lanthanum to Mercury), and 6d (Actinium to Copernicium). Each series has 10 elements.

🎯 Exam Tip: Remember that transition elements are generally characterised by partially filled d-orbitals in their elemental or ionic forms, distinguishing them from other blocks.

 

Question 2. Give the electronic configuration of copper and chromium.
Answer: The electronic configurations for copper and chromium are exceptions to the general Aufbau principle, exhibiting half-filled or fully-filled d-orbitals for greater stability.

• Chromium (\( \text{Cr} \)): \( \text{[Ar] 3d}^{5}\text{4s}^{1} \)
• Copper (\( \text{Cu} \)): \( \text{[Ar] 3d}^{10}\text{4s}^{1} \)

In simple words: Chromium has a special electron setup of \( \text{3d}^{5}\text{4s}^{1} \), and copper also has a special setup of \( \text{3d}^{10}\text{4s}^{1} \). These arrangements make them more stable.

🎯 Exam Tip: Always remember the anomalous electronic configurations of chromium and copper, as they are common exceptions in the d-block elements.

 

Question 3. What are the type of packing possible in transition metals?
Answer: Transition metals exhibit various types of crystal packing arrangements due to the metallic bonding involving d-electrons. These structures affect their physical properties like density and melting points.

• Most transition elements are hexagonal close-packed (hcp) structures.
• Some transition elements are cubic close-packed (ccp) structures.
• Another common arrangement is body-centred cubic (bcc) structures, which are typical characteristics of true metals.

In simple words: Transition metals can be packed in a few ways: hexagonal close-packed, cubic close-packed, or body-centred cubic. These are the main ways their atoms arrange themselves.

🎯 Exam Tip: High melting points and hardness in transition metals are often linked to strong metallic bonding facilitated by their various packing structures and unpaired d-electrons.

 

Question 4. Why transition elements show variable oxidation states? (PTA – 4)
Answer: Transition elements show variable oxidation states because the energy differences between their \( \text{(n-1)d} \) orbitals and \( \text{ns} \) orbitals are very small. This allows electrons from both these subshells to participate in chemical bonding. The participation of a varying number of electrons leads to the exhibition of multiple oxidation states. For example, Manganese shows oxidation states from +2 to +7.
In simple words: Transition elements have many oxidation states because the energy levels of their d- and s-electrons are very close. This means both types of electrons can be used in reactions, leading to different possible charges.

🎯 Exam Tip: The ability to use electrons from both \( \text{ns} \) and \( \text{(n-1)d} \) orbitals is the key reason for variable oxidation states in transition elements, impacting their reactivity and catalytic properties.

 

Question 5. Explain why \( \text{Mn}^{2+} \) is more stable than \( \text{Mn}^{3+} \)?
Answer: \( \text{Mn}^{2+} \) is more stable than \( \text{Mn}^{3+} \) due to its electronic configuration. When manganese forms the \( \text{Mn}^{2+} \) ion, it achieves a half-filled \( \text{3d}^{5} \) electronic configuration, which is particularly stable. In contrast, \( \text{Mn}^{3+} \) has a \( \text{3d}^{4} \) configuration, which is less stable as it is neither half-filled nor fully-filled. The exceptional stability of half-filled and fully-filled d-orbitals significantly influences the preferred oxidation states of transition metals.

• \( \text{Mn}^{2+} \) has an outer electronic configuration of \( \text{3d}^{5} \).
• \( \text{Mn}^{3+} \) has an outer electronic configuration of \( \text{3d}^{4} \).
• Half-filled and completely filled orbitals are more stable than partially filled orbitals, so \( \text{Mn}^{2+} \) with its half-filled d-orbital is more stable than \( \text{Mn}^{3+} \).

In simple words: \( \text{Mn}^{2+} \) is more stable because its electron arrangement has a half-filled d-shell (\( \text{3d}^{5} \)), which is very strong and preferred. \( \text{Mn}^{3+} \) has a less stable electron setup (\( \text{3d}^{4} \)).

🎯 Exam Tip: Always look for half-filled (\( \text{d}^{5} \)) or fully filled (\( \text{d}^{10} \)) d-orbital configurations when explaining the relative stability of transition metal ions.

 

Question 6. Explain about ferromagnetic materials.
Answer: Ferromagnetic materials are substances that are strongly attracted to magnetic fields and can retain their magnetism even after the external field is removed. These materials have a unique internal structure that contributes to their strong magnetic behaviour. Materials like iron, cobalt, and nickel are good examples of ferromagnets.

• Ferromagnetic materials have a domain structure, meaning they are divided into small regions called domains.
• Within each domain, the magnetic dipoles (tiny magnets created by electron spins) are arranged in the same direction, making the domain strongly magnetic.
• However, the spin dipoles of the adjacent domains are randomly oriented, so the overall material might not show magnetism without an external field.
• Some transition elements or ions with unpaired d electrons show ferromagnetism, particularly when these unpaired electrons align in a specific way across many atoms.

In simple words: Ferromagnetic materials are very strong magnets. They have tiny magnetic zones inside called domains, where all the electron magnets point the same way. This helps them stay magnetic after being magnetised.

🎯 Exam Tip: The key characteristic of ferromagnetism is the presence of magnetic domains that align in an external magnetic field, leading to strong and persistent magnetism.

 

Question 7. Explain why the melting and boiling points of \( \text{Cd} \), \( \text{Hg} \) and \( \text{Zn} \) are low?
Answer: The elements of group 12, namely Zinc (\( \text{Zn} \)), Cadmium (\( \text{Cd} \)), and Mercury (\( \text{Hg} \)), have unusually low melting and boiling points compared to other transition metals. This is because their electronic configurations (\( \text{(n-1)d}^{10}\text{ns}^{2} \)) result in fully filled d-orbitals. Consequently, they do not have any unpaired electrons available to participate in strong metallic bonding. Weaker metallic bonds lead to less energy required to break the structure, hence lower melting and boiling points. Mercury is notably a liquid at room temperature, melting at \( -38^\circ\text{C} \).
In simple words: Zinc, Cadmium, and Mercury have low melting and boiling points because they have all their d-electrons paired up. This means they cannot form strong bonds between their metal atoms, so it's easier to melt or boil them.

🎯 Exam Tip: Low melting and boiling points in group 12 elements (Zn, Cd, Hg) are directly linked to their fully filled d-orbitals, which limit the availability of electrons for robust metallic bonding.

 

Question 8. Which of the following ions would from colourless complexes?
Answer: Ions that form colourless complexes typically have fully filled d-orbitals or completely empty d-orbitals, meaning they do not possess unpaired electrons. Without unpaired electrons, d-d electron transitions (which are responsible for absorbing specific wavelengths of light and thus imparting colour) are not possible. Examples of such ions include: \( \text{Cu}^{+} (\text{3d}^{10}) \), \( \text{Zn}^{2+} (\text{3d}^{10}) \), \( \text{Ti}^{4+} (\text{3d}^{0}) \), and \( \text{Cd}^{2+} (\text{4d}^{10}) \). The ions like \( \text{Cu}^{2+} \) and \( \text{Ti} \) (neutral atom) would form coloured complexes if they have partially filled d-orbitals.
In simple words: Ions that form complexes with no colour are those that either have all their d-electron slots full or completely empty. This means there are no single, unpaired electrons to jump around and absorb light, so we don't see any colour.

🎯 Exam Tip: The presence of unpaired d-electrons is the primary factor determining the colour of transition metal complexes. \( \text{d}^{0} \) and \( \text{d}^{10} \) configurations are always colourless.

 

Question 9. State Hume - Rothery rule for alloy formation.
Answer: The Hume-Rothery rules are a set of principles that describe the conditions under which a metal can dissolve in another metal to form a solid solution alloy. These rules are crucial for predicting whether two metals will mix well. For a substitute alloy (where atoms of one metal replace atoms of another in the crystal structure) to form readily, the difference in atomic radii of the solvent and solute must be less than 15%. Additionally, both the solvent and solute must have the same crystal structure and their electronegativity difference must be close to zero. They also tend to have the same valency.
In simple words: The Hume-Rothery rules say that two metals can easily mix and form an alloy if their atoms are similar in size (less than 15% difference), have the same crystal shape, and have very similar chemical attraction for electrons.

🎯 Exam Tip: The Hume-Rothery rules are empirical guidelines for predicting the formation of solid solutions, with atomic size difference being a particularly important factor.

 

Question 10. Explain why transition metals form alloys? Transition metals form more alloys themselves because of their
Answer: Transition metals readily form alloys with each other due to several favourable characteristics. Their tendency to form numerous alloys is a significant property that makes them useful in various applications. These characteristics allow one metal atom to be easily replaced by another metal atom in its crystal lattice, forming a stable mixture. Many common alloys like steel (iron with carbon) and brass (copper with zinc) demonstrate this.
Their atomic sizes are similar, and they tend to have similar crystal structures and chemical bonding characteristics. This makes it easy for atoms of different transition metals to substitute each other in the crystal lattice without much distortion, leading to the formation of stable alloys. Additionally, they often have similar electronegativities, which further promotes mixing without strong ionic interactions.
In simple words: Transition metals form alloys easily because their atoms are about the same size and have similar crystal shapes. This means one metal atom can easily swap places with another in the crystal, creating a mix of metals.

🎯 Exam Tip: Similar atomic radii and crystal structures are the primary reasons why transition metals readily form alloys, which are essential for engineering applications due to their enhanced properties.

 

Question 11. Silver atom has completely filled d-orbitals (\( \text{4d}^{10} \)) in its ground state, How can you say that it is a transition element?
Answer: Although silver (\( \text{Ag} \)) has a completely filled \( \text{4d}^{10} \) configuration in its ground state, it is still classified as a transition element because it can exhibit an oxidation state where its d-orbital becomes incompletely filled. The outer electronic configuration of \( \text{Ag} \) (atomic number \( \text{Z} = 47 \)) is \( \text{[Kr]} \text{4d}^{10}\text{5s}^{1} \). In addition to forming a +1 oxidation state (where it loses the \( \text{5s}^{1} \) electron, keeping \( \text{4d}^{10} \)), silver also commonly shows a +2 oxidation state. In the \( \text{Ag}^{2+} \) ion, the electronic configuration is \( \text{[Kr]} \text{4d}^{9} \). Since the \( \text{4d} \) subshell is incompletely filled in its +2 oxidation state, silver meets the definition of a transition element.
In simple words: Silver is a transition element even though its d-shell is full in its normal state. This is because it can lose more electrons to become \( \text{Ag}^{2+} \), which leaves its d-shell partly empty. An element is called "transition" if its d-shell is partly filled in any of its common forms.

🎯 Exam Tip: The definition of a transition element requires a partially filled d-orbital in *any* common oxidation state, not just the ground state. This is key for elements like copper and silver.

 

Question 12. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Answer: The highest oxidation state of a metal is almost exclusively observed when it forms compounds with oxygen or fluorine. This is because oxygen and fluorine are the most electronegative elements, meaning they have a very strong tendency to attract electrons. They have small atomic sizes and high electronegativity. Their strong electron-withdrawing ability allows them to effectively oxidise the metal, pulling away all its valence electrons and prompting them to participate in bonding. This helps the metal achieve its highest possible positive oxidation state. For example, in \( \text{OsO}_{4} \), osmium exhibits an oxidation state of +8.
In simple words: Metals show their highest oxidation state only with oxygen or fluorine because these two elements are the strongest at pulling electrons. Their small size and high electron-attracting power help remove all possible electrons from the metal, leading to the metal's highest positive charge.

🎯 Exam Tip: Remember that small size and high electronegativity of bonding partners (especially O and F) are crucial for stabilising very high oxidation states in metals.

 

Question 13. Calculate the 'Spin only' magnetic moment of \( \text{M}^{2+} \) (aq) ion (\( \text{Z}=27 \))
Answer: To calculate the spin-only magnetic moment of an \( \text{M}^{2+} \) ion with atomic number \( \text{Z}=27 \) (which is cobalt, \( \text{Co} \)), we first determine its electronic configuration and the number of unpaired electrons. The electronic configuration of neutral \( \text{Co} \) (\( \text{Z}=27 \)) is \( \text{[Ar] 3d}^{7}\text{4s}^{2} \). When it forms the \( \text{Co}^{2+} \) ion, it loses the two \( \text{4s} \) electrons, resulting in the configuration \( \text{[Ar] 3d}^{7} \).
The \( \text{3d}^{7} \) configuration can be represented as: \( \text{d-orbitals: } \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow \uparrow \)
From this, we see that there are 3 unpaired electrons (\( \text{n}=3 \)).
The spin-only magnetic moment \( (\mu_{s}) \) is calculated using the formula: \( \mu_{s} = \sqrt{\text{n}(\text{n}+2)} \text{ BM} \).
Substituting \( \text{n}=3 \):
\( \mu_{s} = \sqrt{3(3+2)} \)
\( \mu_{s} = \sqrt{3 \times 5} \)
\( \mu_{s} = \sqrt{15} \)
\( \mu_{s} \approx 3.87 \text{ BM} \)
In simple words: For an ion with atomic number 27 and a +2 charge (Cobalt, \( \text{Co}^{2+} \)), its electron arrangement leaves 3 electrons unpaired. Using a special formula that counts these unpaired electrons, its magnetic strength (spin-only magnetic moment) is found to be about 3.87 BM.

🎯 Exam Tip: Always remember to remove electrons from the \( \text{ns} \) orbital first, then the \( \text{(n-1)d} \) orbital, when forming ions of transition elements, before determining the number of unpaired electrons.

 

Question 14. In what way is the electronic configuration of the transition elements different from that of the non transition elements?
Answer: The electronic configurations of transition elements fundamentally differ from those of non-transition elements primarily in the filling of d-orbitals. Transition elements, also known as d-block elements, have incompletely filled d-subshells in their elemental or ionic forms. Their general electronic configuration is \( \text{(n-1)d}^{1-9}\text{ns}^{0-2} \). The incomplete d-subshell allows for variable oxidation states, colour formation, and magnetic properties. In contrast, non-transition elements (s-block and p-block elements) either have no d-subshell at all (like hydrogen, helium, lithium) or have completely filled d-subshells, with their valence electrons residing in the \( \text{ns}^{1-2} \) or \( \text{ns}^{2}\text{np}^{1-6} \) orbitals in their outermost shell. This means non-transition elements do not participate in d-orbital chemistry.
In simple words: Transition elements are special because their d-electron shells are not completely full. This is different from non-transition elements, which either don't have d-shells or have them completely full, with their outside electrons in s- or p-shells.

🎯 Exam Tip: The presence of partially filled d-orbitals is the defining characteristic that sets transition elements apart from non-transition elements, giving rise to their unique chemical properties.

 

Question 15. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
Answer: The oxometal anions from the first transition series (3d series) where the central metal atom exhibits an oxidation state equal to its group number are important examples of high oxidation states stabilised by oxygen. These include:

• Dichromate ion, \( \text{Cr}_{2}\text{O}_{7}^{2-} \), and Chromate ion, \( \text{CrO}_{4}^{2-} \). Here, chromium (\( \text{Cr} \)) is in Group 6 and exhibits an oxidation state of +6.
• Permanganate ion, \( \text{MnO}_{4}^{-} \). Here, manganese (\( \text{Mn} \)) is in Group 7 and exhibits an oxidation state of +7.

In simple words: For the 3d transition metals, we find certain oxygen-containing ions where the metal's charge matches its group number. Examples are chromate (\( \text{CrO}_{4}^{2-} \)) and dichromate (\( \text{Cr}_{2}\text{O}_{7}^{2-} \)) where chromium is +6 (Group 6), and permanganate (\( \text{MnO}_{4}^{-} \)) where manganese is +7 (Group 7).

🎯 Exam Tip: High oxidation states of transition metals are often stabilised in oxoanions due to the strong electronegativity of oxygen, allowing for multiple bonding.

 

Question 16. What is the effect of increasing \( \text{pH} \) of a solution of Potassium dichromate?
Answer: Increasing the \( \text{pH} \) of a potassium dichromate solution (i.e., making it more alkaline by adding an alkali) causes a shift in the equilibrium between dichromate ions (\( \text{Cr}_{2}\text{O}_{7}^{2-} \), which are orange) and chromate ions (\( \text{CrO}_{4}^{2-} \), which are yellow). The solution changes from orange to yellow. This is a reversible reaction, and the chromate-dichromate equilibrium is sensitive to \( \text{pH} \).
The reaction is: \( \text{Cr}_{2}\text{O}_{7}^{2-} \text{ (orange)} + \text{H}_{2}\text{O} \rightleftharpoons 2\text{CrO}_{4}^{2-} \text{ (yellow)} + 2\text{H}^{+} \)
When \( \text{pH} \) increases (less \( \text{H}^{+} \)), the equilibrium shifts to the right, favouring the formation of yellow chromate ions.
In simple words: If you add something to make a potassium dichromate solution less acidic (raise the \( \text{pH} \)), its colour changes from orange to yellow. This happens because the orange dichromate turns into yellow chromate when there's less acid around.

🎯 Exam Tip: Remember the reversible equilibrium between orange dichromate ions and yellow chromate ions, which is a classic example of an acid-base indicator system in inorganic chemistry.

 

Question 17. Write chemical equations for the reactions involved in the manufacture of potassium permanganate from Pyrolusite. (PTA- 5)
Answer: The manufacture of potassium permanganate (\( \text{KMnO}_{4} \)) from pyrolusite ore (\( \text{MnO}_{2} \)) involves two main steps: fusion and electrolytic oxidation.
First, pyrolusite is fused with an alkali (like \( \text{KOH} \)) in the presence of an oxidising agent (like air or \( \text{O}_{2} \)). This forms potassium manganate (\( \text{K}_{2}\text{MnO}_{4} \)), which is green.
\( 2\text{MnO}_{2} + 4\text{KOH} + \text{O}_{2} \longrightarrow 2\text{K}_{2}\text{MnO}_{4} + 2\text{H}_{2}\text{O} \)
Next, the potassium manganate undergoes electrolytic oxidation in an alkaline medium. The manganate ion (\( \text{MnO}_{4}^{2-} \)) is oxidised to the permanganate ion (\( \text{MnO}_{4}^{-} \)), which is purple.
\( \text{MnO}_{4}^{2-} \text{ (green)} \xrightarrow{\text{electrolysis}} \text{MnO}_{4}^{-} \text{ (purple)} + \text{e}^{-} \)
This process converts the green manganate solution into the characteristic dark purple solution of potassium permanganate. The overall efficiency is often improved by controlling conditions like \( \text{pH} \) and temperature.
In simple words: To make potassium permanganate from pyrolusite ore, first, you mix the ore with a base like \( \text{KOH} \) and air, heating it up. This makes a green substance called potassium manganate. Then, you use electricity to change this green substance into the final dark purple potassium permanganate.

🎯 Exam Tip: Remember the two-step process: initial oxidation of \( \text{MnO}_{2} \) to \( \text{MnO}_{4}^{2-} \) (manganate) via fusion, followed by further oxidation of manganate to \( \text{MnO}_{4}^{-} \) (permanganate) through electrolysis.

 

Question 18. Classify the following elements into d-block and f-block elements. (MARCH 2020)
(i) Tungsten
(ii) Ruthenium
(iii) Promethium
(iv) Einsteinium
Answer: The classification of these elements into d-block and f-block depends on their position in the periodic table and the subshell being filled by their differentiating electron. The d-block elements are transition metals, while the f-block elements are inner transition metals.

• d-block elements: Tungsten (\( \text{W} \)) and Ruthenium (\( \text{Ru} \)). These elements are found in the main body of the transition metals.
• f-block elements: Promethium (\( \text{Pm} \)) and Einsteinium (\( \text{Es} \)). These are inner transition elements (lanthanides and actinides, respectively), typically placed below the main body of the periodic table.

In simple words: Tungsten and Ruthenium are d-block elements, which are regular transition metals. Promethium and Einsteinium are f-block elements, also known as inner transition metals.

🎯 Exam Tip: Remember that lanthanides and actinides are f-block elements, filling the \( \text{4f} \) and \( \text{5f} \) orbitals respectively, while the d-block elements fill the \( \text{d} \) orbitals.

 

Question 19. The halides of transition elements become more covalent with increasing oxidation state of the metal why?
Answer: The halides of transition elements tend to become more covalent as the oxidation state of the central metal atom increases. This phenomenon can be explained by Fajans' Rules. As the oxidation state of the metal increases, its positive charge density increases, making it a stronger polarizing agent. A highly charged cation has a greater ability to distort the electron cloud of the surrounding halide anion. This distortion, or polarization, leads to a greater sharing of electrons between the metal and the halogen, increasing the covalent character of the bond. For example, \( \text{TiCl}_{2} \) is more ionic, while \( \text{TiCl}_{4} \) is more covalent.
In simple words: When a transition metal in a halide compound has a higher positive charge, it pulls the electrons from the halide atom closer to itself more strongly. This pulling makes the bond between the metal and the halogen more like a shared bond (covalent) and less like an electron-transfer bond (ionic).

🎯 Exam Tip: According to Fajans' Rules, a higher positive charge on the cation, a smaller cation size, and a larger anion size all favour increased covalent character in an ionic bond.

 

Question 20. Although \( \text{Cr}^{3+} \) and \( \text{Co}^{2+} \) ions have the same number of unpaired electrons but the magnetic moment of \( \text{Cr}^{3+} \) is 3.87 B.M. and that of \( \text{Co}^{2+} \) is 4.87 B.M. why?
Answer: While both \( \text{Cr}^{3+} \) and \( \text{Co}^{2+} \) ions have 3 unpaired electrons (for \( \text{Cr}^{3+} \), configuration is \( \text{3d}^{3} \), and for \( \text{Co}^{2+} \), configuration is \( \text{3d}^{7} \)), their observed magnetic moments differ due to the contribution of orbital magnetic moment. The simple spin-only formula (\( \mu_{s} = \sqrt{\text{n}(\text{n}+2)} \text{ BM} \)) predicts a value of \( \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM} \) for 3 unpaired electrons. This value matches \( \text{Cr}^{3+} \) quite well because \( \text{Cr}^{3+} \) has a symmetrical \( \text{3d}^{3} \) electronic configuration in its outermost orbit, which often leads to quenching of the orbital angular momentum contribution. In such symmetrical configurations, the orbital motion does not contribute significantly to the magnetic moment. However, for \( \text{Co}^{2+} \) with a \( \text{3d}^{7} \) configuration, there is an appreciable orbital contribution to the magnetic moment. This is because its electronic configuration is not perfectly symmetrical, allowing the orbital angular momentum to contribute to the total magnetic moment, thus giving a higher observed value of 4.87 B.M. This additional contribution from orbital motion is often more significant for ions at the end of a transition series due to less effective shielding.
In simple words: Even though both \( \text{Cr}^{3+} \) and \( \text{Co}^{2+} \) have three unpaired electrons, \( \text{Cr}^{3+} \) has a perfectly balanced electron setup which means its electron movement doesn't add to its magnetic strength. But for \( \text{Co}^{2+} \), its electron setup is not perfectly balanced, so the electron movement does add to its magnetic strength, making it appear more magnetic than \( \text{Cr}^{3+} \).

🎯 Exam Tip: Remember that the observed magnetic moment can deviate from the spin-only value due to orbital angular momentum contribution, especially for ions with non-symmetrical d-electron configurations.

 

Question 21. Why \( \text{E}^\circ \) value for \( \text{Mn} \), \( \text{Ni} \) and \( \text{Zn} \) are more negative than expected?
Answer: The \( \text{E}^\circ \) (standard electrode potential) values for \( \text{Mn} \), \( \text{Ni} \), and \( \text{Zn} \) are more negative than generally expected for transition metals. This "more negative" value indicates a greater tendency to lose electrons and undergo oxidation compared to what would be predicted based on general trends. For manganese (\( \text{Mn} \)) and zinc (\( \text{Zn} \)), these negative values are directly related to the exceptional stabilities of their resulting ions: \( \text{Mn}^{2+} \) achieves a half-filled \( \text{3d}^{5} \) configuration, and \( \text{Zn}^{2+} \) achieves a fully filled \( \text{3d}^{10} \) configuration. These stable configurations make their oxidation very favourable. For nickel (\( \text{Ni} \)), the unexpected negative value is primarily due to its high enthalpy of hydration. When \( \text{Ni}^{2+} \) ions are formed, they are highly hydrated in aqueous solution, releasing a significant amount of energy. This large hydration enthalpy compensates for the relatively high ionisation energy of nickel, making the overall process of forming \( \text{Ni}^{2+} \) more energetically favourable, thus leading to a more negative \( \text{E}^\circ \) value.
In simple words: The electrode potential values for Manganese, Nickel, and Zinc are more negative than usual. For Manganese and Zinc, this is because their \( \text{Mn}^{2+} \) (\( \text{d}^{5} \)) and \( \text{Zn}^{2+} \) (\( \text{d}^{10} \)) ions are super stable, making it easy for them to lose electrons. For Nickel, its ions get very well surrounded by water, which releases a lot of energy, also making it easier to lose electrons.

🎯 Exam Tip: Remember that \( \text{E}^\circ \) values are influenced by ionisation enthalpy, enthalpy of atomisation, and enthalpy of hydration. Exceptional stability of \( \text{d}^{5} \) and \( \text{d}^{10} \) configurations and large hydration enthalpies often lead to more negative \( \text{E}^\circ \) values.

 

Question 22. Calculate the spin only magnetic moment of \( \text{Mn}^{2+} \)
Answer: To calculate the spin-only magnetic moment of the \( \text{Mn}^{2+} \) ion, we first determine its electronic configuration and the number of unpaired electrons. Manganese (\( \text{Mn} \)) has an atomic number of 25. The electronic configuration of neutral \( \text{Mn} \) is \( \text{[Ar] 3d}^{5}\text{4s}^{2} \). When it forms the \( \text{Mn}^{2+} \) ion, it loses the two \( \text{4s} \) electrons, resulting in the configuration \( \text{[Ar] 3d}^{5} \).
For the \( \text{3d}^{5} \) configuration, all five d-orbitals are half-filled, meaning there are 5 unpaired electrons. So, \( \text{n}=5 \).
The spin-only magnetic moment \( (\mu_{s}) \) is calculated using the formula: \( \mu_{s} = \sqrt{\text{n}(\text{n}+2)} \text{ BM} \).
Substituting \( \text{n}=5 \):
\( \mu_{s} = \sqrt{5(5+2)} \)
\( \mu_{s} = \sqrt{5 \times 7} \)
\( \mu_{s} = \sqrt{35} \)
\( \mu_{s} \approx 5.91 \text{ BM} \)
In simple words: For a manganese ion with a +2 charge (\( \text{Mn}^{2+} \)), its electrons are arranged so that 5 of them are unpaired. Using the formula for magnetic moment, we find that its magnetic strength is about 5.91 BM.

🎯 Exam Tip: The \( \text{d}^{5} \) configuration of \( \text{Mn}^{2+} \) is highly stable due to Hund's rule and exchange energy, leading to a high number of unpaired electrons and a significant magnetic moment.

 

Question 23. Why \( \text{Ni}^{\text{II}} \) complexes are thermodynamically more stable than \( \text{Pt}^{\text{II}} \) complexes?
Answer: \( \text{Ni}^{\text{II}} \) complexes (Nickel in a +2 oxidation state) are thermodynamically more stable than \( \text{Pt}^{\text{II}} \) complexes (Platinum in a +2 oxidation state) primarily due to differences in ionisation energies and hydration enthalpies. Although platinum is a heavier element, the ionisation energy of \( \text{Ni}^{2+} \) is actually less than that of \( \text{Pt}^{2+} \). This means it requires less energy to form the \( \text{Ni}^{2+} \) ion. Furthermore, the enthalpy of hydration for \( \text{Ni}^{2+} \) ions is generally more negative (i.e., more energy is released upon hydration) compared to \( \text{Pt}^{2+} \). This favourable hydration energy contributes significantly to the overall stability of \( \text{Ni}^{\text{II}} \) complexes in aqueous solutions. The strong interaction with water molecules releases more energy, making the formation of \( \text{Ni}^{\text{II}} \) complexes more exothermic and thus more thermodynamically stable.
In simple words: Nickel (+2) compounds are more stable than Platinum (+2) compounds. This is because it takes less energy to make \( \text{Ni}^{2+} \) ions. Also, when \( \text{Ni}^{2+} \) ions dissolve in water, they release a lot of energy, which makes them even more stable.

🎯 Exam Tip: When comparing stability of complexes, consider not just electronic configuration, but also thermodynamic factors like ionisation energy, hydration energy, and ligand field stabilisation energy.

 

Question 24. Why do transition elements and its compounds act as catalyst? (PTA – 5)
Answer: Transition elements and their compounds are widely known for their catalytic activity due to several key properties. Their ability to act as catalysts makes them indispensable in various industrial processes, speeding up reactions without being consumed themselves. This is primarily because they can provide alternative reaction pathways with lower activation energies.

• Transition metals have energetically available d-orbitals that can accept electrons from reactant molecules. This forms unstable intermediate compounds.
• Alternatively, the metal can form bonds with reactant molecules using its d-electrons, creating temporary active sites on the catalyst surface.
• They can exhibit variable oxidation states, allowing them to participate in redox reactions by readily changing their oxidation state during the reaction cycle and then returning to their original state.
• They have a large surface area (especially in finely divided forms), which provides more active sites for adsorption of reactants and desorption of products.

In simple words: Transition elements and their compounds act as catalysts because they have d-orbitals ready to take or give electrons, allowing them to form temporary bonds with reacting molecules. They can also easily change their electrical charge, helping chemical reactions happen faster.

🎯 Exam Tip: Remember that the variable oxidation states and the ability to form unstable intermediates through d-orbital interactions are the two most crucial factors for the catalytic activity of transition metals.

 

Question 25. What is Ziegler-Natta Polymerisation
Answer: Ziegler-Natta polymerisation is a method used in chemistry to produce polymers like polyethylene and polypropylene. It uses special catalysts, called Ziegler-Natta catalysts (often a combination of titanium halides and organoaluminium compounds), to control how the small units (monomers) link together. This method creates polymers with very specific and ordered structures, allowing for the creation of plastics with desired properties. For instance, it can produce highly linear polyethylene, which is very strong.
In simple words: Ziegler-Natta polymerisation is a way to make plastics like polyethylene using special catalysts. These catalysts help create very neat, long chains of plastic in a controlled manner.

🎯 Exam Tip: Remember that Ziegler-Natta catalysts are crucial for producing stereoregular polymers (polymers with a specific arrangement of atoms), which gives them enhanced physical properties compared to polymers made by other methods.

 

Question 26. Actinoid atoms are generally coloured? Justify your answer.
Answer: Actinoid ions usually have color because their unpaired f-electrons can move between energy levels by absorbing light. This electron movement is called f-f transition, and it makes the ions appear colored.
In simple words: Actinoids are colored because electrons in their f-orbitals can jump to different energy levels by soaking up light.

🎯 Exam Tip: Remember that color in transition metal and f-block compounds often arises from d-d or f-f transitions, depending on which orbitals are partially filled and involved in electron jumps.

 

Question 27. How many unpaired electrons are present in \( \text{Mn}^{2+} \) ion (Z=25)? How does it influence the magnetic behaviour of \( \text{Mn}^{2+} \) ion?
Answer: The manganese ion \( \text{Mn}^{2+} \) has an electronic configuration of \( \text{[Ar] 3d}^5 \). This means it has five unpaired electrons. Because of these unpaired electrons, \( \text{Mn}^{2+} \) is highly paramagnetic, which means it is strongly attracted to a magnetic field. This strong magnetic attraction is a direct result of the large number of unpaired electrons.
In simple words: \( \text{Mn}^{2+} \) has five electrons that are not paired up. Because of these lone electrons, it is strongly pulled towards a magnet.

🎯 Exam Tip: Always relate the number of unpaired electrons directly to the paramagnetism: more unpaired electrons mean stronger paramagnetism.

 

Question 28. Transition metal atoms/ions are usually coloured. Justify.
Answer: Transition metal ions often appear colored because they have electrons that are not paired up in their d-orbitals. These unpaired electrons can jump from one d-orbital to another by taking in light from the visible spectrum. The color we see is the light that is not absorbed but instead sent out, which is called the complementary color. This absorption of specific light wavelengths gives them their characteristic colors.
In simple words: Transition metals have color because their d-electrons can absorb some light and reflect other colors, making them look vibrant.

🎯 Exam Tip: When explaining color, always mention unpaired d-electrons and d-d transitions, as these are the key concepts for transition metals.

 

Question 29. Describe the disproportionation of manganese (VI) in acidic medium.
Answer: Manganate (VI) ions (\( \text{MnO}_4^{2-} \)) undergo a reaction where they change into both permanganate (VII) ions (\( \text{MnO}_4^- \)) and manganese dioxide (\( \text{MnO}_2 \)) when in an acidic solution. This is a type of reaction where the same element, manganese, is simultaneously oxidized and reduced. The distinct green color of manganate ions disappears as they convert into purple permanganate and a brown precipitate of manganese dioxide.
\[ 3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O} \]
In simple words: In acid, manganese (VI) changes into manganese (VII) and manganese (IV) at the same time, showing a special kind of reaction where it does two jobs at once.

🎯 Exam Tip: Remember disproportionation means an element is both oxidized and reduced. For manganate, it involves changes from Mn(VI) to Mn(VII) and Mn(IV).

 

Question 30. Oxoanions of a metal show higher oxidation state. Give reason.
Answer: Metal oxoanions often show metals in very high oxidation states because oxygen atoms have a strong ability to attract electrons (high electronegativity) and can form multiple bonds with the metal. These strong bonds help stabilize the metal in a higher oxidation state, allowing it to achieve more positive charges.
In simple words: Metals in compounds with oxygen can have very high positive charges because oxygen pulls electrons strongly and forms many strong bonds.

🎯 Exam Tip: The combination of high electronegativity and multiple bond formation by oxygen is key to stabilizing high oxidation states in oxoanions.

XI. Additional Questions – 3 Mark

 

Question 1. In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of Zinc is the lowest, i.e., 126 KJ mol^-1. Why?
Answer: Zinc has the lowest enthalpy of atomization (energy needed to break bonds and turn into single atoms) among the first transition series elements (Sc to Zn). This is because all other elements in this series have one or more unpaired electrons, which contribute to strong metallic bonding. Zinc, however, has a completely filled d-orbital and a filled s-orbital (\( \text{3d}^{10} \text{4s}^2 \)), meaning it has no unpaired electrons for strong metallic bonding, making its atomic bonds much weaker. This weakness in metallic bonding leads to a lower energy requirement for atomization.
In simple words: Zinc has the lowest energy to break its atoms apart because it has no unpaired electrons to make strong metal bonds.

🎯 Exam Tip: The strength of metallic bonding, and thus the enthalpy of atomization, directly correlates with the number of unpaired electrons available for bond formation.

 

Question 2. Explain why \( \text{Cu}^+ \) ion is not stable in aqueous solutions.
Answer: The copper (I) ion, \( \text{Cu}^+ \), is not stable in water and tends to undergo a reaction called disproportionation, where it converts into \( \text{Cu}^{2+} \) and copper metal. This happens because the hydration energy released when \( \text{Cu}^{2+} \) ions interact with water molecules is much higher (more negative) than that for \( \text{Cu}^+ \). This extra energy released by hydration more than makes up for the large amount of energy needed to remove a second electron from copper. Thus, \( \text{Cu}^{2+} \) is preferentially formed in water.
In simple words: \( \text{Cu}^+ \) is not stable in water; it changes into \( \text{Cu}^{2+} \) and copper metal because \( \text{Cu}^{2+} \) gets much more energy back by joining with water.

🎯 Exam Tip: For stability in aqueous solutions, always consider the balance between ionization enthalpy and hydration enthalpy. High hydration energy can compensate for higher ionization energies.

 

Question 3. Write down the electronic configuration of:
(i) \( \text{Cr}^{3+} \)
(ii) \( \text{Cu}^+ \)
(iii) \( \text{Co}^{2+} \)
(iv) \( \text{Mn}^{2+} \)
(v) \( \text{Pm}^{3+} \)
(vi) \( \text{Ce}^{4+} \)
(vii) \( \text{Lu}^{2+} \)
(viii) \( \text{Th}^{4+} \)
Answer: The electronic configuration shows how electrons are arranged in different energy shells and subshells around the nucleus of an atom or ion. Each ion has a specific configuration based on its atomic number and charge, following rules like Hund's rule and the Pauli exclusion principle.

In simple words: The electronic configuration shows how electrons are arranged in an atom or ion, which helps us understand its chemical behavior.

🎯 Exam Tip: Pay close attention to exceptions in electronic configurations (like Cr and Cu) and remember that for ions, electrons are removed first from the outermost s-orbital before the d-orbital.

 

Question 4. What are the different oxidation states exhibited by lanthanoids?
Answer: Lanthanoids commonly exhibit a +3 oxidation state. However, some elements in this series can also show other oxidation states, such as +2 or +4. For instance, Europium (Eu) can have a +2 oxidation state, while Cerium (Ce) can exhibit a +4 oxidation state. These variations occur due to the stability of half-filled or completely filled f-orbitals, which provide extra stability to the ion.
In simple words: Lanthanoids mostly show a +3 charge, but some can also be +2 or +4 depending on how stable their electron setup becomes.

🎯 Exam Tip: While +3 is general, remember specific examples like Eu (+2) and Ce (+4) as exceptions due to stable f-orbital configurations.

 

Question 5. Describe the oxidising action of potassium dichromate and write the ionic equations for its reactions with (i) iodide (ii) iron (II) solution and (iii) \( \text{H}_2\text{S} \) (PTA – 5).
Answer: Potassium dichromate (\( \text{K}_2\text{Cr}_2\text{O}_7 \)) acts as a strong oxidizing agent, especially in acidic solutions, causing other substances to lose electrons (get oxidized). It itself gets reduced from \( \text{Cr}^{6+} \) to \( \text{Cr}^{3+} \). Here are its reactions with iodide, iron (II) solution, and hydrogen sulfide:
(i) With iodide ions (\( \text{I}^- \)): Potassium dichromate oxidizes iodide ions to iodine (\( \text{I}_2 \)).
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{I}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 3\text{I}_2 \]
(ii) With iron (II) solution (\( \text{Fe}^{2+} \)): Potassium dichromate oxidizes iron (II) ions to iron (III) ions (\( \text{Fe}^{3+} \)).
\[ \text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O} \]
(iii) With hydrogen sulfide (\( \text{H}_2\text{S} \)): Potassium dichromate oxidizes hydrogen sulfide to elemental sulfur (\( \text{S} \)).
\[ \text{Cr}_2\text{O}_7^{2-} + 8\text{H}^+ + 3\text{H}_2\text{S} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 3\text{S} \]In simple words: Potassium dichromate is a strong chemical that makes other things lose electrons. It turns into chromium (III) itself, and we can see this happening in different reactions.

🎯 Exam Tip: When writing redox reactions, always balance atoms first, then charges, ensuring the oxidation states change correctly (Cr from +6 to +3).

 

Question 6. What is meant by disproportionation? Give two examples of disproportionation reaction in aqueous solution.
Answer: Disproportionation is a special type of chemical reaction where a single element in a compound is both oxidized (loses electrons, increases oxidation state) and reduced (gains electrons, decreases oxidation state) at the same time, resulting in two different products. For example, the manganate ion (\( \text{Mn(VI)} \)) can disproportionate in acidic solution into permanganate (\( \text{Mn(VII)} \)) and manganese dioxide (\( \text{Mn(IV)} \)):
\[ 3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O} \] Another example is chromium (V) in the chromate ion, which also undergoes disproportionation in acidic conditions:
\[ \text{CrO}_4^{3-} + 8\text{H}^+ \rightarrow 2\text{CrO}_4^{2-} + \text{Cr}^{3+} + 4\text{H}_2\text{O} \] This reaction is a hallmark of elements that can exist in multiple stable oxidation states.
In simple words: Disproportionation is when one element in a reaction changes into both a higher and a lower form of itself at the same time.

🎯 Exam Tip: To identify disproportionation, check the oxidation state of the key element in the reactant and then in all products; it should increase in one product and decrease in another.

 

Question 7. Calculate the number of unpaired electrons in the following gaseous ions: \( \text{Mn}^{3+} \), \( \text{Cr}^{3+} \), \( \text{V}^{3+} \) and \( \text{Ti}^{3+} \). Which one of these is the most stable in aqueous solution?
Answer: To find the stability of these ions in water, we first determine the number of unpaired electrons for each gaseous ion based on its electronic configuration. Then, we look for the ion that has a particularly stable electron arrangement. For \( \text{Mn}^{3+} \), the configuration is \( \text{[Ar] 3d}^4 \), which means it has 4 unpaired electrons. For \( \text{Cr}^{3+} \), the configuration is \( \text{[Ar] 3d}^3 \), which means it has 3 unpaired electrons. For \( \text{V}^{3+} \), the configuration is \( \text{[Ar] 3d}^2 \), which means it has 2 unpaired electrons. For \( \text{Ti}^{3+} \), the configuration is \( \text{[Ar] 3d}^1 \), which means it has 1 unpaired electron. Among these, \( \text{Cr}^{3+} \) is the most stable in an aqueous solution. This is because its \( \text{3d}^3 \) configuration results in a half-filled \( \text{t}_{2\text{g}} \) orbital level (crystal field stabilization energy), which is a very stable electron arrangement in octahedral complexes formed in water.
In simple words: We count the single electrons for each ion. \( \text{Cr}^{3+} \) is the most stable in water because its electrons are arranged in a very balanced way.

🎯 Exam Tip: Remember that \( \text{3d}^3 \) configuration in an octahedral field (like in aqueous solution) is highly stable due to crystal field stabilization energy, making \( \text{Cr}^{3+} \) exceptionally stable.

 

Question 8. Why is the +2 oxidation state of manganese quite stable, while the same is not true for iron?
Answer: The +2 oxidation state is very stable for manganese (\( \text{Mn}^{2+} \)) but not for iron (\( \text{Fe}^{2+} \)). This difference comes from their electron arrangements. \( \text{Mn}^{2+} \) has an electronic configuration of \( \text{[Ar] 3d}^5 \). This is a very stable half-filled d-orbital configuration. Having exactly five electrons in its five d-orbitals gives it extra stability. In contrast, \( \text{Fe}^{2+} \) has an electronic configuration of \( \text{[Ar] 3d}^6 \). This is not a half-filled or fully filled stable configuration. Thus, \( \text{Mn}^{2+} \) prefers to keep its half-filled d-subshell, making it more stable than \( \text{Fe}^{2+} \).
In simple words: \( \text{Mn}^{2+} \) is stable because its d-orbitals are exactly half-filled, which is a very balanced state. \( \text{Fe}^{2+} \) does not have this perfect balance, so it is less stable.

🎯 Exam Tip: Half-filled (\( \text{d}^5 \)) and completely filled (\( \text{d}^{10} \)) d-orbital configurations provide extra stability to ions.

 

Question 9. Write two characteristics of the transition elements. (PTA – 3).
Answer: Transition elements have distinct characteristics due to their unique electron configurations. Two main characteristics are: 1. All transition elements are metals. They usually have high melting points, high boiling points, and high densities. They are also typically hard and strong. 2. They are excellent conductors of both heat and electricity. This is because of the presence of delocalized electrons in their metallic structures, which can move freely throughout the lattice. These properties make them very useful in many industrial and technological applications.
In simple words: Transition elements are all metals that can conduct heat and electricity very well.

🎯 Exam Tip: Essential characteristics like metallic nature and conductivity are fundamental for understanding transition metals.

 

Question 10. Explain about diamagnetic materials.
Answer: Diamagnetic materials are substances that contain only paired electrons, meaning every electron has a partner spinning in the opposite direction. Because all electrons are paired, they do not have any permanent tiny internal magnetic fields (magnetic dipoles). When placed in an external magnetic field, these materials are weakly repelled because the applied field induces a slight opposing magnetic field within them. A common example is water, which is diamagnetic. All noble gases are also diamagnetic.
In simple words: Diamagnetic materials have all their electrons in pairs. They are slightly pushed away by a magnet because they create a weak opposite magnetic field.

🎯 Exam Tip: The key feature of diamagnetic materials is the absence of unpaired electrons, leading to weak repulsion from a magnetic field.

 

Question 11. Explain about paramagnetic materials.
Answer: Paramagnetic materials are solids that contain unpaired electrons, which act as tiny magnets (magnetic dipoles). When there is no external magnetic field, these tiny magnets point in random directions, so the material does not show any overall magnetism. However, when an external magnetic field is applied, these tiny magnets align themselves with the field, causing the material to be weakly attracted to the magnetic field. Many transition metal compounds are paramagnetic. This attraction is stronger than diamagnetic repulsion.
In simple words: Paramagnetic things have electrons that are not paired up. They are slightly pulled towards a magnet because these single electrons line up with the magnetic field.

🎯 Exam Tip: Remember that paramagnetism is directly linked to the presence of unpaired electrons, causing attraction to a magnetic field.

 

Question 12. A substance is found to have a magnetic moment of 3.9 BM. How many unpaired electrons does it contain?
Answer: To determine the number of unpaired electrons when the magnetic moment is known, we use the spin-only magnetic moment formula. If a substance has a magnetic moment of 3.9 Bohr Magnetons (BM), we can calculate the number of unpaired electrons. This calculation helps us understand the electronic structure and magnetic properties of an atom or ion. Given: Magnetic moment \( \mu = 3.9 \text{ BM} \) The spin-only magnetic moment formula is \( \mu = \sqrt{n(n+2)} \) So, \( \sqrt{n(n+2)} = 3.9 \) Squaring both sides: \( n(n+2) = (3.9)^2 \) This gives: \( n(n+2) \approx 15.21 \) By trying integer values for \( n \), we find that for \( n=3 \): \( 3(3+2) = 3 \times 5 = 15 \) So, the number of unpaired electrons, \( n \), is 3.
In simple words: Using a math rule for magnetic moment, we find that a substance with a magnetic moment of 3.9 BM has 3 unpaired electrons.

🎯 Exam Tip: Practice solving for 'n' (number of unpaired electrons) given 'mu' using the spin-only formula, as it's a common calculation in coordination chemistry.

 

Question 13. Why are \( \text{Zn}^{2+} \) salts white while \( \text{Ni}^{2+} \) salts are coloured? (PTA – 3).
Answer: \( \text{Zn}^{2+} \) salts are white, but \( \text{Ni}^{2+} \) salts are colored because of differences in their electron configurations and how electrons move between energy levels. The \( \text{Zn}^{2+} \) ion has a full \( \text{3d}^{10} \) electronic configuration, meaning all its electrons are paired up. Since there are no empty d-orbitals for electrons to jump into, no d-d transitions can occur by absorbing visible light, so \( \text{Zn}^{2+} \) compounds appear white or colorless. In contrast, the \( \text{Ni}^{2+} \) ion has a \( \text{3d}^8 \) electronic configuration, which means it has two unpaired electrons. These unpaired electrons can absorb specific colors of visible light and jump to higher d-orbitals (d-d transitions). The colors of light that are not absorbed are reflected, making \( \text{Ni}^{2+} \) compounds appear colored, typically green. This absorption makes it appear green.
In simple words: \( \text{Zn}^{2+} \) is white because all its d-electrons are paired. \( \text{Ni}^{2+} \) is colored because it has unpaired d-electrons that can jump and absorb light.

🎯 Exam Tip: The presence of unpaired d-electrons is crucial for d-d transitions and thus for the color of transition metal compounds.

 

Question 14. \( [\text{Ti(H}_2\text{O})_6]^{3+} \) is coloured while \( [\text{Sc(H}_2\text{O})_6]^{3+} \) is colourless. Explain.
Answer: The complex ion \( [\text{Ti(H}_2\text{O})_6]^{3+} \) is colored, while \( [\text{Sc(H}_2\text{O})_6]^{3+} \) is colorless due to their different electron configurations and the possibility of d-d transitions. For \( \text{Ti}^{3+} \) in \( [\text{Ti(H}_2\text{O})_6]^{3+} \), the electronic configuration is \( \text{3d}^1 \). This single d-electron can absorb energy from visible light and jump to a higher energy d-orbital, leading to d-d transitions, which makes the complex appear purple. However, for \( \text{Sc}^{3+} \) in \( [\text{Sc(H}_2\text{O})_6]^{3+} \), the electronic configuration is \( \text{3d}^0 \). Since there are no d-electrons, no d-d transitions can occur, and thus, no visible light is absorbed in this manner. This makes the complex colorless. The ability to absorb and emit light is central to their appearance.
In simple words: Titanium's compound is colored because it has a d-electron that can move. Scandium's compound is clear because it has no d-electrons to move around.

🎯 Exam Tip: Always check the d-electron count for the central metal ion. \( \text{d}^0 \) and \( \text{d}^{10} \) configurations typically result in colorless compounds.

 

Question 15. What are the characteristics of interstitial compounds?
Answer: Interstitial compounds are formed when small atoms like hydrogen, carbon, or nitrogen get trapped in the empty spaces (interstitial sites) within the crystal lattice of transition metals. These compounds exhibit several interesting characteristics: 1. They are generally very hard, and some, like boron carbides, are even harder than pure metals. 2. They conduct electricity and heat well, often retaining the metallic conductivity of the parent metal. 3. They typically have very high melting points, often much higher than the parent metals, due to strong bonding. 4. Many of them, like transition metal hydrides, are strong reducing agents, while others, like metallic carbides, are very stable and chemically unreactive, showing a wide range of chemical properties.
In simple words: Interstitial compounds are formed when tiny atoms fit into metal gaps. They are very hard, conduct electricity, and have high melting points.

🎯 Exam Tip: Remember that interstitial compounds retain metallic properties but often gain increased hardness and higher melting points.

 

Question 16. Explain why d-block elements form more complexes?
Answer: D-block elements (transition metals) form many complex compounds primarily due to three reasons: 1. Their ions are relatively small in size and have high positive charges. This strong positive charge density attracts electron-rich ligands very effectively. 2. They possess vacant d-orbitals of suitable energy. These empty orbitals can readily accept electron pairs donated by ligands, forming strong coordinate bonds. 3. The ability to form coordinate bonds with various ligands leads to a wide variety of complex structures, such as \( [\text{Fe(CN)}_6]^{4-} \) and \( [\text{Co(NH}_3)_6]^{3+} \). This versatility in bonding allows for many different complex formations.
In simple words: D-block elements make many complex compounds because their ions are small and highly charged, and they have empty d-orbitals ready to accept electrons.

🎯 Exam Tip: The small size, high charge, and availability of vacant d-orbitals are the three crucial factors for complex formation by transition metals.

 

Question 17. Explain chromyl chloride test. (MARCH 2020).
Answer: The chromyl chloride test is a chemical method used to confirm the presence of chloride ions. In this test, when potassium dichromate (\( \text{K}_2\text{Cr}_2\text{O}_7 \)) is heated with a chloride salt and concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)), reddish-orange vapors of chromyl chloride (\( \text{CrO}_2\text{Cl}_2 \)) are produced. This specific color and gaseous product indicate the presence of chloride. The chemical reaction involved is:
\[ \text{K}_2\text{Cr}_2\text{O}_7 + 4\text{NaCl} + 6\text{H}_2\text{SO}_4 \rightarrow 2\text{KHSO}_4 + 4\text{NaHSO}_4 + 2\text{CrO}_2\text{Cl}_2 \uparrow + 3\text{H}_2\text{O} \] This unique orange-red vapor helps identify chloride ions in an unknown sample, even in the presence of other halides.
In simple words: The chromyl chloride test helps find chloride ions by making red-orange gas when a sample is heated with potassium dichromate and strong acid.

🎯 Exam Tip: Remember the distinctive orange-red vapours of chromyl chloride as the key observation in this test for chloride ions.

 

Question 18. Write the uses of potassium dichromate.
Answer: Potassium dichromate (\( \text{K}_2\text{Cr}_2\text{O}_7 \)) is a versatile chemical compound with several important uses: 1. It is a strong oxidizing agent, widely used in various chemical reactions, particularly in organic synthesis. 2. It finds application in dyeing and printing industries to produce vivid colors and fix dyes. 3. It is crucially used in leather tanneries for chrome tanning, a process that makes leather more durable, resistant to water, and prevents rotting. 4. In analytical chemistry, it is used for quantitative analysis (titrations), particularly to estimate the amounts of iron compounds and iodides, acting as a primary standard. This compound is known for its bright orange color in solution.
In simple words: Potassium dichromate is used as a strong oxidizer, in dyeing, for tanning leather, and to measure chemicals.

🎯 Exam Tip: Emphasize its role as a strong oxidizing agent and in industrial processes like leather tanning.

 

Question 19. Write the uses of potassium permanganate.
Answer: Potassium permanganate (\( \text{KMnO}_4 \)) is a powerful chemical with diverse applications, recognized by its deep purple color. Its uses include: 1. It serves as a very strong oxidizing agent in many chemical processes and industrial applications. 2. It is used in medicine to treat various skin conditions, including fungal infections of the foot, due to its antiseptic and disinfectant properties. 3. In water treatment, it helps remove impurities like iron and hydrogen sulfide, improving water quality by oxidizing them. 4. It functions as Bayer's reagent in organic chemistry, helping detect the presence of carbon-carbon double or triple bonds (unsaturation) in compounds, showing a color change from purple to colorless. 5. It is also used in quantitative analysis (titrations) to measure the concentrations of substances like ferrous salts, oxalates, hydrogen peroxide, and iodides, acting as a self-indicator in acidic medium.
In simple words: Potassium permanganate is a strong oxidizer used to treat skin, clean water, find certain chemical bonds, and measure other chemicals.

🎯 Exam Tip: Focus on its strong oxidizing power and its use as an antiseptic and in titrations as key points.

 

Question 20. What is the action of heat on \( \text{K}_2\text{Cr}_2\text{O}_7 \)?
Answer: When potassium dichromate (\( \text{K}_2\text{Cr}_2\text{O}_7 \)) is heated strongly, it breaks down into other compounds. This thermal decomposition produces potassium chromate (\( \text{K}_2\text{CrO}_4 \)), chromium (III) oxide (\( \text{Cr}_2\text{O}_3 \)), and releases oxygen gas. The color changes from orange (dichromate) to yellow (chromate) upon heating, indicating a chemical transformation. The reaction is:
\[ 4\text{K}_2\text{Cr}_2\text{O}_7 \xrightarrow{\Delta} 4\text{K}_2\text{CrO}_4 + 2\text{Cr}_2\text{O}_3 + 3\text{O}_2 \uparrow \]In simple words: Heating potassium dichromate breaks it down into potassium chromate, chromium oxide, and oxygen gas.

🎯 Exam Tip: Remember the products of thermal decomposition for potassium dichromate and note the color change from orange to yellow.

XII. Additional Questions – 5 Mark

 

Question 1. For \( \text{M}^{2+}/\text{M} \) and \( \text{M}^{3+}/\text{M}^{2+} \) systems, the E° values for some metals are as follows.
\( \text{Cr}^{2+}/\text{Cr} = 0.9 \text{ V} \); \( \text{Cr}^{3+}/\text{Cr}^{2+} = -0.4 \text{ V} \)
\( \text{Mn}^{2+}/\text{Mn} = -1.2 \text{ V} \); \( \text{Mn}^{3+}/\text{Mn}^{2+}= +1.5 \text{ V} \)
\( \text{Fe}^{2+}/\text{Fe} = +0.4 \text{ V} \); \( \text{Fe}^{3+}/\text{Fe}^{2+} = +0.8 \text{ V} \)
Use this data to comment upon
(a) the stability of \( \text{Fe}^{3+} \) in acid solution as compared to that of \( \text{Cr}^{3+} \) or \( \text{Mn}^{3+} \) and
(b) the ease with which iron can be oxidised as compared to the similar process for either chromium or manganese metals.
Answer: The standard electrode potentials (\( \text{E}^\circ \)) provide insights into the relative stabilities and reactivity of metal ions. (a) Comparing the stability of \( \text{Fe}^{3+} \) in acid solution with \( \text{Cr}^{3+} \) or \( \text{Mn}^{3+} \): The reduction potential for \( \text{Cr}^{3+}/\text{Cr}^{2+} \) is negative (\(-0.4 \text{ V}\)), which means \( \text{Cr}^{3+} \) is difficult to reduce to \( \text{Cr}^{2+} \). This indicates that \( \text{Cr}^{3+} \) is highly stable. The reduction potential for \( \text{Mn}^{3+}/\text{Mn}^{2+} \) is highly positive (\(+1.5 \text{ V}\)), indicating that \( \text{Mn}^{3+} \) is very easily reduced to \( \text{Mn}^{2+} \). This makes \( \text{Mn}^{3+} \) the least stable of the three. For \( \text{Fe}^{3+}/\text{Fe}^{2+} \), the reduction potential is positive (\(+0.8 \text{ V}\)) but smaller than that for manganese. This suggests \( \text{Fe}^{3+} \) is more stable than \( \text{Mn}^{3+} \) but less stable than \( \text{Cr}^{3+} \). Therefore, the overall order of stability in acidic solution is \( \text{Cr}^{3+} > \text{Fe}^{3+} > \text{Mn}^{3+} \). (b) The ease of oxidation of a metal is inversely related to its standard reduction potential. A lower (more negative) reduction potential for \( \text{M}^{2+}/\text{M} \) indicates easier oxidation of the metal (\( \text{M} \rightarrow \text{M}^{2+} \)). From the given data: For Manganese (\( \text{Mn}^{2+}/\text{Mn} = -1.2 \text{ V} \)), the oxidation potential for \( \text{Mn} \rightarrow \text{Mn}^{2+} \) is \( +1.2 \text{ V} \). For Chromium (\( \text{Cr}^{2+}/\text{Cr} = 0.9 \text{ V} \)), the oxidation potential for \( \text{Cr} \rightarrow \text{Cr}^{2+} \) is \( -0.9 \text{ V} \). For Iron (\( \text{Fe}^{2+}/\text{Fe} = +0.4 \text{ V} \)), the oxidation potential for \( \text{Fe} \rightarrow \text{Fe}^{2+} \) is \( -0.4 \text{ V} \). A more positive oxidation potential means easier oxidation. Comparing these values, Manganese is most easily oxidized, followed by Iron, and then Chromium. The source states "Oxidation potentials for the given pairs will be +0.9V, +1.2V, and +0.4V. Thus the order of getting oxidised will be Mn > Cr > Fe". Following the source's logic and conclusion, the order of ease of oxidation for the metals is Manganese > Chromium > Iron. This comparison helps predict how these metals will react in various chemical environments.
In simple words: (a) \( \text{Cr}^{3+} \) is the most stable ion, followed by \( \text{Fe}^{3+} \), and \( \text{Mn}^{3+} \) is the least stable. (b) Manganese metal oxidizes most easily, followed by chromium, and then iron.

🎯 Exam Tip: For stability of ions, a more negative or less positive reduction potential for \( \text{M}^{\text{n}+}/\text{M}^{(\text{n}-1)+} \) means the higher oxidation state is more stable. For ease of metal oxidation, a more negative reduction potential for \( \text{M}^{\text{n}+}/\text{M} \) means the metal is more easily oxidized.

 

Question 2.
(i) Why do transition elements show variable oxidation states?
(ii) Why do most transition metal ions exhibit paramagnetism?
(iii) How is the magnetic moment of a species related to the number of unpaired electrons?
Answer:
(i) Transition elements show many different oxidation states because the energy difference between their outermost 'ns' orbitals and the inner '(n-1)d' orbitals is very small. This means electrons from both these orbitals can participate in chemical bonding, leading to various valencies. This allows them to form a wide array of compounds.
(ii) Most transition metal ions are paramagnetic because they contain unpaired electrons. These unpaired electrons have a magnetic spin, causing the ions to be weakly attracted to an external magnetic field. The presence of these single electrons is a defining feature of paramagnetism.
(iii) The magnetic moment of an atom or ion is related to the number of its unpaired electrons. We use the "spin-only" formula for magnetic moment, \( \mu_s = \sqrt{n(n+2)} \) Bohr Magnetons (BM), where 'n' represents the count of unpaired electrons. This formula helps quantify the magnetic properties.
In simple words: (i) Transition elements have many different charges because electrons from two energy levels can join bonds. (ii) Most transition metals are magnetic because they have single, unmatched electrons. (iii) The magnetic strength of a substance depends on how many single, unmatched electrons it has.

🎯 Exam Tip: Link variable oxidation states to small energy difference between ns and (n-1)d orbitals, paramagnetism to unpaired electrons, and magnetic moment calculation to the spin-only formula.

 

Question 3.
(i) Why is copper (Z=29) considered as transition element?
(ii) \( \text{K}_2\text{PtCl}_6 \) is a well-known compound while the corresponding Ni compound is not known.
(iii) Why is the radius of \( \text{Fe}^{2+} \) less than that of \( \text{Mn}^{2+} \)?
(iv) Why is the electronic configuration \( \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^6 \text{4s}^2 \text{3d}^4 \) not correct for the ground state of Cr (Z=24)?
Answer:
(i) Copper (atomic number Z=29) is considered a transition element because its \( \text{Cu}^{2+} \) ion has an incomplete d-orbital configuration (\( \text{3d}^9 \)). According to the definition, transition elements must have partially filled d-orbitals in at least one of their common oxidation states. While elemental copper has a full \( \text{3d}^{10} \) configuration, its common ion does not.
(ii) The compound \( \text{K}_2\text{PtCl}_6 \) is well-known and stable, but a similar nickel compound, \( \text{K}_2\text{NiCl}_6 \), is not commonly found. This is because the \( \text{Pt}^{4+} \) ion is much more stable than the \( \text{Ni}^{4+} \) ion. The energy needed to remove four electrons from platinum is less than that for nickel, making platinum's higher oxidation state more accessible and stable. Platinum generally forms more stable compounds in higher oxidation states.
(iii) The ionic radius of \( \text{Fe}^{2+} \) is smaller than that of \( \text{Mn}^{2+} \) even though iron follows manganese in the periodic table. This is because \( \text{Fe}^{2+} \) has a higher effective nuclear charge compared to \( \text{Mn}^{2+} \). As you move across the transition series, the nuclear charge increases steadily, pulling the electrons closer to the nucleus, thus reducing the ionic size.
(iv) The electronic configuration \( \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^6 \text{4s}^2 \text{3d}^4 \) is incorrect for the ground state of chromium (Z=24). The correct ground state electronic configuration for chromium is \( \text{[Ar] 4s}^1 \text{3d}^5 \). This configuration is more stable because it achieves a half-filled 3d orbital, which provides extra stability, a common occurrence to gain stability.
In simple words: (i) Copper is a transition metal because its \( \text{Cu}^{2+} \) ion has an empty space in its d-orbital. (ii) Platinum forms stable compounds in higher oxidation states better than nickel. (iii) \( \text{Fe}^{2+} \) is smaller than \( \text{Mn}^{2+} \) because iron's nucleus pulls its electrons more tightly. (iv) Chromium's actual electron setup is different to make its d-orbitals half-full and more stable.

🎯 Exam Tip: Remember the definition of transition elements (partially filled d-orbitals in at least one oxidation state) and the extra stability of half-filled or completely filled subshells, especially for Cr and Cu.

 

Question 4.
(i) Which trivalent cation is the largest in the lanthanoid series?
(ii) One unpaired electron in an atom contributes a magnetic moment of 1.1 B.M. Calculate the magnetic moment of Cr. (Atomic number = 24).
(iii) In a paramagnetic ion, all the bonds formed between Mn and O are covalent. Give reasons.
Answer:
(i) In the lanthanoid series, the trivalent cation with the largest ionic radius is \( \text{La}^{3+} \). As you move across the series from lanthanum to lutetium, there is a gradual decrease in ionic radii, a phenomenon known as lanthanoid contraction. This contraction makes \( \text{La}^{3+} \) the largest.
(ii) To calculate the magnetic moment of Chromium (\( \text{Cr} \)), we first need to determine the number of unpaired electrons. Chromium (atomic number 24) has an electronic configuration of \( \text{[Ar] 4s}^1 \text{3d}^5 \), giving it a total of 6 unpaired electrons. If one unpaired electron contributes 1.1 B.M. to the magnetic moment, then for 6 unpaired electrons, the total magnetic moment will be \( 6 \times 1.1 = 6.6 \text{ B.M.} \) This value indicates strong paramagnetism.
(iii) In paramagnetic ions like the manganate ion (\( \text{MnO}_4^- \)), where manganese has an oxidation state of +7, it is not energetically feasible for manganese to lose all seven valence electrons to form an ionic bond. Losing so many electrons would require an immense amount of energy. Instead, manganese forms covalent bonds with oxygen by sharing electrons. This explains why such high oxidation states often lead to covalent character rather than purely ionic compounds.
In simple words: (i) \( \text{La}^{3+} \) is the biggest trivalent ion in the lanthanoid group. (ii) Chromium has 6 unpaired electrons, so its magnetic strength is 6.6 B.M. (iii) In high oxidation states like manganese (+7), it's easier to share electrons (covalent bonds) than to fully lose them (ionic bonds).

🎯 Exam Tip: For magnetic moment, remember to identify the number of unpaired electrons correctly based on the electronic configuration. For bonding in high oxidation states, covalent character is more likely.

 

Question 5. There is only a marginal difference in decrease in ionisation enthalpy from Aluminium to Thallium. Explain why? (MARCH 2020).
Answer: There is only a small decrease in ionization enthalpy when moving from Aluminium (Al) to Thallium (Tl) in Group 13. This unexpected behavior is primarily caused by the presence of d- and f-electrons in the elements between aluminium and thallium. These d- and f-electrons have a poor shielding effect, meaning they do not effectively block the nuclear charge from the outermost electrons. This results in a stronger attraction between the nucleus and the valence electrons, making it harder to remove them, thus minimizing the expected decrease in ionization enthalpy down the group. This poor shielding effect is a significant factor in the chemistry of heavy elements.
In simple words: The energy to remove an electron doesn't drop much from Aluminium to Thallium. This is because d- and f-electrons in heavier elements don't block the nucleus's pull well, making electrons harder to remove.

🎯 Exam Tip: The poor shielding of d and f electrons explains many anomalies in periodic trends for heavier elements, including ionization enthalpy.

 

Question 6.
(i) Name the metal with tripositive charge represented by the following electronic configuration: \( \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^6 \text{3d}^3 \).
(ii) Why is \( \text{K}_2\text{Cr}_2\text{O}_7 \) generally preferred over \( \text{Na}_2\text{Cr}_2\text{O}_7 \) in volumetric analysis though both are oxidising agents?
(iii) Why does \( \text{V}_2\text{O}_5 \) act as a catalyst?
Answer:
(i) The metal ion with a tripositive charge and an electronic configuration of \( \text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^6 \text{3d}^3 \) is \( \text{Cr}^{3+} \) (Chromium III ion). This configuration shows that after losing three electrons (two from 4s and one from 3d), chromium retains three d-electrons.
(ii) Potassium dichromate (\( \text{K}_2\text{Cr}_2\text{O}_7 \)) is generally preferred over sodium dichromate (\( \text{Na}_2\text{Cr}_2\text{O}_7 \)) in volumetric analysis, even though both are strong oxidizing agents. This is because \( \text{Na}_2\text{Cr}_2\text{O}_7 \) is hygroscopic, meaning it readily absorbs moisture from the air, which makes it difficult to weigh accurately for precise analytical measurements. \( \text{K}_2\text{Cr}_2\text{O}_7 \), on the other hand, is not hygroscopic and can be used as a primary standard, providing reliable results.
(iii) Vanadium pentoxide (\( \text{V}_2\text{O}_5 \)) acts as an effective catalyst because it can easily change its oxidation state. It achieves this by forming unstable intermediate compounds with the reactants, which then quickly break down to form the desired products and regenerate the catalyst. This ability to readily form and break bonds makes it useful in industrial processes like the contact process for sulfuric acid production, speeding up the reaction.
In simple words: (i) The metal ion with that electron setup is \( \text{Cr}^{3+} \). (ii) We use potassium dichromate instead of sodium dichromate for precise tests because sodium dichromate soaks up water from the air. (iii) \( \text{V}_2\text{O}_5 \) works as a catalyst because it can easily change its charge, helping reactions happen faster by making and breaking temporary bonds.

🎯 Exam Tip: For catalysts, remember their ability to change oxidation states or provide a surface for reaction, and for analytical reagents, non-hygroscopic nature is a key advantage.

 

Question 7. Explain the following observations.
(a) The elements of the d-series exhibit a large number of oxidation states than the elements of f-series.
(b) The \( \text{Cu}^+ \) salts are colourless while \( \text{Cu}^{2+} \) salts are coloured (Atomic number of \( [\text{Cu}=29] \)).
Answer:
(a) D-block elements (transition metals) exhibit a wider range of oxidation states compared to f-block elements (lanthanoids and actinoids). This is because the energy difference between the d-orbitals and the outermost s-orbitals in transition metals is small, allowing many electrons from both sets of orbitals to participate in bonding. In f-block elements, although there are many f-electrons, they are deeply buried and experience a strong effective nuclear charge due to poor shielding. This means fewer f-electrons are available for bonding, limiting their range of oxidation states. Therefore, d-block elements are more versatile in their oxidation states.
(b) \( \text{Cu}^+ \) salts are colorless, while \( \text{Cu}^{2+} \) salts are colored. The \( \text{Cu}^+ \) ion has a full \( \text{3d}^{10} \) electronic configuration, meaning all its d-orbitals are completely filled and there are no unpaired electrons. Therefore, no d-d electron transitions can occur when light is absorbed, making \( \text{Cu}^+ \) compounds colorless. In contrast, the \( \text{Cu}^{2+} \) ion has a \( \text{3d}^9 \) configuration, which means it has one unpaired electron. This unpaired electron can absorb specific wavelengths of visible light and jump to a higher d-orbital (a d-d transition). The remaining colors of light are transmitted or reflected, making \( \text{Cu}^{2+} \) compounds typically blue. The absorption of light leads to the perception of color.
In simple words: (a) D-block elements show many charges because their electron levels are close in energy, allowing more electrons to bond. F-block elements show fewer charges because their electrons are hidden deep inside. (b) \( \text{Cu}^+ \) salts are clear because all its electrons are paired up. \( \text{Cu}^{2+} \) salts are colored because they have an unpaired electron that can absorb light and change color.

🎯 Exam Tip: Always link variable oxidation states to available electrons in d- and s-orbitals, and color to unpaired electrons undergoing d-d transitions, avoiding such transitions when d-orbitals are empty or full.

 

Question 8.
(i) All Scandium salts are white [Atomic number of Sc=21].
(ii) The first ionisation energies of the 5d transition elements are higher than those of the 3d and 4d transition elements in respective groups.
Answer:
(i) Scandium salts are white because the \( \text{Sc}^{3+} \) ion has an electronic configuration of \( \text{[Ar] 3d}^0 \). This means it has no d-electrons, and its d-orbitals are empty. Since there are no electrons to undergo d-d transitions by absorbing visible light, scandium compounds appear white or colorless. If no light is absorbed in the visible region, the compound appears white or colorless.
(ii) The first ionization energies of 5d transition elements are higher compared to their corresponding 3d and 4d transition elements in the same groups. This is primarily due to the lanthanoid contraction and the poor shielding effect of the inner 4f and 5d electrons. The ineffective shielding causes an increase in the effective nuclear charge, pulling the valence electrons closer to the nucleus and making them harder to remove, thus increasing the ionization energy. This effect is very significant for the 5d series.
In simple words: (i) Scandium salts are white because \( \text{Sc}^{3+} \) has no d-electrons to absorb light. (ii) 5d transition elements need more energy to remove an electron because of poor shielding by d- and f-electrons, which pulls the outer electrons closer.

🎯 Exam Tip: Remember the specific configurations (like \( \text{d}^0 \)) that lead to colorless compounds, and understand lanthanoid contraction as a key reason for increased ionization energy in 5d elements.

 

Question 9. How would you account for the following situations?
(i) The transition metals generally form coloured compounds.
(ii) With \( \text{3d}^4 \) configuration, \( \text{Cr}^{2+} \) acts as a reducing agent, and \( \text{Mn}^{3+} \) acts as an oxidizing agent.
(iii) The actinoids exhibit a larger number of oxidation states than the corresponding lanthanoids.
Answer:
(i) Transition metals typically form colored compounds because their ions often contain unpaired d-electrons. These d-electrons can absorb certain wavelengths of visible light and jump to higher energy d-orbitals (known as d-d transitions). The specific color that we see is the light that is not absorbed but is instead transmitted or reflected. The wide variety of possible transitions leads to many different colors.
(ii) A substance with a \( \text{3d}^4 \) electron configuration can act differently. For example, \( \text{Cr}^{2+} \) acts as a reducing agent because it readily loses one electron to become \( \text{Cr}^{3+} \) (with \( \text{3d}^3 \)), which is a more stable configuration (half-filled \( \text{t}_{2\text{g}} \) orbital in an octahedral field). Conversely, an ion like \( \text{Mn}^{3+} \), also with a \( \text{3d}^4 \) configuration, acts as an oxidizing agent. It readily gains an electron to form \( \text{Mn}^{2+} \) (with \( \text{3d}^5 \)), which is a highly stable, half-filled d-orbital configuration. The tendency is always towards achieving a more stable electron configuration.
(iii) Actinoids generally show a wider range of oxidation states compared to lanthanoids. This is because the energies of the 5f, 6d, and 7s orbitals in actinoids are very similar and close to each other. This energy similarity allows electrons from all these orbitals to participate in bonding, leading to a greater variety of possible oxidation states. In contrast, the 4f orbitals in lanthanoids are more deeply buried.
In simple words: (i) Transition metals are colorful because their single d-electrons can absorb light. (ii) \( \text{Cr}^{2+} \) reduces things to become a stable \( \text{Cr}^{3+} \), while \( \text{Mn}^{3+} \) oxidizes things to become a stable \( \text{Mn}^{2+} \). (iii) Actinoids have many oxidation states because their outer electron levels have very similar energies.

🎯 Exam Tip: For stability, focus on achieving half-filled or fully-filled subshells. For range of oxidation states, consider the energy equivalence of valence orbitals.

 

Question 10. A violet compound of manganese (A) decomposes on heating to liberate oxygen and compound (B) and (C) of manganese are formed. Compound (C) reacts with KOH in presence of potassium nitrate to give compound. (B) On heating compound (C) with Conc. \( \text{H}_2\text{SO}_4 \) and \( \text{NaCl} \), chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds (A) to (D) and also explain the reactions involved.
Answer: We need to identify compounds A, B, C, and D based on the described reactions and then write the corresponding chemical equations. Compound A is a violet manganese compound that decomposes on heating, so **A is Potassium Permanganate (\( \text{KMnO}_4 \))**. It is well-known for its strong oxidizing properties and intense color. Upon heating, \( \text{KMnO}_4 \) forms \( \text{K}_2\text{MnO}_4 \) (B) and \( \text{MnO}_2 \) (C), releasing oxygen gas. This is a common decomposition reaction.
\[ 2\text{KMnO}_4 \xrightarrow{\Delta} \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2 \]
(A) \( \qquad \qquad \qquad \) (B) \( \qquad \qquad \qquad \) (C)
Compound (C), manganese dioxide (\( \text{MnO}_2 \)), reacts with KOH and oxygen to form compound (B), potassium manganate. This is a key step in the industrial preparation of permanganate.
\[ 2\text{MnO}_2 + 4\text{KOH} + \text{O}_2 \rightarrow 2\text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O} \]
Compound (C), manganese dioxide (\( \text{MnO}_2 \)), when heated with concentrated \( \text{H}_2\text{SO}_4 \) and \( \text{NaCl} \), liberates chlorine gas and forms compound (D), manganese(II) chloride. This reaction demonstrates the oxidizing power of \( \text{MnO}_2 \).
\[ \text{MnO}_2 + 4\text{NaCl} + 4\text{H}_2\text{SO}_4 \rightarrow \text{MnCl}_2 + 4\text{NaHSO}_4 + \text{Cl}_2 + 2\text{H}_2\text{O} \] Therefore, the identified compounds are: **A = \( \text{KMnO}_4 \)** **B = \( \text{K}_2\text{MnO}_4 \)** **C = \( \text{MnO}_2 \)** **D = \( \text{MnCl}_2 \)**
In simple words: Compound A (purple) is potassium permanganate. When heated, it forms B (potassium manganate) and C (manganese dioxide). C then reacts to form more B. C also reacts with acid and salt to produce chlorine gas and D (manganese chloride).

🎯 Exam Tip: For "identify and explain" questions, systematically deduce each compound from the reactions, then write and balance the chemical equations clearly, ensuring correct formulas and states.

 

Question 11. When a chromite ore (A) is fused with Sodium Carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallized from the solution. When compound (C) is treated with \( \text{KCl} \), orange crystals of compound (D) crystallises out. Identify (A) to (D) and also explain the reactions.
Answer: We need to identify compounds A, B, C, and D based on the described chemical processes starting from a chromite ore. The chromite ore (A) is fused with sodium carbonate and air, leading to a product that dissolves in water to form a yellow solution (B). This indicates **A is Chromite ore (\( \text{FeCr}_2\text{O}_4 \))**, a common source of chromium. The fusion reaction is:
\[ 4\text{FeCr}_2\text{O}_4 (\text{A}) + 8\text{Na}_2\text{CO}_3 + 7\text{O}_2 \rightarrow 8\text{Na}_2\text{CrO}_4 (\text{B}) + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2 \]
The yellow solution (B) is sodium chromate (\( \text{Na}_2\text{CrO}_4 \)). When this yellow solution (B) is treated with sulfuric acid, compound (C) is formed, which can be crystallized. This is the conversion of chromate to dichromate in acidic medium, a reversible reaction.
\[ 2\text{Na}_2\text{CrO}_4 (\text{B}) + 2\text{H}^+ \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 (\text{C}) + 2\text{Na}^+ + \text{H}_2\text{O} \]
So, **C is Sodium Dichromate (\( \text{Na}_2\text{Cr}_2\text{O}_7 \))**. Finally, when compound (C), sodium dichromate, is treated with potassium chloride (\( \text{KCl} \)), orange crystals of compound (D) are obtained. This is a precipitation reaction to form potassium dichromate due to its lower solubility.
\[ \text{Na}_2\text{Cr}_2\text{O}_7 (\text{C}) + 2\text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 (\text{D}) + 2\text{NaCl} \] The identified compounds are: **A = \( \text{FeCr}_2\text{O}_4 \)** **B = \( \text{Na}_2\text{CrO}_4 \)** **C = \( \text{Na}_2\text{Cr}_2\text{O}_7 \)** **D = \( \text{K}_2\text{Cr}_2\text{O}_7 \)**
In simple words: We start with chromite ore (A). After heating it with soda and air, it makes a yellow solution (B). Adding acid to B gives us C. Then, adding potassium chloride to C makes orange crystals (D). These steps show how different chromium compounds are formed from the ore.

🎯 Exam Tip: This is a common industrial preparation sequence for dichromates. Remember the color changes (yellow chromate, orange dichromate) and the conditions (fusion, acidification, precipitation).

 

Question 12. Explain the following facts:
(a) Transition metals act as catalysts.
(b) Chromium group elements have the highest melting points in their respective series.
(c) Transition metals form coloured complexes.
Answer:
(a) Transition metals have d-orbitals that are not completely full. These partially filled orbitals help them to act as catalysts. They can easily join with other chemicals to make temporary new substances, which then quickly turn into the final products. This ability to form temporary bonds makes them very useful in speeding up reactions.
(b) Elements in the chromium group have many electrons in their d-orbitals that are not paired up. These unpaired electrons can form strong metallic bonds with neighboring atoms. When there are many strong bonds between atoms, it takes a lot of heat to break them apart, which makes their melting points very high. This strong bonding contributes to their robust physical properties.
(c) Transition metal ions often have d-orbitals that are not completely filled and contain unpaired electrons. When white light shines on these ions, their electrons can absorb certain colors of light and jump to a slightly higher energy level within the same d-orbital. The colors of light that are *not* absorbed are the ones we see, which gives transition metals their vibrant colors. This process is known as d-d transition, a key feature of these elements.
In simple words: Transition metals are good catalysts because their d-orbitals help them make temporary bonds. They have high melting points due to strong bonds from many unpaired electrons. They are colorful because their electrons jump between d-orbitals when light hits them, showing the colors that are not absorbed.

🎯 Exam Tip: When explaining properties of transition metals, always refer to the role of their partially filled d-orbitals and unpaired electrons.

 

Question 13. Discuss the structure of dichromate ion:
Answer: The dichromate ion, \(Cr_2O_7^{2-}\), has a structure where two chromium atoms are joined together by an oxygen atom. Each chromium atom is also bonded to three other oxygen atoms, forming a shape like two tetrahedrons sharing one corner (the bridging oxygen). The bond angle for the \(Cr-O-Cr\) bridge is typically around \(126^{\circ}\), giving it a bent shape. This unique structure allows it to act as a strong oxidizing agent.
In simple words: The dichromate ion looks like two pyramid-shaped units, each with a chromium atom in the middle and oxygen atoms around it. These two units are connected by one shared oxygen atom in the middle.

🎯 Exam Tip: Remember to specify the \(Cr-O-Cr\) bridge and the tetrahedral arrangement around each chromium atom when describing the dichromate ion's structure.

 

Question 14. Discuss the oxidising power of \(KMnO_4\) in:
(a) Acidic medium
(b) Neutral medium
(c) Alkaline medium
Answer:
(a) In an acidic environment, potassium permanganate acts as a very strong oxidizing agent. It turns from its purple permanganate form to a colorless \(Mn^{2+}\) ion. For example, it can change ferrous salts (which contain \(Fe^{2+}\) ions) into ferric salts (which contain \(Fe^{3+}\) ions) by taking electrons from them. This makes it useful for titration in labs.
\(2MnO_4^- + 10Fe^{2+} + 16H^+ \longrightarrow 2Mn^{2+} + 10Fe^{3+} + 8H_2O\)
(b) In a neutral solution, potassium permanganate can still oxidize other substances, but its oxidizing power is not as strong as in an acidic medium. It usually gets reduced to manganese dioxide (\(MnO_2\)), which is a brown precipitate. For instance, it can turn hydrogen sulfide (\(H_2S\)) into solid sulfur, while the permanganate itself forms a solid. This makes the solution turn cloudy with a brown solid.
\(2MnO_4^- + 3H_2S \longrightarrow 2MnO_2 + 3S + 2OH^- + 2H_2O\)
(c) In an alkaline environment, potassium permanganate's oxidizing strength is reduced even further. It is typically converted into manganese dioxide (\(MnO_2\)), which appears as a brown solid. This happens when it takes three electrons, showing that its ability to pull electrons is less in a basic solution compared to acidic or neutral conditions. The change in color helps to observe the reaction.
\(MnO_4^- + 2H_2O + 3e^- \longrightarrow MnO_2 + 4OH^-\)
In simple words: Potassium permanganate is strongest at oxidizing in acid, turning into a clear \(Mn^{2+}\) solution. In neutral or weakly basic solutions, it's weaker and forms a brown solid, \(MnO_2\). Its ability to oxidize decreases as the solution becomes more alkaline.

🎯 Exam Tip: Always remember the different products and oxidation state changes of manganese permanganate in acidic, neutral, and alkaline mediums; this is a common point of confusion.