CBSE Class 12 Chemistry The d and f Block Elements Notes and Questions

Important Questions for NCERT Class 12 Chemistry The d and f Block Elements

Question. General electronic configuration of lanthanides is
(a) (n – 2) f^{1 – 14} (n – 1) s²p⁶d^{0 – 1} ns²
(b) (n – 2) f^{10 – 14} (n – 1) d^{0 – 1} ns²
(c) (n – 2) f ^{0 – 14} (n – 1) d¹⁰ ns²
(d) (n – 2) d^{0 – 1} (n – 1) f^{1 – 14} ns² 
Answer    B

Question. Which of the following statement is not correct?
(a) La(OH)₃ is less basic than Lu(OH)₃.
(b) In lanthanide series ionic radius of Ln⁺³ ion decreases.
(c) La is actually an element of transition series rather lanthanides.
(d) Atomic radius of Zn and Hf are same because of lanthanide contraction.

Answer    A

Question. The lanthanide contraction is responsible for the fact that
(a) Zr and Hf have about the same radius
(b) Zr and Zn have the same oxidation state
(c) Zr and Y have about the same radius
(d) Zr and Nb have similar oxidation state. 

Answer    A

Question. Which of the following statements concerning lanthanide elements is false?
(a) All lanthanides are highly dense metals.
(b) More characteristic oxidation state of lanthanide elements is +3.
(c) Lanthanides are separated from one another by ion exchange method.
(d) Ionic radii of trivalent lanthanides steadily increases with increase in the atomic number.

Answer    D

Question. The reason for greater range of oxidation states in actinoids is attributed to
(a) actinoid contraction
(b) 5f, 6d and 7s levels having comparable energies
(c) 4f and 5d levels being close in energies
(d) the radioactive nature of actinoids.

Answer    B

Question. Which of the following exhibits only +3 oxidation state?
(a) U
(b) Th
(c) Ac
(d) Pa 

Answer    C

Question. The chief ore of Hg is: 
(a) Pyrolusite
(b) Bauxite
(c) Galena
(d) Cinnabar

Question. The transition element which shows the highest oxidation state is: 
(a) Iron
(b) Vanadium
(c) Manganese
(d) Chromium

Question. Gun metal is: 
(a) Cu + Zn
(b) Cu + Sn + Zn
(c) Cu + Sn
(d) Zn + Sn

Question. Transition elements form coloured ions due to :
(a) d-d transition 
(b) fully filled d-orbitals
(c) smaller atomic radii
(d) availability of s-electrons

Question. CuSO₄ and KCN react to produce : 
(a) CuCN₂
(b) CuCN
(c) K₃[Cu(CN)₄]
(d) K₄[Cu(CN)₆ ]

Question. Which of the following is greatest paramagnetic?
(a) Cu+
(b) Fe2+ 
(c) Fe3+
(d) Cu2+

Question. The Mohr’s salt is shown by: 
(a) FeSO₄ (NH₄)₂SO₄.6H₂O
(b) FeSO₄ (NH₃)2SO4.6H₂O
(c) K2SO₄Al₂ (SO₄ )3.24H₂O
(d) FeSO₃(NH₂ )4SO₄.6H₂O

Question. German silver is an alloy of: 
(a) Fe, Cr, Ni
(b) Ag, Cu Au
(c) Cu, Zn, Ni
(d) Cu, Zn, Sn

Question. The composition of duralumin is :
(a) Al 94%, Mg 6%
(b) Cu 56%, Zn 24%, Ni 20%
(c) Cu 95%, Al 5%
(d) Al 95%, Cu 4%, Mn 0.5%, Mg 0.5%

Question. The number of water molecules in Mohr's salt is :
(a) 2
(b) 4
(c) 6
(d) 8

Question. Philosopher's wool on heating with BaO at 1100°C produces : 
(a) Ba + ZnCl₂
(b) BaCdO₂
(c) BaZnO₂
(d) BaO2 + Zn

Question. Which of the following is not an ore of iron?
(a) limonite
(b) casiterite 
(c) magnetite
(d) none of these

Question. The colourless species is : 
(a) VCl₃
(b) VOSO₄
(c) Na₃VO₄
(d) [V(H₂O)₆]SO₄. H₂O

Question. Lanthanide for which +II and +III oxidation states are common is : 
(a) La
(b) Pr
(c) Ce
(d) Eu 

POINTS TO BE REMEMBERED:

1. The elements of the periodic table belonging to group 3 to 12 are known as d-Block elements.
2. The general electronic configuration of these elements is \((n - 1)d^{1-10} ns^{1-2}\).
3. d-Block elements are collectively known as Transition Elements because properties of these elements vary in between s-Block and p-Block elements.
4. A transition element should have partially filled \((n - 1)\) d orbital.
5. Group 12 elements i.e. Zn, Cd, Hg have completely filled \((n - 1)\) d-orbital in atomic and ionic state & thus these elements are considered as Typical Transition Elements.
6. All these elements are metals. They are less electropositive than s-block elements & more electropositive than p-block elements.
7. The atomic radii decreases from group 3 to 6 (i.e. Sc to Cr) because of increase in effective nuclear charge gradually.
8. The atomic radii of group 7, 8 & 9 elements (i.e. Fe, Co, Ni) is almost same because pairing of electrons take place in \((n - 1)d\) orbital causing repulsion i.e. shielding of \((n - 1)d\) orbital.
9. Group 11 & 12 elements i.e. Cu & Zn have bigger size due to strong shielding of completely filled \((n - 1)d\) orbital.
10. The transition elements show variable oxidation state due to small energy difference between \((n - 1)d\) & \(ns\) orbital as a result both \((n - 1)d\) & \(ns\) electrons take part in bond formation.
11. The highest oxidation state of an element is equal to number of unpaired electrons present in \((n - 1)d\) & \(ns\) orbital.
12. Transition elements have high enthalpy of atomization/sublimation Because of large number of unpaired electrons in their atoms, they have stronger interatomic interaction and hence strong metallic bonding is present between atoms.
13. Most of transition elements are paramagnetic due to presence of unpaired electrons in \((n - 1)d\) orbital.
14. Most of transition elements are used as catalyst. It is due to (i) partially filled \((n - 1)d\) orbital (ii) Variable oxidation state (iii) Ability to change oxidation state frequently.
15. Most of transition elements form coloured compounds due to presence of unpaired electrons in \((n - 1)d\) orbital & thus they can undergo d-d transition.
16. Most of transition elements form complex compounds due to (i) small size (ii) high charge (iii) presence of vacant d-orbital of suitable energy.
17. Transition elements have lower value of Reduction Potential due to high ionization potential, high heat of sublimation & low enthalpy of hydration.
18. Transition elements form interstitial compounds because size of interstitial voids is similar to size of non-metals C, N, O, B.
19. Transition elements form alloys due to similar ionic radii.
20. The oxides of transition metals in lower oxidation state are BASIC, intermediate oxidation state are AMPHOTERIC, highest oxidation state are ACIDIC.

LANTHANOIDS:

1. The 14 elements after Lanthanum having atomic number 58 to 71 are collectively known as Lanthanoids.
2. The general electronic configuration of these elements is \([Xe] 4f^{1-14}, 5d^{0-1}, 6s^2\).
3. Most common oxidation state of these elements is \(+3\), but Ce shows \(+4\), Eu \(+2\), because they acquire stable configuration.
4. The size of Lanthanoids and its trivalent ion decreases from La to Lu due to poor shielding of 4f electrons. It is known as lanthanoids contraction.

ACTINOIDS:

1. The 14 elements after Actinium having atomic number 90 to 113 are collectively known as Actinoids.
2. The general electronic configuration of these elements is \([Rn] 5f^{1-14}, 6d^{0-1}, 7s^2\).
3. The size of actinoids and its trivalent ion decreases from Ac to Lw due to poor shielding of 5f electrons. It is known as actinoids contraction.
4. The elements after U (92) are man made known as transuranic elements.

POTASSIUM DICHROMATE:

Preparation: It takes place in three steps -
(i) Conversion of chromite ore to sodium chromate.
(ii) Conversion of sodium chromate to sodium dichromate.
(iii) Conversion of sodium dichromate to potassium dichromate
Following reaction take place:
\(4\text{FeCr}_2\text{O}_4 + 4\text{Na}_2\text{CO}_3 + 7\text{O}_2 \longrightarrow 2\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2\)
\(2\text{Na}_2\text{CrO}_4 + 2\text{H}^+ \longrightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{Na}^+ + \text{H}_2\text{O}\)
\(\text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \longrightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2\text{NaCl}\)

The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution.
A) In Acidic medium (pH < 7)
\(2\text{CrO}_4^{2-} + 2\text{H}^+ \longrightarrow \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}\)
B) In basic medium (pH > 7)
\(\text{Cr}_2\text{O}_7^{2-} + 2\text{OH}^- \longrightarrow 2\text{CrO}_4^{2-} + \text{H}_2\text{O}\)

POTASSIUM PERMANGANATE:

Preparation:
It takes place in two steps:
(i) Conversion of pyrolusite ore into potassium magnate
(ii) Conversion of potassium magnate to potassium permanganate
Following reactions take place:
\(2\text{MnO}_2 + 4\text{KOH} + \text{O}_2 \longrightarrow 2\text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O}\)
\(3\text{MnO}_4^{2-} + 4\text{H}^+ \longrightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}\)

QUESTION ANSWERS

(TWO MARK QUESTIONS)

Question 1. Explain briefly how +2 oxidation state becomes more and more stable in the first half of the first row transition elements with increasing atomic number.
Answer: In \(M^{2+}\) ions, 3d-orbitals get occupied gradually as the atomic number increases. Since, the number of empty d-orbitals decreases, the stability of cations increases from \(\text{Sc}^{2+}\) to \(\text{Mn}^{2+}\). \(\text{Mn}^{2+}\) is most stable as all d-orbitals are singly occupied.
In simple words: Half-filled d-orbitals in manganese provide extra stability.

Exam Tip: Mention the \(3d^5\) half-filled configuration to secure full marks.

Question 2. Explain why transition elements have many irregularities in their electronic configurations?
Answer: In the transition elements, there is a little difference in the energy of \((n-1)\) d-orbitals and \(ns\)-orbitals. Thus, incoming electron can occupy either of shell. Hence, transition elements exhibit many irregularities in their electronic configurations.
In simple words: Very small energy gaps between orbitals cause electrons to shift easily.

Exam Tip: Always state that \((n-1)d\) and \(ns\) subshells have comparable energies.

Question 3. What are different oxidation states exhibited by Lanthanides?
Answer: The common stable oxidation state of lanthanides is \(+3\). However some members also show oxidation states of \(+2\) & \(+4\).
In simple words: Lanthanides mostly show +3, with a few showing +2 or +4.

Exam Tip: Give examples like Ce(\(+4\)) and Eu(\(+2\)).

Question 4. How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.
Answer: The transition elements use its \((n-1)d\), \(ns\) and \(np\) orbital and the successive oxidation states differ by unity. For example, Mn shows all the oxidation states from \(+2\) to \(+7\). On other hand non transition elements use its \(ns\), \(np\) and \(nd\) orbitals and the successive oxidation states differ by two units e.g. \(\text{Sn}^{2+}\), \(\text{Sn}^{4+}\) etc.
In simple words: Transition metals differ by 1 unit, while non-transition metals differ by 2 units.

Exam Tip: Highlight that d-electrons are available for bonding in transition metals.

Question 5. Why do transition elements show variable oxidation states?
Answer: The transition elements show variable oxidation state due to small energy difference between \((n-1)d\) & \(ns\) orbital as a result both \((n-1)d\) & \(ns\) electrons take part in bond formation.
In simple words: Both inner d and outer s electrons participate in bonding.

Exam Tip: Use the phrase "comparable energies of \((n-1)d\) and \(ns\) orbitals".

Question 6. Why are \(\text{Mn}^{2+}\) compounds more stable than \(\text{Fe}^{2+}\) compounds towards oxidation to +3 state?
Answer: The electronic configuration of \(\text{Mn}^{2+}\) is \([\text{Ar}] 3d^5\), i.e. all five d-orbitals are singly occupied. Thus this is stable electronic configuration and further loss of electron requires high energy. On other hand side the electronic configuration of \(\text{Fe}^{2+}\) is \([\text{Ar}] 3d^6\), i.e. Loss of one electron requires low energy.
In simple words: Manganese has a stable half-filled d-shell, unlike iron.

Question 7. To what extend do the electronic configuration decide the stability of oxidation state in the first series of the transition elements? Illustrate your answer with an example.
Answer: In a transition series, the oxidation state which lead to exactly half filled or completely filled orbitals are more stable e.g. the electronic configuration of Fe is \([\text{Ar}] 3d^6, 4s^2\). It shows various oxidation state but Fe(III) is more stable than Fe(II).
In simple words: Half-filled and fully filled subshells dictate extra stability.

Exam Tip: Mention electronic configurations explicitly in terms of stability.

Question 8. What is meant by disproportionation? Give two examples.
Answer: Those reactions in which same substance undergoes oxidation as well as reduction are called disproportionation reactions. e.g.
\(2\text{Cu}^+ \longrightarrow \text{Cu}^{2+} + \text{Cu}\)
\(3\text{MnO}_4^{2-} + 4\text{H}^+ \longrightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}\)
In simple words: A single species gets both oxidized and reduced simultaneously.

Exam Tip: Always check oxidation states of the element on both reactant and product sides.

Question 9. Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
Answer: Copper with configuration \([\text{Ar}] 3d^{10} 4s^1\) exhibits \(+1\) oxidation state. Copper loses \(4s^1\) electron easily and achieved a stable configuration \(3d^{10}\) by forming \(\text{Cu}^+\).
In simple words: Losing one electron gives copper a completely full 3d shell.

Exam Tip: Note that \(\text{Cu}^{2+}\) is actually more stable in aqueous solution due to high hydration enthalpy.

Question 10. What are inner transition elements?
Answer: The f-block elements in which the last electron accommodated on \((n-2)\) f-subshell are called inner transition elements. These include atomic numbers 58 to 71 and from 90 to 103.
In simple words: Elements where electrons fill the inner f-orbitals.

Exam Tip: Clearly state \((n-2)f\) filling.

Question 11. The paramagnetic character in 3d-transition series elements increases upto Mn and then decreases. Explain why?
Answer: In the 3d-transition series as we move from Sc (21) to Mn (25) the number of unpaired electrons increases and hence paramagnetic character increases. After Mn, the pairing of electrons in the d-orbital starts and the number of unpaired electrons decreases and hence, paramagnetic character decreases.
In simple words: Unpaired electrons peak at manganese and then drop off as pairing begins.

Exam Tip: Paramagnetism depends directly on the number of unpaired electrons.

Question 12. Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition metal.
Answer: The following points justify that the given statement is true:
(i) Ionization enthalpies of heavier transition elements are higher than the elements of 3d series. Consequently, heavier transition elements are less reactive in comparison to 3d-elements.
(ii) Melting points of heavier transition elements are higher than 3d-elements.
(iii) Higher oxidation states of heavier transition elements are stable whereas lower oxidation states are stable in 3d-elements.
In simple words: Heavier transition metals have stronger bonds and higher ionization energies.

Exam Tip: Mention 4d/5d vs 3d differences clearly.

Question 13. What are transition elements? Which d-block elements are not regarded as transition elements and why?
Answer: An element which has partially filled \((n-1)\) d orbital is known as transition element. Group 12 elements i.e. Zn, Cd, Hg have completely filled \((n-1)\) d-orbital in atomic & ionic state & thus these elements are not considered as Transition Elements.
In simple words: Zinc, cadmium, and mercury have full d-shells, so they are not true transition metals.

Exam Tip: Emphasize both atomic and ionic states having filled d orbitals.

Question 14. What are interstitial compounds? Why are such compounds well known for transition metal?
Answer: Compounds of transition metal with relatively smaller non-metals are known as interstitial compounds. These compounds are well known for transition metals because size of C, N, O, and B is similar to size of interstitial voids of transition metal.
In simple words: Small atoms fit snugly into the empty spaces of transition metal lattices.

Exam Tip: List the typical trapped non-metal atoms: H, C, N, B.

Question 15. For the first row of transition metals the \(E^0\) values are:

Explain the irregularity in the above values.
Answer: The \(E^0(M^{2+}/M)\) values are not regular which can be explained from the irregular variation of ionization energy and sublimation energy of Mn due to half-filled orbitals.
In simple words: Irregularities stem from varying hydration, sublimation, and ionization energies.

Exam Tip: Reference the sum of ionization enthalpies \((\Delta_i H_1 + \Delta_i H_2)\).

(THREE MARK QUESTIONS)

Question 1. Decide giving reason which one of the following pairs exhibits the property indicated:
(i) \(\text{Sc}^{3+}\) or \(\text{Cr}^{3+}\) exhibits paramagnetism
(ii) V or Mn exhibits more number of oxidation states
(iii) \(\text{V}^{4+}\) or \(\text{V}^{5+}\) exhibits colour
Answer: (i) \(\text{Sc} = [\text{Ar}] 3d^1 4s^2\); \(\text{Sc}^{3+} = [\text{Ar}]\); it has no unpaired electron so diamagnetic.
\(\text{Cr} = [\text{Ar}] 3d^5 4s^1\); \(\text{Cr}^{3+} = [\text{Ar}] 3d^3\); it has three unpaired electrons, hence paramagnetic.
(ii) \(\text{V} = [\text{Ar}] 3d^3 4s^2\), \(\text{Mn} = [\text{Ar}] 3d^5 4s^2\). Thus V exhibits oxidation states of \(+2, +3, +4, +5\), whereas Mn exhibits oxidation states of \(+2\) to \(+7\).
(iii) \(\text{V}^{4+} = [\text{Ar}] 3d^1 \rightarrow\) coloured, \(\text{V}^{5+} = [\text{Ar}] \rightarrow\) colourless.
In simple words: Presence of unpaired d-electrons determines color and paramagnetism.

Exam Tip: Count unpaired electrons carefully for each ion.

Question 2.(a) Describe the general trends in the following properties of the first series of the transition elements:
(i) Stability of +2-oxidation state
(ii) Formation of oxometal ions
(b) Write steps involved in the preparation of \(\text{KMnO}_4\) from \(\text{K}_2\text{MnO}_4\)
Answer: (a) (i) The elements of first transition series show decreasing tendency to form divalent cation as we move left to right in the series. This trend is due to general increase in the first and second ionization energy. The greater stability of \(\text{Mn}^{2+}\) is due to half filled \(d^5\) configuration and that of zinc is due to \(d^{10}\) configuration.
(ii) All metal except Sc form oxide of type MO which are basic. The highest oxidation number in all oxide, coincide with the group number and is attain in \(\text{Sc}_2\text{O}_3\) to \(\text{Mn}_2\text{O}_7\). Formation of oxoanions is due to high electro negativity and small size of oxygen atom.
(b) It takes place in two steps:
Conversion of pyrolusite ore into potassium mangnate:
\(2\text{MnO}_2 + 4\text{KOH} + \text{O}_2 \longrightarrow 2\text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O}\)
Conversion of potassium mangnate to potassium permanganate:
\(3\text{MnO}_4^{2-} + 4\text{H}^+ \longrightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}\)
In simple words: Divalent stability drops across the series, while high electronegativity of oxygen favors high oxometal ions.

Exam Tip: Specify the acidic/neutral oxidation methods for manganate conversion.

Question 3.(a) Write the steps involve in the preparation of \(\text{K}_2\text{Cr}_2\text{O}_7\) from chromite ore.
(b) What is the effect of pH on dichromate ion solution?
Answer: (a) It takes place in three steps:
(iv) Conversion of chromite ore to sodium chromate:
\(4\text{FeCr}_2\text{O}_4 + 4\text{Na}_2\text{CO}_3 + 7\text{O}_2 \longrightarrow 2\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2\)
(v) Conversion of sodium chromate to sodium dichromate:
\(2\text{Na}_2\text{CrO}_4 + 2\text{H}^+ \longrightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{Na}^+ + \text{H}_2\text{O}\)
(vi) Conversion of sodium dichromate to potassium dichromate:
\(\text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \longrightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2\text{NaCl}\)
(b) Dichromate ion is orange in acidic solution (pH < 7) and turns yellow in basic solution. It is due to interconversion of dichromate ion to chromate ion. Following reactions take place:
\(2\text{CrO}_4^{2-} (\text{yellow}) + 2\text{H}^+ \longrightarrow \text{Cr}_2\text{O}_7^{2-} (\text{orange}) + \text{H}_2\text{O}\)
\(\text{Cr}_2\text{O}_7^{2-} (\text{orange}) + 2\text{OH}^- \longrightarrow 2\text{CrO}_4^{2-} (\text{yellow}) + \text{H}_2\text{O}\)
In simple words: Acid turns chromate to orange dichromate; base turns dichromate to yellow chromate.

Exam Tip: Remember: Orange in acid, Yellow in base.

Question 4.(a) What is lanthanide contraction? What effect does it have on the chemistry of the elements, which follow lanthanoids?
(b) The chemistry of actinoid elements is not so much smooth as that of lanthanoids. Justify these statements by giving some examples from the oxidation state of these elements.
Answer: (a) The size of Lanthanoids and its trivalent ion decreases from La to Lu. It is known as lanthanoids contraction.
Cause: It is due to poor shielding of 4f electrons.
Consequences of lanthanide contraction: (i) Basic strength of hydroxide decreases from \(\text{La(OH)}_3\) to \(\text{Lu(OH)}_3\). (ii) Because of similar chemical properties lanthanides are difficult to separate.
(b) Lanthanoids show limited number of oxidation states i.e. \(+2, +3, +4\) (out of which \(+3\) is most common). This is because of a large energy gap between 4f, 5d and 6s subshell. The dominant oxidation state of actinides is also \(+3\) but they show a number of other oxidation state also e.g. \(+4, +5,\) and \(+7\). This is due to small energy difference between 5f, 6d and 7s orbitals.
In simple words: Poor f-orbital shielding shrinks radii across the period, while smaller energy gaps give actinides diverse oxidation states.

Exam Tip: Link lanthanide contraction to the atomic radii of 4d and 5d transition series.

Question 5. Give examples and suggest reasons for the following features of the transition metal chemistry:
(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.
(iii) Of the \(d^4\) species, \(\text{Cr}^{2+}\) is strongly reducing while manganese(III) is strongly oxidizing.
Answer: (i) The oxide of transition metals in lower oxidation states are generally basic while those in the higher oxidation states are acidic. Acidic character increases with increase in oxidation state is due to decrease in size of metal ion and increase in charge density. e.g. \(\text{MnO}\) (basic), \(\text{Mn}_3\text{O}_4\) (amphoteric), \(\text{Mn}_2\text{O}_7\) (acidic).
(ii) A transition metal exhibits higher oxidation states in oxides and fluorides because oxygen and fluorine are the most electronegative elements and thus easily can unpair electrons of metal atom.
(iii) Because oxidizing and reducing property depends on \(E^0\) value. Since \(E^0\) value of \(\text{Cr}^{3+}/\text{Cr}^{2+}\) is negative while that of \(\text{Mn}^{3+}/\text{Mn}^{2+}\) is positive, as a result \(\text{Cr(II)}\) act as reducing agent and \(\text{Mn(III)}\) is strong oxidizing.
In simple words: Higher oxidation makes metal oxides more acidic, and fluorine/oxygen stabilize high oxidation states due to high electronegativity.

Exam Tip: Use oxidation state changes to explain acid-base properties of oxides.

Question 6. For \(M^{2+}/M\) and \(M^{3+}/M^{2+}\) systems, the \(E^0\) 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:
(i) the stability of \(\text{Fe}^{3+}\) in acid solution as compared to \(\text{Cr}^{3+}\) or \(\text{Mn}^{3+}\) and
(ii) the ease with which iron can be oxidized as compared to a similar process for either chromium or manganese metal.
Answer: (i) \(E^0\) for \(\text{Cr}^{3+}/\text{Cr}^{2+}\) is -0.4V i.e. negative, this means \(\text{Cr}^{3+}\) ions in the solution cannot be reduced to \(\text{Cr}^{2+}\) easily i.e. \(\text{Cr}^{3+}\) is stable. As \(\text{Mn}^{3+}/\text{Mn}^{2+}\) is +1.5V i.e positive means \(\text{Mn}^{3+}\) can easily reduced to \(\text{Mn}^{2+}\) ions in comparison to \(\text{Fe}^{3+}\) ions. Thus relatively stability of these ions is: \(\text{Mn}^{3+} < \text{Fe}^{3+} < \text{Cr}^{3+}\)
(ii) The oxidation potentials for the given pairs will be +0.9V, +1.2V and +0.4V. Thus, the order of their getting oxidized will be in the order \(\text{Mn} > \text{Cr} > \text{Fe}\).
In simple words: Negative reduction potential implies stable oxidized states, while positive potential implies easy reduction.

Exam Tip: High positive \(E^0\) means strong oxidizing power for the oxidized species.

Question 7. Account for the following statements:
(i) Cobalt (II) is stable in aqueous solution but in the presence of strong ligands and air, it can be oxidized to Co (III).
(ii) The \(d^1\) configuration is very unstable in ions.
(iii) One among the lanthanides, Ce (III) can be easily oxidized to Ce (IV).
Answer: (i) Strong ligands force cobalt (II) to lose one more electron from 3d-subshell and thereby induced \(d^2sp^3\) hybridisation.
(ii) The ion with \(d^1\) configuration try to lose the only electron in order to acquire inert gas configuration.
(iii) The configuration of Ce is \([\text{Xe}] 4f^1 5d^1 6s^2\). There is no much difference between the energy of 4f, 5d and 6s orbitals and thus, Ce can utilize electrons present in these orbitals and hence oxidation state of +4.
In simple words: Ligand field splitting and noble gas configurations drive these transitions.

Question 8. Compare the chemistry of actinides with that of the lanthanoids with special reference to:
(i) electronic configuration
(ii) atomic and ionic sizes
(iii) oxidation state
(iv) chemical reactivity
Answer:

In simple words: Actinides show more variable oxidation states and higher reactivity than lanthanides due to smaller 5f-6d-7s energy gaps.

Exam Tip: Present comparisons in a tabular format for maximum clarity.

Question 9.(a) What is actinides contraction? What effect does it have on the chemistry of the elements, which follow actinides?
(b) Name an important alloy, which contains some of the lanthanide metals. Mention its uses.
Answer: (a) The size of actinoid and its trivalent ion decreases from Ac to Lw. It is known as actinides contraction.
Cause: It is due to poor shielding of 5f electrons.
Consequences of actinides contraction: (i) Basic strength of hydroxide decreases from \(\text{Ac(OH)}_3\) to \(\text{Lw(OH)}_3\). (ii) Because of similar chemical properties actinides are difficult to separate.
(b) An important alloy containing lanthanoid metals is mischmetal, which contains 95% lanthanide metal and 5% Fe along with traces of S, C, Ca and Al. It is used in Mg-based alloy to produce bullets, shells and lighter flint.
In simple words: Actinide contraction mirrors lanthanide contraction, shrinking ionic radii across the series.

Exam Tip: Remember the composition of mischmetal: 95% lanthanide + 5% iron.

Question 10. Complete following reactions:
(i) \(\text{MnO}_4^- + \text{H}^+ + \text{Fe}^{2+} \longrightarrow \dots + \dots + \dots\)
(ii) \(\text{MnO}_4^- + \text{C}_2\text{O}_4^{2-} + \text{H}^+ \longrightarrow \dots + \dots + \dots\)
(iii) \(\text{MnO}_4^- + \text{OH}^- + \text{I}^- \longrightarrow \dots + \dots + \dots\)
Answer: (i) \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \longrightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\)
(ii) \(2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^+ \longrightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}\)
(iii) \(2\text{MnO}_4^- + \text{H}_2\text{O} + \text{I}^- \longrightarrow 2\text{MnO}_2 + 2\text{OH}^- + \text{IO}_3^-\)
In simple words: Permanganate oxidizes iron(II), oxalate, and iodide in respective acidic or alkaline media.

Exam Tip: Memorize stoichiometric balancing coefficients for permanganate titrations.

(FIVE MARK QUESTIONS)

Question 1. Explain giving reasons:
(i) Transition metals and many of their compounds show paramagnetic behaviour.
(ii) The enthalpies of atomisation of the transition metals are high.
(iii) The transition metals generally form coloured compounds.
(iv) Transition metals and their many compounds act as good catalyst.
(v) Transition metals have a strong tendency to form complexes.
Answer: (i) Transition metals and many of their compounds show paramagnetic behaviour due to presence of unpaired electrons in \((n-1)\) d orbital.
(ii) The enthalpies of atomisation of the transition metals are high Because of large number of unpaired electrons in their atoms, they have stronger interatomic interaction and hence strong metallic bonding is present between atoms.
(iii) The transition metals generally form coloured compounds due to presence of unpaired electrons in \((n-1)\) d orbital & thus they can undergo d-d transition.
(iv) Transition metals and their many compounds act as good catalyst It is due to (i) partially filled \((n-1)\) d orbital (ii) Variable oxidation state (iii) Ability to change oxidation state frequently.
(v) Transition metals have a strong tendency to form complexes Most of transition elements form complex compounds due to (i) small size (ii) high charge (iii) presence of vacant d-orbital of suitable energy.
In simple words: Unpaired d-electrons and vacant orbitals account for paramagnetism, color, catalysis, and complexation.

Exam Tip: Group these five properties under characteristic features of d-block elements.

Question 2. Give reasons for the following:
(i) Fe has higher melting point than Cu.
(ii) \([\text{Ti(H}_2\text{O)}_6]^{3+}\) is coloured while \([\text{Sc(H}_2\text{O)}_6]^{3+}\) is colourless.
(iii) The 4d and 5d series of transition metals have more frequent metal-metal bonding in their compound than do the 3d metals.
(iv) Transition metals some time exhibit very low oxidation state such as +1 and 0.
(v) Hg is not considered a transition metal.
Answer: (i) This is because Fe (\(3d^6, 4s^2\)) has four unpaired electrons in 3d-subshell. While Cu (\(3d^{10}, 4s^1\)) only one unpaired electron in 4s shell. Hence metallic bonding is stronger in Fe than those in Cu.
(ii) The oxidation state of Ti in \([\text{Ti(H}_2\text{O)}_6]^{3+}\) is \(+3\) and its configuration is \([\text{Ar}] 3d^1\) i.e one unpaired electron and hence it is coloured. Whereas the oxidation state of Sc in \([\text{Sc(H}_2\text{O)}_6]^{3+}\) is \(+3\) and its configuration is \([\text{Ar}] 3d^0\) i.e no unpaired electron and hence it is colourless.
(iii) In the same group of d-block elements, the 4d and 5d transition elements are larger size than that of 3d elements. Hence, the valence electrons are less tightly held and form metal-metal bond more frequently.
(iv) +1 oxidation state is shown by elements like Cu because after loss of one electron, it acquire stable configuration. Zero oxidation state is shown in metal carbonyl, because \(\pi\)-electrons donated by CO are accepted into the empty orbital.
(v) The characteristic properties of transition metal are due to partially filled d-orbitals. Hg has completely filled d-orbital, as a result it doesn't show properties of transition metals and hence is not considered as transition metal.
In simple words: Metallic strength, d-electron occupancy, and orbital sizes explain these specific chemical behaviours.

Exam Tip: D-d transitions require unfilled d-orbitals with at least one electron.

Question 3.(a) write electronic configuration of element having atomic number 101.
(b) Which element show maximum oxidation state in 3d transition series?
(c) What is mischmetal?
(d) Explain why \(\text{Cu}^+\) ion is not stable in aqueous solution?
(e) Name the transition metal which is well known to exhibit +4 oxidation state?
Answer: (a) \([\text{Rn}] 5f^{13} 6d^0 7s^2\)
(b) Mn, Which shows \(+7\) oxidation state in \(\text{KMnO}_4\).
(c) It is an important alloy, which contains 95% lanthanide metal and 5% Fe along with traces of S, C, Ca and Al. It is used in Mg-based alloy to produce bullets, shells and lighter flint.
(d) Water is a good complexing agent and thus Cu forms complex by losing one more electron from 3d orbital.
(e) Cerium (\(Z=58\))
In simple words: Element 101 is mendelevium, manganese shows the highest oxidation state, and hydration stabilizes copper(II).

Exam Tip: Mention hydration enthalpy compensating for second ionization energy in copper solutions.

Question 4.(a) Write the preparation of potassium dichromate from iron chromite. What happens when potassium dichromate reacts with (i) Hydrogen sulphide (ii) \(\text{FeSO}_4\)?
(b) Why do Zr and Hf exhibit almost similar properties?
(c) Why is \(\text{La(OH)}_3\) stronger base than \(\text{Lu(OH)}_3\).
Answer: (a) Preparation: It takes place in three steps:
(i) Conversion of chromite ore to sodium chromate.
(ii) Conversion of sodium chromate to sodium dichromate.
(iii) Conversion of sodium dichromate to potassium dichromate
Following reaction takes place:
\(4\text{FeCr}_2\text{O}_4 + 4\text{Na}_2\text{CO}_3 + 7\text{O}_2 \longrightarrow 2\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2\)
\(2\text{Na}_2\text{CrO}_4 + 2\text{H}^+ \longrightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{Na}^+ + \text{H}_2\text{O}\)
\(\text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \longrightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2\text{NaCl}\)
Reactions:
(i) \(\text{Cr}_2\text{O}_7^{2-} + 8\text{H}^+ + 3\text{H}_2\text{S} \longrightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 3\text{S}\)
(ii) \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \longrightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{Fe}^{3+}\)
(b) Because both have similar ionic size.
(c) Due to lanthanoid contraction size of \(\text{La}^{3+}\) is smaller than \(\text{Lu}^{3+}\) as a result \(\text{Lu-O}\) bond will stronger than \(\text{La-O}\) bond.
In simple words: Zirconium and hafnium match in size due to lanthanoid contraction, and larger La-O size makes lanthanum hydroxide more basic.

Exam Tip: Mention the diagonal-like relationship caused by lanthanide contraction for Zr and Hf.

Question 5. Give reasons for the following:
(i) Transition metals have high enthalpy of hydration.
(ii) Zn, Cd and Hg are not regarded as transition metal.
(iii) d block elements exhibit a large number of oxidation state than f block elements.
(iv) The second and third members in each group of transition element have similar atomic radii.
(v) \(\text{K}_2[\text{PtCl}_6]\) is well known compound whereas the corresponding Ni compound is not known.
Answer: (i) Transition metal ions are smaller and have higher charge, therefore have high enthalpy of hydration.
(ii) Group 12 elements i.e. Zn, Cd, Hg have completely filled \((n-1)\) d-orbital in atomic & ionic state & thus these elements are not considered as transition elements.
(iii) The difference in the energy of \((n-1)\) d orbital and ns orbital is very small and thus both sub-shells are used for bond formation. Whereas in f block elements \((n-2)f\) orbitals lie underneath and hence are not available for bond formation.
(iv) The second and third members in each group of transition element have similar atomic radii due to lanthanoid contraction. It arises due to poor shielding of d and f electron.
(v) The oxidation state of Pt is \(+4\) which is stable for Pt. The \(+4\) oxidation state for Ni is very difficult to achieve because the sum of first four ionization energies is very high. Hence, the corresponding Ni(II) compound is known.
In simple words: High charge density, full d-shells, and lanthanide contraction explain these chemical trends.

Exam Tip: Highlight that \(\text{Pt}^{4+}\) is accessible while \(\text{Ni}^{4+}\) requires prohibitive ionization energy.