RBSE Solutions Class 12 Chemistry Chapter 8 d and f Block Elements

RBSE Class 12 Chemistry Chapter 8 Text Book Type Questions

RBSE Class 12 Chemistry Chapter 8 Multiple Choice Questions

 

Question 1. Which of the following shows highest oxidation state of +7?
(a) Co
(b) Cr
(c) Mn
(d) v
Answer: (c) Mn
In simple words: Manganese (Mn) has an electron configuration that allows it to lose or share seven electrons, reaching an oxidation state of +7. This is the highest possible oxidation state among the given options due to its d-electron availability.

🎯 Exam Tip: Remember the common highest oxidation states for 3d transition metals. Manganese is notable for reaching +7 due to its half-filled d-subshell and two s-electrons, which are all available for bonding.

 

Question 2. The number of unpaired electrons in Fe²⁺ are:
(a) 4
(b) 5
Answer: (a) 4
In simple words: Iron (Fe) has an atomic number of 26. Its electron configuration is [Ar]3d⁶4s². When it forms an Fe²⁺ ion, it loses the two 4s electrons, leaving [Ar]3d⁶. With 6 electrons in the 3d subshell, 4 of them are unpaired according to Hund's rule, and one pair is formed.

🎯 Exam Tip: To find the number of unpaired electrons, first write the electron configuration for the neutral atom, then remove electrons from the outermost s-orbital first (for cations) and fill the d-orbitals according to Hund's rule.

 

Question 4. Which of the following has maximum magnetic moment ?
(a) V³⁺
(b) Cr³⁺
(c) Fe³⁺
(d) C³⁺
Answer: (c) Fe³⁺
In simple words: Magnetic moment is directly related to the number of unpaired electrons. Fe³⁺ has the most unpaired electrons (5 unpaired electrons from its 3d⁵ configuration) compared to the other ions, so it will have the largest magnetic moment.

🎯 Exam Tip: Calculate the number of unpaired electrons for each ion. The ion with the highest number of unpaired electrons will have the maximum magnetic moment according to the spin-only formula.

 

Question 5. What is the common oxidation state of lanthanoid series?
(a) + 1
(b) + 4
(c) + 2
(d) + 3
Answer: (d) + 3
In simple words: Most lanthanoids readily form ions with a +3 charge. This is because losing two 6s electrons and one 5d or 4f electron gives them a stable configuration.

🎯 Exam Tip: Remember that +3 is the most stable and common oxidation state for lanthanoids, although some elements can show +2 or +4 in specific compounds.

 

Question 6. Lanthanoid contraction is due to increase in:
(a) effective nuclear charge
(b) atomic number
(c) size of 4 f - orbital
(d) none of these
Answer: (a) effective nuclear charge
In simple words: Lanthanoid contraction means that the atomic and ionic sizes of elements get smaller across the lanthanoid series. This happens because the 4f electrons do not shield the nuclear charge very well, causing the outer electrons to be pulled closer to the nucleus.

🎯 Exam Tip: The poor shielding effect of 4f electrons is the key reason for lanthanoid contraction. This leads to an increased effective nuclear charge experienced by the outer electrons, pulling them in tighter.

 

Question 7. One of the member of lanthanoid series shows +4 oxidation state is :
(a) Ce
(b) Lu
(c) Eu
(d) Pm
Answer: (a) Ce
In simple words: While +3 is common, Cerium (Ce) is known for exhibiting a stable +4 oxidation state. This is because it can achieve a stable noble gas-like electron configuration by losing four electrons.

🎯 Exam Tip: Be aware of exceptions to the common +3 oxidation state in lanthanoids. Cerium (Ce) commonly shows +4, and Europium (Eu) often shows +2, due to achieving more stable f⁰, f⁷, or f¹⁴ configurations.

 

Question 9. Which of the following has maximum first ionisation enthalpy ?
(a) Ti
(b) Mn
(c) Fe
(d) Ni
Answer: (d) Ni
In simple words: As you move across the 3d transition series, the atomic size generally decreases slightly, and the effective nuclear charge increases. This makes it harder to remove the first electron. Among the given options, Nickel (Ni) has the highest first ionization enthalpy.

🎯 Exam Tip: First ionization enthalpy generally increases across a period due to increasing nuclear charge and decreasing atomic size, making electrons harder to remove. Exceptions exist due to stable electronic configurations.

 

Question 10. In which of the following ions, all the electrons are paired ?
(a) Cr²⁺
(b) Cu²⁺
(c) Cu⁺
(d) Ni²⁺
Answer: (c) Cu⁺
In simple words: An ion has all its electrons paired when its d-subshell is completely full (d¹⁰). Copper (Cu) as a neutral atom is [Ar]3d¹⁰4s¹. When it forms Cu⁺, it loses the 4s electron, resulting in [Ar]3d¹⁰, where all d-electrons are paired.

🎯 Exam Tip: To determine if all electrons are paired, write the electron configuration of each ion. Look for ions with completely filled subshells (like d¹⁰) as these will have no unpaired electrons.

RBSE Class 12 Chemistry Chapter 8 Very Short Answer Type Questions

 

Question 1. Zn is not considerd as transition element. Explain why ?
Answer: Zinc (Z=30) has the electron configuration [Ar]3d¹⁰4s². It does not have any partially filled d-subshell in its neutral atomic form or in its common oxidation state, Zn²⁺ (which is 3d¹⁰). Because transition elements are defined by having partially filled d-orbitals, Zinc does not fit this definition. Zinc's filled d-subshell prevents it from showing typical transition metal properties like variable oxidation states and color. Therefore, it is not considered a transition element.
In simple words: Zinc is not a transition element because its d-orbitals are always full, both in its normal state and as an ion. Transition elements must have d-orbitals that are only partly filled.

🎯 Exam Tip: The key criterion for being a transition element is the presence of incompletely filled d-orbitals in its ground state or in any of its common oxidation states.

 

Question 2. Ti⁴⁺ ion is colourless. Give reason.
Answer: The Ti⁴⁺ ion has an electron configuration of (3s²3p⁶) which is like a noble gas. This means it does not contain any unpaired electrons in its d-subshell. Color in transition metal compounds often comes from d-d electronic transitions, where an electron moves between different d-orbitals. Since Ti⁴⁺ has no d-electrons at all, these transitions cannot occur. Therefore, Ti⁴⁺ compounds are colourless. This makes them appear white or clear in solutions.
In simple words: Ti⁴⁺ ions are colourless because they have no electrons in their d-orbitals. Without d-electrons, they cannot absorb visible light to show color.

🎯 Exam Tip: Remember that transition metal ions are typically coloured if they have partially filled d-orbitals, allowing for d-d transitions. Ions with d⁰ or d¹⁰ configurations are usually colourless.

RBSE Class 12 Chemistry Chapter 8 Short Answer Type Questions

 

Question 1. What is lanthanoid contraction ? Explain.
Answer: Lanthanoid contraction refers to the steady decrease in the atomic and ionic radii of lanthanoid elements as you move from Lanthanum (La³⁺) to Lutetium (Lu³⁺) across the series. This decrease happens because the 4f electrons, which are added one by one across the series, provide very poor shielding of the nuclear charge. The 4f orbitals are diffuse and buried deep within the atom, so they are not very effective at blocking the positive charge of the nucleus from the outer electrons. This leads to a stronger pull of the nucleus on the outer electrons, causing the atomic and ionic sizes to shrink more than expected. This contraction has significant effects on the chemistry of the elements that follow the lanthanoids.
In simple words: Lanthanoid contraction is the gradual shrinking in size of atoms and ions as we go across the lanthanoid series. It happens because the inner 4f electrons are not good at shielding the outer electrons from the strong pull of the nucleus.

🎯 Exam Tip: When explaining lanthanoid contraction, always mention the "poor shielding effect of 4f electrons" and its consequence: "increased effective nuclear charge" on the outer electrons.

 

Question 2. What are alloys ? Give one use.
Answer: An alloy is a homogeneous mixture of two or more metallic elements, or a metallic element combined with one or more non-metallic elements. Alloys are created to achieve properties that are superior to those of the individual pure metals. For example, steel is an alloy of iron and carbon, known for its strength. A notable alloy involving lanthanoid metals is misch metal. Misch metal is composed of about 95% lanthanoid metal (mainly cerium) and about 5% iron, along with traces of sulfur, carbon, calcium, or aluminium. One important use of misch metal is in magnesium-based alloys for producing bullets, shells, and lighter flints due to its pyrophoric properties.
In simple words: Alloys are mixtures made by combining two or more metals, or a metal with a non-metal, to make them stronger or better. An example is misch metal, used in making lighter flints.

🎯 Exam Tip: Define an alloy as a homogeneous mixture and state its purpose (improved properties). Always include at least one specific example and its application.

 

Question 3. Write electronic configuration of Cu²⁺. Calculate its magnetic moment.
Answer: The atomic number of Copper (Cu) is 29. Its ground state electron configuration is [Ar]3d¹⁰4s¹. When copper forms the Cu²⁺ ion, it loses one electron from the 4s orbital and one electron from the 3d orbital. Therefore, the electronic configuration of Cu²⁺ will be [Ar]3d⁹. This configuration means there is one unpaired electron in the 3d subshell (9 electrons in 5 orbitals, so 4 are paired and 1 is unpaired).
The magnetic moment (\(\mu\)) is given by the spin-only formula: \( \mu = \sqrt{n(n+2)} \) B.M.
Where n = number of unpaired electrons.
For Cu²⁺, n = 1 (one unpaired electron).
So, \( \mu = \sqrt{1(1+2)} = \sqrt{3} \)
\( \mu \approx 1.73 \) B.M. (Bohr Magnetons). The magnetic moment tells us how strongly a substance is attracted to a magnetic field.
In simple words: Cu²⁺ has an electron setup of [Ar]3d⁹, meaning it has one electron that is not paired up. Its magnetic strength, called magnetic moment, is about 1.73 Bohr Magnetons.

🎯 Exam Tip: Remember to remove electrons from the 4s orbital first, then from the 3d orbital, when forming transition metal ions. Magnetic moment calculations depend solely on the number of unpaired electrons.

 

Question 6. Explain the following:
(a) The size of 5d transition elements is almost similar to 4d transition elements.
(b) Transition elements forms coordinate bonds
Answer:
(a) The size of 5d transition elements is almost similar to 4d transition elements in the same group. For example, in group 3, there's a regular increase in size from Scandium (Sc) to Yttrium (Y) and then to Lanthanum (La). However, after Lanthanum, the 5d series elements follow the lanthanoids. The lanthanoid contraction occurs just before the 5d series begins, causing a significant decrease in atomic radii across the lanthanoid series. This effect carries over, leading to the 5d transition elements having nearly the same size as their corresponding 4d counterparts. This similarity in size leads to similar chemical properties for elements in the 4d and 5d series of the same group.
(b) Transition elements readily form coordinate bonds because they have vacant d-orbitals available. These empty d-orbitals can accept lone pairs of electrons from ligands (molecules or ions that can donate electrons), forming coordinate covalent bonds. The d-orbitals are energetically accessible, allowing them to act as Lewis acids. The tendency to form coordinate compounds generally increases with a decrease in the atomic size of the metal ion and an increase in its positive charge, as this enhances its ability to attract electron pairs from ligands. This is why many transition metals form brightly coloured complex ions.
In simple words: (a) Elements in the 5d series are almost the same size as those in the 4d series because of the "lanthanoid contraction" that happens before the 5d series. (b) Transition elements can form special bonds called coordinate bonds because they have empty d-orbitals that can accept electrons from other atoms or molecules.

🎯 Exam Tip: For (a), link the similar sizes of 4d and 5d elements directly to the lanthanoid contraction. For (b), emphasize the availability of vacant d-orbitals as the primary reason for coordinate bond formation.

 

Question 7. Write four differences between lanthanoids and actinoids.
Answer: The lanthanoids and actinoids are two series of f-block elements, each with distinct properties. Here are four key differences between them:

Lanthanoids and actinoids both belong to the f-block but differ significantly in their electronic configurations and chemical behaviors due to the nature of 4f and 5f orbitals. Actinoids are generally more reactive and complex than lanthanoids.
In simple words: Lanthanoids are mostly not radioactive, usually have a +3 charge, and form less basic compounds. Actinoids are all radioactive, can have many different charges, form more basic compounds, and make special ions called oxo ions.

🎯 Exam Tip: Focus on distinguishing features like radioactivity, range of oxidation states, basicity of hydroxides, and tendency to form complexes/oxo-ions when comparing lanthanoids and actinoids.

 

Question 8. The atomic radius of Zr (57) and Hf (72) is almost same. Give reason.
Answer: Zirconium (Zr, Z=57) and Hafnium (Hf, Z=72) are elements in the same group (Group 4) of the periodic table, with Zr being in the 4d series and Hf in the 5d series. Normally, atomic radius increases down a group. However, the atomic radius of Hf is almost the same as that of Zr. This unexpected similarity in size is a direct consequence of the lanthanoid contraction. The lanthanoid series (elements 58-71) comes just before Hafnium in the periodic table. As electrons fill the 4f orbitals across the lanthanoid series, the poor shielding effect of these 4f electrons leads to a significant increase in the effective nuclear charge. This stronger nuclear pull causes the elements following the lanthanoids, including Hf, to have smaller atomic radii than would be expected. This contraction effectively counteracts the normal increase in size down a group, making Hf similar in size to Zr. This phenomenon explains why these elements have very similar chemical properties.
In simple words: Zirconium and Hafnium have almost the same atomic size because of lanthanoid contraction. The elements between them cause the outer electrons to be pulled in tighter, making Hafnium smaller than it should be.

🎯 Exam Tip: The similar atomic radii of 4d and 5d elements in the same group (e.g., Zr and Hf, Nb and Ta) is a classic example demonstrating the impact of lanthanoid contraction.

 

Question 9. The ionisation potential of Au (79) and Ag (47) are almost same. Explain.
Answer: Silver (Ag, Z=47) is in the 4d series, and Gold (Au, Z=79) is in the 5d series, both belonging to Group 11. While the atomic size of Au and Ag are relatively similar due to relativistic effects and the lanthanoid contraction (which impacts Au more significantly by making it smaller than expected), their ionization enthalpies are not identical. The statement "ionisation potential of Au (79) and Ag (47) are almost same" as given in the question is not entirely accurate. In reality, the ionization enthalpy of Gold (Au) is significantly *higher* than that of Silver (Ag). This is because the effective nuclear charge (Zeff) of Au is much greater than that of Ag, primarily due to the poor shielding effect of the f-orbitals (lanthanoid contraction) that precedes gold, along with relativistic effects. These effects cause the electrons in gold to be held much more tightly by the nucleus, making it harder to remove them. Therefore, the ionization enthalpy of Au (79) is actually higher than Ag (47). This difference in electron binding energy contributes to gold's greater nobility.
In simple words: The ionization potential, which is how much energy is needed to remove an electron, for Gold (Au) is actually higher than Silver (Ag). This is because Gold's electrons are held more strongly by its nucleus, partly due to the f-orbitals not shielding well.

🎯 Exam Tip: Be cautious about statements claiming "almost same" properties for 4d and 5d elements. While sizes can be similar due to lanthanoid contraction, ionization enthalpies often differ due to cumulative effects of effective nuclear charge and relativistic effects, especially for heavier elements like Gold.