Chapter 4 d and f Block Elements – NCERT Class 12 Chemistry

Question 4.1

Write down the electronic configuration of:
(i) $Cr^{3+}$ (ii) $Pm^{3+}$ (iii) $Cu^{+}$ (iv) $Ce^{4+}$ (v) $Co^{2+}$ (vi) $Lu^{2+}$ (vii) $Mn^{2+}$ (viii) $Th^{4+}$

(i) $Cr^{3+} (Z=24)$: $[Ar] 3d^3$
(ii) $Pm^{3+} (Z=61)$: $[Xe] 4f^4$
(iii) $Cu^{+} (Z=29)$: $[Ar] 3d^{10}$
(iv) $Ce^{4+} (Z=58)$: $[Xe]$ (Inert gas configuration)
(v) $Co^{2+} (Z=27)$: $[Ar] 3d^7$
(vi) $Lu^{2+} (Z=71)$: $[Xe] 4f^{14} 5d^1$
(vii) $Mn^{2+} (Z=25)$: $[Ar] 3d^5$
(viii) $Th^{4+} (Z=90)$: $[Rn]$ (Inert gas configuration)

Question 4.2

Why are $Mn^{2+}$ compounds more stable than $Fe^{2+}$ towards oxidation to their +3 state?

Electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$, which is a stable half-filled configuration. It resists losing another electron.

Electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$. By losing one electron, it forms $Fe^{3+} ([Ar] 3d^5)$, gaining extra stability. Thus, $Fe^{2+}$ oxidizes easily, while $Mn^{2+}$ does not.

Question 4.3

Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

As the atomic number increases from Sc to Mn, the nuclear charge increases, and the 3d orbitals are progressively filled. However, the removal of the two 4s electrons becomes energetically favorable because the increasing nuclear charge pulls the d-electrons closer, making them harder to remove (to form higher oxidation states). Particularly, $Mn^{2+}$ achieves a stable half-filled $d^5$ configuration, making the +2 state very stable at the midpoint.

Question 4.4

To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Stability is strongly influenced by $d^0, d^5$ (half-filled), and $d^{10}$ (fully filled) configurations.

$Sc^{3+} (d^0)$: Highly stable (Noble gas config). Sc does not show +2.
$Mn^{2+} (d^5)$: Very stable due to half-filled subshell.
$Zn^{2+} (d^{10})$: Very stable due to fully filled subshell. Zn does not show variable oxidation states.

Question 4.5

What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: $3d^3, 3d^5, 3d^8, 3d^4$?

$3d^3 4s^2$ (Vanadium): +5 (loss of all electrons), +4, +3, +2.
$3d^5 4s^1$ (Chromium): +3, +6.
$3d^5 4s^2$ (Manganese): +2, +7.
$3d^8 4s^2$ (Nickel): +2.
$3d^4$ (Does not exist in ground state, usually becomes $3d^5$). If $Cr$, +3, +6.

Question 4.6

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.

Permanganate ion ($MnO_4^-$): Mn is in +7 state (Group 7).
Chromate ion ($CrO_4^{2-}$): Cr is in +6 state (Group 6).
Vanadate ion ($VO_4^{3-}$): V is in +5 state (Group 5).

Question 4.7

What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Definition

The steady decrease in the atomic and ionic radii of lanthanoid elements with increasing atomic number is called Lanthanoid Contraction. This is due to the poor shielding effect of 4f electrons.

Consequences

Similarity in sizes of 2nd and 3rd transition series: E.g., Zr (160 pm) and Hf (159 pm) have almost identical radii.
Difficulty in separation: Lanthanoids have very similar chemical properties due to similar sizes.
Basicity differences: $La(OH)_3$ is more basic than $Lu(OH)_3$ due to the decrease in size from La to Lu (Covalent character increases).

Question 4.8

What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

Characteristics: Variable oxidation states, formation of coloured ions, paramagnetic behaviour, catalytic activity, formation of complexes and alloys.

Why: They represent a transition in properties from highly electropositive s-block metals to electronegative p-block elements. They have incompletely filled d-orbitals in ground or stable oxidation states.

Exceptions: Zn, Cd, and Hg are d-block elements but not regarded as typical transition elements because they have fully filled $d^{10}$ configuration in their ground state as well as in their common oxidation states.

Question 4.9

In what way is the electronic configuration of the transition elements different from that of the non-transition elements?

Transition elements involve the filling of the penultimate shell (n-1)d orbitals, while non-transition elements (representative elements) involve the filling of the valence shell ns or np orbitals.

General Config: Transition $(n-1)d^{1-10} ns^{1-2}$ vs Non-transition $ns^{1-2}$ or $ns^2 np^{1-6}$.

Question 4.10

What are the different oxidation states exhibited by the lanthanoids?

The most common and stable oxidation state is +3. However, some elements show +2 and +4 states to attain stable configurations:

+2: Eu, Yb (due to $f^7, f^{14}$).
+4: Ce, Tb (due to $f^0, f^7$).

Question 4.11

Explain giving reasons:
(i) Transition metals show paramagnetic behaviour.
(ii) Enthalpies of atomisation are high.
(iii) Form coloured compounds.
(iv) Act as good catalysts.

(i) Paramagnetism: Due to the presence of unpaired electrons in (n-1)d orbitals.
(ii) Enthalpy of Atomisation: High due to strong metallic bonding arising from the participation of both (n-1)d and ns electrons in bonding (large number of unpaired electrons).
(iii) Colour: Due to d-d transitions. Electrons absorb energy from visible light to jump from lower d-orbital to higher d-orbital; the complementary colour is transmitted.
(iv) Catalysis: Due to variable oxidation states (ability to form intermediates) and large surface area (provides sites for reactants). Example: V2O5, Fe.

Question 4.12

What are interstitial compounds? Why are such compounds well known for transition metals?

Interstitial compounds are formed when small atoms like H, C, or N are trapped inside the crystal lattice of metals.

Why: Transition metals have crystal lattices with large voids (interstices) that can easily accommodate these small atoms without changing the chemical properties significantly, but altering physical properties (hardness, melting point).

Question 4.13

How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.

In transition metals, oxidation states differ by 1 unit (e.g., $Fe^{2+}, Fe^{3+}$; $Cu^+, Cu^{2+}$) because (n-1)d and ns energies are close.

In non-transition metals (p-block), oxidation states differ by 2 units due to the inert pair effect (e.g., $Pb^{2+}, Pb^{4+}$; $Sn^{2+}, Sn^{4+}$).

Question 4.14

Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Preparation Steps

Fusion: Chromite ore ($FeCr_2O_4$) is fused with sodium carbonate in excess air.
$$4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \to 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$$
Acidification: Yellow sodium chromate is acidified to orange sodium dichromate.
$$2Na_2CrO_4 + 2H^+ \to Na_2Cr_2O_7 + 2Na^+ + H_2O$$
Conversion: Treatment with KCl forms $K_2Cr_2O_7$ (less soluble).
$$Na_2Cr_2O_7 + 2KCl \to K_2Cr_2O_7 + 2NaCl$$

Effect of pH

Increasing pH (adding base) converts orange Dichromate ($Cr_2O_7^{2-}$) to yellow Chromate ($CrO_4^{2-}$).

$$Cr_2O_7^{2-} + 2OH^- \to 2CrO_4^{2-} + H_2O$$

Question 4.15

Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with: (i) iodide (ii) iron(II) solution and (iii) H2S.

In acidic medium, $Cr_2O_7^{2-}$ accepts 6 electrons to become $2Cr^{3+}$.

$$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$$

(i) Iodide: $6I^- \to 3I_2 + 6e^-$. Net: $Cr_2O_7^{2-} + 14H^+ + 6I^- \to 2Cr^{3+} + 3I_2 + 7H_2O$.
(ii) Iron(II): $6Fe^{2+} \to 6Fe^{3+} + 6e^-$. Net: $Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \to 2Cr^{3+} + 6Fe^{3+} + 7H_2O$.
(iii) H2S: $3H_2S \to 6H^+ + 3S + 6e^-$. Net: $Cr_2O_7^{2-} + 8H^+ + 3H_2S \to 2Cr^{3+} + 3S + 7H_2O$.

Question 4.16

Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) SO2 and (iii) oxalic acid?

Preparation

From Pyrolusite ($MnO_2$): Fused with KOH/O2 to form green Manganate, then electrolytically oxidised to purple Permanganate.

$$2MnO_2 + 4KOH + O_2 \to 2K_2MnO_4 + 2H_2O$$ $$MnO_4^{2-} \xrightarrow{\text{electrolysis}} MnO_4^- + e^-$$

Reactions (Acidic Medium)

Reduction Half: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$

(i) Iron(II): $5Fe^{2+} \to 5Fe^{3+} + 5e^-$. Net: $MnO_4^- + 8H^+ + 5Fe^{2+} \to Mn^{2+} + 5Fe^{3+} + 4H_2O$.
(ii) SO2: $5SO_2 + 10H_2O \to 5SO_4^{2-} + 20H^+ + 10e^-$. (Requires 2 MnO4). Net: $2MnO_4^- + 5SO_2 + 2H_2O \to 2Mn^{2+} + 5SO_4^{2-} + 4H^+$.
(iii) Oxalic Acid: $5C_2O_4^{2-} \to 10CO_2 + 10e^-$. Net: $2MnO_4^- + 16H^+ + 5C_2O_4^{2-} \to 2Mn^{2+} + 10CO_2 + 8H_2O$.

Question 4.17

Use electrode potentials to comment on: (i) stability of $Fe^{3+}$ vs $Cr^{3+}/Mn^{3+}$ (ii) ease of oxidation of Fe vs Cr/Mn.

Data Analysis

$E^\circ (Cr^{3+}/Cr^{2+}) = -0.4V$, $E^\circ (Mn^{3+}/Mn^{2+}) = +1.5V$, $E^\circ (Fe^{3+}/Fe^{2+}) = +0.8V$.

(i) Stability of +3 State: A negative or low reduction potential from +3 to +2 indicates stability of +3.
- $Cr^{3+}$ is most stable (-0.4V means it resists reduction to +2).
- $Mn^{3+}$ is strong oxidising agent (+1.5V, wants to reduce to $Mn^{2+}$).
- $Fe^{3+}$ is moderately stable but $Fe^{2+}$ tends to oxidize.
(ii) Ease of Oxidation (M to M2+): Look at $M^{2+}/M$.
- $Mn (-1.2V) < Cr (-0.9V) < Fe (-0.4V)$.
- Lower potential means easier oxidation. Order of ease: Mn > Cr > Fe.

Question 4.18

Predict coloured ions in aqueous solution: $Ti^{3+}, V^{3+}, Cu^+, Sc^{3+}, Mn^{2+}, Fe^{3+}, Co^{2+}$.

Colour depends on the presence of unpaired d-electrons.

$Ti^{3+} (d^1)$: Coloured.
$V^{3+} (d^2)$: Coloured.
$Cu^+ (d^{10})$: Colourless (No unpaired e-).
$Sc^{3+} (d^0)$: Colourless (No d electrons).
$Mn^{2+} (d^5)$: Coloured (Pale pink).
$Fe^{3+} (d^5)$: Coloured (Yellow/Brown).
$Co^{2+} (d^7)$: Coloured (Pink).

Question 4.21

Account for: (i) $Cr^{2+}$ is reducing, $Mn^{3+}$ is oxidising (both $d^4$). (ii) $Co(II)$ is stable but oxidised in complexes. (iii) $d^1$ configuration unstable.

(i) $Cr^{2+} \to Cr^{3+} (d^3)$. $d^3$ has stable half-filled $t_{2g}^3$ config. Hence reducing.
$Mn^{3+} \to Mn^{2+} (d^5)$. $d^5$ is stable half-filled. Hence oxidising.
(ii) Strong field ligands cause pairing in $Co^{3+} (d^6)$ to form stable $t_{2g}^6$ (low spin), reducing the energy and making oxidation of $Co(II) \to Co(III)$ favourable.
(iii) $d^1$ ions easily lose the single electron to form stable noble gas configuration or disproportionate. Hydration energy often compensates for ionisation.

Question 4.22

What is ‘disproportionation’? Give two examples.

Disproportionation: A reaction where a particular element in one oxidation state is simultaneously oxidised and reduced.

Examples:

$2Cu^+ \to Cu^{2+} + Cu$ (Cu(I) to Cu(II) and Cu(0)).
$3MnO_4^{2-} + 4H^+ \to 2MnO_4^- + MnO_2 + 2H_2O$ (Mn(VI) to Mn(VII) and Mn(IV)).

Question 4.26

Indicate steps in preparation of: (i) K2Cr2O7 (ii) KMnO4.

(Refer to detailed steps in Q4.14 and Q4.16 solutions above).

(i) K2Cr2O7: Chromite $\to$ Na-Chromate $\to$ Na-Dichromate $\to$ K-Dichromate.
(ii) KMnO4: Pyrolusite $\to$ K-Manganate $\to$ K-Permanganate.

Question 4.28

Decide atomic numbers of inner transition elements: 29, 59, 74, 95, 102, 104.

Inner transition elements are f-block elements:

Lanthanoids: Z = 58 to 71.
Actinoids: Z = 90 to 103.

Result: 59 (Pr), 95 (Am), 102 (No).

Question 4.31

Derive electronic configuration of $Ce^{3+}$ and calculate magnetic moment.

Ce (Z=58): $[Xe] 4f^1 5d^1 6s^2$.
$Ce^{3+}$: Remove 2 from 6s and 1 from 5d. Config: $[Xe] 4f^1$.

Unpaired electrons $n = 1$.

$$\mu = \sqrt{n(n+2)} = \sqrt{1(3)} = \sqrt{3} \approx 1.73 \text{ BM}$$

Question 4.38

Inferred from magnetic moments: $K_4[Mn(CN)_6]$ (2.2), $[Fe(H_2O)_6]^{2+}$ (5.3), $K_2[MnCl_4]$ (5.9).

Analysis

$K_4[Mn(CN)_6]$: $Mn^{2+} (d^5)$. $\mu=2.2 \implies n=1$. Strong field CN causes pairing ($t_{2g}^5$). Low spin octahedral.
$[Fe(H_2O)_6]^{2+}$: $Fe^{2+} (d^6)$. $\mu=5.3 \implies n=4$. Weak field H2O, no pairing ($t_{2g}^4 e_g^2$). High spin octahedral.
$K_2[MnCl_4]$: $Mn^{2+} (d^5)$. $\mu=5.9 \implies n=5$. Weak field Cl, no pairing. High spin tetrahedral.

NCERT Solutions