Chapter 4 d- and f-Block Elements

Frequently Asked Questions

NCERT Solutions Class 12 Chemistry Chapter 4: d- and f-block Elements

NCERT Solutions for Class 12 Chemistry Chapter 4 (d- and f-Block Elements) play a crucial role in helping students build a strong conceptual foundation for both CBSE board exams and competitive exams like JEE Main, JEE Advanced, and NEET. This chapter gives students the theoretical and chemical foundations for understanding the Transition and Inner-transition elements. Understanding these concepts is essential for students as they lead to many of the systems that utilize this technology, including industrial catalysis and the manufacturing of high-strength alloys and magnets.

The NCERT Solutions for Class 12 Chemistry Chapter 4 will help students learn how to use Electronic Configurations to explain why transition metals exhibit variable oxidation states and form colored ions. Well-explained NCERT solutions enable students to strengthen fundamentals, practice important reasoning-based questions, and perform better in term exams, board exams, and national-level competitive tests.

1.0Class 12 Chemistry Chapter 4 d- and f-block Elements: Key Concepts

d- and f-Block Elements is an important chapter in Class 12 Chemistry. It covers the electronic configuration, physical properties, oxidation states, magnetic properties, chemical reactions, and important compounds of transition and inner-transition elements.

d-Block and Transition Elements: Learn the electronic configuration of transition elements and understand why Zn, Cd, and Hg are not considered transition elements.
Physical Properties: Study the important physical properties of transition elements, including their metallic character, high melting points, hardness, and high enthalpy of atomisation.
Oxidation States: Understand why transition elements show variable oxidation states. Manganese is an important example because it shows several oxidation states.
Magnetic Properties: Learn how to calculate the magnetic moment of transition metal ions using the formula:
μ = √[n(n + 2)] BM
where n represents the number of unpaired electrons.
Colour of Transition Metal Ions: Understand why many transition metal ions and compounds are coloured and how d-d transitions cause the absorption of certain wavelengths of light.
Lanthanoid Contraction: Learn about the gradual decrease in the atomic and ionic radii of lanthanoids and its effect on the properties of elements that follow them.
Actinoids: Study the electronic configuration, oxidation states, and chemical properties of actinoids and understand their similarities and differences with lanthanoids.
Potassium Dichromate (K₂Cr₂O₇): Learn its preparation, properties, important reactions, and uses.
Potassium Permanganate (KMnO₄): Study its preparation, properties, oxidising action, important reactions, and uses.
Important Trends and Reactions: Revise important trends in atomic size, ionic size, oxidation states, magnetic behaviour, colour, and reactivity of d- and f-block elements.

2.0NCERT Solutions Class 12 Chemistry Chapter 4d- and f-block: Detailed Solutions

INTEXT QUESTIONS

1. Silver atom has completely filled d orbitals $(4 d^{10})$ in its ground state. How can you say that it is a transition element? Sol. Ag has a completely filled 4d-orbital ( $4 d^{10} 5 s^{1}$ ) in its ground state. Now, silver displays two oxidation states (+1 and +2). In the +1 oxidation state, an electron is removed from the s-orbital. However, in the +2 oxidation state, an electron is removed from the d-orbital. Thus, the d-orbital now becomes incomplete $(4 d^{9})$ . Hence, it is a transition element.

In the series $Sc (Z = 21)$ to $Zn (Z = 30)$ , the enthalpy of atomization of zinc is the lowest, i.e., $126 kJ mol^{- 1}$ . Why? Sol. The extent of metallic bonding an element undergoes decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals (except Zn, electronic configuration: $3 d^{10} 4 s^{2}$ ), there are some unpaired electrons that account for their stronger metallic bonding. Due to the absence of these unpaired electrons, the interatomic electronic bonding is the weakest in Zn and as a result, it has the least enthalpy of atomization.
Which of the 3d-series of the transition metals exhibits the largest number of oxidation states and why ? Sol. $Mn (Z = 25) = 3 d^{5} 4 s^{2}$ Mn has the maximum number of unpaired electrons present in the d-subshell (5 electrons). Hence, Mn exhibits the largest number of oxidation states, ranging from +2 to +7.
The $E^{θ} (M^{2 +} / M)$ value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high energy and low energy) Sol. The $E^{\circ} (M^{2 +} / M)$ value of a metal depends on the energy changes involved in the following: (i) Sublimation : The energy required for converting one mole of an atom from the solid state to the gaseous state.

$M_{(s)} \to M_{(g)}$

$Δ_{s} H$ (Sublimation energy) (ii) Ionization : The energy required to take out electrons from one mole of atoms in the gaseous state.

$M_{(g)} \to M^{2 +}$

$Δ_{i} H$ (Ionization energy) (iii) Hydration : The energy released when one mole of ions are hydrated.

$M_{(g)}^{2 +} \to M_{(aq)}^{2 +}$

$Δ_{hyd} H$ (Hydration energy) Now, copper has a high energy of atomization and low hydration energy. Hence, the $E^{\circ} (M^{2 +} / M)$ value for copper is positive.

5. How would you account for the irregular variation of ionization enthalpies (first and second) in the first series of the transition elements? Sol. Ionization enthalpies are found to increase in the given series due to a continuous filling of the inner d-orbitals. The irregular variations of ionization enthalpies can be attributed to the extra stability of configurations such as $d^{0}, d^{5}$ , $d^{10}$ . Since these states are exceptionally stable, their ionization enthalpies are very high. In case of first ionization energy, Cr has low ionization energy. This is because after losing one electron, it attains the stable configuration $(3 d^{5})$ .

On the other hand, Zn has exceptionally high first ionization energy as an electron has to be removed from stable and fully-filled orbitals ( $3 d^{10} 4 s^{2}$ ). Second ionization energies are higher than the first since it becomes difficult to remove an electron when an electron has already been taken out. Also, elements like Cr and Cu have exceptionally high second ionization energies as after losing the first electron, they have attained the stable configuration ( $Cr^{+} : 3 d^{5}$ and $Cu^{+} : 3 d^{10}$ ). Hence, taking out one electron more from this stable configuration will require a lot of energy.

Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only? Sol. Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they are able to oxidize the metal to its highest oxidation state.
Which is a stronger reducing agent $Cr^{2 +}$ or $Fe^{2 +}$ and why? Sol. The following reactions are involved when $Cr^{2 +}$ and $Fe^{2 +}$ act as reducing agents.

$Cr^{2 +} ⟶ Cr^{3 +} Fe^{2 +} ⟶ Fe^{3 +}$

The $E_{Cr^{3 +} / Cr^{2 +}}^{o}$ value is -0.41 V and $E_{Fe^{3 +} / Fe^{2 +}}^{o}$ is +0.77 V. This means that $Cr^{2 +}$ can be easily oxidized to $Cr^{3 +}$ , but $Fe^{2 +}$ does not get oxidized to $Fe^{3 +}$ easily. Therefore, $Cr^{2 +}$ is a better reducing agent that $Fe^{3 +}$ .

8. Calculate the 'spin only' magnetic moment of $M_{(aq)}^{2 +}$ ion $(Z = 27)$ . Sol. $Z = 27$

$\Rightarrow [Ar] 3 d^{7} 4 s^{2} ∴ M^{2 +} = [Ar] 3 d^{7} 3 d^{7} = 11111$

i.e., 3 unpaired electrons

$∴ n = 3$

$\Rightarrow n (n + 2) = μ$ $\Rightarrow 3 (3 + 2) = μ$ $\Rightarrow 15 = μ$ $\Rightarrow μ \approx 4 BM$

9. Explain why $Cu^{+}$ ion is not stable in aqueous solutions? Sol. In an aqueous medium, $Cu^{2 +}$ is more stable than $Cu^{+}$ . This is because although energy is required to remove one electron from $Cu^{+}$ to $Cu^{2 +}$ , high hydration energy of $Cu^{2 +}$ compensates for it. Therefore, $Cu^{+}$ ion in an aqueous solution is unstable. It disproportionate to give $Cu^{2 +}$ and Cu .

10. Actinoid contraction is greater from element to element than lanthanoide contraction. Why? Sol. In actinoids, 5f orbitals are filled. These 5f orbitals have a poorer shielding effect than 4f orbitals (in lanthanoides). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinides is much more that that experienced by lanthanoides. Hence, the size contraction in actinides is greater as compared to that in lanthanoides.

3.0NCERT EXERCISE

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) $Th^{4 +}$ Sol. (i) $Cr^{3 +} : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{3}$ (ii) $Pm^{3 +} : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2}$ $5 p^{6} 4 f^{4}$ (iii) $Cu^{+ 1} : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10}$

(iv) $Ce^{4 +} : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2}$ $5 p^{6}$ (v) $Co^{2 +} : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7}$ (vi) $Lu^{2 +} : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2}$ $5 p^{6} 4 f^{14} 5 d^{1}$ (vii) $Mn^{2 +} : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5}$ (viii) $Th^{4 +} : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 4 f^{14}$ $5 s^{2} 5 p^{6} 5 d^{10} 6 s^{2} 6 p^{6}$

2. Why are $Mn^{2 +}$ compounds more stable than $Fe^{2 +}$ towards oxidation to their +3 state ? Sol. Electronic configuration of $Mn^{2 +}$ is $[Ar]^{18} 3 d^{5}$ . (half-filled stability) Electronic configuration of $Fe^{2 +}$ is [Ar] $^{18} 3 d^{6}$ . (After losing one electron gain half-filled stability).

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 ? Sol. The oxidation states displayed by the first half of the first row of tranition metals are given in the table below. Oxidation state Sc Ti V Cr Mn

$+ 2 + 2 + 2 + 2 + 3 + 3 + 3 + 3 + 3 + 4 + 4 + 4 + 4 + 5 + 5 + 6 + 6 + 7$

It can be easily observed that except Sc, all others metals display +2 oxidation state. Also, on moving from Sc to Mn, the atomic number increases from 21 to 25. This means the number of electron in the 3d-orbital also increase from 1 to 5.

$Sc (+ 2) = d^{1} Ti (+ 2) = d^{2} V (+ 2) = d^{3} Cr (+ 2) = d^{4} Mn (+ 2) = d^{5}$

+2 oxidation state is attained by the loss of the two 4s electrons by these metals. Since the number of d electrons in (+2) state also increase from $Ti (+ 2)$ to $Mn (+ 2)$ , the stability of +2 state increases (as d-orbital is becoming more and more half-filled). $Mn (+ 2)$ has $d^{5}$ electrons (that is half-filled d shell, which is highly stable).

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. Sol. The elements in the first half of the transition series exhibit many oxidation states with Mn exhibiting maximum number of oxidation states (+2 to +7). The stability of +2 oxidation state increases with the increase in atomic number. This happens as more electrons are getting filled in the d-orbital.

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 : $3 d^{3}, 3 d^{5}, 3 d^{8}$ and $3 d^{4}$ ?

Sol.

	Electronic configuration in ground state	Stable oxidation states
(i)	$3 d^{3}$ (Vanadium)	+2, +3, +4 and +5
(ii)	$3 d^{5}$ (Chromium)	+3, +4, +6
(iii)	$3 d^{3}$ (Manganese)	+2, +4, +6, +7
(iv)	$3 d^{8}$ (Cobalt)	+2, +3
(v)	$3 d^{4}$	There is no 3 deconfiguration in ground state

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.

Sol.

(i) Vanadate, $VO_{3}^{-}$ Oxidation state of V is +5. (ii) Chromate, $CrO_{4}^{2 -}$ Oxidation state of Cr is +6 (iii) Permanganate, $MnO_{4}^{-}$ Oxidation state of Mn is +7.

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

Sol. A regular decrease (contraction) in the atomic and ionic radii of lanthanoides with increasing atomic number is known as lanthanoid contraction.

Consequences of lanthanoid contraction

(i) There is similarity in the properties of second and third transition series. (ii) Separation of lanthanoids is possible due to lanthanoide contraction. (iii) It is due to lanthanoide contraction that there is variation in the basic strength of lanthanoide hydroxides. (Basic strength decreases from La $(OH)_{3}$ to $Lu (OH)_{3}$ )

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? Sol. Transition elements are those elements in which the atoms or ions (in stable oxidation state) contain partially filled d-orbital. These elements lie in the d-block and show a transition of properties between s-block and p-block. Therefore, these are called transition elements. Elements such as Zn, Cd and Hg cannot be classified as transition elements because these have completely filled d-subshell.

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

Sol. Transition metals have a partially filled d-orbital. Therefore, the electronic configuration of transition elements is $(n - 1) d^{1 - 10} ns^{0 - 2}$ . The non-transition elements either do not have a d-orbital or have a fully filled d-orbital. Therefore, the electronic configuration of nontransition elements is $ns^{1 - 2}$ or $ns^{2} np^{1 - 6}$ .

10. What are the different oxidation states exhibited by the lanthanoids? Sol. In the lanthanoide series, +3 oxidation state is most common i.e., Ln (III) compounds are predominant. However, +2 and +4 oxidation states can also be found in the solution or in solid compounds.

11. 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) Transition metals and their many compounds act as good catalyst.

Sol.

(i) Transition metals show paramagnetic behaviour. Paramagnetism arises due to the presence of unpaired electrons with each electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum. (ii) Transition elements have high effective nuclear charge and a large number of valence electrons. Therefore, they form very strong metallic bonds. As a result, the enthalpy of atomization of transition metals is high. (iii) The catalytic activity of the transition elements can be explained by two basic facts. (a) Show variable oxidation states and form complexes (b) Provide a suitable surface

What are interstitial compounds? Why are such compounds well known for transition metals? Sol. Transition metals are large in size and contain lots of interstitial sites. Transition elements can trap atoms of other elements (that have small atomic size), such as H, C, N, in the interstitial sites of their crystal lattices. The resulting compounds are called interstitial compounds.
How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples. Sol. In transition elements, the oxidation state can vary from +1 to the highest oxidation state by removing all its valence electrons. Also, in transition elements, the oxidation states differ by

$1 (Fe^{2 +} and Fe^{3 +}; Cu^{+} and Cu^{2 +}) .$

In non-transition elements, the oxidation states differ by 2 ,

$for example, + 2 and + 4 or + 3 and + 5, etc.$

Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate ? Sol. Potassium dichromate is prepared from chromite ore $(FeCr_{2} O_{4})$ in the following steps. Step (1) : Preparation of sodium chromate

$4 FeCr_{2} O_{4} + 16 NaOH + 7 O_{2} \to 8 Na_{2} CrO_{4} + 2 Fe_{2} O_{3} + 8 H_{2} O$

Step (2) : Conversion of sodium chromate into sodium dichromate

$2 Na_{2} CrO_{4} + conc. H_{2} SO_{4} ⟶ Na_{2} Cr_{2} O_{7} + Na_{2} SO_{4} + H_{2} O$

Step (3) : Conversion of sodium dichromate to potassium dichromate

$Na_{2} Cr_{2} O_{7} + 2 KCl ⟶ K_{2} Cr_{2} O_{7} + 2 NaCl$

Potassium dichromate being less soluble than sodium dichromate is obtained in the form of orange coloured crystals and can be removed by filtration.The dichromate ion $(Cr_{2} O_{7}^{2 -})$ exists in equilibrium with chromate $(CrO_{4}^{2})$ ion at pH 4 . However, by changing the pH, they can be interconverted

$Chromate ion(Yellow) 2 CrO_{4}^{2 -} + 2 H^{+} pH = 4 Cr_{2} O_{7}^{2 -} + H_{2} O$

On increasing pH,

$Cr_{2} O_{7}^{2 -} + 2 OH^{-} ⟶ 2 CrO_{4}^{2 -} + H_{2} O$

Describe the oxidation action of potassium dichromate and write the ionic equations for its reaction with : (i) iodide (ii) iron(II) solution and (iii) $H_{2} S$ Sol. $K_{2} Cr_{2} O_{7}$ acts as a very strong oxidising agent in the acidic medium.

$K_{2} Cr_{2} O_{7} + 4 H_{2} SO_{4} ⟶ K_{2} SO_{4} + Cr_{2} (SO_{4})_{3} + 4 H_{2} O + 3 [O]$

$K_{2} Cr_{2} O_{7}$ takes up electrons to get reduced and acts as an oxidising agent. The reaction of $K_{2} Cr_{2} O_{7}$ with iodide ion, iron (II) solution and $H_{2} S$ are given below. (i) $K_{2} Cr_{2} O_{7}$ oxidizes iodide to iodine.

$Cr_{2} O_{7}^{2 -} + 6 I^{-} + 14 H^{+} \to 2 Cr^{3 +} + 3 I_{2} + 7 H_{2} O$

(ii) $K_{2} Cr_{2} O_{7}$ oxidizes iron (II) solution to iron (III) solution i.e., ferrous ions to ferric ions.

$Cr_{2} O_{7}^{2 -} + 14 H^{+} + 6 Fe^{2 +} \to 2 Cr^{3 +} + 6 Fe^{+ 3} + 7 H_{2} O$

(iii) $K_{2} Cr_{2} O_{7}$ oxidizes $H_{2} S$ to sulphur.

$Cr_{2} O_{7}^{2 -} + 3 H_{2} S + 8 H^{+} \to 2 Cr^{3 +} + 3 S + 7 H_{2} O$

Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron (II) ions (ii) $SO_{2}$ (iii) oxalic acid? Write the ionic equations for the reactions.

Sol. Potassium permanganate can be prepared from pyrolusite $(MnO_{2})$ . The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as $KNO_{3}$ or $KClO_{4}$ , to give $K_{2} MnO_{4}$ .

$2 MnO_{2} + 4 KOH + O_{2} heat 2 K_{2} MnO_{4} + 2 H_{2} O$

The green mass can be extracted with water and then oxidized either electrolytically or by passing chlorine/ozone into the solution. Electrolytic oxidation

$K_{2} MnO_{4} ⟷ 2 K^{+} + MnO_{4}^{2 -} H_{2} O ⟷ H^{+} + OH^{-}$

At anode, manganate ions are oxidized to permanganate ions.

$Green MnO_{4}^{2 -} ⟷ Purple MnO_{4}^{-} + e^{-}$

Oxidation by chlorine

$2 K_{2} MnO_{4} + Cl_{2} \to 2 KMnO_{4} + 2 KCl 2 MnO_{4}^{2 -} + Cl_{2} \to 2 MnO_{4}^{-} + 2 Cl^{-}$

Oxidation by ozone

$2 K_{2} MnO_{4} + O_{3} + H_{2} O \to 2 KMnO_{4} + 2 KOH + O_{2} 2 MnO_{4}^{2 -} + O_{3} + H_{2} O \to 2 MnO_{4}^{2 -} + 2 OH^{-} + O_{2}$

(i) Acidified $KMnO_{4}$ solution oxidizes Fe (II) ions to Fe (III) ions i.e., ferrous to ferric ions.

$MnO_{4}^{-} + 5 Fe^{2 +} + 8 H^{+} \to Mn^{2 +} + 5 Fe^{3 +} + 4 H_{2} O$

(ii) Acidified potassium permanganate oxidizes $SO_{2}$ to sulphuric acid.

$2 MnO_{4}^{-} + 10 SO_{2} + 5 O_{2} + 4 H_{2} O ⟶ 2 Mn^{2 +} + 10 SO_{4}^{2 -} + 8 H^{+}$

(iii) Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide.

$+ 2 MnO_{4}^{-} + 5 C_{2} O_{7}^{2 -} + 16 H^{+} ⟶ 2 Mn^{2 +} 10 CO_{2} + 8 H_{2} O$

For $M^{2 +} / M$ and $M^{3 +} / M^{2 +}$ systems, the $E^{⊖}$ values for some metals are as follows :

$Cr^{2 +} / Cr - 0.9 V Cr^{3} / Cr^{2 +} - 0.4 V Mn^{2 +} / Mn^{2 +} - 1.2 V Mn^{3 +} / Mn^{2 +} + 1.5 V Fe^{2 +} / Fe^{2 +} - 0.4 V Fe^{3 +} / Fe^{2 +} + 0.8 V$

Use this data to comment upon : (i) The stability of $Fe^{3 +}$ in acid solution as compared to that of $Cr^{3 +}$ and $Mn^{3 +}$ and (ii) The ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Sol.

(i) The $E^{Θ}$ value for $Fe^{3 +} / Fe^{2 +}$ is higher than that for $Cr^{3 +} / Cr^{2 +}$ and lower than that for $Mn^{3 +} / Mn^{2 +}$ . So, the reduction of $Fe^{3 +}$ to $Fe^{2 +}$ is easier than the reduction of $Cr^{3 +}$ to $Cr^{2 +}$ , but not as easy as the reduction of $Mn^{3 +}$ to $Mn^{2 +}$ . Hence, $Fe^{3 +}$ is more stable than $Mn^{3 +}$ , but less stable than $Cr^{3 +}$ . These metal ions can be arranged in the increasing order of their stability as: $Mn^{3 +} < Fe^{3 +} < Cr^{3 +}$ (ii) The reduction potentials for the given pairs increase in the following order.

$Mn^{2 +} / Mn < Cr^{2 +} / Cr < Fe^{2 +} / Fe$

So, the oxidation of Fe to $Fe^{2 +}$ is not a easy as the oxidation of Cr to $Cr^{2 +}$ and the oxidation of Mn to $Mn^{2 +}$ . Thus these metals can be arranged in the increasing order of their ability to get oxidised :

$Fe < Cr < Mn$

Predict which of the following will be coloured in aqueous solution? $Ti^{3 +}, V^{3 +}, Cu^{+}, Sc^{3 +}, Mn^{2 +}, Fe^{3 +}$ and $Co^{2 +}$ . Give reasons for each.

Sol. Only the ions that have unpaired electrons in d-orbital will be coloured. The ions in which d-orbital is empty or fully filled will be colourless.

Element	Atomic Number	Ionic State	Electronic Configuration in ionic state
Ti	22	$Ti^{3 +}$	$[Ar] 3 d^{2}$
V	23	$V^{3 +}$	$[Ar] 3 d^{2}$
Cu	29	$Cu^{3 +}$	$[Ar] 3 d^{10}$
Sc	21	$Sc^{3 +}$	[Ar]
Mn	25	$Mn^{3 +}$	$[Ar] 3 d^{5}$
Fe	26	$Fe^{3 +}$	$[Ar] 3 d^{5}$
Co	27	$Co^{3 +}$	$[Ar] 3 d^{7}$

19. Compare the stability of +2 oxidation state for the elements of the first transition series. Sol.

Sc			+3
Ti	+1	+2	+3	+4
V	+1	+2	+3	+4	+5
Cr	+1	+2	+3	+4	+5	+6
Mn	+1	+2	+3	+4	+5	+6	+7
Fe	+1	+2	+3	+4	+5	+6
Co	+1	+2	+3	+4	+5
Ni	+1	+2	+3	+4
Cu	+1	+2
Zn		+2

The relative stability of the +2 oxidation state increases on moving from top to bottom. This is because on moving from top to bottom, it become more and more difficult to remove the third electron from the d-orbital.

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

Sol.

(i) Electronic configuration : The general electronic configuration for lanthanoids is $[Xe]^{54} 4 f^{1 - 14} 5 d^{0 - 1} 6 s^{2}$ and that for actinoids is $[Rn]^{86} 5 f^{1 - 14} 6 d^{0 - 1} 7 s^{2}$ . (ii) Atomic and Ionic sizes : Similar to lanthanoids, actinoids also exhibit actinoid contraction (overall decrease in atomic and ionic radii). The contraction is greater due to the poor shielding effect of 5f orbitals. (iii) Oxidation states : The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of +2 and +4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d and 7s levels are of comparable energies. (iv) Chemical reactivity : In the lanthanoide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanoides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided.

21. How would you account for the following : (i) Of the $d^{4}$ species, $Cr^{2 +}$ is strongly reducing while manganese(III) is strongly oxidising. (ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. (iii) The $d^{1}$ configuration is very unstable in ions.

Sol.

(i) $Cr^{2 +}$ is strongly reducing in nature. It has a $d^{4}$ configuration. While acting as a reducting agent, it gets oxidized to $Cr^{3 +}$ (electronic configuration, $d^{3}$ ). This $d^{3}$ configuration can be written as $t_{2 g}^{3}$ configuration, which is a more stable configuration.

In the case of $Mn^{3 +} (d^{4})$ , it acts as an oxidizing agent and gets reduced to $Mn^{2 +} (d^{5})$ . This has an exactly half- filled d-orbital and is highly stable.

(ii) Co(II) is stable is aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidized to Co(III). Although the $3^{rd}$ ionization energy for Co is high, but the higher amount of crystal field stabilization energy (CFSE) released in the presence of strong field ligands overcomes this ionization energy. (iii) The ions in $d^{1}$ configuration tend to lose one more electron to get into stable $d^{0}$ configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the d-orbital of these ions. Therefore, they act as reducing agents.

22. What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution. Sol. It is found that sometimes a relatively less stable oxidation state undergoes an (oxidation -reduction)reactioninwhichit is simultaneously oxidised and reduced. This is called disproportionation. For example, (1) $Cr (V) 2 CrO_{4}^{3 -} + 8 H^{+} ⟶ Cr (VI) 2 CrO_{4}^{2 -} + Cr (III) Cr^{3 +} + 4 H_{2} O$ Cr(V) is oxidized to Cr(VI) and reduced to Cr(III). (2) $Mn (VI) MnO_{4}^{2 -} + 4 H^{+} ⟶ Mn (VII) 2 MnO_{4}^{-} + Mn (IV) MnO_{2} + 2 H_{2} O$ Mn (VI) is oxidized to Mn (VII) and reduced to Mn (IV).

23. Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why? Sol. In the first transition series, Cu exhibits +1 oxidation state very frequently. It is because Cu (+1) has an electronic configuration of [Ar] $3 d^{10}$ . The completely filled d-orbital makes it highly stable.

24. Calculate the number of unpaired electrons in the following gaseous ions: $Mn^{3 +}, Cr^{3 +}, V^{3 +}$ and $Ti^{3 +}$ . Which one of these is the most stable in aqueous solution ? Sol.

	Gaseous ions	Number of Unpaired electrons
(i)	$Mn^{3 +}, [Ar] 3 d^{4}$	4
(ii)	$Cr^{3 +}, [Ar] 3 d^{3}$	3
(iii)	$V^{3 +}, [Ar] 3 d^{2}$	2
(iv)	$Ti^{3 +} [Ar] 3 d^{1}$	1

$Cr^{3 +}$ is the most stable in aqueous solutions showing to a $t_{2 g}^{3}$ half filled configuration.

25. 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) The highest oxidation state is exhibited in oxoanions of a metal.

Sol.

(i) In the case of a lower oxide of a transition metal, the metal atom has a low oxidation state. This means that some of the valence electrons of the metal atom are not involved in bonding. As a result, it can donate electrons and behave as a base. On the other hand, in the case of a higher oxide of a transition metal, the metal atom has a high oxidation state. This means that the valence electrons are involved in bonding and so, they are unavailable. There is also a high effective nuclear charge. As a result, it can accept electrons and behave as an acid. For example, $Mn^{II} O$ is basic and $Mn_{2}^{VII} O_{7}$ is acidic.

(ii) Oxygen and fluorine act as strong oxidising agents because of their high electronegatives and small sizes. Hence, they bring out the highest oxidation states from the transition metals. In other words, a transition metal exhibits higher oxidation states in oxides and fluorides. For example, in $OsF_{6}$ and $V_{2} O_{5}$ , the oxidation states of Os and V are +6 and +5 respectively.

(iii) Oxygen is a strong oxidising agent due to its high electronegativity and small size. So, oxoanions of a metal have the highest oxidation state. For example, in $MnO_{4}^{-}$ , the oxidation state of Mn is +7.

26. What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses. Sol. An alloy is a solid of two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys are usually found to possess different physical properties than those of the component elements. An important alloy of lathanoidsis Mischmetal. It contains lanthanoids (94-95%), iron (5%), and traces of S, C, Si, Ca and Al.

Uses:

(1) Mischmetal is used in cigarettes and gas lighters. (2) It is used in flame throwing tanks. (3) It is used in tracer bullets and shells.

27. What are inner transition elements ? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements : 29, 59, 74, 95, 102, 104. Sol. Inner transition elements are those elements in which the last electron enters the f-orbital. The elements in which the 4 f and and 5f orbitals are progressively filled are called f-block elements. Among the given atomic numbers, the atomic numbers of the inner transition elements are 59, 95 and 102.

28. The chemistry of the actinoid elements is not so smooth as that of the Lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements. Sol. Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d and 6s orbitals is quite large. On the other hand, the energy difference between 5f, 6d, and 7s orbitals is very less. Hence actinoids display a large number of oxidation states. For example, uranium and plutonium display +3, +4, +5, and +7. The most common oxidation state in case of actinoids is also +3.

29. Which is the last elements in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element. Sol. The last element in the actinoid series is lawrencium, Lr. its atomic number is 103 and its electronic configuration is $[Rn] 5 f^{14} 6 d^{1} 7 s^{2}$ . The most common oxidation state displayed by it is +3; because after losing 3 electrons it attains stable $f^{14}$ configuration.

30. Use Hund's rule to derive the electronic configuration of $Ce^{3 +}$ ion and calculate its magnetic moment on the basis of 'spin - only' formula. Sol. $Ce^{3 +} : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2}$ $5 p^{6} 4 f^{1}$ Magnetic moment can be calculated as :

$μ = n (n + 2)$

Where, $n =$ number of unpaired electrons. In $Ce^{3 +}, n = 1$ Therefore,

$μ = 1 (1 + 2) = 3 = 1.732 BM$

Name the members of the lanthanoid series which exhibit +4 oxidation state and those which exhibit +2 oxidation state. Try to correlate this type of behaviour with the electronic configuration of these elements. Sol. The lanthanoides that exhibit + 2 and +4 states are shown in the given table . The atomic number of these elements are given in the parenthesis.

+2	+4
Nd (60)	Ce (58)
Sm (62)	Pr (59)
Eu (63)	Nd (60)
Tm (69)	Tb (65)
Yb (70)	Dy (66)

Ce after forming $Ce^{4 +}$ attains a stable electronic configuration of [Xe]. Tb after forming $Tb^{4 +}$ attains a stable electronic configuration of $[Xe] 4 f^{7}$ Eu after forming $Eu^{2 +}$ attains a stable electronic configuration of $[Xe] 4 f^{7}$ Yb after forming $Yb^{2 +}$ attains a stable electronic configuration of [Xe] 4f $^{14}$

32. Write the electronic configuration of the elements with the atomic numbers 61, 91, 101 and 109. Sol.

Atomic Number	Electronic Configuration
61	$[Xe]^{54} 4 f^{5} 5 d^{0} 6 s^{2}$
91	$[Rn]^{86} 5 f^{2} 6 d^{1} 7 s^{2}$
101	$[Rn]^{86} 5 f^{13} 5 d^{0} 7 s^{2}$
109	$[Rn]^{86} 5 f^{14} 6 d^{7} 7 s^{2}$

33. Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns.

Give special emphasis on the following points :

(i) electronic configurations, (ii) oxidation states, (iii) ionisation enthalpies, and (iv) atomic sizes.

Sol.

(i) In the $1^{st}, 2^{nd}$ and $3^{rd}$ transition series, the $3 d, 4 d$ and 5d orbitals are respectively filled. We know that elements in the same vertical column generally have similar electronic configurations. In the first transition series, two elements show unusual electronic configurations :

$Cr (24) = 3 d^{5} 4 s^{1} Cu (29) = 3 d^{10} 4 s^{1}$

Similarly, there are exceptions in the second transition series. These are :

$Nb (41) = 4 d^{4} 5 s^{1} Mo (42) = 4 d^{5} 5 s^{1} Tc (43) = 4 d^{6} 5 s^{1} Ru (44) = 4 d^{7} 5 s^{1} Rh (45) = 4 d^{8} 5 s^{1} Pd (46) = 4 d^{10} 5 s^{0} Ag (47) = 4 d^{10} 5 s^{1}$

There are some exceptions in the third transition series as well. These are :

$Pt (78) = 5 d^{9} 6 s^{1} Au (79) = 5 d^{10} 6 s^{1}$

As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group are dissimilar. (ii) In each of the three transition series the number of oxidation states shown by the elements is the maximum in the middle and the minimum at the extreme ends. However, +2 and +3 oxidation states are quite stable for all elements present in the first transition series.

All metals present in the first transition series form stable compounds in the +2 and +3 oxidation states. The stability of the +2 and +3 oxidation states decreases in the second and the third transition series, wherein higher oxidation states are more important. For example

$[Fe II (CN)_{6}]^{4 -}, [Co III (NH_{3})_{6}]^{3 +}, [Ti (H_{2} O)_{6}]^{3 +}$

are stable complexes, but no such complexes are known for the second and third transition series such as Mo, W, Rh, In. They form complexes in which their oxidation states are high. For example : $WCl_{6}, ReF_{7}, RuO_{4}$ , etc.

(iii) In each of the transition series, the first ionisation enthalpy increases from left to right. However, there are some exceptions. The first ionisation enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to the poor shielding effect of 4f electrons in the third transition series. Certain elements in the second transition series have higher first ionisation enthalpies than elements corresponding to the same vertical column in the first transition series. There are also elements in the $2^{nd}$ transition series whose first ionisation enthalpies are lower than those of the elements corresponding to the same vertical column in the $1^{st}$ transition series.

(iv) Atomic size generally decreases from left to right across a period. Now, among the three transition series, atomic sizes of the elements in the second transition series are greater than those of the element corresponding to the same vertical column in the first transition series. However, the atomic sizes of the element in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.

34. Write down the number of 3d electrons in each of the following ions:

$Ti^{2 +}, V^{2 +}, Cr^{3 +}, Mn^{2 +}, Fe^{2 +}, Fe^{3 +}, Co^{2 +}, Ni^{2 +}$

and $Cu^{2 +}$ Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).

Sol.

Metal ion	Number of d-electrons	Filling of d-orbitals
$Ti^{2 +}$	2	$t_{2 g}^{2}$
$V^{2 +}$	3	$t_{2 g}^{3}$
$Cr^{3 +}$	3	$t_{2 g}^{3}$
$Mn^{2 +}$	5	$t_{2 g}^{3} e_{g}^{2}$
$Fe^{2 +}$	6	$t_{2 g}^{4} e_{g}^{2}$
$Fe^{3 +}$	5	$t_{2 g}^{3} e_{g}^{2}$
$Co^{2 +}$	7	$t_{2 g}^{5} e_{g}^{2}$
$Ni^{2 +}$	8	$t_{2 g}^{6} e_{g}^{2}$
$Cu^{2 +}$	9	$t_{2 g}^{6} e_{g}^{3}$

35. What can be inferred from the magnetic moment values of the following complex species?

Example	Magnetic Moment (BM)
$K_{4} [Mn (CN)_{6}]$	2.2
$[Fe (H_{2} O)_{6}]^{2 +}$	5.3
$K_{2} [MnCl_{4}]$	5.9

Sol. Magnetic moment $(μ)$ is given as

$μ = n (n + 2) BM$

For value

$n = 1, μ = 1 (1 + 2) = 3 = 1.732 BM$

For value

$n = 2, μ = 2 (2 + 2) = 8 = 2.83 BM$

For value

$n = 3, μ = 3 (3 + 2) = 15 = 3.87 BM$

For value

$n = 4, μ = 4 (4 + 2) = 24 = 4.899 BM$

For value

$n = 5, μ = 5 (5 + 2) = 35 = 5.92 BM$

(i) $K_{4} [Mn (CN)_{6}]$ : For transition metals, the magnetic moment is calculated from the spinonly formula. Therefore, $μ = n (n + 2)$ $= 2.2 BM$ We can see from the above calculation that the given value is closest to $n = 1$ . Also, in this complex, Mn in in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital. Hence, we can say that $CN^{-}$ is a strong field ligand that cause the pairing of electrons. (ii) $[Fe (H_{2} O)_{6}]^{2 +}$

$μ = n (n + 2) = 5.3 BM$

We can see from the above calculation that the given value is closest to $n = 4$ , Also, in this complex, Fe is in the +2 oxidation state. This means that Fe has 6 electrons in the d-orbital. Hence, we can say that $H_{2} O$ is a weak field ligand and does not cause the pairing of electrons. (iii) $K_{2} [MnCl_{4}]$

$μ = n (n + 2) = 5.9 BM$

We can see from the above calculation that the given value is closest to $n = 5$ . Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital. Hence, we can say that $Cl^{-}$ is a weak field ligand and does not cause the pairing of electrons.

4.0NCERT Solutions Class 12 Chemistry | Chapter-wise Links

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Correct Electronic Configurations: Learn how to write the electronic configurations of d-block elements, including the special cases of Chromium (Cr) and Copper (Cu).
Important Periodic Trends: Understand important trends in ionisation enthalpy, atomic size, ionic size, oxidation states, and standard electrode potential (E°).
Lanthanoids and Actinoids: Learn the differences between lanthanoids and actinoids, including their electronic configurations, oxidation states, atomic sizes, and chemical properties.
Important Compounds and Reactions: Study the preparation, properties, and important reactions of potassium dichromate (K₂Cr₂O₇) and potassium permanganate (KMnO₄).
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NCERT Solutions Class 12 Chemistry Chapter 4: d- and f-block Elements

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Table of Contents

Sc			+3
Ti	+1	+2	+3	+4
V	+1	+2	+3	+4	+5
Cr	+1	+2	+3	+4	+5	+6
Mn	+1	+2	+3	+4	+5	+6	+7
Fe	+1	+2	+3	+4	+5	+6
Co	+1	+2	+3	+4	+5
Ni	+1	+2	+3	+4
Cu	+1	+2
Zn		+2

Sc			+3
Ti	+1	+2	+3	+4
V	+1	+2	+3	+4	+5
Cr	+1	+2	+3	+4	+5	+6
Mn	+1	+2	+3	+4	+5	+6	+7
Fe	+1	+2	+3	+4	+5	+6
Co	+1	+2	+3	+4	+5
Ni	+1	+2	+3	+4
Cu	+1	+2
Zn		+2

Sc			+3
Ti	+1	+2	+3	+4
V	+1	+2	+3	+4	+5
Cr	+1	+2	+3	+4	+5	+6
Mn	+1	+2	+3	+4	+5	+6	+7
Fe	+1	+2	+3	+4	+5	+6
Co	+1	+2	+3	+4	+5
Ni	+1	+2	+3	+4
Cu	+1	+2
Zn		+2