Page No 235: - Chapter 8 D & F Block Elements Exercise Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Question 8.16:

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

Write the ionic equations for the reactions.

Answer:

Potassium permanganate can be prepared from pyrolusite (MnO₂). The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as KNO₃ or KClO₄, to give K₂MnO₄.

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

Electrolytic oxidation

At anode, manganate ions are oxidized to permanganate ions.

Oxidation by chlorine

Oxidation by ozone

(i) Acidified KMnO₄ solution oxidizes Fe (II) ions to Fe (III) ions i.e., ferrous ions to ferric ions.

(ii) Acidified potassium permanganate oxidizes SO₂ to sulphuric acid.

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

Question 8.17:

For M²⁺/M and M³⁺/M²⁺ systems, the   values for some metals are as follows:

Cr²⁺/Cr −0.9V

Cr³/Cr²⁺ −0.4 V

Mn²⁺/Mn −1.2V

Mn³⁺ /Mn²⁺ +1.5 V

Fe²⁺ /Fe −0.4V

Fe³⁺/Fe²⁺ +0.8 V

Use this data to comment upon:

(i) The stability of Fe³⁺ in acid solution as compared to that of Cr³⁺ or Mn³⁺ and

(ii) The ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Answer:

(i) The   value for Fe³⁺/Fe^{2+ ­­} is higher than that for Cr³⁺/Cr²⁺ and lower than that for Mn³⁺/Mn²⁺. So, the reduction of Fe³⁺ to Fe²⁺ is easier than the reduction of Mn³⁺ to Mn²⁺, but not as easy as the reduction of Cr³⁺ to Cr²⁺. Hence, Fe³⁺ is more stable than Mn³⁺, but less stable than Cr³⁺. These metal ions can be arranged in the increasing order of their stability as: Mn³⁺ < Fe³⁺ < Cr³⁺

(ii) The reduction potentials for the given pairs increase in the following order.

Mn²⁺ / Mn < Cr²⁺ / Cr < Fe²⁺ /Fe

So, the oxidation of Fe to Fe²⁺ is not as easy as the oxidation of Cr to Cr²⁺ and the oxidation of Mn to Mn²⁺. Thus, these metals can be arranged in the increasing order of their ability to get oxidised as: Fe < Cr < Mn

Question 8.18:

Predict which of the following will be coloured in aqueous solution? Ti³⁺, V³⁺,

Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺ and Co²⁺. Give reasons for each.

Answer:

Only the ions that have electrons in d-orbital and in which d-d transition is possible will be coloured. The ions in which d-orbitals are empty or completely filled will be colourless as no d-d transition is possible in those configurations.

Question 8.19:From the above table, it can be easily observed that only Sc³⁺ has an empty d-orbital and Cu⁺ has completely filled d-orbitals. All other ions, except Sc³⁺ and Cu⁺, will be coloured in aqueous solution because of d−d transitions.

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

Answer:

Question 8.20:From the above table, it is evident that the maximum number of oxidation states is shown by Mn, varying from +2 to +7. The number of oxidation states increases on moving from Sc to Mn. On moving from Mn to Zn, the number of oxidation states decreases due to a decrease in the number of available unpaired electrons. 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 becomes more and more difficult to remove the third electron from the d-orbital.

Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

(i) electronic configuration (iii) oxidation state

(ii) atomic and ionic sizes and (iv) chemical reactivity.

Answer:

(i) Electronic configuration

The general electronic configuration for lanthanoids is [Xe]⁵⁴ 4f^0-14 5d^0-1 6s² and that for actinoids is [Rn]⁸⁶ 5f^1-14 6d^0-1 7s². Unlike 4f orbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent.

(ii) 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. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.

(iii) Atomic and lonic 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.

4. Chemical reactivity

In the lanthanide 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 lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of the formation of a protective oxide layer).

Question 8.21:

How would you account for the following:

(i) Of the d⁴species, Cr²⁺ 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 d¹ configuration is very unstable in ions.

Answer:

(i) Cr²⁺ is strongly reducing in nature. It has a d⁴ configuration. While acting as a reducing agent, it gets oxidized to Cr³⁺ (electronic configuration, d³). This d³ configuration can be written as   configuration, which is a more stable configuration. In the case of Mn³⁺ (d⁴), it acts as an oxidizing agent and gets reduced to Mn²⁺ (d⁵). This has an exactly half-filled d-orbital and is highly stable.

(ii) Co(II) is stable in 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 d¹ configuration tend to lose one more electron to get into stable d⁰ 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.

Question 8.22:

What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.

Answer:

It is found that sometimes a relatively less stable oxidation state undergoes an oxidation­−reduction reaction in which it is simultaneously oxidised and reduced. This is called disproportionation.

For example,

Cr(V) is oxidized to Cr(VI) and reduced to Cr(III).

Mn (VI) is oxidized to Mn (VII) and reduced to Mn (IV).

Question 8.23:

Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?

Answer:

In the first transition series, Cu exhibits +1 oxidation state very frequently. It is because Cu ( +1) has an electronic configuration of [Ar] 3d¹⁰. The completely filled d-orbital makes it highly stable.

Question 8.24:

Calculate the number of unpaired electrons in the following gaseous ions: Mn³⁺, Cr³⁺, V³⁺ and Ti³⁺. Which one of these is the most stable in aqueous solution?

Answer:

Question 8.25:Cr³⁺ is the most stable in aqueous solutions owing to a  configuration.

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.

Answer:

(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,  is basic and  is acidic.

(ii) Oxygen and fluorine act as strong oxidising agents because of their high electronegativities 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 OsF₆ and V₂O₅, 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, oxo-anions of a metal have the highest oxidation state. For example, in , the oxidation state of Mn is +7.

Question 8.26:

Indicate the steps in the preparation of:

(i) K2Cr2O7 from chromite ore.

(ii) KMnO4 from pyrolusite ore.

Answer:

(i)

Potassium dichromate ( ) is prepared from chromite ore  in the following steps.

Step (1): Preparation of sodium chromate

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

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

Potassium chloride being less soluble than sodium chloride is obtained in the form of orange coloured crystals and can be removed by filtration.

The dichromate ion  exists in equilibrium with chromate  ion at pH 4. However, by changing the pH, they can be interconverted.

(ii)

Potassium permanganate ( ) can be prepared from pyrolusite (MnO₂). The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as KNO₃ or KClO₄, to give K₂MnO₄.

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

Electrolytic oxidation

At anode, manganate ions are oxidized to permanganate ions.

Oxidation by chlorine

Oxidation by ozone

Question 8.27:

What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

Answer:

An alloy is a solid solution 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 lanthanoids is 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.

Question 8.28:

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.

Answer:

Inner transition metals are those elements in which the last electron enters the f-orbital. The elements in which the 4f and the 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.

Question 8.29:

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.

Answer:

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 +6 oxidation states while neptunium displays +3, +4, +5, and +7. The most common oxidation state in case of actinoids is also +3.

Question 8.30:

Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

Answer:

The last element in the actinoid series is lawrencium, Lr. Its atomic number is 103 and its electronic configuration is . The most common oxidation state displayed by it is +3; because after losing 3 electrons it attains stable f¹⁴ configuration.