Chemistry Class 12 NCERT Solutions: Chapter 8 The d and f Block Elements Part 6

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Q: 24. Calculate the number of unpaired electrons in the following gaseous . Which one of these is the most stable in aqueous solution?

Answer:

Q_24_The Number of Unpaired Electrons in the Gaseous
Q_24_The Number of Unpaired Electrons in the Gaseous

Gaseous ions

Number of unpaired electrons

(i)

4

(ii)

3

(iii)

2

(vi)

1

is the most stable in aqueous solutions owing to a configuration

Q: 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 oxyanions 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, osmium shows an oxidation states of + 6 in and vanadium shows an oxidation states of + 5 in .

(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 .

Q: 26. Indicate the steps in the preparation of:

(i) from chromite ore

(ii) from Pyrolusite ore

Answer:

(i)

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

Step(1): Preparation of sodium chromate

Step(3): 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 . However, by changing the , they can be interconverted

Potassium permanganate can be prepared from Pyrolusite . The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as

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, manganite ions are oxidized to permanganate ions.

Oxidation by chlorine

Oxidation by ozone