Q. 165.0( 2 Votes )

Complete the foll

Answer :


In the above equation,

MnO4- will convert to MnO2 and thiosulphate will convert to Sulphate ion and Water will dissociate into hydroxyl(-OH) ion.

MnO4- MnO2

3e- +Mn+7 Mn+4 ----------(i)

In the above,

Mn+7 will convert into Mn4+ ion,

Oxidation No. changes by 3.

S2O32- SO42-

2S+2 2S+6 + 8e- -----------(ii)

Oxidation No. changes by +4.

Multiplying equation(i) by 8 and equation(ii) by 3 we get as below,

(ii) Cr2O72-(aq) + Fe2+(aq) + H+(aq)

In the above equation,

By the above reaction we get following,

6e- +Cr2+12 2Cr3+ ------------(i)

Fe2+ Fe3+ + e- --------------(ii)

Multiplying equation (i) by 1 and equation (ii) by 6, we get as below;


(i) Hydration energy of Cu2+ is high so, in an aqueous medium, Cu2+ is more stable than Cu+. Although more energy is required to remove one electron from Cu+ to Cu2+, high hydration energy of Cu2+ compensates for it.

Therefore, Cu+ ion in an aqueous solution is unstable. It disproportionates to give Cu2+and Cu.

(ii) Unlike Cr3+, Mn2+, Fe3+ and the subsequent other M2+ ions of the 3d series of elements, the 4d and the 5d series metals generally do not form stable cationic species. This is due to Lanthanoid contraction as expected increase in size does not occur.

The lanthanoid contraction is the ‘greater than expected decrease’ in ionic radii of the elements in the lanthanide series from atomic number 57 - lanthanum, to 71 - lutetium, which results in smaller than otherwise expected ionic radii for the subsequent elements. This is because atomic radii of 4d and 5d transition elements are nearly same and this similarity in size is due to weak shielding of d-electrons.

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