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The p-Block Elements

Short Answer Type

591. Knowing the electron gain enthalpy values for O

\to

O^– and O

\to

O²^–as -141 and 702 kJ mol^–1 respectively, how can you account for the formation of a large number of oxides having O^2-species and not O^–?

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Long Answer Type

592. Using VSEPR theory, predict the structures of SO₃,²^– IF₇^–, XeF₂, ClO₄^–, ICl₄^– and IBr₂^–.

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Short Answer Type

593.

Assign a reason for each of the following statement:
Ammonia is a stronger base than phosphine.

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Long Answer Type

594. Draw the structures of the following molecules (i) SF₄, (ii) XeF₄.

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One Word Answers

595. Answer the following:
Of Bi(V) and Sb(V) which may be a stronger oxidising agent and why?

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Short Answer Type

596. Assign a reason for the following statement?
Phosphorus shows marked tendency for catenation but nitrogen shows little tendency for catenation.

Nitrogen has a very little tendency to show catenation as N–N bond is very weak because of the repulsion in the electron pairs on the nitrogen atom. Nitrogen forms a chain of two atoms as N≡ N, hydrazine H₂N–NH₂ and of three atoms as in azide ion, N₃^–. The change in behaviour in phosphorus and nitrogen is due to weakness of the N–N single bond which in turn may be due to repulsion between lone pair of electrons on the two adjacent nitrogen atoms.
The lesser tendency of nitrogen to show catenation in comparison to phosphorus is their law (M–M) bond dissociation energy.
Bond N–N P–P
Bond energy (KJ mol^-1) 163.8 201.6

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597. Assign a reason for the following statement?
The electron gain enthalpy with negative sign for oxygen (–141'k] mol–¹) is less that for sulphur (–200 kj mol^{– 1}).

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