p-Block Elements — JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 Question with Solution

There are some deposits of nitrates and phosphates in earth's crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions. Hybridization easily explains the ease of sigma donation capability of

NH_{3}

and

PH_{3}

. Phosphine is a flammable gas and is prepared from white phosphorous.

White phosphorus on reaction with NaOH gives

PH_{3}

as one of the products. This is a
(A) dimerization reaction
(B) disproportionation reaction
(C) condensation reaction
(D) precipitation reaction

(A)dimerization reaction
(B)disproportionation reaction✓ Correct
(C)condensation reaction
(D)precipitation reaction

Answer: (B)

Solution

The reaction is:

P_{4} + 3 NaOH + 3 H_{2} O ⟶ 3 NaH_{2} PO_{2} + PH_{3}

In white phosphorus (

P_{4}

), phosphorus has an oxidation state of 0. In the products:

In $NaH_{2} PO_{2}$ (sodium hypophosphite), phosphorus has an oxidation state of +1
In $PH_{3}$ (phosphine), phosphorus has an oxidation state of -3

Since the same element (phosphorus) is simultaneously oxidized and reduced, this is a disproportionation reaction.

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · p-Block Elements

Solution

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