Match List I with List II

1-d; 2-e; 3-a; 4-c; 5-b

The correct answer is option 1. 1-d; 2-e; 3-a; 4-c; 5-b.

Let us break down the reasoning for the bond angles in the hydrides of Group 15 elements \((NH_3, PH_3, AsH_3, SbH_3, BiH_3)\).

Group 15 elements form hydrides of the type EH3 (where E is N, P, As, Sb, or Bi). These hydrides have a trigonal pyramidal shape due to the presence of a lone pair of electrons on the central atom. However, the bond angles between the hydrogen atoms vary significantly due to changes in atomic size and lone pair repulsion as we move down the group.

Bond Angles and Factors Influencing Them

1. Ammonia \((NH_3)\): Bond angle: d. 107.8º

Nitrogen is a small atom with a high electronegativity. In \(NH_3\), the lone pair of electrons is close to the nucleus and exerts a strong repulsive force on the bonding pairs, resulting in a bond angle slightly less than the tetrahedral angle (109.5º).

2. Phosphine \((PH_3)\): Bond angle: e. 93.6º

Phosphorus is larger than nitrogen, and its lone pair of electrons is less tightly held (more diffuse). This reduces the repulsion between the lone pair and the bonding pairs, resulting in a smaller bond angle compared to \(NH_3\).

3. Arsine \((AsH_3)\): Bond angle: a. 91.8º
   - Arsenic is even larger than phosphorus, with the lone pair being more diffuse. Consequently, the bond angle further decreases.

4. Stibine \((SbH_3)\): Bond angle: c. 91.3º

Antimony is larger than arsenic, and the lone pair repulsion is further reduced, leading to a bond angle close to 91.3º.

5. Bismuthine \((BiH_3)\): Bond angle: b. 90º

Bismuth, being the largest in the group, has its lone pair farthest from the nucleus and the least effective repulsion on the bonding pairs, resulting in the bond angle being almost 90º.

Thus, the correct option is:1. 1-d; 2-e; 3-a; 4-c; 5-b