Choose the correct statements about the hydrides of group 15 elements.
A. The stability of the hydrides decreases in the order NH₃>PH₃>AsH₃> SbH₃>BiH₃.
B. The reducing ability of the hydride increases in the order NH₃<PH₃<AsH₃ <SbH₃<BiH₃.
C. Among the hydrides, NH₃ is strong reducing agent while BiH₃ is mild reducing agent.
D. The basicity of the hydrides increases in the order NH₃<PH₃<AsH₃< SbH₃<BiH₃.
Choose the most appropriate from the options given below :

Solution

Chemistry ma’am · 4 steps

Everything here follows from one fact: the E–H bond gets weaker as the central atom gets bigger.

1. Bond strength. Going down from N to Bi, the central atom gets bigger, so its overlap with the small H 1s orbital gets poorer. The E–H bond becomes weaker and longer.
2. Stability and reducing power. A weaker E–H bond breaks more easily on heating, so thermal stability decreases down the group. The same easy breaking lets the hydride give up hydrogen to other species, so reducing power increases down the group.
3. Basicity. On small N, the lone pair sits in a compact orbital and is readily donated to H⁺. On bigger atoms it is spread over a larger volume and is less available, so basicity decreases from NH₃ to BiH₃.

   Statement	What it claims	Actual trend	Verdict
   A	Stability: NH₃ highest, BiH₃ lowest	Weakening E–H bond, stability falls	Correct
   B	Reducing ability: NH₃ lowest, BiH₃ highest	Easier E–H breaking, reducing power rises	Correct
   C	NH₃ strong, BiH₃ mild reducing agent	Reversed: BiH₃ is the strongest	Incorrect
   D	Basicity: NH₃ lowest, BiH₃ highest	Reversed: NH₃ is the most basic	Incorrect

4. So. Only statements A and B are correct, which is option (c).

The idea to keep. For group 15 hydrides, going down the group: stability falls and basicity falls, while reducing power rises. A bigger central atom means a weaker E–H bond and a more spread-out lone pair.

Answer from the NTA final answer key. Solution by PrepWiser.