Question

A group 15 element forms \( d\pi - d\pi \) bond with transition metals. It also forms a hydride, which is the strongest base among the hydrides of other group members that form \( d\pi - d\pi \) bonds. The atomic number of the element is ________________________.

Hint:

Phosphorus forms strong \( d\pi - d\pi \) bonds and has strong basic hydrides.

Answer 1

Correct Answer: 15

The element must be from group 15 and form \( d\pi - d\pi \) bonds. Possible candidates: N, P, As.

• Phosphorus (\( P \)) forms \( d\pi - d\pi \) bonds. 
• \( PH_3 \) is a strong base compared to \( AsH_3 \).

Atomic number of Phosphorus = 15.

Answer 2

Step 1: Understanding the question.
We are asked to identify a Group 15 element that forms \( d\pi - d\pi \) bonds with transition metals and whose hydride is the strongest base among the hydrides of the other group members that form such bonds.

Step 2: Recall the elements of Group 15.
The Group 15 elements are:
N (Z = 7), P (Z = 15), As (Z = 33), Sb (Z = 51), Bi (Z = 83).

Step 3: \( d\pi - d\pi \) bonding capability.
For forming \( d\pi - d\pi \) bonds with transition metals, the element must have vacant \( d \)-orbitals. - Nitrogen (N) does not have \( d \)-orbitals (only 2s and 2p). - Phosphorus (P), Arsenic (As), and Antimony (Sb) have vacant \( d \)-orbitals and can form \( d\pi - d\pi \) bonds. Hence, possible elements: P, As, Sb.

Step 4: Basicity of their hydrides.
Hydrides: PH₃, AsH₃, SbH₃.
Basicity decreases down the group because the size of the central atom increases, and the lone pair availability decreases:
\[ \text{PH}_3 > \text{AsH}_3 > \text{SbH}_3. \] Thus, PH₃ (phosphine) is the strongest base among these hydrides.

Step 5: Conclusion.
The element that forms \( d\pi - d\pi \) bonds with transition metals and whose hydride is the strongest base is **Phosphorus (P)**.
The atomic number of phosphorus is 15.

Final Answer:
\[ \boxed{15} \]