JEE Chemistry 2025 Question with Solution

Given: The element belongs to group $15$, forms $d\pi - d\pi$ bonds with transition metals, and its hydride is the strongest base among the hydrides of the group members that can form such bonds.

Find: The atomic number of the element.

Elements of group $15$ are $N, P, As, Sb, Bi$.

For formation of $d\pi - d\pi$ bonds with transition metals, the element must be able to participate through available vacant $d$-orbitals. Hence, nitrogen is excluded, and the relevant candidates are $P, As, Sb$.

Now compare the basicity of their hydrides:

$$\text{PH}_3 > \text{AsH}_3 > \text{SbH}_3$$

So, phosphine $\text{PH}_3$ is the strongest base among these hydrides.

Therefore, the required element is phosphorus.

Atomic number of phosphorus is $15$.

Therefore, the answer is $15$.

Given: A group $15$ element forms $d\pi - d\pi$ bond with transition metals, and its hydride is the strongest base among the hydrides of the other eligible group members.

Find: The atomic number of the element.

1. Group $15$ elements are:

$$N \,(Z=7),\; P \,(Z=15),\; As \,(Z=33),\; Sb \,(Z=51),\; Bi \,(Z=83)$$

2. Nitrogen does not have vacant $d$-orbitals, so it is not considered for forming $d\pi - d\pi$ bonds in this context.
3. The possible elements are therefore:

$$P,\; As,\; Sb$$

4. Their hydrides are:

$$\text{PH}_3,\; \text{AsH}_3,\; \text{SbH}_3$$

5. Basicity decreases down the group:

$$\text{PH}_3 > \text{AsH}_3 > \text{SbH}_3$$

Hence, the strongest base is $\text{PH}_3$, so the element is phosphorus.

The atomic number of phosphorus is $15$.

Therefore, the final answer is $15$.