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

Elements of group 15 are: Nitrogen (N, $$Z = 7$$), Phosphorus (P, $$Z = 15$$), Arsenic (As, $$Z = 33$$), Antimony (Sb, $$Z = 51$$) and Bismuth (Bi, $$Z = 83$$).

Step 1 – Condition for $$d\pi-d\pi$$ bonding
To form $$d\pi-d\pi$$ bonds with transition metals, the p-block element must have empty d-orbitals that can overlap with filled d-orbitals of the metal. Nitrogen has no vacant d-orbitals (only $$2s$$ and $$2p$$ levels), so it cannot participate in $$d\pi-d\pi$$ back-bonding. Phosphorus and all heavier congeners (As, Sb, Bi) possess vacant $$3d$$/$$4d$$/$$5d$$/$$6d$$ orbitals respectively and therefore can form $$d\pi-d\pi$$ bonds.

Hence the eligible elements are P, As, Sb and Bi.

Step 2 – Basicity trend of their hydrides
The hydrides are $$PH_3$$, $$AsH_3$$, $$SbH_3$$ and $$BiH_3$$. Basic strength in this series decreases down the group because:
  • Electronegativity falls, so the lone pair becomes less concentrated.
  • Atomic size increases, causing poorer orbital overlap with a proton.
Therefore, $$\text{basicity order} : PH_3 \gt AsH_3 \gt SbH_3 \gt BiH_3$$.

The question asks for the element whose hydride is the strongest base among those elements that can form $$d\pi-d\pi$$ bonds. From the order above, $$PH_3$$ is the strongest base.

Step 3 – Identifying the element
The hydride $$PH_3$$ corresponds to Phosphorus, whose atomic number is $$15$$.

Final answer: $$15$$