Chemistry - CHEMICAL BONDING Question with Solution | TestHub

The orbitals used for $\pi$-bond formation, assuming one atom (the central atom) uses its $np$ orbital.

(P) SOF₄: Sulfur is in its +6 oxidation state (3rd period, $n=3$). It uses its $3d$ orbitals for $\pi$-bonding with oxygen ($2p$). Thus, it matches with (3) $(n+1)d$ (if $n=2$ for Oxygen).

(Q) CS₃²⁻: Carbon ($n=2$) and Sulfur ($n=3$) both use $p$-orbitals for $\pi$-bonding. It matches with (2) $np$ (both in $p$-block).

(R) CO: Carbon ($n=2$) and Oxygen ($n=2$) both use $2p$ orbitals for the $\pi$-bond. It matches with (1) $np$.

(S) BrO₃⁻: Bromine (4th period, $n=4$) uses $4d$ orbitals for $\pi$-bonding with oxygen ($2p$). If $n=2$ for Oxygen, $4d$ is $(n+2)d$, matching (4).