Concept: Hybridization involves the mixing of atomic orbitals to form new hybrid orbitals suitable for bonding. The type of hybridization depends on the number of sigma bonds and lone pairs around the central atom (steric number). For elements to exhibit ${\mathrm{sp}}^3\mathrm{d}$ hybridization, they must have vacant d-orbitals available for participation in bonding.

Why (B) is correct:

• ${\mathrm{sp}}^3\mathrm{d}$ hybridization requires the involvement of one s, three p, and one d orbital. This means the central atom must have accessible d-orbitals in its valence shell.
• Carbon (C) is in Period 2 and has the electronic configuration $[He]2s^{2}2p^2$. It does not have any d-orbitals in its valence shell (n=2). Therefore, carbon cannot undergo ${\mathrm{sp}}^3\mathrm{d}$ hybridization.
• Boron (B) is also in Period 2 and has the electronic configuration $[He]2s^{2}2p^1$. Like carbon, it lacks d-orbitals in its valence shell. Thus, boron cannot exhibit ${\mathrm{sp}}^3\mathrm{d}$ hybridization.
• Phosphorus (P) is in Period 3 and has the electronic configuration $[Ne]3s^{2}3p^3$. It has vacant 3d-orbitals, which can be utilized for hybridization, allowing it to form compounds like ${\mathrm{PCl}}_5$ (where P is ${\mathrm{sp}}^3\mathrm{d}$ hybridized).
• Chlorine (Cl) is also in Period 3 and has the electronic configuration $[Ne]3s^{2}3p^5$. It has vacant 3d-orbitals and can exhibit ${\mathrm{sp}}^3\mathrm{d}$ hybridization in compounds like ${\mathrm{ClF}}_3$ (${\mathrm{sp}}^3\mathrm{d}$ hybridized) or ${\mathrm{ClF}}_5$ (${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridized).

Therefore, Carbon and Boron cannot exhibit ${\mathrm{sp}}^3\mathrm{d}$ hybridization.

Option Analysis:

• (a) C: Incorrect, as Carbon cannot exhibit ${\mathrm{sp}}^3\mathrm{d}$ hybridization.
• (b) P: Incorrect, as Phosphorus can exhibit ${\mathrm{sp}}^3\mathrm{d}$ hybridization.
• (c) Cl: Incorrect, as Chlorine can exhibit ${\mathrm{sp}}^3\mathrm{d}$ hybridization.
• (d) B: Incorrect, as Boron cannot exhibit ${\mathrm{sp}}^3\mathrm{d}$ hybridization.

The question asks which elements cannot exhibit ${\mathrm{sp}}^3\mathrm{d}$ hybridization. These are C and B.

Correct Answer: (B)