WBJEE 2026 Physics and Chemistry Paper

Concept:Tetrahedral geometry arises from $sp^3$ hybridization, which depends on the metal's $d$-electron count and the ligand field strength.Explanation:A: $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$: Ni is in $+2$ oxidation state with $d^8$ configuration. $\mathrm{CN}^-$ is a strong field ligand, so pairing occurs and hybridization is $dsp^2$. Hence the structure is square planar, not tetrahedral.B: $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$: Ni is in $0$ oxidation state with $d^{10}$ configuration. $\mathrm{CO}$ is a neutral ligand and no pairing is needed. Hybridization is $sp^3$, giving a tetrahedral structure.C: $\left[\mathrm{NiCl}_{4}\right]^{2-}$: Ni is in $+2$ oxidation state with $d^8$ configuration. $\mathrm{Cl}^-$ is a weak field ligand, so no pairing occurs. Hybridization is $sp^3$, giving a tetrahedral structure.D: $\mathrm{CrO}_{4}^{2-}$: Cr is in $+6$ oxidation state with $d^0$ configuration. The chromate ion has $sp^3$ hybridization and a tetrahedral structure.Answer:B, C, D