JEE Advanced 2024 · Paper 2, Chemistry question 12

Clearly tetrahedral (4):

• $\ce{[Co(CO)3(NO)]}$: isoelectronic with $\ce{Ni(CO)4}$, $sp^3$.
• $\ce{[PCl4]+}$: four bond pairs, $sp^3$.
• $\ce{[Cu(CN)4]^3-}$: $\ce{Cu+}$ is $d^{10}$, so $sp^3$.
• $\ce{P4}$: a tetrahedron of P atoms.

Not tetrahedral:

• $\ce{XeF4}$ and $\ce{[ICl4]-}$: four bond pairs and two lone pairs, square planar.
• $\ce{[PdCl4]^2-}$: $d^8$, $dsp^2$, square planar.

Ambiguous. $\ce{[Co(CN)4]^4-}$ contains Co(0), $d^9$. Whether its geometry is tetrahedral is not settled at this level.

So the total is $4$, or $5$ if $\ce{[Co(CN)4]^4-}$ is counted. The official key accepts 4 or 5.