Among $[Co(CN)_4]^{4-}$, $[Co(CO)_3(NO)]$, $XeF_4$, $[PCl_4]^+$, $[PdCl_4]^{2-}$, $[ICl_4]^-$, $[Cu(CN)_4]^{3-}$ and $P_4$ the total number of species with tetrahedral geometry is _____....

The correct numerical answer is 4. Click to view the complete step-by-step calculation.

JEE Advanced 2024 Paper 2 Chemistry Question 12 Solution

$[Co(CO)_3(NO)]$ : Cobalt follows the 18-electron rule here, geometry is tetrahedral.
$XeF_4$ : $sp^3d^2$ hybridization with 2 lone pairs, square planar.
$[PCl_4]^+$ : $sp^3$ hybridization, tetrahedral.
$[PdCl_4]^{2-}$ : $4d^8$ configuration with strong field-like behavior from $Cl^-$ in $4d$ series, square planar ( $dsp^2$ ).
$[ICl_4]^-$ : $sp^3d^2$ hybridization with 2 lone pairs, square planar.
$[Cu(CN)_4]^{3-}$ : $Cu^+$ is $d^{10}$ , $sp^3$ hybridization, tetrahedral.
$P_4$ : Each Phosphorus atom is $sp^3$ hybridized in a tetrahedral cage, geometry is tetrahedral.
$[Co(CN)_4]^{4-}$ : Co(0) $d^9$ is generally not considered tetrahedral in standard JEE contexts or is highly unstable. The species with tetrahedral geometry are: $[Co(CO)_3(NO)]$ , $[PCl_4]^+$ , $[Cu(CN)_4]^{3-}$ , and $P_4$ . Total = 4.