Match Column I (Hybrid bond orbitals ) with Column II (species) and select the correct answer.

Codes:

$3\; 4\; 2\; 1\;$ ;$ICl_4$ is $sp^3d^2 , TeCl_4$ is $sp^3d , Ni(CN)_4^{2-}$ is $dsp^2 , MnO_4^-$ is $d^3s$
Hence (d) is the correct answer.