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

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$3\; 4\; 2\; 1\;$ ;$ 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.
answered Dec 9, 2013 by sreemathi.v
 

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