Question

Select correct statements. (I) Hybridisation of $\text{ClO}_4^-$ is $dsp^3$ (II) $[\text{Ni}(\text{CN})_4]^{2-}$ is tetrahedral (III) $[\text{Co}(\text{H}_2\text{O})_6]^{2+}$ has $sp^3d^2$ hybridisation (IV) $[\text{Mn}(\text{CN})_6]^{4-}$ has $sp^3d^2$ hybridisation

• A. II and III and
• B. III only
• C. II, III and IV only
• D. I, II, III and IV

Hint:

For $d^8$ square planar complexes like $[\text{Ni}(\text{CN})_4]^{2-}$, the hybridization is $dsp^2$. If the complex is $d^5$ low spin octahedral, the hybridization is $d^2sp^3$ (inner orbital).

Answer 1

The Correct Option is B

(I) $\text{ClO}_4^-$: $sp^3$ (Tetrahedral). Listed $dsp^3$. (I) is Incorrect.
(II) $[\text{Ni}(\text{CN})_4]^{2-}$: $dsp^2$ (Square Planar). Listed Tetrahedral. (II) is Incorrect.
(III) $[\text{Co}(\text{H}_2\text{O})_6]^{2+}$: $\text{Co}^{2+}$ ($d^7$), $\text{H}_2\text{O}$ ($\text{WF}$). High spin $sp^3d^2$. (III) is Correct.
(IV) $[\text{Mn}(\text{CN})_6]^{4-}$: $\text{Mn}^{2+}$ ($d^5$), $\text{CN}^-$ ($\text{SF}$). Low spin $d^2sp^3$. Listed $sp^3d^2$. (IV) is Incorrect.
Only statement (III) is correct.