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
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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).
The question involves determining the incorrect statements regarding the hybridization and geometry of the given chemical species. Let's analyze each option one by one:
\(ClO_4^−\) Hybridization: The structure of \(ClO_4^−\) (perchlorate ion) is tetrahedral, with a \(sp^3\) hybridization. The central chlorine atom forms sigma bonds with four oxygen atoms, resulting in a tetrahedral geometry not \(dsp^3\). Hence, statement I is incorrect.
\([\text{Ni}(\text{CN})_4]^{2-}\) Geometry: The coordination number of Ni in this complex is 4, and it's a square planar, not a tetrahedral complex. Given the electronic configuration of Ni and the strong field ligand \(\text{CN}^−\), this leads to a low-spin complex with \(dsp^2\) hybridization. Hence, statement II is incorrect.
\([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) Hybridization: In this coordination complex, cobalt is in +2 oxidation state, with a coordination number of 6, leading to an octahedral geometry. The hybridization of Co in an octahedral complex is \(sp^3d^2\). Hence, statement III is correct.
\([\text{Mn}(\text{CN})_6]^{4-}\) Hybridization: This is a high-spin octahedral complex due to the presence of \(\text{CN}^−\) as a strong field ligand. However, the hybridization is \(d^2sp^3\) following the octahedral and low-spin nature of the complex. Thus, statement IV is also incorrect.
Based on the analysis above, only statement III is correct. Hence, the correct option is: III only.
