JEE MAIN - Chemistry (2025 - 8th April Evening Shift - No. 10)
The number of species from the following that are involved in sp3d2 hybridization is :
[Co(NH3)6]3+, SF6, [CrF6]3−, [CoF6]3−, [Mn(CN)6]3−, and [MnCl6]3−
Explanation
In $\left[\mathrm{Co}\left(\mathrm{NH}_3\right]_6\right]^{3+}, \mathrm{Co}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^6, \mathrm{NH}_3$ is S.F.L
Hybridisation state of $\mathrm{Co}^{3+}$ is $\mathrm{d}^2 \mathrm{sp}^3$
In $\mathrm{SF}_6$, Hybridisation state of sulphur is $\mathrm{sp}^3 \mathrm{~d}^2$
In $\left[\mathrm{CrF}_6\right]^{3-}, \mathrm{Cr}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^3$
Hybridisation state of $\mathrm{Cr}^{3+}$ is $\mathrm{d}^2 \mathrm{sp}^3$
$\left[\mathrm{CoF}_6\right]^{3-}, \mathrm{Co}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^6 \mathrm{~F}^{-}$is W.F.L
Hybridisation state of $\mathrm{Co}^{3+}$ is $\mathrm{sp}^3 \mathrm{~d}^2$
$\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}, \mathrm{Mn}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^4 \mathrm{CN}^{-}$is S.F.L
Hybridisation state of $\mathrm{Mn}^{3+}$ is $\mathrm{d}^2 \mathrm{sp}^3$
$\left[\mathrm{MnCl}_6\right]^{3-}, \mathrm{Mn}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^4 \mathrm{Cl}^{-}$is W.F.L
Hybridisation state of $\mathrm{Cl}^{-}$is $\mathrm{sp}^3 \mathrm{~d}^2$
Total number of $\mathrm{sp}^3 \mathrm{~d}^2$ hybridized molecules is 3
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