Q10 / 75 · JEE Main 2025 (8 Apr Shift 2) · Chemistry

- A$5$
- B$6$
- C$4$
- Correct answerD$3$
Solution
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
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
Hence, the correct answer is option: 'D'.