Q6 / 90 · JEE Main 2024 (05 Apr Shift 2) · Chemistry

The number of complexes from the following with no electrons in the \(t_2\) orbital is \(\qquad\) \(\mathrm{TiCl}_4,\left[\mathrm{MnO}_4\right]^{-},\left[\mathrm{FeO}_4\right]^{2-},\left[\mathrm{FeCl}_4\right]^{-},\left[\mathrm{CoCl}_4\right]^{2-}\)

  • A
    1
  • B
    4
  • Correct answer
    C
    3
  • D
    2

Solution

$\begin{array}{ll}\mathrm{TiCl}_4 \Rightarrow \mathrm{Ti}^{+4} & \mathrm{e}^0 \mathrm{t}_2^0 \\ \mathrm{MnO}_4^{-} \Rightarrow \mathrm{Mn}^{+7} & \mathrm{e}^0 \mathrm{t}_2^0 \\ \mathrm{FeO}_4^{2-} \Rightarrow \mathrm{Fe}^{+6} & \mathrm{e}^2 \mathrm{t}_2^0 \\ \mathrm{FeCl}_4^{2-} \Rightarrow \mathrm{Fe}^{+2} & \mathrm{e}^3 \mathrm{t}_2^3 \\ \mathrm{CoCl}_4^{2-} \Rightarrow \mathrm{Co}^{+2} & \mathrm{e}^4 \mathrm{t}_2^3\end{array}$

Hence, the correct answer is option: 'C'.

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