JEE MAIN - Chemistry (2023 - 10th April Evening Shift - No. 9)
Match List I with List II
| List - I Complex |
List - II Crystal Field splitting energy ($$\Delta_0$$) |
||
|---|---|---|---|
| A. | $${[Ti{({H_2}O)_6}]^{2 + }}$$ | I. | $$-1.2$$ |
| B. | $${[V{({H_2}O)_6}]^{2 + }}$$ | II. | $$-0.6$$ |
| C. | $${[Mn{({H_2}O)_6}]^{3 + }}$$ | III. | 0 |
| D. | $${[Fe{({H_2}O)_6}]^{3 + }}$$ | IV. | $$-0.8$$ |
Choose the correct answer from the options given below:
A-IV, B-I, C-II, D-III
A-II, B-IV, C-III, D-I
A-II, B-IV, C-I, D-III
A-IV, B-I, C-III, D-II
Explanation
$$
\begin{aligned}
& \text{(A)} ~~ {\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \\\\
& \mathrm{Ti}^{2+} \Rightarrow 3 \mathrm{~d}^2 4 \mathrm{~s}^0 \\\\
& \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=2 \\\\
& \mathrm{e}_{\mathrm{g}} \mathrm{e}^{-}=0 \\\\
& \mathrm{CFSE}=[-0.4 \times 2+0.6 \times 0] \Delta_0 \\\\
& \quad=-0.8 \Delta
\end{aligned}
$$
$$ \begin{aligned} & \text{(B)} ~~ {\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \\\\ & \mathrm{V}^{2+} \Rightarrow 3 \mathrm{~d}^3 4 \mathrm{~s}^0 \\\\ & \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=3 \\\\ & \mathrm{e}_{\mathrm{g}} \mathrm{e}^{-}=0 \\\\ & \mathrm{CFSE}=[-0.4 \times 3+0.6 \times 0] \Delta_0 \\\\ & =-1.2 \Delta_0 \end{aligned} $$
$$ \begin{aligned} & \text{(C)} ~~{\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}} \\\\ & \mathrm{Mn}^{3+} \Rightarrow 3 \mathrm{~d}^4 4 \mathrm{~s}^0 \\\\ & \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=3 \\\\ & \mathrm{e}_{\mathrm{g}} \mathrm{e}^{-}=1 \\\\ & \mathrm{CFSE}=[-0.4 \times 3+0.6 \times 1] \Delta_0 \\\\ & =-0.6 \Delta_0 \end{aligned} $$
$$ \begin{aligned} & \text{(D)} ~~{\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}} \\\\ & \mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{~d}^5 4 \mathrm{~s}^0 \\\\ & \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=3 \quad \mathrm{e}_{\mathrm{g}}=2 \\\\ & \mathrm{CFSE}=[-0.4 \times 3+0.6 \times 2] \Delta_0 \\\\ & \quad=0 \Delta_0 \end{aligned} $$
$$ \begin{aligned} & \text{(B)} ~~ {\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \\\\ & \mathrm{V}^{2+} \Rightarrow 3 \mathrm{~d}^3 4 \mathrm{~s}^0 \\\\ & \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=3 \\\\ & \mathrm{e}_{\mathrm{g}} \mathrm{e}^{-}=0 \\\\ & \mathrm{CFSE}=[-0.4 \times 3+0.6 \times 0] \Delta_0 \\\\ & =-1.2 \Delta_0 \end{aligned} $$
$$ \begin{aligned} & \text{(C)} ~~{\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}} \\\\ & \mathrm{Mn}^{3+} \Rightarrow 3 \mathrm{~d}^4 4 \mathrm{~s}^0 \\\\ & \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=3 \\\\ & \mathrm{e}_{\mathrm{g}} \mathrm{e}^{-}=1 \\\\ & \mathrm{CFSE}=[-0.4 \times 3+0.6 \times 1] \Delta_0 \\\\ & =-0.6 \Delta_0 \end{aligned} $$
$$ \begin{aligned} & \text{(D)} ~~{\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}} \\\\ & \mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{~d}^5 4 \mathrm{~s}^0 \\\\ & \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=3 \quad \mathrm{e}_{\mathrm{g}}=2 \\\\ & \mathrm{CFSE}=[-0.4 \times 3+0.6 \times 2] \Delta_0 \\\\ & \quad=0 \Delta_0 \end{aligned} $$
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