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:
Solution
$$
\begin{aligned}
& \text{(A)} ~~ {\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \\\\
& \mathrm{Ti}^{2+} \Rightarrow 3 \mathrm{~d}^2 4 \mathrm{~s}^0 \\\\
& \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=2 \\\\
& \mathrm{e}_{\mathrm{g}} \mathrm{e}^{-}=0 \\\\
& \mathrm{CFSE}=[-0.4 \times 2+0.6 \times 0] \Delta_0 \\\\
& \quad=-0.8 \Delta
\end{aligned}
$$
<br/><br/>
$$
\begin{aligned}
& \text{(B)} ~~ {\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}} \\\\
& \mathrm{V}^{2+} \Rightarrow 3 \mathrm{~d}^3 4 \mathrm{~s}^0 \\\\
& \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=3 \\\\
& \mathrm{e}_{\mathrm{g}} \mathrm{e}^{-}=0 \\\\
& \mathrm{CFSE}=[-0.4 \times 3+0.6 \times 0] \Delta_0 \\\\
& =-1.2 \Delta_0
\end{aligned}
$$
<br/><br/>
$$
\begin{aligned}
& \text{(C)} ~~{\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}} \\\\
& \mathrm{Mn}^{3+} \Rightarrow 3 \mathrm{~d}^4 4 \mathrm{~s}^0 \\\\
& \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=3 \\\\
& \mathrm{e}_{\mathrm{g}} \mathrm{e}^{-}=1 \\\\
& \mathrm{CFSE}=[-0.4 \times 3+0.6 \times 1] \Delta_0 \\\\
& =-0.6 \Delta_0
\end{aligned}
$$<br/><br/>
$$
\begin{aligned}
& \text{(D)} ~~{\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}} \\\\
& \mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{~d}^5 4 \mathrm{~s}^0 \\\\
& \mathrm{t}_{2 \mathrm{~g}} \mathrm{e}^{-}=3 \quad \mathrm{e}_{\mathrm{g}}=2 \\\\
& \mathrm{CFSE}=[-0.4 \times 3+0.6 \times 2] \Delta_0 \\\\
& \quad=0 \Delta_0
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Coordination Compounds · Topic: Ligands and Coordination Number
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