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QJEEM23Ch4f45: JEE Main 2023 Chemistry - Class 12 JEEM23Ch4f45 Solution: Match List I with List II .tg…

Match List I with List II

List - I
Complex		 List - II
Crystal Field splitting energy ( Δ0\Delta_0 )
A. [Ti(H2O)6]2+{[Ti{({H_2}O)_6}]^{2 + }} I. −1.2-1.2
B. [V(H2O)6]2+{[V{({H_2}O)_6}]^{2 + }} II. −0.6-0.6
C. [Mn(H2O)6]3+{[Mn{({H_2}O)_6}]^{3 + }} III. 0
D. [Fe(H2O)6]3+{[Fe{({H_2}O)_6}]^{3 + }} IV. −0.8-0.8

Choose the correct answer from the options given below:

Multiple Choice Options

  • A A-IV, B-I, C-II, D-III
  • B A-II, B-IV, C-III, D-I
  • C A-II, B-IV, C-I, D-III
  • D A-IV, B-I, C-III, D-II

Step-by-Step Verified Solution

amp;(A)  [Ti(H2O)6]2+amp;Ti2+⇒3 d24 s0amp;t2 ge−=2amp;ege−=0amp;CFSE=[−0.4×2+0.6×0]Δ0amp;=−0.8Δ\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}

amp;(B)  [V(H2O)6]2+amp;V2+⇒3 d34 s0amp;t2 ge−=3amp;ege−=0amp;CFSE=[−0.4×3+0.6×0]Δ0amp;=−1.2Δ0\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}

amp;(C)  [Mn(H2O)6]3+amp;Mn3+⇒3 d44 s0amp;t2 ge−=3amp;ege−=1amp;CFSE=[−0.4×3+0.6×1]Δ0amp;=−0.6Δ0\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}

amp;(D)  [Fe(H2O)6]3+amp;Fe3+⇒3 d54 s0amp;t2 ge−=3eg=2amp;CFSE=[−0.4×3+0.6×2]Δ0amp;=0Δ0\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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