JEE MAIN - Chemistry (2023 - 8th April Evening Shift - No. 1)
Match List I with List II
| LIST I Coordination Complex |
LIST II Number of unpaired electrons |
||
|---|---|---|---|
| A. | $$\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{3-}$$ | I. | 0 |
| B. | $$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$$ | II. | 3 |
| C. | $$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$$ | III. | 2 |
| D. | $$\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$$ | IV. | 4 |
Choose the correct answer from the options given below:
A-IV, B-III, C-II, D-I
A-II, B-I, C-IV, D-III
A-II, B-IV, C-I, D-III
A-III, B-IV, C-I, D-II
Explanation
For option (A)
$$ \begin{aligned} & \mathrm{Cr}^{+3}: 3 \mathrm{~d}^3 \\\\ & \mathrm{CN}^{-} \rightarrow \mathrm{SFL} \end{aligned} $$
No. of unpaired electrons $=3$
For option (B)
$$ \begin{aligned} & \mathrm{Fe}^{+2}: 3 \mathrm{~d}^6 \\\\ & \mathrm{H}_2 \mathrm{O}: \mathrm{WFL} \end{aligned} $$
No. of unpaired electrons $=4$
For option (C)
$$ \begin{aligned} & \mathrm{Co}^{+3}: 3 \mathrm{~d}^6 \\\\ & \mathrm{NH}_3: \mathrm{SFL} \end{aligned} $$
No. of unpaired electrons $=0$
For option (D)
$$ \begin{aligned} & \mathrm{Ni}^{+2}: 3 \mathrm{~d}^8 \\\\ & \mathrm{NH}_3: \mathrm{SFL} \end{aligned} $$
No. of unpaired electrons $=2$
$$ \begin{aligned} & \mathrm{Cr}^{+3}: 3 \mathrm{~d}^3 \\\\ & \mathrm{CN}^{-} \rightarrow \mathrm{SFL} \end{aligned} $$
No. of unpaired electrons $=3$
For option (B)
$$ \begin{aligned} & \mathrm{Fe}^{+2}: 3 \mathrm{~d}^6 \\\\ & \mathrm{H}_2 \mathrm{O}: \mathrm{WFL} \end{aligned} $$
No. of unpaired electrons $=4$
For option (C)
$$ \begin{aligned} & \mathrm{Co}^{+3}: 3 \mathrm{~d}^6 \\\\ & \mathrm{NH}_3: \mathrm{SFL} \end{aligned} $$
No. of unpaired electrons $=0$
For option (D)
$$ \begin{aligned} & \mathrm{Ni}^{+2}: 3 \mathrm{~d}^8 \\\\ & \mathrm{NH}_3: \mathrm{SFL} \end{aligned} $$
No. of unpaired electrons $=2$
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