JEE MAIN - Chemistry (2024 - 9th April Evening Shift - No. 15)
Match List I with List II
| LIST I | LIST II | ||
|---|---|---|---|
| A. | $$\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]$$ | I. | $$sp^3$$ |
| B. | $$\left[\mathrm{Ni}(\mathrm{CO})_4\right]$$ | II. | $$sp^3d^2$$ |
| C. | $$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$$ | III. | $$dsp^2$$ |
| D. | $$\mathrm{Na}_3\left[\mathrm{CoF}_6\right]$$ | IV. | $$d^2sp^3$$ |
Choose the correct answer from the options given below:
Explanation
$$\begin{aligned} & \stackrel{+2}{\text { (A) } \mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]} \\ & \mathrm{Ni}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^{\circ},\left(\mathrm{CN}^{-} \text {is } \mathrm{S} . \mathrm{F} . \mathrm{L}\right) \\ & \text { Pre hybridization state of } \mathrm{Ni}^{+2} \end{aligned}$$
_9th_April_Evening_Shift_en_15_1.png)
(B) $$[\mathrm{Ni}(\mathrm{CO})_4]$$
$$\mathrm{Ni}:[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^2$$
$$\mathrm{CO}$$ is S.F.L , so pairing occur
Pre hybridization state of $$\mathrm{Ni}$$
_9th_April_Evening_Shift_en_15_2.png)
(C) $$[\mathrm{Co}(\mathrm{NH}_3)_6] \mathrm{Cl}_3$$
$$\mathrm{Co}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^6 4 \mathrm{~s}^0$$
With $$\mathrm{Co}^{3+}, \mathrm{NH}_3$$ act as S.F.L
_9th_April_Evening_Shift_en_15_3.png)
(D) $$\mathrm{Na}_3[\mathrm{CoF}_6]$$
$$\mathrm{Co}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^6\left(\mathrm{~F}^{\ominus}: \text { W.F.L }\right)$$
_9th_April_Evening_Shift_en_15_4.png)
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