Question

Match each set of hybrid orbitals from LIST-I with complex(es) given in LIST-II.

| LIST-I | LIST-II |
|--------|--------|
| P. $\mathrm{dsp}^{2}$ | 1. $\left[\mathrm{FeF}_{6}\right]^{4-}$ |
| Q. $\mathrm{sp}^{3}$ | 2. $\left[\mathrm{Ti}(\mathrm{H}_{2}\mathrm{O})_{3}\mathrm{Cl}_{3}\right]$ |
| R. $\mathrm{sp}^{3}\mathrm{d}^{2}$ | 3. $\left[\mathrm{Cr}(\mathrm{NH}_{3})_{6}\right]^{3+}$ |
| S. $\mathrm{d}^{2}\mathrm{sp}^{3}$ | 4. $\left[\mathrm{FeCl}_{4}\right]^{2-}$ |
| | 5. $\mathrm{Ni}(\mathrm{CO})_{4}$ |
| | 6. $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ |

The correct option is

Accepted Answer

(C) $\mathbf{P} \rightarrow 6; \mathbf{Q} \rightarrow 4,5; \mathbf{R} \rightarrow 1; \mathbf{S} \rightarrow 2,3$

Answer

(A) $\mathbf{P} \rightarrow 5; \mathbf{Q} \rightarrow 4,6; \mathbf{R} \rightarrow 2,3; \mathbf{S} \rightarrow 1$

Answer

(B) $\mathbf{P} \rightarrow 5,6; \mathbf{Q} \rightarrow 4; \mathbf{R} \rightarrow 3; \mathbf{S} \rightarrow 1,2$

Answer

(D) $\mathbf{P} \rightarrow 4,6; \mathbf{Q} \rightarrow 5,6; \mathbf{R} \rightarrow 1,2; \mathbf{S} \rightarrow 3$

LIST-I	LIST-II
P. $dsp^{2}$	1. $[FeF_{6}]^{4 -}$
Q. $sp^{3}$	2. $[Ti (H_{2} O)_{3} Cl_{3}]$
R. $sp^{3} d^{2}$	3. $[Cr (NH_{3})_{6}]^{3 +}$
S. $d^{2} sp^{3}$	4. $[FeCl_{4}]^{2 -}$
	5. $Ni (CO)_{4}$
	6. $[Ni (CN)_{4}]^{2 -}$

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Coordination Compounds

Solution

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