JEE Advance - Chemistry (2022 - Paper 1 Online - No. 17)

LIST-I contains metal species and LIST-II contains their properties.

List-I	List-II
(I) $\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-}$	(P) $t_{2 \mathrm{g}}$ orbitals contain 4 electrons
(II) $\left[\mathrm{RuCl}_{6}\right]^{2-}$	(Q) $\mu$ (spin-only $)=4.9 \mathrm{BM}$
(III) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$	(R) low spin complex ion
(IV) $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$	(S) metal ion in $4+$ oxidation state
	(T) $d^{4}$ species

[Given: Atomic number of $\mathrm{Cr}=24, \mathrm{Ru}=44, \mathrm{Fe}=26$ ]

Match each metal species in LIST-I with their properties in LIST-II, and choose the correct option

I $\rightarrow$ R, T; II $\rightarrow$ P, S; III $\rightarrow$ Q, T; IV $\rightarrow$ P, Q

I $\rightarrow$ R, S; II $\rightarrow$ P, T; III $\rightarrow$ P, Q; IV $\rightarrow$ Q, T

I $\rightarrow$ P, R; II $\rightarrow$ R, S; III $\rightarrow$ R, T; IV $\rightarrow$ P, T

I $\rightarrow$ Q, T; II $\rightarrow$ S, T; III $\rightarrow$ P, T; IV $\rightarrow$ Q, R

(I) $${[Cr{(CN)_6}]^{4 - }}$$

$$C{r^{ + 2}} = [Ar]\,3{d^4}\,4{s^0}$$

It is d²sp³ hybridised as CN^$$-$$ is a strong field ligand.

(II) $${[RuC{l_6}]^{2 - }}$$

$$R{u^{ + 4}} = [Kr]\,4{d^4}\,5{s^0}$$

t_2g set contains 4 electron.

(III) $${[Cr{({H_2}O)_6}]^{2 + }}$$

$$C{r^{ + 2}} = [Ar]\,3{d^4}\,4{s^0}$$

It has 4 unpaired e^$$-$$ as H₂O is weak field ligand.

So, its $$\mu$$ = 4.9 B.M.

(IV) $${[Fe{({H_2}O)_6}]^{2 + }}$$

$$F{e^{ + 2}} = [Ar]\,3{d^6}\,4{s^0}$$

$$ = t_{2g}^4\,e_g^2$$

It has 4 unpaired e^$$-$$, its $$\mu$$ = 4.9 B.M.