Assertion (A) $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}_6\right)\right] \mathrm{Cl}_2$ and $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2$ are reducing in nature.
Reason (R) Unpaired electrons are present in their d-orbitals.
Assertion (A) Linkage isomerism arises in coordination compounds containing ambidentate ligand.
Reason (R) Ambidentate ligand has two different donor atoms.
Assertion (A) Complexes of $M X_6$ and $M X_5 L$ type ( $X$ and $L$ are unidentate) do not show geometrical isomerism.
Reason (R) Geometrical isomerism is not shown by complexes of coordination number 6.
Assertion $(\mathrm{A})\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ ion shows magnetic moment corresponding to two unpaired electrons.
Reason (R) Because it has $d^2 s p^3$ type hybridisation.
Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following
(a) $\left[\mathrm{CoF}_6\right]^{3-},\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$
(b) $\mathrm{FeF}_6{ }^{3-},\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
(a) $\left[\mathrm{CoF}_6\right]^{3-}$.
$\mathrm{F}^{-}$is a weak field ligand.
Configuration of $\mathrm{Co}^{3+}=3 d^6$ (or $t_{2 g}^4 e_g^2$)
Number of unpaired electrons $(n)=4$
Magnetic moment $(\mu)=\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24}=4.9 \mathrm{BM}$
$\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$, $\mathrm{H}_2 \mathrm{O}$ is a weak field ligand.
Configuration of $\mathrm{Co}^{2+}=3 d^7\left(\right.$ or $\left.t_{2 g}^5 e_g^2\right)$
Number of unpaired electrons $(n)=3$
$$\mu=\sqrt{3(3+2)}=\sqrt{15}=3.87 \mathrm{BM}$$
$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$ i.e., $\mathrm{Co}^{3+}$
$\because \quad \mathrm{CN}$ is strong field ligand.
$$\mathrm{Co}^{3+}=3 \mathrm{~d}^6\left(\mathrm{or} t_{2 g}^6 \mathrm{e}_g^0\right)$$
There is no unpaired electron, so it is diamagnetic.
$$\mu=0$$
(b) $\left[\mathrm{FeF}_6\right]^{3-}$,
$\mathrm{Fe}^{3+}=3 d^5\left(\right.$ or $\left.t_{2 g}^3 e_g^2\right)$
Number of unpaired electrons, $n=5$
$$\begin{aligned} \mu & =\sqrt{5(5+2)} \\ & =\sqrt{35}=5.92 \mathrm{BM} \\ {\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} } & \end{aligned}$$
$$\mathrm{Fe}^{2+}=3 d^6\left(\operatorname{or} t_{2 g}^4 e_g^2\right)$$
Number of unpaired electrons, $n=4$
$$\begin{aligned} \mu & =\sqrt{4(4+2)} \\ Z & =\sqrt{24} \\ & =4.98 \mathrm{BM} \\ {\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} } & \end{aligned}$$
Since, $\mathrm{CN}^{-}$is a strong field ligand, all the electrons get paired.
$$\mathrm{Fe}^{2+}=3 d^6\left(\operatorname{or} t_{2 g}^6 e_g^0\right)$$
Because there is no unpaired electron, so it is diamagnetic in nature.