Answer

$\mathrm{Ni}$ is in the +2 oxidation state i.e., in $\mathrm{d}^{8}$ configuration.

There are $4 \mathrm{CN}^{-}$ ions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since $\mathrm{CN}^{-}$ ion is a strong field ligand, it causes the pairing of unpaired $3 d$ electrons.

It now undergoes $\mathrm{dsp}^{2}$ hybridization. Since all electrons are paired, it is diamagnetic.

In case of $\left[\mathrm{NiCl_4}\right]^{2-}, $ chloride ion is a weak field ligand. Therefore, it does not lead to the pairing of unpaired $3 d$ electrons. Therefore, it undergoes $s p^{3}$ hybridization.

Since there are 2 unpaired electrons in this case, it is paramagnetic in nature.

Answer

Though both $\left[\mathrm{NiCl_4}\right]^{2-}$ and $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. $\mathrm{Cl}^{-}$ is a weak field ligand and it does not cause the pairing of unpaired $3 d$ electrons. Hence, $\left[\mathrm{NiCl_4}\right]^{2-}$ is paramagnetic.

In $\mathrm{Ni}(\mathrm{CO})_{4}, \mathrm{Ni}$ is in the zero oxidation state i.e., it has a configuration of $3 d^{8} 4 s^{2}$

But $\mathrm{CO}$ is a strong field ligand. Therefore, it causes the pairing of unpaired $3 d$ electrons. Also, it causes the $4 \mathrm{~s}$ electrons to shift to the $3 d$ orbital, thereby giving rise to $s p^{3}$ hybridization. Since no unpaired electrons are present in this case, $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$ is diamagnetic.

Answer

In both $[\mathrm{Fe(H_2O)_6}]^{3+}$ and $[\mathrm{Fe}\mathrm{(CN)_6}]^{3-}$, Fe exists in the +3 oxidation state i.e., in $d^{5}$ configuration.

Since $\mathrm{CN}^-$ is a strong field ligand, it causes the pairing of unpaired electrons. Therefore, there is only one unpaired electron left in the $d$-orbital.

Therefore,

$ \begin{aligned} \mu & =\sqrt{n(n+2)} \\ & =\sqrt{1(1+2)} \\ & =\sqrt{3} \\ & =1.732\ \mathrm{BM} \end{aligned} $

On the other hand, $\mathrm{H_2} \mathrm{O}$ is a weak field ligand. Therefore, it cannot cause the pairing of electrons. This means that the number of unpaired electrons is 5 .

Therefore,

$ \begin{aligned} \mu & =\sqrt{n(n+2)} \\ & =\sqrt{5(5+2)} \\ & =\sqrt{35} \\ & 3.873\ \mathrm{BM} \end{aligned} $

Thus, it is evident that $[\mathrm{Fe(H_2 O)_6}]^{3+}$ is strongly paramagnetic, while $[\mathrm{Fe(CN)_6}]^{3-}$ is weakly paramagnetic.

Answer

$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$	$\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}$
Oxidation state of cobalt $=+3$	Oxidation state of $\mathrm{Ni}=+2$
Electronic configuration of cobalt $=d^6$	Electronic configuration of nickel $=d^8$
$\mathrm{NH}_3$ being a strong field ligand for $CO^{3+}$ causes the pairing. Therefore, Cobalt can undergo $d^2 s p^3$ hybridization.	If $\mathrm{NH}_3$ causes the pairing, then only one $3 d$ orbital is empty. Thus, it cannot undergo $d^s s p^3$ hybridization. Therefore, it undergoes $s p^3 d^2$ hybridization.
Hence, it is an inner orbital complex.	Hence, it forms an outer orbital complex.

Answer

$\left[\mathrm{Pt}(\mathrm{CN})_{4}\right]^{2-}$

In this complex, $\mathrm{Pt}$ is in the +2 state. It forms a square planar structure. This means that it undergoes $d p^{2}$ hybridization. Now, the electronic configuration of $\mathrm{Pd}(+2)$ is $5 d^{8}$.

$\mathrm{CN}^{-}$ being a strong field ligand causes the pairing of unpaired electrons. Hence, there are no unpaired electrons in $\left[\operatorname{Pt}(\mathrm{CN})_{4}\right]^{2-}$.

Answer

Hence, hexaaquo manganese (II) ion has five unpaired electrons, while hexacyano ion has only one unpaired electron.