Unit 9 Coordination Compounds (Intext Questions-3)
Intext Questions
9.5 Explain on the basis of valence bond theory that $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ ion with square planar structure is diamagnetic and the $\left[\mathrm{NiCl_4}\right]^{2-}$ ion with tetrahedral geometry is paramagnetic.
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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 $\left[\mathrm{NiCl_4}\right]^{2}, \mathrm{Cl}$ 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.
9.6 $\left[\mathrm{NiCl_4}\right]^{2-}$ is paramagnetic while $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ is diamagnetic though both are tetrahedral. Why?
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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.
9.7 $\left[\mathrm{Fe}\left(\mathrm{H_2} \mathrm{O}\right)_6\right]^{3+}$ is strongly paramagnetic whereas $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ is weakly paramagnetic. Explain.
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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.
$ \begin{array}{|c|c|c|c|c|} \hline \uparrow \downarrow & \uparrow \downarrow &\uparrow & & \\ \hline \end{array} $
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} \\ & \simeq 6 \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.
9.8 Explain $\left[\mathrm{Co}\left(\mathrm{NH_3}\right)_6\right]^{3+}$ is an inner orbital complex whereas $\left[\mathrm{Ni}\left(\mathrm{NH_3}\right)_6\right]^{2+}$ is an outer orbital complex.
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9.9 Predict the number of unpaired electrons in the square planar $\left[\mathrm{Pt}(\mathrm{CN})_{4}\right]^{2-}$ ion.
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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 $5 d^{8}$.
$\underset{3d^8}{\begin{array}{|c|c|c|c|c|}\hline \uparrow \downarrow & \uparrow \downarrow &\uparrow \downarrow & \uparrow & \uparrow \\ \hline \end{array}}$
$\mathrm{CN}^{-}$being a strong field ligand causes the pairing of unpaired electrons. Hence, there are no unpaired electrons in $\left[\operatorname{Pt}(\mathrm{CN})_{4}\right]^{2-}$.
9.10 The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.
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Answer
| $\left[\mathrm{Mn}\left(\mathrm{H_2} \mathrm{O}\right)_{6}\right]^{2+}$ | $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ |
|---|---|
| $\mathrm{Mn}$ is in the +2 oxidation state. | $\mathrm{Mn}$ is in the +2 oxidation state. |
| The electronic configuration is $\mathrm{d}^{5}$ | The electronic configuration is $\mathrm{d}^{5}$ |
| The crystal field is octahedral. Water is a weak field ligand. Therefore, the arrangement of the electrons in $\left[\mathrm{Mn}\left(\mathrm{H_2} \mathrm{O}\right)_{6}\right]^{2+}$ is t2g3eg2. | The crystal field is octahedral. Cyanide is a strong field ligand. Therefore, the arrangement of the electrons in $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}$ is T2g5eg0. |
Hence, hexaaquo manganese (II) ion has five unpaired electrons, while hexacyano ion has only one unpaired electron.
