(1)

A) In ${\left[{\mathrm{NiCl}}_4\right]}^{2-}$, there are two unpaired electrons because ${\mathrm{Cl}}^{-}$ is a weak field ligand. Therefore, it is paramagnetic in nature.

[IMAGE 26]------------------------

B) In $\left[\mathrm{Ni}{\left(\mathrm{CO}\right)}_4\right]$

     ${\mathrm{Ni}}^0 (\mathrm{Z}=28):3d^{8}4s^2$

     Since CO is a strong field ligand, it forces electrons to pair up. Therefore, it is diamagnetic in nature.

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C) In ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$, there is no unpaired electron because ${\mathrm{CN}}^{-}$ is a strong field ligand. Therefore, it is diamagnetic in nature.

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D) Oxidation state of Ni in ${\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}=+2$

     ${\mathrm{Ni}}^{2+} (\mathrm{Z}=28):3d^8$

     ${\mathrm{H}}_2\mathrm{O}$ is a weak field ligand hence pairing of electrons does not take place and compound is paramagnetic on account of two unpaired electrons.

E) In $\left[\mathrm{Ni}{\left({\mathrm{PPh}}_3\right)}_4\right]$, oxidation state of Ni = 0

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       It is diamagnetic in nature.

(4)

Stronger is the strength of the ligand, greater is the splitting and higher is the energy absorbed by the complex.

As  $E=\frac{hc}{\lambda } \text{or} E\propto \frac{1}{\lambda }$

So, wavelength of light absorbed is inversely proportional to the strength of the ligand.

(1)

In ${\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}$:

Hybridisation – $d^{2}s{p}^3$

Structure – Octahedral

Nature – Diamagnetic

In ${\left[{\mathrm{CoF}}_6\right]}^{3-}$:

Hybridisation – $s{p}^3{d}^2$

Structure – Octahedral

Nature – Paramagnetic

(3)

Chelating ligand increases the stability of complex compound and higher the number of chelating ligands, higher will be the stability. Stronger is the strength of ligand, greater is the energy absorbed by the complex. Hence, the order is :  

$\underset{C}{{\left[\mathrm{Ni}{\left(\mathrm{en}\right)}_3\right]}^{2+}} > \underset{A}{{\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_2{\left(\mathrm{en}\right)}_2\right]}^{2+}} > \underset{B}{{\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4\right(\mathrm{en}\left)\right]}^{2+}}$

(1)

${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}$

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Spin only magnetic moment = $\sqrt{n(n+2)}$

where $n$ = number of unpaired electrons.

Hence, $n=1$

$\mu =\sqrt{1(1+2)}=\sqrt{3}=1.73$ B.M.

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$n=5$,   $\mu =\sqrt{5(5+2)}=\sqrt{35}=5.92$ B.M.

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$n=0,$   $\mu =\sqrt{0}=0$ B.M.

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$n=4,$   $\mu =\sqrt{4(4+2)}=\sqrt{24}=4.90$ B.M.

(1)

According to spectrochemical series, order of increasing field strength is:

${\text{SCN}}^{-}<{\text{F}}^{-}<{\text{C}}_2{\text{O}}_4^{2-}<{\text{CN}}^{-}$

(3)

In ${\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$ complex, Fe is in +2 oxidation state.

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As ${\mathrm{CN}}^{-}$ is a strong field ligand, it causes pairing of electrons. Therefore, electronic configuration of ${\mathrm{Fe}}^{2+}$ in ${\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$ is $t_{2g}^6{e}_g^0$

(1)

(4)

${\Delta }_t=\frac{4}{9}{\Delta }_o=\frac{4}{9}\times 18000=8000$ ${\text{cm}}^{-1}$

(2)

Ni(28) : [Ar] $3d^{8}4s^2$

$\because$  CO is a strong field ligand, so, unpaired electrons get paired.

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Thus, the complex is $s{p}^3$ hybridised with tetrahedral geometry and diamagnetic in nature.