Find the oxidation state and d-electron count of each metal ion

(a) $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$

• $\mathrm{NH}_3$ is a neutral ligand.
• So oxidation state of Co is $+3$.
• Cobalt: $Z=27$, electronic configuration of Co is $[\mathrm{Ar}]\,3d^7 4s^2$.
• Therefore, $\mathrm{Co}^{3+}$ is: $3d^6$

(b) $[\mathrm{NiCl}_4]^{2-}$

• Let oxidation state of Ni be $x$.
• Each $\mathrm{Cl}^-$ contributes $-1$.
• So, $x+4(-1)=-2$ $x-4=-2$ $x=+2$
• Nickel: $Z=28$, electronic configuration is $[\mathrm{Ar}]\,3d^8 4s^2$.
• Therefore, $\mathrm{Ni}^{2+}$ is: $3d^8$

Determine geometry and spin state

(a) $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$

• Coordination number $=6$, so geometry is octahedral.
• For $\mathrm{Co}^{3+}$, even with $\mathrm{NH}_3$ (reasonably strong ligand), the complex is typically low spin.
• Octahedral low-spin $d^6$ configuration: $t_{2g}^6 e_g^0$
• All electrons are paired.
• Number of unpaired electrons = 0.

(b) $[\mathrm{NiCl}_4]^{2-}$

• Coordination number $=4$.
• $\mathrm{Cl}^-$ is a weak-field ligand.
• $[\mathrm{NiCl}_4]^{2-}$ is tetrahedral.
• For tetrahedral $d^8$, the complex is high spin.
• Electron arrangement gives 2 unpaired electrons.

Add the unpaired electrons

$0+2=2$

Final Answer

Total number of unpaired electrons is: $\boxed{2}$