Question

Total number of unpaired electrons in the complex ion [Co(NH$_3$)$_6$]$^{3+}$ and [NiCl$_4$]$^{2-}$ is

Answer 1

Correct Answer: 2

To determine the total number of unpaired electrons in the complex ions [Co(NH₃)₆]³⁺ and [NiCl₄]²⁻, we must analyze each complex individually:

[Co(NH₃)₆]³⁺:

• Cobalt (Co) has an atomic number of 27, giving it an electron configuration of [Ar]3d⁷4s².
• In the 3+ oxidation state, Co loses 3 electrons, resulting in an electron configuration of [Ar]3d⁶.
• NH₃ is a strong field ligand, causing pairing of electrons in the 3d orbitals.
• The 3d orbital configuration is t_2g⁶e_g⁰, indicating that all electrons are paired.
• Thus, there are 0 unpaired electrons in this complex.

[NiCl₄]²⁻:

• Nickel (Ni) has an atomic number of 28, giving it an electron configuration of [Ar]3d⁸4s².
• In the 2− oxidation state, Ni loses 2 electrons, resulting in an electron configuration of [Ar]3d⁸.
• Cl⁻ is a weak field ligand, so there is no electron pairing in the 3d orbitals.
• The 3d orbital configuration is e_g⁴t_2g⁴, with 2 unpaired electrons.

Combining both results, the total number of unpaired electrons in the complex ions is 0 (from Co) + 2 (from Ni) = 2.

This result fits the provided range: 2,2.

Answer 2

[Co(NH$_3$)$_6$]$^{3+}$
-Oxidation state of Co: $+3$
-Electronic configuration of Co$^{3+}$: $3d^6$
-NH$_3$ is a strong field ligand, causing pairing of electrons in the $t_{2g}$ orbitals.
-Distribution: $t_{2g}^6 e_g^0$ 
-Unpaired electrons: 0
[NiCl$_4$]$^{2-}$
-Oxidation state of Ni: $+2$
-Electronic configuration of Ni$^{2+}$: $3d^8$
-Cl$^-$ is a weak field ligand, causing no pairing of electrons.
-Distribution: $e_g^2 t_{2g}^6$
-Unpaired electrons: 2
Total unpaired electrons: $0 + 2 = 2$