Determining Magnetic Moment for Hexacyanidomanganate(II)

The question asks for the spin-only magnetic moment of the Hexacyanidomanganate(II) ion. First, we need to determine the formula and the oxidation state of manganese (Mn).

• The ion is stated as Hexacyanidomanganate(II).
• "Hexa" indicates 6 cyanide ligands ($CN^-$).
• "Manganate(II)" indicates manganese in the +2 oxidation state ($Mn^{2+}$).
• Each $CN^-$ ligand has a charge of -1. The total charge from 6 $CN^-$ ligands is $6 \times (-1) = -6$.
• The overall charge of the complex ion is the sum of the manganese charge and the ligand charges: $(+2) + (-6) = -4$.
• Therefore, the complex ion is $[Mn(CN)_6]^{4-}$.

Electronic Configuration and Unpaired Electrons

Next, we find the electronic configuration of the central metal ion, $Mn^{2+}$.

• Manganese (Mn) has atomic number 25. Its ground state electronic configuration is $[Ar] 3d^5 4s^2$.
• The $Mn^{2+}$ ion is formed by losing two electrons from the 4s orbital, resulting in the configuration: $[Ar] 3d^5$.
• The cyanide ligand ($CN^-$) is a strong field ligand. In an octahedral crystal field, strong field ligands cause pairing of electrons in the lower energy $t_{2g}$ orbitals before occupying the higher energy $e_g$ orbitals.
• For a $d^5$ configuration in an octahedral field with a strong ligand, the electron arrangement is $t_{2g}^5 e_g^0$.
• In the $t_{2g}^5$ configuration, there are two paired electrons and one unpaired electron.
• Thus, the number of unpaired electrons ($n$) is 1.

Calculating Spin-Only Magnetic Moment

The spin-only magnetic moment ($\mu_s$) is calculated using the formula:

$ \mu_s = \sqrt{n(n+2)} \, \text{B.M.} $

Where '$n$' is the number of unpaired electrons.

• Substituting $n=1$ into the formula:
• $ \mu_s = \sqrt{1(1+2)} $
• $ \mu_s = \sqrt{1 \times 3} $
• $ \mu_s = \sqrt{3} $
• $ \mu_s \approx 1.73 \, \text{B.M.} $

The spin-only magnetic moment of the Hexacyanidomanganate(II) ion is approximately 1.73 B.M.