The spin-only magnetic moment of Hexacyanomanganate(II) ion is _____BM.

The correct option is 5.90.

To find the spin-only magnetic moment of the Hexacyanomanganate(II) ion, we follow these steps:
First, we identify the chemical formula of the Hexacyanomanganate(II) ion, which is [Mn(CN)₆]^4-.

Next, we determine the oxidation state of the central metal atom, Manganese (Mn):
Let the oxidation state of Mn be x. Since each cyanide (CN^-) ligand carries a charge of -1, we have:
x + 6(-1) = -4
x - 6 = -4
x = +2
Thus, manganese is in the +2 oxidation state (Mn²⁺).

The atomic number of manganese is 25. Its ground-state electronic configuration is:
Mn: [Ar] 3d⁵ 4s²
For the Mn²⁺ ion, the electronic configuration is formed by removing two electrons from the outer 4s orbital:
Mn²⁺: [Ar] 3d⁵

In the high-spin configuration of the 3d⁵ system, all 5 d-orbitals contain one electron each. Therefore, the number of unpaired electrons (n) is 5.

The formula for the spin-only magnetic moment ($\mu$) in Bohr Magnetons (BM) is given by:

$$\mu =\sqrt{n(n+2)}$$

Substituting n = 5 into the formula:

$$\mu =\sqrt{5(5+2)}=\sqrt{35}\approx 5.92\text{BM}$$

This calculated value matches the option of 5.90 BM.