In hexacyanidomanganate(II) ion the Mn atom assumes `d^(2)sp^(3)`- hybrid sates. Then the number of unpaired electrons in the complex is .

Explanation

To determine the number of unpaired electrons in the hexacyanidomanganate(II) ion, we can follow these steps:

### Step 1: Identify the oxidation state of manganese in the complex
The hexacyanidomanganate(II) ion can be represented as [Mn(CN)₆]²⁻. Here, the cyanide (CN⁻) is a negatively charged ligand, and since there are six cyanide ligands, the total negative charge contributed by the ligands is -6. Let the oxidation state of manganese be \( x \).

Setting up the equation for the overall charge:
\[ x + 6(-1) = -2 \]
\[ x - 6 = -2 \]
\[ x = +4 \]

However, since we are dealing with the manganese(II) ion, we have:
\[ x = +2 \]

### Step 2: Determine the electron configuration of manganese in the +2 oxidation state
The atomic number of manganese (Mn) is 25. The ground state electron configuration of Mn is:
\[ \text{Mn: } [Ar] 4s^2 3d^5 \]

For Mn²⁺, we remove two electrons from the 4s orbital:
\[ \text{Mn}^{2+}: [Ar] 3d^5 \]

### Step 3: Analyze the effect of the ligand
Cyanide (CN⁻) is a strong field ligand, which causes pairing of electrons in the d-orbitals. In the case of Mn²⁺ with a 3d⁵ configuration, the presence of a strong field ligand will lead to the pairing of electrons.

The d-orbital configuration after pairing will be:
- 3d: ↑↓ ↑ ↑ ↑ ↑ (3 paired and 2 unpaired)

### Step 4: Determine the hybridization
The complex [Mn(CN)₆]²⁻ has six ligands, which requires the formation of six equivalent hybrid orbitals. The hybridization involved is \( d^2sp^3 \), which corresponds to an octahedral geometry.

### Step 5: Count the number of unpaired electrons
From the paired and unpaired electron configuration:
- There are 5 d-electrons, and after pairing due to the strong field ligand, we find that there is only **1 unpaired electron** remaining.

### Final Answer
The number of unpaired electrons in the hexacyanidomanganate(II) ion is **1**.

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