In the following compounds of manganese what is the distribution of electrons on d-orbitals of manganese ?
(i) `[Mn(H_2O)_6]^(2+)`
(ii) `[Mn(CN)_6]^(4-)`

Options

A. ` t _ 2 ^ 3 e _ g ^ 2 ` in both
B. ` t _ ( 2g ) ^ 5 e _ g^ o ` in both
C. ` t_ (2g ) ^3 e _ g ^2 ` in (i) and ` t _ (2g ) ^ 5 e _ g ^ o ` in (ii)
D. ` t _ (2g ) ^ 3 e _ g ^ 2 ` in ( i) and ` t _ (2g ) ^ 3 e _ g ^ 2 ` (ii)

Answer

To determine the distribution of electrons on the d-orbitals of manganese in the given complexes, we need to analyze each compound step by step.

### Step 1: Determine the oxidation state of manganese in each complex.

**For (i) `[Mn(H2O)6]^(2+)`:**
- The overall charge of the complex is +2.
- Water (H2O) is a neutral ligand, contributing 0 to the oxidation state.
- Therefore, the oxidation state of manganese (Mn) must be +2.

**For (ii) `[Mn(CN)6]^(4-)`:**
- The overall charge of the complex is -4.
- Each cyanide ion (CN) has a charge of -1, and there are 6 cyanide ions.
- Therefore, the total contribution from the cyanide ligands is -6.
- Let the oxidation state of manganese be x. We can set up the equation:
\( x + (-6) = -4 \)
Solving this gives \( x = +2 \).

### Step 2: Write the electronic configuration of manganese.

- The atomic number of manganese (Mn) is 25.
- The ground state electronic configuration of Mn is:
\[ \text{Mn: } [Ar] 4s^2 3d^5 \]

### Step 3: Adjust the electronic configuration for the oxidation state.

**For (i) `[Mn(H2O)6]^(2+)`:**
- Since Mn is in the +2 oxidation state, we remove 2 electrons from the outermost shell (4s before 3d).
- The electronic configuration becomes:
\[ \text{Mn}^{2+}: [Ar] 3d^5 \]

**For (ii) `[Mn(CN)6]^(4-)`:**
- Again, Mn is in the +2 oxidation state, so the electronic configuration remains:
\[ \text{Mn}^{2+}: [Ar] 3d^5 \]

### Step 4: Determine the splitting of d-orbitals based on the ligands.

**For (i) `[Mn(H2O)6]^(2+)`:**
- H2O is a weak field ligand, which means it does not cause significant pairing of electrons.
- The 3d electrons will occupy the d-orbitals singly before pairing occurs.
- The d-orbitals split into two sets: T2G (lower energy) and EG (higher energy).
- The distribution will be:
- T2G: 3 electrons
- EG: 2 electrons
- Therefore, the distribution is:
- T2G: 3
- EG: 2

**For (ii) `[Mn(CN)6]^(4-)`:**
- CN is a strong field ligand, which causes pairing of electrons.
- The 3d electrons will pair up in the lower energy T2G orbitals before occupying the EG orbitals.
- The distribution will be:
- T2G: 5 electrons (all paired)
- EG: 0 electrons
- Therefore, the distribution is:
- T2G: 5
- EG: 0

### Final Answer:
- For `[Mn(H2O)6]^(2+)`: T2G = 3, EG = 2
- For `[Mn(CN)6]^(4-)`: T2G = 5, EG = 0