The shape of `MnO_(4)^(-)` ion and the hybridisation of Mn in `MnO_(4)^(-)` is

Options

A. tetrahedral, `sp^(3)`
B. tetrahedral, `d^(3)s`
C. square planar , `dsp^(2)`
D. square planar , `sp^(3)`

Answer

To determine the shape of the `MnO4^(-)` ion and the hybridization of manganese (Mn) in this ion, we can follow these steps:

### Step 1: Determine the oxidation state of manganese in `MnO4^(-)`.
- Manganese (Mn) in `MnO4^(-)` has an oxidation state of +7. This is calculated by considering that each oxygen (O) has an oxidation state of -2. With four oxygens, the total contribution from oxygen is -8. To balance this with the overall charge of -1, manganese must be +7.

### Step 2: Count the total number of valence electrons.
- Manganese has 7 valence electrons (as it is in group 7 of the periodic table).
- Each oxygen contributes 6 valence electrons, and with four oxygens, that gives us 4 × 6 = 24 valence electrons.
- The total number of valence electrons in `MnO4^(-)` is thus 7 (from Mn) + 24 (from O) + 1 (for the negative charge) = 32 valence electrons.

### Step 3: Determine the number of bonding pairs and lone pairs.
- In `MnO4^(-)`, manganese forms four sigma bonds with the four oxygen atoms. This accounts for 4 bonding pairs.
- Since all valence electrons are used in bonding (7 from Mn and 24 from O), there are no lone pairs on the manganese.

### Step 4: Determine the hybridization of manganese.
- The hybridization can be determined by the number of sigma bonds and lone pairs. Here, we have 4 sigma bonds and 0 lone pairs.
- The hybridization corresponding to 4 sigma bonds is `sp^3`.

### Step 5: Determine the shape of the `MnO4^(-)` ion.
- With 4 bonding pairs and no lone pairs, the molecular geometry is tetrahedral.

### Final Answer:
- The shape of `MnO4^(-)` ion is tetrahedral, and the hybridization of Mn in `MnO4^(-)` is `sp^3`.

---