The hybridisation of Nitrogen in Nitrate ion is

Options

A. sp
B. `sp^2`
C. `sp^3`
D. `sp^3d`

Explanation

To determine the hybridization of nitrogen in the nitrate ion (NO3^-), we can follow these steps:

### Step 1: Identify the Lewis Structure of the Nitrate Ion
- The nitrate ion consists of one nitrogen atom and three oxygen atoms.
- In the Lewis structure, nitrogen is at the center, bonded to three oxygen atoms. One of these bonds is a double bond while the other two are single bonds.
- The overall charge of the ion is -1, which is distributed among the oxygen atoms.

### Step 2: Count the Number of Sigma Bonds
- In the nitrate ion, there are:
- One double bond (N=O), which contributes one sigma bond and one pi bond.
- Two single bonds (N-O), each contributing one sigma bond.
- Therefore, the total number of sigma bonds is:
- 1 (from the double bond) + 2 (from the single bonds) = 3 sigma bonds.

### Step 3: Determine the Hybridization
- The number of hybrid orbitals formed corresponds to the number of sigma bonds.
- Since there are 3 sigma bonds, nitrogen must have 3 hybrid orbitals.
- The hybridization that corresponds to 3 hybrid orbitals is **sp²** hybridization (1 s orbital and 2 p orbitals mix to form 3 sp² hybrid orbitals).

### Step 4: Conclusion
- Therefore, the hybridization of nitrogen in the nitrate ion (NO3^-) is **sp²**.

### Final Answer:
The hybridization of nitrogen in the nitrate ion (NO3^-) is **sp²**.

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