`underset("has garlic smell")underset("white solid")underset("highly reactive")((A)) overset("burns in")underset("excess air")rarr underset("agent")underset("dehydrating")underset("strong")((B)) overset("conc".HNO_(3))rarr(F)+underset("solid")underset("colourless")((C)) overset(I_(2))rarr underset("solid")((D)) +E(g)`
When `(C )` reacts with `NaCl` the state of hybridisation of `N-`atom in the products

Options

A. `sp^(2)only`
B. `sp^(2)`and`sp^(3)`
C. `sp^(3)`only
D. `sp` and `sp^(2)`

Answer

To solve the problem, let's break down the reactions step by step and determine the hybridization of the nitrogen atom in the products formed when C reacts with NaCl.

### Step 1: Identify Compound A
- **A** is described as a white solid with a garlic smell and is highly reactive.
- This corresponds to **white phosphorus (P4)**.

### Step 2: Reaction of A with Excess Air
- When **P4** burns in excess air (oxygen), it forms **phosphorus pentoxide (P4O10)**.
- The reaction can be written as:
\[ P_4 + 5O_2 \rightarrow P_4O_{10} \]
- Here, **B** is **P4O10**, which is a strong dehydrating agent.

### Step 3: Reaction of B with Concentrated HNO3
- **B (P4O10)** reacts with concentrated nitric acid (HNO3) to produce:
\[ P_4O_{10} + 4HNO_3 \rightarrow 2N_2O_5 + 4HPO_3 \]
- Here, **C** is **N2O5**, which is a colorless solid, and **F** is **HPO3**.

### Step 4: Reaction of C with NaCl
- Now, we need to consider the reaction of **C (N2O5)** with sodium chloride (NaCl):
\[ N_2O_5 + NaCl \rightarrow NaNO_2 + NaOCl \]
- In this reaction, **NaNO2** and **NaOCl** are formed.

### Step 5: Determine Hybridization of Nitrogen in Products
1. **For NaNO2**:
- The nitrite ion (NO2-) has nitrogen in a +3 oxidation state.
- The hybridization can be calculated as follows:
- Valency of nitrogen = 5
- Negative charge = +1
- Number of monatomic ions around nitrogen = 0
\[ \text{Hybridization} = \frac{(5 + 1 + 0)}{2} = \frac{6}{2} = 3 \quad \text{(sp}^2\text{)} \]

2. **For NaOCl**:
- The nitrogen in the nitrosyl ion (NO+) has nitrogen in a +2 oxidation state.
- The hybridization can be calculated as follows:
- Valency of nitrogen = 5
- Positive charge = -1
- Number of monatomic ions around nitrogen = 0
\[ \text{Hybridization} = \frac{(5 - 1 + 0)}{2} = \frac{4}{2} = 2 \quad \text{(sp)} \]

### Conclusion
- The hybridization of nitrogen in **NaNO2** is **sp²**.
- The hybridization of nitrogen in **NaOCl** is **sp**.

### Final Answer
The state of hybridization of the nitrogen atom in the products formed when C reacts with NaCl is **sp² and sp**.

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