The oxygen atoms in `H_2O_2` undergone___hybridisation.

Options

A. (a) `sp^(3)`
B. (b) `sp^(2)`
C. (c) `sp`
D. (d) `sp^(3)d^(2)`

Answer

To determine the hybridization of the oxygen atoms in hydrogen peroxide (H₂O₂), we can follow these steps:

### Step-by-Step Solution:

1. **Draw the Structure of H₂O₂**:
- The molecular structure of hydrogen peroxide consists of two oxygen atoms connected by a single bond (peroxy linkage) and each oxygen atom is also bonded to a hydrogen atom.
- The structure can be represented as:
```
H H
\ /
O - O
/ \
H H
```

2. **Count the Valence Electrons**:
- Each oxygen atom has 6 valence electrons. In H₂O₂, each oxygen atom is involved in two bonds (one with another oxygen and one with a hydrogen atom).
- Therefore, for each oxygen atom, 2 electrons are used for bonding, leaving 4 electrons (6 - 2 = 4) which will be present as lone pairs.

3. **Determine Lone Pairs and Bond Pairs**:
- Each oxygen atom has 2 lone pairs (4 electrons) and forms 2 sigma bonds (one with hydrogen and one with the other oxygen).
- Thus, for each oxygen atom:
- Number of sigma bonds = 2
- Number of lone pairs = 2

4. **Calculate the Total Number of Electron Pairs**:
- The total number of electron pairs around each oxygen atom is the sum of the number of sigma bonds and the number of lone pairs:
- Total = Number of sigma bonds + Number of lone pairs
- Total = 2 (sigma bonds) + 2 (lone pairs) = 4

5. **Determine the Hybridization**:
- The hybridization can be determined using the total number of electron pairs:
- 4 electron pairs correspond to sp³ hybridization.
- Therefore, the oxygen atoms in H₂O₂ undergo sp³ hybridization.

### Conclusion:
The oxygen atoms in H₂O₂ undergo **sp³ hybridization**.

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