The electrovalent linkage is present in

Options

A. `O_2` molecule
B. `"CCl"_4` molecule
C. `CHCl_3` molecule
D. NaBr molecule

Explanation

To determine where the electrovalent (ionic) linkage is present, we need to analyze the bonding in the given compounds. Electrovalent or ionic bonds are formed through the transfer of electrons between atoms, typically between metals and non-metals.

### Step-by-Step Solution:

1. **Identify the Compounds**: We need to analyze the bonding in the following compounds: O2, CCl4, CHCl3, and NaBr.

2. **Analyze O2**:
- O2 is formed by the sharing of electrons between two oxygen atoms.
- Each oxygen atom has 6 valence electrons, and they share electrons to achieve a stable configuration.
- **Conclusion**: O2 has a covalent bond, not an ionic bond.

3. **Analyze CCl4**:
- CCl4 (carbon tetrachloride) consists of one carbon atom and four chlorine atoms.
- Carbon has 4 valence electrons, and each chlorine has 7 valence electrons. They share electrons to form covalent bonds.
- **Conclusion**: CCl4 has covalent bonds, not ionic bonds.

4. **Analyze CHCl3**:
- CHCl3 (chloroform) consists of one carbon atom, one hydrogen atom, and three chlorine atoms.
- Similar to CCl4, carbon shares electrons with hydrogen and chlorine to form covalent bonds.
- **Conclusion**: CHCl3 has covalent bonds, not ionic bonds.

5. **Analyze NaBr**:
- NaBr (sodium bromide) consists of sodium (Na) and bromine (Br).
- Sodium has 1 valence electron, and bromine has 7 valence electrons. Sodium transfers its electron to bromine to achieve a stable electronic configuration.
- This transfer of electrons leads to the formation of Na+ and Br- ions, resulting in an ionic bond.
- **Conclusion**: NaBr has an electrovalent (ionic) bond.

### Final Answer:
The electrovalent linkage is present in **NaBr**.