Ionic bond is present in which of the following species :

Options

A. `O_2`
B. `CHCl_3`
C. `NaBr`
D. `C Cl_4`

Explanation

To determine which species contains an ionic bond, we need to analyze the nature of the bonds formed between the elements in each option. Ionic bonds typically form between metals and non-metals through the transfer of electrons. Let's evaluate each option step by step:

### Step 1: Identify the species
We are given several species to analyze for the presence of ionic bonds. Let's denote them as follows:
1. O2 (Oxygen)
2. HCl (Hydrogen Chloride)
3. NaBr (Sodium Bromide)
4. CCl4 (Carbon Tetrachloride)

### Step 2: Analyze O2
- **Composition**: O2 consists of two oxygen atoms.
- **Nature of Bond**: Both oxygen atoms are non-metals. They share electrons to achieve a full outer shell.
- **Conclusion**: O2 forms a **covalent bond**, not an ionic bond.

### Step 3: Analyze HCl
- **Composition**: HCl consists of hydrogen and chlorine.
- **Nature of Bond**: Hydrogen is a non-metal, and chlorine is also a non-metal. They share electrons.
- **Conclusion**: HCl forms a **covalent bond**, not an ionic bond.

### Step 4: Analyze NaBr
- **Composition**: NaBr consists of sodium and bromine.
- **Nature of Bond**: Sodium is a metal, and bromine is a non-metal. Sodium tends to lose one electron to achieve a full outer shell (octet), while bromine gains that electron.
- **Conclusion**: NaBr forms an **ionic bond** due to the transfer of electrons from sodium to bromine.

### Step 5: Analyze CCl4
- **Composition**: CCl4 consists of carbon and chlorine.
- **Nature of Bond**: Carbon is a non-metal, and chlorine is also a non-metal. They share electrons to achieve a full outer shell.
- **Conclusion**: CCl4 forms a **covalent bond**, not an ionic bond.

### Final Conclusion
Among the given species, the only one that contains an ionic bond is **NaBr (Sodium Bromide)**.

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