Successive ionisation potentials of an element M are `8.3,25.1,37.9,259.3` and 340.1 ev. The formula of its bromide is

Options

A. `MBr_(5)`
B. `MBr_(4)`
C. `MBr_(3)`
D. `MBr_(2)`

Answer

To determine the formula of the bromide of the element M, we need to analyze the given successive ionization potentials (IP) and identify how many electrons can be removed from the element M before a significant increase in energy is observed.

### Step-by-Step Solution:

1. **List the Successive Ionization Potentials**:
- First IP = 8.3 eV
- Second IP = 25.1 eV
- Third IP = 37.9 eV
- Fourth IP = 259.3 eV
- Fifth IP = 340.1 eV

2. **Analyze the Changes in Ionization Potentials**:
- The first three ionization potentials (8.3 eV, 25.1 eV, and 37.9 eV) show relatively small increases in energy.
- However, there is a significant jump between the third (37.9 eV) and the fourth ionization potential (259.3 eV). This indicates that removing the fourth electron requires a much larger amount of energy.

3. **Determine the Number of Valence Electrons**:
- The small increases in the first three ionization potentials suggest that the element M can easily lose three electrons.
- The large increase in the fourth ionization potential indicates that after losing three electrons, the element M reaches a stable electronic configuration, making it very difficult to remove a fourth electron.

4. **Conclude the Formula of the Bromide**:
- Since M can lose three electrons to form a stable ion, it will combine with bromine (Br), which typically forms a -1 ion.
- Therefore, the formula of the bromide will be MBr₃, as one M ion will combine with three Br ions to balance the charges.

### Final Answer:
The formula of the bromide of element M is **MBr₃**.