Find the formula of halide of a metal whose successive ionization enthalpies are x, 2x, 5x, 100 x kJ `"mol"^(-1)` respectively

Options

A. MX
B. `MX_(2)`
C. `MX_(3)`
D. `M_(2)X`

Answer

To find the formula of the halide of a metal whose successive ionization enthalpies are given as x, 2x, 5x, and 100x kJ/mol, we can follow these steps:

### Step 1: Analyze the Ionization Enthalpies
The successive ionization enthalpies are:
- First ionization enthalpy = x
- Second ionization enthalpy = 2x
- Third ionization enthalpy = 5x
- Fourth ionization enthalpy = 100x

Notice that there is a significant jump between the third and fourth ionization enthalpies (from 5x to 100x). This indicates that removing the fourth electron requires much more energy than removing the first three electrons.

### Step 2: Determine the Stability of Oxidation States
The large increase in the fourth ionization enthalpy suggests that the metal is particularly stable in the +3 oxidation state. This is because after removing three electrons, the metal achieves a stable electronic configuration, and removing a fourth electron becomes significantly more difficult.

### Step 3: Predict the Halide Formula
Halogens typically have a valency of -1. When a metal with a +3 oxidation state reacts with halogens, it will likely form a halide with a formula that reflects this oxidation state. Since the metal can lose three electrons, it will combine with three halide ions (each with a -1 charge).

Thus, the formula of the halide will be:
\[ \text{MX}_3 \]
where M is the metal and X is the halogen.

### Conclusion
The formula of the halide of the metal is:
\[ \text{MX}_3 \]