What would be the formula of metal halide if successive ionisation energies of metal are 5eV, 50eV, 68eV, 90 eV and 120eV?

M X

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M X_{2}

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M_{2} X_{2}

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M X_{4}

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Solution

The correct option is A $M X$
There is a large difference between $I E_{1}$ and $I E_{2}$ . So more chances of forming $M^{+}$ ion.

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