The $I_{1}, I_{2}, I_{3}, I_{4}$ values of an element $M$ are $120$ kJ/mole, $600$ kJ/mole, $1000$ kJ/mole and $8000$ kJ/mole. Then the formula of its sulphate is:

M S O_{4}

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M_{2} (S O_{4})_{3}

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M_{2} S O_{4}

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M_{3} (S O_{4})_{2}

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Solution

The correct option is B $M_{2} (S O_{4})_{3}$
The given element has high fourth ionization potential. Thus it should have three valence electrons. Therefore the formula of its sulphate will be $M_{2} (S O_{4})_{3}$ .

$M_{2} {(S O_{4})}_{3} \to 2 M^{3 + |} + 2 {S O}_{4}^{3 -}$

Hence, the correct option is $B$

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