If the consecutive ionization energies of an element A are 496, 4564, 6918, 9542 kJ/mole respectively, the formula of the oxide formed by A is:

A_{2} O_{3}

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A O

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A O_{2}

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A_{2} O

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Solution

The correct option is D $A_{2} O$
In element A there is a large difference between I and II ionization energies so it can form stable ion $A^{+}$ .

The oxide of A would be $A_{2} O$ .

Option D is correct.

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If the consecutive ionization energies of an element A are 496, 4564, 6918, 9542 kJ/mole respectively, the formula of the oxide formed by A is:

The correct option is D A2OIn element A there is a large difference between I and II ionization energies so it can form stable ion A+. The oxide of A would be A2O.Option D is correct.

The correct option is D $A_{2} O$
In element A there is a large difference between I and II ionization energies so it can form stable ion $A^{+}$ .

The oxide of A would be $A_{2} O$ .

Option D is correct.