1.

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

Answer»

(a) `A_(2)O_(3)`
(B) AO
(C) `AO_(2)`
(d) `A_(2)O`

Solution :In element A there is a large DIFFERENCE between I
and II ionisation energies, so, it can form the STABLE
ION `A^(+)` the oxide of A Would be `A_(2)O.`


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