First slide

Periodic Trends

Question

First ionisation enthalpy of an element is 756 kJ/mole. The change in enthalpy for the process $M (g) \to M^{2 +} (g)$ will be

Moderate

A

900 kJ
B

2196 kJ
C

1071 kJ
D

1512 kJ

Solution

Given
I.P₁ of ‘M’ = + 756 kJ/mole ;
I.P₂ of ‘M’ must be > 756 kJ/mole ;
Thus sum of I.P₁ and I.P₂ will be > 1512 kJ
Thus
$M (g) \to M^{2 +} (g), ∆ H > 1512 kJ;$
By inspection, the only option which is greater than 1512 kJ is Option (2) ;

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