Consider the process, Mg→M2+g , ∆H=270 kJ ;If the IP1 of M(g) is 100 kJ/mole then the electron gain enthalpy of M2+(g) will be

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+270 kJ

-170 kJ

+170 kJ

-370 kJ

answer is B.

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Detailed Solution

Mg→+100 kJM+g→+170 kJM2+g ; Thus M2+g+e-→M+g , ∆egH =-170 kJ ;

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