Xg+ →Xg → Xg−;ΔH for the entire process -1600 kJ/mole.If IP of ‘X' is 1260 kJ/mole then Electron gain enthalpy of ‘X’ would be

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+2860 kJ/mole

+340 kJ/mole

–340kJ/mole

–2860 kJ/mole

answer is C.

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Detailed Solution

Given Xg → Xg+, ΔH = +1260kJ∴Xg+ → Xg ΔH = −1260kJ/moleXg+ →−1260Xg →ΔH2 Xg−–1260+∆H2=–1600 ΔH2 = −340kJ/mole

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