Q.

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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a

+2860 kJ/mole

b

+340 kJ/mole

c

–340kJ/mole

d

–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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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