question_answer

Given                 Reaction                                Energy Change (in KJ)                 \[\operatorname{L}\operatorname{i}(\operatorname{s})\to \operatorname{Li}(g)\]                                           161                 \[\operatorname{L}\operatorname{i}(g)\to {{\operatorname{Li}}^{+}}(g)\]                             520                 \[\frac{1}{2}{{\operatorname{F}}_{2}}(g)\to \operatorname{F}(g)\]                          77\[{{\operatorname{F}}_{2}}(g)+{{e}^{-}}\to {{\operatorname{F}}^{-}}(g)\]               (Electron gain                                                                                 enthalpy)                 \[\operatorname{L}{{\operatorname{i}}^{+}}(g)+{{\operatorname{F}}^{-}}(g)\to Li\,\,\operatorname{F}(s)\]   -1047                 \[\operatorname{L}\operatorname{i}(\operatorname{s})+\frac{1}{2}{{\operatorname{F}}_{2}}(g)\to \operatorname{Li}\operatorname{F}(s)\]       -617                 Based on data provided, the value of electron gain enthalpy of fluorine would be:     JEE Main  Online Paper (Held On 22 April 2013)

A) \[-300\operatorname{k}\operatorname{J}{{\operatorname{mol}}^{-1}}\]              

B) \[-350\operatorname{k}\operatorname{J}{{\operatorname{mol}}^{-1}}\]

C) \[-328\operatorname{k}\operatorname{J}{{\operatorname{mol}}^{-1}}\]              

D) \[-228\operatorname{k}\operatorname{J}{{\operatorname{mol}}^{-1}}\]