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IE(1) and IE(2) of Mg are 178 and 348 kc...

`IE_(1)` and `IE_(2)` of `Mg` are `178` and `348 kcal mol^(-1)`. The energy required for the reaction `Mg rarr Mg^(2+)+2e^(-)` is

A

`+170 kcal`

B

`+526 kcal`

C

`-170 kcal`

D

`-526 kcal`

Text Solution

Verified by Experts

The correct Answer is:
B
(i) `Mg_((g)) rarr Mg_((g))^(o+) + e^(-), IE_(1) = 178 kcal mol^(-1)`
(ii) `Mg_((g))^(o+) rarr Mg_((g))^(2+) + e^(-) " " IE_(2) = 348 kcal mol^(-1)`
Adding (i) and (ii), we have
`ul(bar(Mg_((g)) rarr Mg_((g))^(2+) + 2e^(-)))`
`:. E = IE_(2)+IE_(2) = 178 + 348 = 526 kcal`
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