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N(0)//2 atoms of X(g) are converted into...

`N_(0)//2` atoms of X(g) are converted into `X^(+)` (g) by energy `E_(1) . N_(0)//2` atoms of X(g) are converted into `X^(-)` (g) by the energy `E_(2)` . Hence ionisation potential and electron affinity of X(g) are :

A

`I.E. = (2E_(1))/(N_(0)), Delta_(eq)H =- (E_(2))/(2N_(0))`

B

`I.E. =- (E_(2))/(2N_(0)). Delta_(eq)H = (2E_(1))/(N_(0))`

C

`I.E. = (E_(1))/(2N_(0)), Delta_(eq)H =- (E_(2))/(2N_(0))`

D

`I.E. = (N_(0))/(2E_(0)). Delta_(eq)H =- (2N_(0))/(E_(2))`

Text Solution

Verified by Experts

The correct Answer is:
A
`X(g) rarr X^(+) (g)+e^(-)`
If `I.E.` is ionisation enthalpy, then
`:. (N_(0))/(2) (I.E.) = E_(1)`
`:. I.E. = (2E_(1))/(N_(0))`
`X(g) +e^(-) rarr X^(-)(g)`
If `Delta_(eg)H` is electron gain enthalpy, then
`:. 2N_(0) (E.A) =- E_(2)`
`:. Delta_(eg)H =- (E_(2))/(2N_(0))`.
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