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Periodic Classification and Periodic Properties - Result Question 13
10. The first ionisation potential of $Na$ is $5.1 eV$. The value of electron gain enthalpy of $Na^{+}$will be
(a) $-2.55 eV$
(b) $-5.1 eV$
(c) $-10.2 eV$
(d) $+2.55 eV$
(2013 Main)
Show Answer
Answer:
Correct Answer: 10. (b)
Solution:
$ \begin{aligned} Na & \longrightarrow Na^{+}+e^{-} \text {First IE } \\ Na^{+}+e^{-} & \longrightarrow Na \end{aligned} $
Electron gain enthalpy of $Na^{+}$is reverse of (IE)
Because reaction is reverse so $\Delta H(e q)=-5.1 eV$
BETA
