Periodic Classification and Periodic Properties 2 Question 10
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10. The first ionisation potential of $\mathrm{Na}$ is $5.1 \mathrm{eV}$. The value of electron gain enthalpy of $\mathrm{Na}^{+}$will be
======= ####10. The first ionisation potential of $\mathrm{Na}$ is $5.1 \mathrm{eV}$. The value of electron gain enthalpy of $\mathrm{Na}^{+}$will be
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $-2.55 \mathrm{eV}$
(b) $-5.1 \mathrm{eV}$
(c) $-10.2 \mathrm{eV}$
(d) $+2.55 \mathrm{eV}$
(2013 Main)
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Answer:
Correct Answer: 10. (b)
Solution:
$$ \begin{aligned} \mathrm{Na} & \longrightarrow \mathrm{Na}^{+}+e^{-} \text {First IE } \ \mathrm{Na}^{+}+e^{-} & \longrightarrow \mathrm{Na} \end{aligned} $$
Electron gain enthalpy of $\mathrm{Na}^{+}$is reverse of (IE)
Because reaction is reverse so $\Delta H(e q)=-5.1 \mathrm{eV}$
