Question

The first ionisation potential of $Na$ is $5.1eV$. The value of electron gain enthalpy of $N{a}^{+}$ will be

• A. - 2.55 eV
• B. - 5.1 eV
• C. - 10.2 eV
• D. + 2.55 eV

Answer 1

Answer: (B)

$Na⟶N{a}^{+}+e^{-}$ First $IE$
$N{a}^{+}+e^{-}⟶Na$
Electron gain enthalpy of $N{a}^{+}$ is reverse of (IE) Because reaction is reverse so
$\Delta H(eq)=-5.1eV$