Question

If the ionization potential of $Na$ is $495.8Kjmo{l}^{-1}$, the electron affinity of $CI$ is $349.4kJmo{l}^{-1}$ and lattice energy is $776kJmo{l}^{-1}$, the true statement for the formation of ionic bond in $NaCl$ is :

• A. total energy released per mole of NaCl formed is $626kJ$
• B. the ionic bond formed is unstable
• C. the total energy absorbed per mole of $NaCl$ formed is $1125.4kJ$
• D. net energy released in this process is $629.6J$

Answer 1

The correct option is A total energy released per mole of NaCl formed is $626kJ$

$Na\to {Na}^{+}+e^{-},IE=495.8kJ{mol}^{-1}$

$Cl+e^{-}\to {Cl}^{-},EA=-349.4kJ{mol}^{-1}$

$Na\left(s\right)+\frac{1}{2}{Cl}_2\left(g\right)\to NaCl\left(s\right)$

${Na}^{+}+{Cl}^{-}\left(g\right)\to NaCl\left(s\right)$

$\Delta H_{f^0}=108+496+122-349-788=-411kJ{mol}^{-1}$

Total energy required is $(108+496+122)=626kJ{mol}^{-1}$