Among ${\mathrm{Me}}_3\mathrm{N}, {\mathrm{C}}_5{\mathrm{H}}_5\mathrm{N} \mathrm{and} \mathrm{MeCN}$ ($\mathrm{Me}$ = methyl group), the electronegativity of $\mathrm{N}$ is in the order

The electronic configuration of two elements $\mathrm{X}$ and $\mathrm{Y}$ are given below:

$$\begin{gathered} \mathrm{X} = 1{\mathrm{s}}^2 2{\mathrm{s}}^2 2{\mathrm{p}}^6 3{\mathrm{s}}^2 3{\mathrm{p}}^6 4{\mathrm{s}}^2 ; \\ \mathrm{Y} = 1{\mathrm{s}}^2 2{\mathrm{s}}^2 2{\mathrm{p}}^6 3{\mathrm{s}}^2 3{\mathrm{p}}^5 \end{gathered}$$

The formula of the ionic compound that can be formed between these elements is

The first ionization energy for $\mathrm{Li}$ is $5.4 \mathrm{eV}$ and electron affinity of $\mathrm{Cl}$ is $3.61 \mathrm{eV}$. The $∆\mathrm{H} (\mathrm{in} \mathrm{kJ}/\mathrm{mol})$ for the reaction,

$\mathrm{Li} \left(\mathrm{g}\right) + \mathrm{Cl} \left(\mathrm{g}\right) \to {\mathrm{Li}}^{+} \left(\mathrm{g}\right) + {\mathrm{Cl}}^{-} \left(\mathrm{g}\right)$,

if the resulting ions do not combine with each other is $(1 \mathrm{eV}=1.6 \mathrm{x} {10}^{-19} \mathrm{J})$.