An element  ${ }_{11}{\text{X}}^{23}$ forms an ionic compound with another element Y then the charge on the ion formed by X is

${ }_{11}{\text{X}}^{23}\to \text{\hspace{0.17em}metal, Na}$

Electronic configuration of Na is $\frac{1{\text{s}}^2}{\text{K}}\frac{2{\text{s}}^{2}2{\text{p}}^6}{\text{L}}\frac{3{\text{s}}^1}{\text{M}}$

Based on valency shell configuration , sodium Na lose one electron to the element Y. Therefore, the amount of charge formed by ‘X’ is  $+1$