Question

The first ionization enthalpy $\left({\Delta }_{\text{t}}\text{H}\right)$ values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ mol${}^{-1}$. Predict whether the first ${\Delta }_{\text{t}}\text{H}$ value for Al will be more close to 575 or 760 kJ mol${}^{\text{- 1}}$? Justify your answer.

Answer 1

The electronic configuration are as follows;

${}_{1}1Na\left(3s{)}^1{,}_{1}2Mg\right(3s{)}^2{,}_{1}3Al\left(3s{)}^2\right(3p{)}^1{,}_{1}4Si\left(3s{)}^2\right(3p{)}^2$

The ionisation energy of $Mg$ will be larger than that of $Na$ due to fully filled configuration $(3s{)}^2$

The ionization of $Al$ will be smaller than that of $Mg$ due to one electron extra than stable configuration but smaller than $Si$ due to increase in effective nuclear charge of $Si$.

$Na<Mg>Al<Si$

So, the first ${\Delta }_{t}H$ value of $Al$ will be more close to $575Kj/mol$. The value for $Al$ should be lower than that of $Mg$ because of effective shielding of $3p$ electron from the nucleus by $3s$ electron.