Question

Ionization energies of elements A are given below in kJ/mol:
$I.E{.}_{1}I.E{.}_{2}I.E{.}_3$
$12043013000$
If A reacts with different elements, which compounds are possible:

• A. $AF$
• B. $A_{2}O$
• C. $A_{3}N$
• D. $A_3{N}_2$

Answer 1

Answer: (D)

$A_3{N}_2$

Third ionization energy ($13000$ kJ/mol) is much higher than the second ionization energy ($430$ kJ/mol). This indicates alkaline earth metal with $2$ valence electrons. When $2$ valence electrons are removed, the stable electronic configuration of inert gas is obtained. When the third electron is lost, the stable electronic configuration of inert gas is broken. This requires a large amount of energy. Hence, the third ionization energy is very high.
The valency of alkaline earths is $+2$. It forms $A^{2+}$ ions.The valence of $N$ is $-3$.It forms $N^{3-}$ ions.Thus, A will react with N to form $A_3{N}_2$.With $F$, A will form $A{F}_2$. With $O$, A will form $AO$ .