Question

The number of valence electrons in the elements A and B are 4 and 6 respectively. The most probable compound formed from A and B is:

• A. $A_2{B}_3$
• B. $A{B}_3$
• C. $A{B}_2$
• D. $A_{2}B$

Answer 1

The correct option is C $A{B}_2$

In a neutral molecule, the sum of bonding valence electrons on each element must be equal to the total number of valence electrons in the entire compound. So, the total charge on the cation has to be equal to the total charge on anion. Let $x$ be the number of A atoms with charge $C_1$ and $y$ be the number of B atoms with charge $C_2$. In ionic compounds, valency gives indication of charge on ion.
So, $C_1$ = 4 since A loses 4 electrons and $C_2$ = 2 as B gains 2 electrons.
$x\times C_1=y\times C_2$
$4\times x=2\times y$
$\frac{x}{y}=\frac{2}{4}=\frac{1}{2}$
$\therefore A{B}_2$