Question

An element A has 3 electrons in the outermost orbit and element B has 6 electrons in the outermost orbit. What will be the formula of the compound formed?

• A. $A{B}_2$
• B. $A_2{B}_3$
• C. $A_2{B}_6$
• D. $A_3{B}_2$

Answer 1

The correct option is B $A_2{B}_3$

In a neutral molecule, the sum of the bonding valence electrons on each element must be equal to the total number of valence electrons in the entire compound.

If we have a 3 valence electrons, the "A" charge will be either +3 or -5 for a full octet, and 6 valence electrons in "B" atoms will most likely result in acquisition of additional electrons. for an octet and relative charge of -2. Thus, "A" will be the positive ion, at +3. Let $x$ be the number of A atoms with charge $C_1$ and $y$ be the number of B atoms with charge $C_2$.
$x\times C_1=y\times C_2$
$3\times x=2\times y$
$\frac{x}{y}=\frac{2}{3}=\frac{1}{2}$
$\therefore A_2{B}_3$ compound will be formed.