Question

An atom $A$ has 2, 8 and 3 electrons in K, L and M sub-shells respectively. Another atom $B$ has 2 and 6 electrons in K and L sub-shells respectively. The formula of the compound formed between $A$ and $B$ is :

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

Answer 1

Answer: (B)

$A_2{B}_3$

For atom $A$,
Electronic configuration: $1s^{2}2s^{2}2p^{6}3s^{2}3p^1$;

The outermost shell has 3 electrons, hence valency is $3$.

For atom $B$,
Electronic configuration: $1s^{2}2s^{2}2p^4$;

The outermost shell has 6 electrons, hence valency $=(8-6)=2.$

Now, the compound formed is:

$A^3{B}^2$ $⟹A_2{B}_3$

So, the compound formed between A and B is $A_2{B}_3$.

Hence, option B is correct.