Question

If the element $A$ has three electrons in an outer shell and $B$ has six electrons in an outer shell, then the formula of the compound formed by these element will be:

• A. $A_3{B}_2$
• B. $A_2{B}_3$
• C. $A{B}_3$
• D. $A_6{B}_3$

Answer 1

Answer: (B)

$A_2{B}_3$

Since ${}^{\prime }A{l}^{\prime }$ $(2,8,3)$ has three valance electron to loose, euhen it looses all $3e^{-}s$ and then it becomes ${Al}^{3+}$ and oxygen has valence electron $2$ less than nobel gas configuration. So, $3^{\prime }{O}^{\prime }$ atom got $6e^{-}$ from $2^{\prime }A{l}^{\prime }$ atom result in ${Al}_2{O}_3$ or $A_2{B}_3$.