# The formula of the compound is A_2B_5. The number of electrons in the outermost orbits of A and B respectively are?

## 6 and 3 5 and 6 5 and 2 2 and 3

Nov 24, 2017

'2. 5 and 6

#### Explanation:

In a neutral (uncharged) molecule, the sum of the bonding valence electrons must be equal. So, the products of the negative element and its charge and the positive element and its charge must be equal.
${C}_{1} \times {N}_{1} = {C}_{2} \times {N}_{2}$

If we have 2 "A" atoms and 5 "B" atoms we have:
${C}_{1} \times 2 = {C}_{2} \times 5$

To be equal, ${C}_{1} = 5$ and ${C}_{2} = 2$

This actually refers to the "charge" of the ion. The "number of electrons in the outer shell" would depend on the specific element, and if the ion form is intended, then the correct value would be the positive ions charge less and the negative ions added to the existing ones.

So, if we have ${\left({A}^{+ 5}\right)}_{2} {\left({B}^{- 2}\right)}_{5}$ we could say that the original number of electrons in B was 6 - it had to gain 2 to balance the shell at 8, with a net charge of -2. Similarly, as already shown, A had to have 5, which it gave up (or shared) to stabilize the shell and end up with a +5 charge.

In that case, the number of electrons in each atoms' shell would be:
$A = 5$
$B = 6$