Question

The formula of the compound is ${{A}_{2}}{{B}_{5}}$.The number of electrons in the outermost orbitals of A and B respectively are:
A) 6 and 3
B) 5 and 6
C) 5 and 2
D) 2 and 3

Answer 1

Hint: The valence electrons are the electrons which are present in the outermost orbital of an element from which the valency of the element can be calculated.
The valency can also be calculated from the molecular formulae of the compound.

Complete step by step answer:
As we know, in the question a compound is given formed from A atoms and B atoms And now we have to say about the number of electrons in the outermost orbitals of A and B or can be simplified and said as we have to find the number of valence electrons in A and B atoms.
We know that a molecular formula is written by criss-cross method. By crisscrossing the oxidation states of the atoms involved we could get the correct molecular formula of the compound.
Here, from the formulae of the compound ${{A}_{2}}{{B}_{5}}$ could be written as ${{A}^{+5}}{{B}^{-2}}$
We know that the cation will have a positive oxidation state and the anion will have a negative oxidation state.
As for A the oxidation state is +5 in${{A}^{+5}}{{B}^{-2}}$, there is 5 electrons in the valence, by losing these five electrons the atom A may attain the stable octet configuration.
If the compound exists as ${{B}^{2-}}$ i.e. it is having a negative oxidation state, it refers that by gaining two more electrons it could complete the octet formation.
So in that case the number of valence electrons present in B should be equal to 6.
Number of valence or outermost electrons = $(8 - valency)$
Number of valence or outermost electrons = $(8- 2) =6$
So the total number of valence electrons in the B atom is six.

Therefore the correct answer for this question is option (B), five valence electrons in atom A and six electrons in atom B.

Note: There are metals that possess variable valencies, mainly these conditions are for transitions metals like $Cu$, $Fe$ etc. In these cases we should assign the oxidation state after calculating the oxidation state of the metal.
For example Fe may possess +3 and +2 oxidation state, so it should be taken care of.
Another point that should be noted is taking into consideration the signs + and –, while solving these type problems.