Question

Two elements P and Q combine to form a compound. If P has 2 and Q has 6 electrons in their outermost shell, what will be the formula of the compound formed?
A. $PQ$
B. ${{P}_{2}}Q$
C. \[{{P}_{2}}{{Q}_{3}}\]
D. $P{{Q}_{2}}$

Answer 1

Hint: The electrons mentioned in the question are the valence electrons of the elements P and Q. Valence electrons are the electrons that participate in bond formation. Hence, using these values we can find the molecular formula of the compound.

Complete step by step answer:
The number of valence electrons mention in the question are:
$2$ in P element and $6$ in Q element. Now we should remember that all compounds desire to find the octet configuration that is, they want to get 8 electrons. For this reason, they undergo bond formation. Be it, covalent bond or ionic bond or others. Now we can produce this octet configuration in these two elements by adding and removing a few electrons.
If we consider element P, we can remove 2 electrons from the outermost orbital and thus it will receive octet configuration in the $+2$ -oxidation state. This can be denoted in the following manner:
$P\to {{P}^{+2}}+2{{e}^{-}}$
For element Q, we need to add two electrons in order to get the octet configuration. Thus, we can add the two electrons that we have taken from P. the change can be represented in this way.
$Q+2{{e}^{-}}\to {{Q}^{-2}}$
This leads to element Q obtaining the $-2$ oxidation state.
Now putting both these variables together we get,
${{P}^{+2}}+{{Q}^{-2}}\to PQ$
Because the two oxidation states being opposite in sign will cancel each other out.

So, the correct answer is Option A.

Note: Always remember that the sign of the oxidation states present on the ions of the elements has to be considered. If the oxidation state of the ions varies in magnitude, they can be multiplied criss cross in this manner:
${{A}^{+2}}+{{B}^{-3}}\to {{A}_{3}}{{B}_{2}}$