Elements A, B and C have atomic numbers 9, 20 and 10 respectively.

(i) State which one is (1) a non-metal; (2) a metal; (3) chemically inert.

(ii) Write down the formula of the compound formed by two of the above elements.

(i) The electronic configurations of the elements are:
    A[9] = [2, 7]
    B[20] = [2, 8, 8, 2]
    C[10] = [2, 8]

(1) A is a non-metal as it will gain electron to attain a stable octet.

(2) B is a metal as it will lose two electrons to attain a stable octet.

(3) C is chemically inert as it's valence shell is completely filled.

(ii) A has 7 electrons in the outermost shell. Therefore, it will try to gain one electron to attain stable octet, hence, A has valency [1-]

As B has 2 in the outermost shell. Therefore, it will try to lose 2 electrons to attain stable octet, hence, B has valency [2+]

Since valency of A is 1- and valency of B is 2+

Formula of the compound:

$\text{B}^{2+} \phantom{\nearrow} \text{A}^{1-} \\[0.5em] \overset{\phantom{2}{2}}{\text{B}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{A}} \Rightarrow \underset{\phantom{1}{1}}{\text{B}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{A}} \\[0.5em]$

So, we get the formula as $\bold{BA}_\bold{2}$