Question

The electronic configuration of four elements are given in brackets

$L\left( 1s^{2},2s^{2}2p^{1} \right);M\left( 1s^{2},2s^{2}2p^{5} \right)$

$Q\left( 1s^{2},2s^{2}2p^{6},3s^{1} \right);R\left( 1s^{2},2s^{2}2p^{2} \right)$

The element that would most readily form a diatomic molecule is

• A. Q
• B. M
• C. R
• D. L

Answer 1

Answer: (B)

M

Answer 2

Non-metals readily form diatomic molecules by sharing of electrons. Element $M(1s^{2}2s^{2}2p^{5})$ has seven electrons in its valence shell and thus needs one more electron to complete its octet. Therefore, two atoms share one electron each to form a diatomic molecule $(M_{2})$

$:\underset{..}{\overset{..}{M}}. + .\overset{..}{\underset{..}{M}}:$ $\rightarrow$ $:\underset{..}{\overset{..}{M}}:\overset{..}{\underset{..}{M}}:$