Question

Match the electronic configurations of the elements given in List - I with their correct characteristic(s) (i.e., properties for given configuration) given in List - II.
$\begin{array}{cccc}& \text{List-I}& & \text{List - II}\\ \text{P.}& 1s^2& 1.& \text{Element shows highest negative oxidation state}\\ \text{Q.}& 1s^{2}2s^{2}2p^5& 2.& \text{Element shows highest first ionisation enthalpy}\\ \text{R.}& 1s^{2}2s^{2}2p^{6}3s^{2}3p^5& 3.& \text{Element shows highest electronegativity on Pauling scale}\\ \text{S.}& 1s^{2}2s^{2}2p^2& 4.& \text{Element shows maximum electron gain enthalpy (most exothermic)}\end{array}$

• A. $P-4,Q-2,R-3,S-1$
• B. $P-2,Q-3,R-4,S-1$
• C. $P-2,Q-1,R-3,S-4$
• D. $P-1,Q-2,R-3,S-4$

Answer 1

The correct option is B $P-2,Q-3,R-4,S-1$

In option (P): The configuration belongs to $He$ which has the highest first ionisation enthalpy amongst all the elements of the periodic table. This is attributed to the stable configuration and small size.

Option (Q) and (R): Group ${17}^{th}$ has $n{s}^{2}n{p}^5$ valence shell electron configuration. They have highest $EN$ values and very high negative electron gain enthalpy because they can attain stable noble gas electronic configuration by picking up an electron. Configuration in option (Q) belongs to fluorine and has highest electronegativity on Pauling scale. configuration in option (R) belongs to chlorine, which has the maximum negative electron gain enthalpy (even greater than fluorine, due to its larger size and lesser interelectronic repulsion).

In option (S) :This configuration belongs to carbon and it shows -4 oxidation state because it attains inert gas configuration of neon by gaining four electrons.