Question

Match the elecgtronic configurations of the elements given in List - I with their correct characteristic(s) (i.e. properties for given configuration) given in List-II and select the correct answer using the code given below the lists.
$\begin{array}{cc}List-I& List-II\\ \left(I\right)1s^2& \text{(P) Element shows highest negative oxidation state.}\\ \left(II\right)1s^{2}2s^{2}2p^5& \text{(Q) Element shows highest first ionisation enthalpy.}\\ \left(III\right)1s^{2}2s^{2}2p^{6}3s^{2}3p^5& \text{(R) Element shows highest electronegativity on Pauling scale.}\\ \left(IV\right)1s^{2}2s^{2}2p^2& \text{(S) Element shows maximum electron gain enthalpy (most}\\ & \text{exothermic).}\\ & \text{(T) Configuration belongs to}Cl\\ & \text{(U) configuration belongs to}C\end{array}$

• A. I - Q; II - R; III - S, T; IV - P, U
• B. I - P; II - Q; III - S, T; IV - P, U
• C. I - Q; II - R; III - S; IV - U
• D. I - P; II - R; III - S, T; IV - P

Answer 1

The correct option is A I - Q; II - R; III - S, T; IV - P, U

(I) This configuration belongs to $He$ which was highest first ionization enthalpy amongst all the elements of the periodic table. This is attributed to stable configuration and it's small size.

(II) configuration belongs to fluorine and $F$ has highest electronegativity on Pauling scale.

(III) configuration belongs to $Cl$, which has the maximum negative electron gain enthalpy (even greater than $F$; due to it's larger size and lesser inter-electronic repulsion).

(IV) This configuration belongs to $C$ and it shows -4 oxidation state because it attains inert gas configuration of neon by gaining four electrons.