Ncert Exercises & Solutions

The electronic configuration of some elements is given in Column-I and their electron gain enthalpies are given in Column-II. Match the electronic configuration with the electron gain enthalpy:

	Column-I (Electron configuration)		Column-II (Electron gain enthalpy/ kJ mol¯¹)
A.	$1 s^2 2 s^2 2 p^6$	(i)	$-53$
B.	$1 s^2 2 s^2 2 p^6 3 s^1$	(ii)	$-328$
C,	$1 s^2 2 s^2 2 p^5 $	(iii)	$-141$
D.	$ 1 s^2 2 s^2 2 p^4 $	(iv)	$+48$

Options:	A	B	C	D
1.	(iv)	(i)	(ii)	(iii)
2.	(i)	(ii)	(iii)	(iv)
3.	(i)	(iv)	(iii)	(ii)
4.	(iv)	(i)	(iii)	(ii)

Hint: Fully filled configuration will have positive electron gain enthalpy.

A. $\to$ (4) B. $\to$ (1) C. $\to$ (2) D. $\to$ (3)

A. This electronic configuration corresponds to the noble gas i.e., neon. Since, noble gases have

+ ∆_{eg} H

values, therefore, electronic configuration (A) corresponds to the

∆_{eg} H = + 48 kJ {mol}^{- 1}

B. This electronic configuration corresponds to the alkali metal i.e., potassium. Alkali metals have small negative

∆_{eg} H

values, hence, electronic configuration (B) correspongds to

∆_{eg} H = - 53 kJ {mol}^{- 1}

C. This electronic configuration corresponds to the halogen i.e., fluorine. Since, halogens have high negative

∆_{eg} H

values, therefore, electronic configuration (C) corresponds to

∆_{eg} H = 328 kJ {mol}^{- 1}

D. This electronic configuration corresponds to the chalcogen i.e., oxygen. Since, chalcogens have

∆_{eg} H

values less negative than those of halogens, therefore, electronic configuration (D) corresponds to

∆_{eg} H = - 141 kJ {mol}^{- 1}

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	Column-I (Electron configuration)		Column-II (Electron gain enthalpy/ kJ mol¯¹)
A.	\(1 s^2 2 s^2 2 p^6\)	(i)	\(-53\)
B.	\(1 s^2 2 s^2 2 p^6 3 s^1\)	(ii)	\(-328\)
C,	\(1 s^2 2 s^2 2 p^5 \)	(iii)	\(-141\)
D.	\( 1 s^2 2 s^2 2 p^4 \)	(iv)	\(+48\)