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

Codes:

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)

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}

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions