Electronic configuration of four elements A, B ,C and D are given below
A) `1s^(2), 2s^(2), 2p^(6)`
B) `1s^(2), 2s^(2), 2p^(4)`
C) `1s^(2), 2s^(2), 2p^(6), 3s^(1)`
D) `1s^(2), 2s^(2), 2p^(5)`
Which of the following is the correct order of increasing tendency to gain electron?

Options

A. `A lt C lt B lt D`
B. `A lt B lt C lt D`
C. `D lt B lt C lt A`
D. `D lt A lt B lt C`

Answer

A

Explanation

Electronic configurations of elements suggest that A is a noble gas (i.e., Ne) B is oxygen (Group 16), C is an alkali metal ( group1) and D is a halogen ( group 17) .
(i) Noble gases have no tendency to gain electrons since all their orbitals are completely filled. Thus element A has hte least electron gain enthalpy.
Now since element D has one electron less and element B has two electrons less than the corresponding noble gas configuration therefore element D has the highest electron gain enthalpy followed by element B.
(iii) Since element C has one electron in the s-orbital and hence needs one more electron to complete it therefore electron gain enthalpy of C is less than that of element B. Combing all the facts given above the electron gain enthalpy of the four elements increases in the order `: A lt C lt b lt D,` i.e., option (a) is correct .