Class 11 Chemistry - Chapter Structure of Atom NCERT Solutions | :(i) Write the electronic configurations

Welcome to the NCERT Solutions for Class 11th Chemistry - Chapter Structure of Atom. This page offers a step-by-step solution to the specific question from Exercise 1, Question 23: i write the electronic configurations of the fo....
Question 23

:(i) Write the electronic configurations of the following ions: (a) H(b) Na+ (c) O2–(d) F

(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1 (b) 2p3 and (c) 3p5?

(iii) Which atoms are indicated by the following configurations?

(a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1.

Answer

Electronic configuration of an atom is defined as the representation of the position of electrons in the various energy shell & subshells.

Now A negative charge on the species indicates the gain of an electron by it & A positive charge denotes the loss of an electron

 

(i) (a) H ion

The electronic configuration of H atom is 1s1.(atomic number = 1)

∴ Electronic configuration of H = 1s2

 

(b) Na+ ion

The electronic configuration of Na atom is 1s2 2s2 2p6 3s1.(atomic number = 11)

∴ Electronic configuration of Na+ = 1s2 2s2 2p6 3s0  Or  1s2 2s2 2p6

 

(c) O2– ion

The electronic configuration of 0 atom is 1s2 2s2 2p4.(atomic number = 8)

∴ Electronic configuration of O2– ion = 1s2 2s2 p6

 

(d) F ion

The electronic configuration of F atom is 1s2 2s2 2p5.(atomic number = 9)

∴ Electron configuration of F ion = 1s2 2s2 2p6

 

(ii) (a) 3s1

Completing the electron configuration of the element as 1s2 2s2 2p6 3s1.

∴ Number of electrons present in the atom of the element

= 2 + 2 + 6 + 1 = 11

∴ Atomic number of the element = 11(sodium)

 

(b) 2p3

Completing the electron configuration of the element as 1s2 2s2 2p3.

∴ Number of electrons present in the atom of the element = 2 + 2 + 3 = 7

∴ Atomic number of the element = 7(nitrogen)

 

(c) 3p5

Completing the electron configuration of the element as 1s2 2s2 2p5.

∴ Number of electrons present in the atom of the element = 2 + 2 + 5 = 9

∴ Atomic number of the element = 9(fluorine)

 

(iii) (a) [He] 2s1

The electronic configuration of the element is [He] 2s1 = 1s2 2s1.

∴ Atomic number of the element = 3 (lithium , a p-block element)

 

(b) [Ne] 3s2 3p3

The electronic configuration of the element is [Ne] 3s2 3p3= 1s2 2s2 2p6 3s2 3p3.

∴ Atomic number of the element = 15(phosphorous, a p block element)

 

(c) [Ar] 4s2 3d1

The electronic configuration of the element is [Ar] 4s2 3d1= 1s2 2s2 2p6 3s2 3p6 4s2 3d1.

∴ Atomic number of the element = 21(scandium , a d block element)

1 Comment(s) on this Question

Write a Comment: