Question

(i) Write the electronic configurations of the following ions: (a) H^– (b) Na⁺ (c) O^2–(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 1

(i) (a) H^– ion

The electronic configuration of H atom is 1s¹.

A negative charge on the species indicates the gain of an electron by it.

∴ Electronic configuration of H^– = 1s²

(b) Na⁺ ion

The electronic configuration of Na atom is 1s² 2s² 2p⁶ 3s¹.

A positive charge on the species indicates the loss of an electron by it.

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

(c) O^2– ion

The electronic configuration of 0 atom is 1s² 2s² 2p⁴.

A dinegative charge on the species indicates that two electrons are gained by it.

∴ Electronic configuration of O^2– ion = 1s² 2s² p⁶

(d) F^– ion

The electronic configuration of F atom is 1s² 2s² 2p⁵.

A negative charge on the species indicates the gain of an electron by it.

∴ 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

(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

(c) 3p⁵

Completing the electron configuration of the element as

$1s^{2}2s^{2}2p^{6}3s^{2}3p^5$

∴ Number of electrons present in the atom of the element = 2 + 2 + 6+2+5 = 17

∴ Atomic number of the element = 17

(iii) (a) [He] 2s¹

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

∴ Atomic number of the element = 3

Hence, the element with the electronic configuration [He] 2s¹ is lithium (Li).

(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

Hence, the element with the electronic configuration [Ne] 3s² 3p³ is phosphorus (P).

(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

Hence, the element with the electronic configuration [Ar] 4s² 3d¹ is scandium (Sc).