Question

Outer electronic configurations of a few elements are given below. Explain them and identify the period, group and block in the periodic table to which they belong.

₂He: 1s², ₅₄Xe: 5s²5p⁶, ₁₆S: 3s²3p⁴, ₇₉Au: 6s¹5d¹⁰

Answer 1

i. ₂He: 1s²

Here, n = 1. Therefore, ₂He belongs to the 1^st period.

The shell n = 1 has only one subshell, namely 1s. The outer electronic configuration 1s² of ‘He’ corresponds to the maximum capacity of 1s, the complete duplet. Therefore, He is placed at the end of the 1^st period in the group 18 of inert gases. So, ‘He’ belongs to p-block.

ii. ₅₄Xe: 5s²5p⁶

Here, n = 5. Therefore, ₅₄Xe belongs to the 5^th period.

The outer electronic configuration. 5s²5p⁶ corresponds to complete octet. Therefore, ₅₄Xe is placed in group 18 and belongs to p-block.

iii. ₁₆S: 3s²3p⁴

Here, n = 3. Therefore, ₁₆S belongs to the 3^rd period. The 3p subshell in ‘S’ is partially filled and short of completion of octet by two electrons. Therefore, ‘S’ belongs to (18 – 2) = 16^th group and p-block.

iv. ₇₉AU: 6s¹5d¹⁰

Here n = 6. Therefore, ‘Au’ belongs to the 6^th period.

The sixth period begins with filling of electron into 6s and then into 5d orbital.

The outer configuration of ‘Au’: 6s¹ 5d¹⁰ implies that (1 + 10) = 11 electrons are filled in the outer orbitals to give ‘Au’. Therefore ‘Au’ belongs to the group 11.

As the last electron has entered ‘d’ orbital ‘Au’ belongs to the d-block.