Question

Given below are some elements of the modern periodic table: (3)4Be, 9Fe, 14Si, 19K, 20Ca(i) Select the element that has one electron in the outermost shell and write its electronic configuration.(ii) Select two elements that belong to the same group. Give reason for your answer.(iii) Select two elements that belong to the same period. Which one of the two has bigger atomic size?

Answer 1

(i) 19K has one electron in the outermost shell and its electronic configuration is2 8 8 1
(ii) 4Be and 20Ca belongs to same group i.e. Group-2.
Electronic configuration:4Be - 2 2
20Ca- 2 8 8 2
The number of electrons in the outermost shell of 4Be and 20Ca is same hence they belong to the same shell.
(iii) 9F and 4Be belongs to the same period,

Period 2. Electronic configuration:9F - 2 74Be - 2 24Be has bigger atomic size then 9F because the atomic radius decreases as we move from left to right due to increase in nuclear charge which tends to pull the electrons closer to the nucleus and hence size of F reduces.

Answer 2

(i) The elements is potassium (K).

(ii) The elements Be and Ca belong to second group since their atoms have two electrons in the valence shell.

(iii) The elements belonging to the same period are Be and F (belong to second period) , K and Ca (belong to fourth period). The element Be has bigger size than F since atomic size decreases along a period. The element Ca has smaller size because of the same reason.