Question

. Given below are some elements of the modern periodic table: 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 you answer. (iii)Select two elements that belong to the same period. Which one of the two has bigger atomic size?

Answer 1

Electronic Config. of elements are
Be= 2,2
F = 2,7
Si= 2,8,4
K =. 2,8,8,1
Ca= 2,8,8,2
I) k has one electron in it's outer orbit
2,8,8,1
ii) Element Be and Ca belong to same group as they have same no. of electron in their outermost shell
iii) Elements K and Ca belong to same period since no.of shells are equal
Size of K is larger as compared to Ca which has larger nuclear force as compared to K so K is larger

Answer 2

Answer:

(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.

Hope it helps you ....