Given below are some elements of the modern periodic table:

_{$4^{B e}, 9^{F e}, 14^{S i}, 19^{K}, 20^{C a}$}

(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?

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Solution

(i) $19^{K}$ has one electron in the outermost shell and its electronic configuration is 2 8 8 1.

(ii) $4^{B e}$ and $20^{C a}$ belongs to same group i.e. Group-2. Electronic configuration :

$4^{B e}$ - 2 2

$20^{C a}$ - 2 8 8 2

The number of electrons in the outermost shell of $4^{B e}$ and $20^{C a}$ is same hence they belong to the same shell

(iii) $9^{F}$ and $4^{B e}$ belongs to the same period, Period 2. Electronic configuration
$9^{F}$ - 2 7
$4^{B e}$ - 2 2

$4^{B e}$ has bigger atomic size then $9^{F}$ 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.

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From the following elements:
$4^{B e}; 9^{F}; 19^{K}; 20^{C a}$
(i) Select the element having one electron is the outermost shell.
(ii) Two elements of the same group.
Write the formula of and mention the nature of the compound formed by the union of 19K and element X (2, 8, 7).

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