From the following elements:
`""_4Be, ""_9F, ""_19K, ""_20Ca`
(i) Select the element having one electron in the outermost shell.
(ii) Two elements of the same group. Write the formula and mention the nature the compound formed by the union of 19K and element X (2,8, 7).

Explanation

(a) the first period has only two elements beginning with Li (Z=3) and ending with Ne (Z= 2)
the second period has also only 8 elements beginning with Li (Z=3) and ending with Ar (Z= 10)
The third period also has 8 elements beginning with Na (Z=11) and ending with Kr (Z=18)
The Fourth period has 18 elements beginnings with K (Z=19) and ending with Kr (Z=36)
Since each period begins with on a element having only one electron in the valence shell therefore out of the four elements listed in the question (i.e., ._(4)Be, ._(9)F, ._(19)K, ._(20)Ca)` only ._(19)K` has one electron in the valence shell.
(b) Since second group has two electrons in the valence shell therefore `._(4)Be " and " ._(20)Ca` belong to the same group.
( c) the element X with electronic configuration 2,8.7 (i.e., , CI ,Z= 17) has one electron less than the nearest inert gas i.e, Ar (Z=18) while K with electronic configuration (2,8,8,1) has on a electron more than the nearest inert gas Ar (2,8,8) . Therefore to achieve the the nearest inert gas configuration , K loses one electron to form `K^(+)` and CI gains one electron to form `CI^(-)` .The two ions then combine to form ionic compound `K^(+) CI^(-)` or simply KCI.