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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Solution

(i) Among the given elements, $_{19} K$ has one electron in the outermost shell. Its electronic configuration is 2, 8, 8, 1.
(ii) The elements $_{4} B e$ and $_{20} C a$ belong to the same group, that is, Group 2. They both have two electrons each in their outermost shells.
Electronic configurationof $4^{B e}$ is 2, 2.
Electronic configurationof $20^{C a}$ is 2, 8, 8, 2.

Atomic number of element X = 17
Electronic configuration of element X is 2, 8, 7.
Number of valence electrons in X = 7
Valency of X = 8 - 7 = 1

Atomic number of K = 19
Electronic configuration of K is 2, 8, 8, 1

Number of valence electrons in K = 1
Valency of K = 1

Therefore, the formula of the compound formed when K and element X combine is KX. The compound KX is of ionic nature.

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From the following elements: 4Be;9F;19K;20Ca (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).

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).