CBSE Class 10 Science 2017 · Paper 531/3 · Question 8

Periodic Classification of Elements 3 marks Medium

Consider the following elements (The atomic numbers are given in parentheses) : Be(4); F(9); K(19); Ca(20)

(a)Select the element having one electron in the outermost shell. Also write its electronic configuration.
(b)Select two elements of the same group.
(c)Write the formula and nature of the compound formed, if K(19) reacts/combines with an element X(2, 8, 7).

Answer

Potassium has one outer electron; beryllium and calcium are in group 2; K and chlorine form the ionic compound $KCl$ .

(a) Potassium, K(19), has one electron in its outermost shell. Its electronic configuration is 2, 8, 8, 1. (b) Beryllium, Be(4), and calcium, Ca(20). Both have two electrons in the outermost shell (2, 2 and 2, 8, 8, 2), so both are in group 2. (c) X (2, 8, 7) has atomic number 17 (chlorine). K loses one electron to form

K^{+}

and X gains it to form

X^{-}

, so the formula is KX, that is

KCl

. It is an ionic (electrovalent) compound, a salt whose solution in water is neutral.

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CBSE Class 10 Science 2017 · Paper 531/3 · Question 8

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