An element X (atomic number 17) reacts with an element Y (atomic number 20) to form a divalent halide
(a) Where in the periodic table are elements X and Y placed ?
(b) Classify X and Y as metal (s), non-metal (s) or metalloid (s)
(c) What will be the nature of oxide of element Y ? Identify the nature of bonding in the compound formed
(d) Draw the electron dot structure of the divalent halide.

Explanation

(a) The electronic configuration of element 'X' with atomic number 17 is 2,8,7. Since, it has 7 valence electrons. Therefore, it lies in group 17 (10 + 7). Further, since in element X, third shell is being filled, it lies in third period. In other wordds, X is chlorine
The electronic configuration of element 'Y' with atomic number 20 is 2,8,8,2. Since, it has 2 valence electrons, it lies in group 2. Further, since in element Y, fourth shell is being filled, it lies in `4^(th)` period. In other words, Y is calcium.
(b) Since, element X (i.e., Cl) has seven electrons in the valence shell and needs one more electron complete its octet. Therefore it is non-metal. Further , since element Y has two eletrons in the valence shell which it can easily lose to achieve the stable electronic configuration of the nearest inert gas, therefore, it is a metal
(c) Since, element Y (i.e., Ca) is a metal, therefore, its oxide (i.e., CaO) must be basic in nature. Further, since metals and non-metals form ionic compounds, therefore, the nature of bonding in calcium oxide is ionic
(d) Electronic configuration of `._(20)Ca=2,8,8,2` [valence electron = 2] electronic configuration of `._(17)Cl = 2,8,7` [ valence electrons = 7]. The electron dot structure of divalent metal halide, i.e., `CaCl_(2)` is