Question

Q 31. An element is placed in the second group and third period of the periodic table and
in presence of Oxygen forms a basic oxide. (3)
(i) Identify the element and write its electronic configuration.
(ii) Write a balanced chemical equation when it burns in the presence of air.
(iii) Draw the electron dot structure for the formation of this oxide.​

Answer 1

(a) Magnesium 

(b)  2, 8, 2 

(c) 2Mg+O2→2MgO

(d) MgO+H2O→Mg(OH)2

(e) Electron dot structure for the formation of magnesium oxide is shown in image: 

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Answer 2

An element is placed in second group , thus its electronic configuration is 2,8,2 because it is placed in second group thus outer most shell will also have only 2 electron and it is 3rd period , thus it will form 3 shell of electrons .

i) The element is magnesium (Mg) and its electronic configuration is 2,8,2

ii) When magnesium is burn is the presence of air the following reaction takes place

2Mg(s) + O2(g)➠2MgO(s)

iii) Magnesium oxide is formed during this reaction and the formation takes place like this

K L M

Mg =2 , 8, 2

O. = 2, 6

So Oxygen will gain 2 electrons from magnesium forming O2- ions and magnesium will be Mg2+ ions

× ➘ • •

Mg O• •

× ➚ • •

Or [Mg]²+ [O]²-

Thus it will form MgO (magnesium oxide).

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