P, Q, R, S are four elements. Their atomic numbers are 8, 17, 12 and 16 respectively. Find the type of chemical bond in these compounds formed by combining the following pairs of elements. Construct and exhibit the type of bonds using different substances (e.g. pearls).

(Electronegativity values: P=3.44, Q=3.16, R=1.31, S=2.5)

1. P, R

2. P, S

3. Q, R

Electronegativity difference between the constituent elements of a compound decide the nature of compound. If the difference is 1.7 or more

than 1.7, then it generally shows ionic character and if its less than 1.7 it shows covalent character. Subtracting the respective electronegativity values of constituents gives difference.

1. P and R

Electronegativity difference = 3.44 – 1.31 = 2.13

Ionic compound

2. P and S

Electronegativity difference = 3.44 – 2.5 = 0.94

Covalent compound

3. Q and R

Electronegativity difference = 3.16 – 1.31 = 1.85

Ionic compound

The electronic configuration of elements and valency decide the chemical formula of the compound. Electronic configuration of –

P is 2,6 ( oxygen )

Q is 2,8,7 ( chlorine )

R is 2,8,2 ( magnesium )

S is 2,8,6 ( sulphur )

P tries to gain 2 electrons and obtain octet.

Q tries to gain 1 electron and obtain octet.

R tries to lose 2 electrons and get stability.

S tries to gain 2 electrons and obtain octet.

1. P and R form the compound PR ( MgO ) by transfer of electrons. It’s a ionic compound.

2. P and S form the compound SP₂ ( SO₂) by sharing of electrons. Its covalent compound.

3. Q and R form the compound RQ₂ ( MgCl₂) by transfer of electrons which is a ionic compound.

The 2 electrons lost by magnesium are gained by 2 chlorine atoms.

Thus, MgCl₂ is formed.