Chapter U2.44, Problem 6E

Interpretation Introduction

Interpretation : Three pairs of atoms with ionic bonds with difference in their electronegativity values are to be written.

Concept Introduction :

The tendency of an atom to attract shared electrons is called electronegativity. A bond between two atoms with different electronegativities is polar covalent.When the difference in electronegativity between two atoms is too large (> 2.1), the bond is no longer covalent but it becomes ionic.

Answer to Problem 6E

Three pairs of atoms with ionic bonds are:

1. NaF- It has an electronegativity difference of 3.05
2. K-F-It has an electronegativity difference of 3.16
3. NaCl- It has an electronegativity difference of 2.23

Explanation of Solution

When the difference in electronegativity between two atoms is more than 2.1, the bond is no longer covalent but it becomes ionic as the electrons get transferred from less electronegative atom to more electronegative atom.

1. NaF- From the electronegativity scale, electronegativity value of sodium is 0.93 and fluorine is 3.98. Difference in electronegativity value $=3.98-0.93=3.05$
2. K-F- From the electronegativity scale, electronegativity value of potassium is 0.82 and fluorine is 3.98. Difference in electronegativity value $=3.98-0.82=3.16$
3. NaCl- From the electronegativity scale, electronegativity value of sodium is 0.93 and chlorine is 3.16. Difference in electronegativity value $=3.98-0.93=2.23$

Conclusion

In a polar molecule one atom is more electronegative than other atom. In ionic molecule the difference in electronegativity value is more than 2.1.