Question

Determine if the individual bonds in each compound are polar or nonpolar.

Include electronegativity values and the difference to show your work.
EXAMPLE: Se=2.4 O= 3.5 3.5-2.4=1.1 Polar
1. SCl2
2. H2S
3. CF4
4. PC15
5. CHA
6. CaBr2
7. SF2
8. CO2
9. NH3
10. Na s

Answer 1

See explanation

Step-by-step explanation:

For SCl2

S = 2.8, Cl = 3.16 difference = 0.4 polar

H2S

H = 2.2 S= 2.8 difference = 0.6 polar

CF4

C = 2.55, F= 3.98 difference = 1.43

This compound is nonpolar irrespective of the difference in electronegativity because the molecule is symmetrical

PCl5

P= 2.19, Cl= 3.16 difference = 0.97

This compound is nonpolar irrespective of the difference in electronegativity because the molecule is symmetrical

CH4

C = 2.55 H = 2.2 difference = 0.35 the molecule is nonpolar

CaBr2

Ca= 1 Br= 2.96 difference = 1.96

The compound is not just polar but ionic in nature due to large electronegativity difference between the bonding atoms

SF2

S = 2.8 F= 3.98 difference = 1.18

The molecule is polar

CO2

C= 2.55 O= 3.44 difference = 0.89

This compound is nonpolar irrespective of the difference in electronegativity because the molecule is symmetrical

NH3

N= 3.04 H = 2.2 difference =0.84

The molecule is polar

NaS

Na= 0.93 S = 2.8 difference = 1.87

The compound is not just polar but ionic in nature due to large electronegativity difference between the bonding