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

See explanation

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

a electronegativity decreases beacuse the size of nucleus is bigger.