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

plasma is the state of matter where the substance is ionized and has no definite shape or volume. the fact of being ionized makes the substance a good electrical conductor.