Arrange the following molecules in order of decreasing polarity of their bonds. a. PBr3
b. SF2
c. H2O
d. NCl3

SF2 > H2O > PBr3 > NCl3

Explanation:

Compare the electronegativity values for the atoms and classify the nature of the bonding based on the electronegativity difference.

P has an electronegativity of 2.1, while Br has an electronegativity of 2.96. The difference is 0.86, indicating that these atoms will form covalent bonds.

S has an electronegativity of 2.58 while F has an electronegativity of 4.0. The difference is 1.42, indicating that these atoms will form polar covalent bonds.

O has an electronegativity of 3.5 while H has an electronegativity of 2.1. The difference is 1.4, indicating that these atoms will form polar covalent bonds.

N has an electronegativity of 3.04, whereas Cl has an electronegativity of 3.5. This difference of 0.46 indicates that these atoms will form covalent bonds.

We know that the greater the electronegativity, the higher the polarity. In decreasing order of polarity, we have:

SF2 > H2O > PBr3 > NCl3

the answer is definitely c.

you need to have a picture of the atom that way we can answer it