Question

Rank the members of each set of compounds in order of decreasing ionic character of their bonds. Use partial charges to indicate the bond polarity of each:
A. PCl3, PBr3, PF3
B. NF3, BF3, CF4
C. SeF4, TeF4, BrF3

Answer 1

Answer:

Ionic character

A.  PF₃ > PBr₃ > PCl₃

B. BF₃ > CF₄ > NF₃

C. TeF₄ > BrF₃ > SeF₄

Explanation:

The most electronegative element is fluorine, followed chlorine, phosphorous nitrogen etc.

• Atoms with high electronegativity tend to form negative ions.

• Ionic compounds formed between elements with high electronegativity difference.

• % ionic character is directly proportional to electronegativity difference.
• According to Pauling Scale E.n for F(4.0), O(3.5), N(3.0), C(2.5), B(2.0), P(2.19), Se(2.55) , Te (2.1), Cl(3.16) and Br(2.96)
• ΔE.N (Electronegativity difference) between( P and F = 4 - 2.19 = 1.81), (P and Br = 2.96 - 2.19 = 0.77) , (P and Cl = 3.16 - 2.96 = 0.2 )

• ΔE.N (Electronegativity difference) between( N and F = 4 - 3 = 1), (B and F = 4 - 2 = 2) , (C and F = 4 - 2.5 = 1.5 )

• ΔE.N (Electronegativity difference) between( Se and F = 4 - 2.55 = 1.45), (F and Te = 4 - 2.1 = 1.9) , (F and Br = 4 - 2.19 = 1.81 )