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
1 answer:
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 )
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