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Chemical BondingMedium

Electronegativity difference and bond type

Determine the bond type (ionic, pure covalent, polar covalent) for Na-Cl (EN: Na=0.93, Cl=3.16), H-O (H=2.20, O=3.44), C-C (C=2.55).
Thresholds: ΔEN < 0.4 → pure cov.; 0.4–1.7 → polar cov.; > 1.7 → ionic.
Given data
EN(Na)=0.93EN(Cl)=3.16EN(H)=2.20EN(O)=3.44EN(C)=2.55
Review the theory: Legami Chimici
Steps
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  1. Calculate ΔEN for Na-Cl.
  2. What bond type is Na-Cl?
  3. Calculate ΔEN for H-O.
  4. What bond type is H-O?
  5. ΔEN for C-C?
  6. What bond type is C-C?
Full worked solution
  1. Calculate ΔEN for Na-Cl.
    3.16−0.933.16 - 0.93
    3.16−0.93=2.23=2.233.16 - 0.93 = 2.23 = \mathbf{2.23}. ΔEN > 1.7.
  2. What bond type is Na-Cl?
    2.23>1.7=2.232.23 > 1.7 = \mathbf{2.23}. Ionic bond.
  3. Calculate ΔEN for H-O.
    3.44−2.203.44 - 2.20
    3.44−2.20=1.24=1.243.44 - 2.20 = 1.24 = \mathbf{1.24}. ΔEN between 0.4 and 1.7.
  4. What bond type is H-O?
    1.241.24 in [0.4,1.7]=1.24[0.4, 1.7] = \mathbf{1.24}. Polar covalent bond.
  5. ΔEN for C-C?
    2.55−2.552.55 - 2.55
    2.55−2.55=0=02.55 - 2.55 = 0 = \mathbf{0}. Same element.
  6. What bond type is C-C?
    0<0.4=00 < 0.4 = \mathbf{0}. Pure covalent bond.
Result:Na-Cl: ionic (ΔEN=2.23) — H-O: polar covalent (ΔEN=1.24) — C-C: pure covalent (ΔEN=0).
TheoryLegami ChimiciFormula sheetChemistry

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