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Determine the electronegativity difference, the probable bond type, and the more-electronegative atom with respect to bonds formed between the following pairs of atoms:

(a) H & I
(b) S & O
(c) K & Br
(d) Si & Cl
(e) H & F
(f) Se & S
(g) C & H

User Gurubelli
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Final answer:

The electronegativity difference between atoms helps determine the bond type, and the atom with the higher electronegativity in each pair is the more-electronegative one. Bond types range from nonpolar covalent, with very small differences, to ionic, with larger differences. Examples include H and I forming a nonpolar covalent bond, S and O forming a polar covalent bond, and K and Br forming an ionic bond.

Step-by-step explanation:

To determine the electronegativity difference, probable bond type, and the more-electronegative atom with respect to the bonds formed between the following pairs of atoms, we apply our understanding of electronegativity values:

  • H & I: H (2.1) & I (2.5), Difference = 0.4, Bond type: Nonpolar Covalent, More-electronegative: I
  • S & O: S (2.5) & O (3.5), Difference = 1.0, Bond type: Polar Covalent, More-electronegative: O
  • K & Br: K (0.8) & Br (2.8), Difference = 2.0, Bond type: Ionic, More-electronegative: Br
  • Si & Cl: Si (1.8) & Cl (3.0), Difference = 1.2, Bond type: Polar Covalent, More-electronegative: Cl
  • H & F: H (2.1) & F (4.0), Difference = 1.9, Bond type: Polar Covalent, More-electronegative: F
  • Se & S: Se (2.4) & S (2.5), Difference = 0.1, Bond type: Nonpolar Covalent, More-electronegative: S
  • C & H: C (2.5) & H (2.1), Difference = 0.4, Bond type: Nonpolar Covalent, More-electronegative: C

User Snehal Ghumade
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