Sure! Let's analyze each bond one by one to determine the electronegativity difference between the two atoms and the type of bond formed.
### Bond: H - Cl
1. Electronegativity of H (Hydrogen): 2.20
2. Electronegativity of Cl (Chlorine): 3.16
3. Difference in Electronegativity (ΔEN):
[tex]\[
\Delta EN = |3.16 - 2.20| = 0.96
\][/tex]
4. Bond Type: Since the electronegativity difference is 0.96, the bond is polar covalent.
### Bond: Li - O
1. Electronegativity of Li (Lithium): 0.98
2. Electronegativity of O (Oxygen): 3.44
3. Difference in Electronegativity (ΔEN):
[tex]\[
\Delta EN = |3.44 - 0.98| = 2.46
\][/tex]
4. Bond Type: Since the electronegativity difference is 2.46, the bond is ionic.
### Bond: H - O
1. Electronegativity of H (Hydrogen): 2.20
2. Electronegativity of O (Oxygen): 3.44
3. Difference in Electronegativity (ΔEN):
[tex]\[
\Delta EN = |3.44 - 2.20| = 1.24
\][/tex]
4. Bond Type: Since the electronegativity difference is 1.24, the bond is polar covalent.
### Bond: Br - Br
1. Electronegativity of both Br (Bromine): 2.96
2. Difference in Electronegativity (ΔEN):
[tex]\[
\Delta EN = |2.96 - 2.96| = 0.0
\][/tex]
4. Bond Type: Since the electronegativity difference is 0.0, the bond is nonpolar covalent.
### Bond: N - Cl
1. Electronegativity of N (Nitrogen): 3.04
2. Electronegativity of Cl (Chlorine): 3.16
3. Difference in Electronegativity (ΔEN):
[tex]\[
\Delta EN = |3.16 - 3.04| = 0.12
\][/tex]
4. Bond Type: Since the electronegativity difference is 0.12, the bond is nonpolar covalent.
### Summary:
1. H - Cl: ΔEN = 0.96, Bond Type = Polar Covalent
2. Li - O: ΔEN = 2.46, Bond Type = Ionic
3. H - O: ΔEN = 1.24, Bond Type = Polar Covalent
4. Br - Br: ΔEN = 0.0, Bond Type = Nonpolar Covalent
5. N - Cl: ΔEN = 0.12, Bond Type = Nonpolar Covalent
