Question

1. draw draw the electron dot structure for each mol-ecule. identify polar covalent bonds by assigning slightly positive (δ+) and slightly negative (δ−) symbols to the appropriate atoms.

a. hooh
c. hbr
b. brcl
d. h₂o

Explanation:

Step1: Determine electron - dot structure for \(HOOH\)

• Oxygen has 6 valence electrons and hydrogen has 1 valence electron.
• In \(HOOH\), the structure is \(H - O - O - H\).
• Each \(O\) has 2 lone pairs. The electron - dot structure is \(H:\ddot{O}:\ddot{O}:H\).
• For the \(O - H\) bonds: Oxygen is more electronegative than hydrogen. So, in \(O - H\) bonds, \(H\) has \(\delta+\) and \(O\) has \(\delta-\). The \(O - O\) bond is non - polar (same atoms).

Step2: Determine electron - dot structure for \(BrCl\)

• Bromine and chlorine each have 7 valence electrons.
• The electron - dot structure is \(\ddot{Br}:\ddot{Cl}\).
• Chlorine (\(Cl\), electronegativity \(= 3.16\)) is more electronegative than bromine (\(Br\), electronegativity \(= 2.96\)). So, \(Br\) has \(\delta+\) and \(Cl\) has \(\delta-\).

Step3: Determine electron - dot structure for \(HBr\)

• Hydrogen has 1 valence electron and bromine has 7 valence electrons.
• The electron - dot structure is \(H:\ddot{Br}\).
• Bromine is more electronegative than hydrogen. So, \(H\) has \(\delta+\) and \(Br\) has \(\delta-\).

Step4: Determine electron - dot structure for \(H_{2}O\)

• Oxygen has 6 valence electrons and hydrogen has 1 valence electron.
• The electron - dot structure is \(H:\ddot{O}:H\).
• Oxygen is more electronegative than hydrogen. So, in \(O - H\) bonds, \(H\) has \(\delta+\) and \(O\) has \(\delta-\).

Answer:

a. Electron - dot structure: \(H:\ddot{O}:\ddot{O}:H\); Polar bonds: \(H\ (\delta+)-O\ (\delta-)\)
b. Electron - dot structure: \(\ddot{Br}:\ddot{Cl}\); Polar bonds: \(Br\ (\delta+)-Cl\ (\delta-)\)
c. Electron - dot structure: \(H:\ddot{Br}\); Polar bonds: \(H\ (\delta+)-Br\ (\delta-)\)
d. Electron - dot structure: \(H:\ddot{O}:H\); Polar bonds: \(H\ (\delta+)-O\ (\delta-)\)