Student fills out a chemistry worksheet on molecular polarity and valence electrons.

Problem Analysis:

The image shows a table with information about molecular structures, including the number of valence electrons, geometry, resonance, bond polarity, and whether the molecule is polar or non-polar. The task involves completing the missing information for the molecules H₂O (Water), CCl₄ (Carbon Tetrachloride), and NO₃⁻ (Nitrate ion).

Step-by-Step Solution:

#### 1. H₂O (Water)
- Molecule: H₂O
- Valence Electrons:
- Oxygen (O): 6 valence electrons
- Hydrogen (H): 1 valence electron each
- Total: \( 6 + 2 \times 1 = 8 \)
- Geometry:
- Water has two bonding pairs (O-H bonds) and two lone pairs on the oxygen atom.
- This results in a bent (V-shaped) geometry.
- Resonance:
- Water does not exhibit resonance because there are no equivalent resonance structures.
- Polar Bonds:
- The O-H bond is polar due to the difference in electronegativity between oxygen (3.5) and hydrogen (2.1):
\[
\text{Electronegativity difference} = 3.5 - 2.1 = 1.4
\]
- Polar Molecule:
- Water is a polar molecule because the bent geometry causes the dipoles of the O-H bonds to add up, resulting in a net dipole moment.

Completed Row for H₂O:
- Valence Electrons: 8
- Geometry: Bent (V-shaped)
- Resonance: No Resonance
- Polar Bonds: \( 3.5 - 2.1 = 1.4 \)
- Polar Molecule: Polar molecule

#### 2. CCl₄ (Carbon Tetrachloride)
- Molecule: CCl₄
- Valence Electrons:
- Carbon (C): 4 valence electrons
- Chlorine (Cl): 7 valence electrons each
- Total: \( 4 + 4 \times 7 = 32 \)
- Geometry:
- Carbon tetrachloride has four bonding pairs (C-Cl bonds) and no lone pairs.
- This results in a tetrahedral geometry.
- Resonance:
- CCl₄ does not exhibit resonance because all C-Cl bonds are identical.
- Polar Bonds:
- The C-Cl bond is polar due to the difference in electronegativity between carbon (2.5) and chlorine (3.0):
\[
\text{Electronegativity difference} = 3.0 - 2.5 = 0.5
\]
- Polar Molecule:
- Despite the polar C-Cl bonds, CCl₄ is a non-polar molecule because the tetrahedral symmetry cancels out the individual bond dipoles.

Completed Row for CCl₄:
- Valence Electrons: 32
- Geometry: Tetrahedral
- Resonance: No Resonance
- Polar Bonds: \( 3.0 - 2.5 = 0.5 \)
- Polar Molecule: Non-polar molecule

#### 3. NO₃⁻ (Nitrate Ion)
- Molecule: NO₃⁻
- Valence Electrons:
- Nitrogen (N): 5 valence electrons
- Oxygen (O): 6 valence electrons each
- Charge: +1 extra electron
- Total: \( 5 + 3 \times 6 + 1 = 24 \)
- Geometry:
- The nitrate ion has three bonding pairs (N-O bonds) and no lone pairs on nitrogen.
- This results in a trigonal planar geometry.
- Resonance:
- Nitrate exhibits resonance because the negative charge is delocalized over the three oxygen atoms, leading to equivalent resonance structures.
- Polar Bonds:
- The N-O bond is polar due to the difference in electronegativity between nitrogen (3.0) and oxygen (3.5):
\[
\text{Electronegativity difference} = 3.5 - 3.0 = 0.5
\]
- Polar Molecule:
- Despite the polar N-O bonds, NO₃⁻ is a non-polar molecule because the trigonal planar symmetry and resonance cancel out the individual bond dipoles.

Completed Row for NO₃⁻:
- Valence Electrons: 24
- Geometry: Trigonal planar
- Resonance: Yes (Resonance)
- Polar Bonds: \( 3.5 - 3.0 = 0.5 \)
- Polar Molecule: Non-polar molecule

Final Answer:

\[
\boxed{
\begin{array}{|c|c|c|c|c|c|c|}
\hline
\text{Molecule} & \text{Valence Electrons} & \text{Geometry} & \text{Resonance?} & \text{Polar Bonds} & \text{Polar Molecule?} \\
\hline
\text{CH₄} & 8 & \text{Tetrahedral} & \text{No Resonance} & 2.5 - 2.1 = 0.4 & \text{Non-polar molecule} \\
\hline
\text{H₂O} & 8 & \text{Bent (V-shaped)} & \text{No Resonance} & 3.5 - 2.1 = 1.4 & \text{Polar molecule} \\
\hline
\text{CCl₄} & 32 & \text{Tetrahedral} & \text{No Resonance} & 3.0 - 2.5 = 0.5 & \text{Non-polar molecule} \\
\hline
\text{NO₃⁻} & 24 & \text{Trigonal planar} & \text{Yes (Resonance)} & 3.5 - 3.0 = 0.5 & \text{Non-polar molecule} \\
\hline
\end{array}
}
\]