💡 9th Grade Other: Lewis Dot Structure, Polarity-Nonpolarity, Naming Compounds, and Intermolecular Forces Practice Questions

Here's how to draw the Lewis dot structure for \( \text{CH}_4 \):

• Step 1: Count Total Valence Electrons.
  Carbon (C) is in Group 14, so it has 4 valence electrons.
  Hydrogen (H) is in Group 1, so it has 1 valence electron.
  Since there are four hydrogen atoms, total valence electrons \( = 4 \text{ (from C)} + (4 \times 1 \text{ (from H)}) = 4 + 4 = 8 \) valence electrons.
• Step 2: Identify the Central Atom.
  Carbon is typically the central atom because it is less electronegative than hydrogen and can form more bonds.
• Step 3: Arrange Atoms and Form Single Bonds.
  Place the carbon atom in the center and arrange the four hydrogen atoms around it.
  Draw a single bond (two electrons) between the central carbon and each hydrogen atom.
  
  \( \text{H} - \text{C} - \text{H} \)
  \( \quad \mid \)
  \( \quad \text{H} \)
  \( \quad \mid \)
  \( \quad \text{H} \)
  This uses \( 4 \times 2 = 8 \) electrons.
• Step 4: Distribute Remaining Electrons (if any).
  We started with 8 valence electrons and used all 8 in the single bonds. There are no remaining electrons to distribute as lone pairs.
• Step 5: Check for Octet Rule and Duet Rule.
  Each hydrogen atom has 2 electrons (duet rule satisfied).
  The carbon atom has \( 4 \times 2 = 8 \) electrons (octet rule satisfied).
  ✅ The Lewis dot structure for \( \text{CH}_4 \) is: \[ \begin{array}{c} \text{H} \\ | \\ \text{H} - \text{C} - \text{H} \\ | \\ \text{H} \end{array} \]

Let's name each compound based on its type:

• a) \( \text{MgCl}_2 \)
  👉 Identify Compound Type: This compound consists of a metal (Magnesium, Mg) and a nonmetal (Chlorine, Cl). This indicates it is an ionic compound.
  👉 Naming Rules for Ionic Compounds:
  • The metal cation is named first, simply by its element name: Magnesium.
  • The nonmetal anion is named second, by taking the root of the element name and adding the suffix "-ide": Chlorine becomes Chloride.
  • Since magnesium typically forms only a \( 2+ \) ion, we don't need to use Roman numerals.
  ✅ Therefore, \( \text{MgCl}_2 \) is named Magnesium chloride.
• b) \( \text{N}_2\text{O}_4 \)
  👉 Identify Compound Type: This compound consists of two nonmetals (Nitrogen, N, and Oxygen, O). This indicates it is a covalent (molecular) compound.
  👉 Naming Rules for Covalent Compounds:
  • Use prefixes to indicate the number of atoms of each element.
  • The first element in the formula is named first, using a prefix if there is more than one atom. (No "mono-" prefix for the first element).
  • The second element is named with a prefix indicating the number of atoms, and its name ends with "-ide".
  • Prefixes: di- (2), tri- (3), tetra- (4), penta- (5), etc.
  👉 Apply Prefixes:
  • For Nitrogen (\( \text{N}_2 \)): "di-" for two nitrogen atoms, so "dinitrogen".
  • For Oxygen (\( \text{O}_4 \)): "tetra-" for four oxygen atoms, so "tetroxide" (we drop the 'a' from tetra- before 'oxide' for easier pronunciation).
  ✅ Therefore, \( \text{N}_2\text{O}_4 \) is named Dinitrogen tetroxide.

To determine if \( \text{NH}_3 \) is polar or nonpolar, we need to consider its Lewis structure, molecular geometry, and bond polarities.

• Step 1: Draw the Lewis Dot Structure for \( \text{NH}_3 \).
  Nitrogen (N) has 5 valence electrons. Each Hydrogen (H) has 1 valence electron.
  Total valence electrons \( = 5 + (3 \times 1) = 8 \).
  Nitrogen is the central atom. It forms three single bonds with hydrogen atoms and has one lone pair of electrons.
  \[ \begin{array}{c} \text{H} \\ | \\ \text{H} - \underset{\centerdot \centerdot}{\text{N}} - \text{H} \end{array} \]
• Step 2: Determine the Molecular Geometry.
  The central nitrogen atom has 3 bonding pairs and 1 lone pair of electrons. According to VSEPR theory, this electron domain arrangement leads to a trigonal pyramidal molecular geometry.
  📌 In a trigonal pyramidal shape, the central atom is at the apex, and the three bonded atoms form the base of a pyramid. The lone pair pushes the bonding pairs downwards.
• Step 3: Assess Bond Polarity.
  Nitrogen (electronegativity \( \approx 3.0 \)) is more electronegative than Hydrogen (electronegativity \( \approx 2.2 \)).
  Therefore, each \( \text{N}-\text{H} \) bond is a polar covalent bond, with the electron density shifted towards the nitrogen atom. This creates a partial negative charge on N and partial positive charges on H atoms.
• Step 4: Determine Overall Molecular Polarity.
  Since the \( \text{NH}_3 \) molecule has a trigonal pyramidal geometry, the bond dipoles (from the polar \( \text{N}-\text{H} \) bonds) do not cancel out. Additionally, the lone pair of electrons on the nitrogen atom contributes significantly to the overall polarity, creating a net dipole moment.
  ✅ Because of the asymmetrical shape and the presence of a lone pair causing an uneven distribution of charge, the ammonia molecule (\( \text{NH}_3 \)) is polar.

To identify the primary intermolecular force in water, we first need to understand its structure and polarity.

• Step 1: Determine the Polarity of the Molecule.
  Water (\( \text{H}_2\text{O} \)) has a bent molecular geometry due to two lone pairs on the central oxygen atom. The oxygen atom is significantly more electronegative than hydrogen, making the \( \text{O}-\text{H} \) bonds polar. Because of its bent shape, the bond dipoles do not cancel out, resulting in a net dipole moment. Therefore, water is a polar molecule.
• Step 2: Identify Possible Intermolecular Forces.
  There are three main types of IMFs:
  • London Dispersion Forces (LDFs): Present in all molecules, arising from temporary fluctuations in electron distribution.
  • Dipole-Dipole Forces: Present in polar molecules, arising from the attraction between permanent dipoles of adjacent molecules.
  • Hydrogen Bonding: A special, strong type of dipole-dipole force that occurs when hydrogen is bonded to a highly electronegative atom (N, O, or F).
• Step 3: Determine the Primary IMF.
  Since water is a polar molecule, it experiences both London Dispersion Forces and Dipole-Dipole Forces. However, because hydrogen is directly bonded to a highly electronegative oxygen atom in \( \text{H}_2\text{O} \), water molecules can form hydrogen bonds with each other.
  📌 Hydrogen bonding is the strongest type of IMF among these three.

✅ Therefore, the primary type of intermolecular force present between molecules of water (\( \text{H}_2\text{O} \)) is hydrogen bonding. This strong IMF is responsible for many of water's unique properties, such as its relatively high boiling point.

The difference in polarity between \( \text{CO}_2 \) and \( \text{H}_2\text{O} \) is crucial and stems from their molecular geometries, which are determined by their Lewis structures.

• 1. Carbon Dioxide (\( \text{CO}_2 \)):
  
  • Lewis Structure: Carbon is the central atom, double-bonded to each oxygen atom. There are no lone pairs on the central carbon atom.
    \[ \centerdot \centerdot \text{O} = \text{C} = \text{O} \centerdot \centerdot \]
  • Molecular Geometry: With two bonding groups and zero lone pairs on the central carbon, \( \text{CO}_2 \) adopts a linear molecular geometry. The oxygen atoms are positioned 180 degrees apart from each other.
  • Bond Polarity: Oxygen is more electronegative than carbon, so each \( \text{C}=\text{O} \) bond is polar. The electron density is pulled towards the oxygen atoms.
  • Overall Molecular Polarity: Due to the linear geometry, the two individual bond dipoles (pulling electrons towards opposite oxygen atoms) are equal in magnitude and point in opposite directions. They effectively cancel each other out. \[ \stackrel{\longleftarrow}{\text{O}} = \text{C} = \stackrel{\longrightarrow}{\text{O}} \] ✅ Therefore, \( \text{CO}_2 \) is a nonpolar molecule.
• 2. Water (\( \text{H}_2\text{O} \)):
  
  • Lewis Structure: Oxygen is the central atom, single-bonded to two hydrogen atoms, and has two lone pairs of electrons.
    \[ \centerdot \centerdot \text{O} \\ \diagup \quad \diagdown \\ \text{H} \quad \text{H} \]
  • Molecular Geometry: With two bonding groups and two lone pairs on the central oxygen, \( \text{H}_2\text{O} \) adopts a bent (or V-shaped) molecular geometry. The lone pairs push the hydrogen atoms closer together.
  • Bond Polarity: Oxygen is more electronegative than hydrogen, so each \( \text{O}-\text{H} \) bond is polar. The electron density is pulled towards the oxygen atom.
  • Overall Molecular Polarity: Due to the bent geometry, the two individual bond dipoles (pulling electrons towards the central oxygen) do not cancel out. Instead, they add up to create a net dipole moment for the entire molecule, with the oxygen side being partially negative and the hydrogen sides being partially positive. The lone pairs also contribute to this uneven charge distribution. \[ \quad \quad \quad \text{H} \\ \quad \quad \quad \diagdown \\ \stackrel{\longleftarrow \longleftarrow}{\text{O}} \\ \quad \quad \quad \diagup \\ \quad \quad \quad \text{H} \] ✅ Therefore, \( \text{H}_2\text{O} \) is a polar molecule.
• Conclusion: The key difference lies in their molecular geometries. \( \text{CO}_2 \)'s linear shape allows its polar bonds to cancel, making it nonpolar. \( \text{H}_2\text{O} \)'s bent shape prevents its polar bonds from canceling, making it polar.

Let's analyze each compound:

Compound 1: \( \text{NaBr} \)

• 1. Identify Bond Type:
  Sodium (Na) is an alkali metal (Group 1). Bromine (Br) is a halogen (nonmetal, Group 17).
  📌 When a metal bonds with a nonmetal, electrons are typically transferred, forming ions. This results in an ionic bond. ✅ Therefore, \( \text{NaBr} \) is an ionic compound.
• 2. Name the Compound:
  👉 Rules for naming ionic compounds:
  • The cation (metal ion) is named first, using its element name: Sodium.
  • The anion (nonmetal ion) is named second, by taking the root of the element name and adding the suffix "-ide": Bromine becomes Bromide.
  • Since Sodium typically forms only a \( 1+ \) ion, Roman numerals are not needed.
  ✅ The name of \( \text{NaBr} \) is Sodium bromide.

Compound 2: \( \text{SF}_6 \)

• 1. Identify Bond Type:
  Sulfur (S) is a nonmetal (Group 16). Fluorine (F) is also a nonmetal (Group 17).
  📌 When two nonmetals bond, they share electrons to achieve stability, forming covalent bonds. ✅ Therefore, \( \text{SF}_6 \) is a covalent (molecular) compound.
• 2. Name the Compound:
  👉 Rules for naming covalent compounds:
  • Use prefixes to indicate the number of atoms of each element.
  • The first element in the formula is named first. If there is only one atom of the first element, the "mono-" prefix is usually omitted.
  • The second element is named with a prefix indicating the number of atoms, and its name ends with "-ide".
  • Prefixes: mono- (1), di- (2), tri- (3), tetra- (4), penta- (5), hexa- (6), etc.
  👉 Apply Prefixes:
  • For Sulfur (S): There is one sulfur atom, so it's simply "Sulfur". (No "monosulfur").
  • For Fluorine (\( \text{F}_6 \)): "hexa-" for six fluorine atoms, so "hexafluoride".
  ✅ The name of \( \text{SF}_6 \) is Sulfur hexafluoride.

The immiscibility (inability to mix) of oil and water is a classic example of how molecular polarity dictates solubility, following the principle of "like dissolves like."

• 1. Polarity of Water:
  Water (\( \text{H}_2\text{O} \)) is a highly polar molecule. Its bent molecular geometry and the significant electronegativity difference between oxygen and hydrogen create a permanent positive end and a permanent negative end within each water molecule. These strong partial charges allow water molecules to form strong hydrogen bonds and dipole-dipole attractions with each other.
• 2. Polarity of Oil:
  Most oils (like vegetable oil or petroleum-based oils) are composed primarily of long chains of carbon and hydrogen atoms (hydrocarbons). The electronegativity difference between carbon and hydrogen is very small, making \( \text{C}-\text{H} \) bonds essentially nonpolar. The overall molecular structure of these long chains is symmetrical enough that any slight bond polarities cancel out.
  ✅ Therefore, oil molecules are generally nonpolar. The primary intermolecular forces between oil molecules are weaker London Dispersion Forces.
• 3. Why They Don't Mix ("Like Dissolves Like"):
  When oil and water are mixed, the water molecules are strongly attracted to each other (due to hydrogen bonding). The oil molecules are attracted to each other (due to London Dispersion Forces). However, there is very little attractive force between the highly polar water molecules and the nonpolar oil molecules.
  The strong attractions between water molecules would have to be broken to allow oil molecules to disperse, and the energy required to do this is not compensated by favorable interactions with the nonpolar oil molecules. It's energetically more favorable for water molecules to stay together and for oil molecules to stay together.
  👉 This leads to the water molecules "excluding" the oil molecules, causing them to separate into distinct layers, with the less dense oil typically floating on top of the water.

The boiling point of a substance is directly related to the strength of the intermolecular forces (IMFs) between its molecules. To boil a liquid, enough energy must be supplied to overcome these attractive forces and allow the molecules to escape into the gas phase.

• 1. Intermolecular Forces in Methane (\( \text{CH}_4 \)):
  
  • Lewis Structure & Geometry: Methane has a central carbon atom bonded to four hydrogen atoms with no lone pairs. It has a tetrahedral geometry.
  • Bond Polarity: The electronegativity difference between carbon and hydrogen is small, making \( \text{C}-\text{H} \) bonds only slightly polar, almost nonpolar.
  • Overall Molecular Polarity: Due to its symmetrical tetrahedral shape, any slight bond dipoles cancel out, making methane a nonpolar molecule.
  • Primary IMF: The only intermolecular forces present between methane molecules are weak London Dispersion Forces (LDFs). These forces arise from temporary, induced dipoles due to the constant movement of electrons.
• 2. Intermolecular Forces in Water (\( \text{H}_2\text{O} \)):
  
  • Lewis Structure & Geometry: Water has a central oxygen atom bonded to two hydrogen atoms and has two lone pairs. It has a bent molecular geometry.
  • Bond Polarity: Oxygen is significantly more electronegative than hydrogen, making the \( \text{O}-\text{H} \) bonds highly polar.
  • Overall Molecular Polarity: Due to its bent shape, the bond dipoles do not cancel out, making water a highly polar molecule.
  • Primary IMF: Because hydrogen is directly bonded to the highly electronegative oxygen atom, water molecules can form strong hydrogen bonds with each other. In addition to hydrogen bonds, water also experiences dipole-dipole forces and London Dispersion Forces. However, hydrogen bonding is by far the strongest of these.
• 3. Conclusion: The Role of IMF Strength:
  Hydrogen bonds are significantly stronger than London Dispersion Forces.
  • To boil methane, only the relatively weak London Dispersion Forces need to be overcome. This requires very little energy, hence its very low boiling point (\( -161.5 \text{%C} \)).
  • To boil water, the much stronger hydrogen bonds (along with dipole-dipole and LDFs) must be overcome. This requires a substantial amount of energy, resulting in its relatively high boiling point (\( 100 \text{%C} \)).
  ✅ The presence of strong hydrogen bonding in water is the primary reason for its unusually high boiling point compared to other molecules of similar size.