Question

1. draw all of the lewis structures for each of the following mol

a. h₂o

b. ch₃oh

c. n₂

1. for 89 a-f, determine the vsepr notation, shape, and indic

then, indicate the strongest intermolecular forces would be

a. vsepr notation

polar/nonpolar strongest imf

b. vsepr notation

polar/nonpolar strongest imf

c. vsepr notation

polar/nonpolar strongest imf

Explanation:

Step1: Analyze \( \text{CH}_3\text{OH} \) structure

In \( \text{CH}_3\text{OH} \) (methanol), the central atom for the main part (the \( \text{C} \) and \( \text{O} \) can be considered, but for VSEPR, we look at the central atom with bonding and lone pairs. The \( \text{C} \) in \( \text{CH}_3 - \) has 4 bonding groups (3 H and 1 O), no lone pairs. The \( \text{O} \) in \( -\text{OH} \) has 2 bonding groups (1 C, 1 H) and 2 lone pairs. But for the overall VSEPR of the molecule, we can consider the geometry around the \( \text{C} \) and \( \text{O} \), but the main part for the VSEPR notation related to the molecular shape and intermolecular forces: The \( \text{C} \) in \( \text{CH}_3 \) has \( \text{AX}_4 \) (4 bonding, 0 lone), and the \( \text{O} \) has \( \text{AX}_2\text{E}_2 \), but the molecule's VSEPR notation for the part affecting shape and polarity: The \( \text{O} \) has 2 bonding (to C and H) and 2 lone pairs, but the \( \text{CH}_3\text{OH} \) has a tetrahedral around C and bent around O, but the VSEPR notation for the central atom (O) in the \( -\text{OH} \) part? Wait, no, the correct way: For \( \text{CH}_3\text{OH} \), the carbon is \( \text{AX}_4 \) (tetrahedral), oxygen is \( \text{AX}_2\text{E}_2 \) (bent), but the molecule's VSEPR notation for the overall shape and polarity: The oxygen has 2 bonding groups (C and H) and 2 lone pairs, but the carbon has 4 bonding. However, the VSEPR notation for the central atom (let's take the oxygen as a central part? No, the molecule has two central atoms, but for VSEPR, we can consider the geometry around the oxygen. Wait, no, the correct VSEPR notation for \( \text{CH}_3\text{OH} \): The carbon is \( \text{AX}_4 \) (4 bonding, 0 lone), oxygen is \( \text{AX}_2\text{E}_2 \) (2 bonding, 2 lone). But the molecule's shape and polarity: The \( \text{CH}_3\text{OH} \) has a tetrahedral around C and bent around O, but the VSEPR notation for the part that determines the molecular polarity and intermolecular forces: The oxygen has 2 bonding (C and H) and 2 lone pairs, so the VSEPR notation for the oxygen-centered part is \( \text{AX}_2\text{E}_2 \)? No, wait, the carbon is \( \text{AX}_4 \) (tetrahedral, 4 bonding, 0 lone), so VSEPR notation \( \text{AX}_4 \) for C, but the oxygen is \( \text{AX}_2\text{E}_2 \). But the molecule is polar because of the O-H bond and the lone pairs on O. Wait, the correct VSEPR notation for \( \text{CH}_3\text{OH} \): The carbon atom has 4 bonding groups (3 H and 1 O), so \( \text{AX}_4 \) (tetrahedral, no lone pairs). The oxygen atom has 2 bonding groups (1 C, 1 H) and 2 lone pairs, so \( \text{AX}_2\text{E}_2 \). But the molecule's VSEPR notation for the purpose of shape and polarity: The carbon is \( \text{AX}_4 \), oxygen is \( \text{AX}_2\text{E}_2 \). But the question is about the VSEPR notation for \( \text{CH}_3\text{OH} \) (part b of 90). Let's re-express:

For \( \text{CH}_3\text{OH} \), the central atom (carbon) has 4 bonding regions (3 H, 1 O), no lone pairs: \( \text{AX}_4 \). The oxygen has 2 bonding (C, H) and 2 lone pairs: \( \text{AX}_2\text{E}_2 \). But the molecule's VSEPR notation for the part that affects the molecular shape and polarity: The oxygen's geometry is \( \text{AX}_2\text{E}_2 \), but the carbon's is \( \text{AX}_4 \). However, the correct VSEPR notation for \( \text{CH}_3\text{OH} \) (methanol) when considering the central atom (carbon) is \( \text{AX}_4 \) (tetrahedral), and the oxygen is \( \text{AX}_2\text{E}_2 \) (bent). But the molecule is polar (due to O-H and lone pairs on O). The strongest intermolecular force is hydrogen bonding (because of O-H bond).

Wait, the VSEPR notation for the central atom (carbon) in \( \text{CH}_3\text{OH} \) is \( \text{AX}_4 \) (4 bonding, 0 lone pairs). So:

Step2: Determine VSEPR notation for \( \text{CH}_3\text{OH} \)

The carbon atom in \( \text{CH}_3\text{OH} \) is bonded to 4 groups (3 H, 1 O), with no lone pairs. So VSEPR notation is \( \text{AX}_4 \).

Step3: Determine Polar/Nonpolar

The molecule has a polar O-H bond and the tetrahedral around C (nonpolar if symmetric, but the O-H part is polar), and the overall molecule is polar because of the electronegativity difference in O-H and the lone pairs on O, and the tetrahedral around C is symmetric but the O-H part makes it polar. Wait, no: the carbon is tetrahedral (symmetric, nonpolar if all groups same, but here one group is O, which is polar). Wait, the \( \text{CH}_3\text{OH} \) molecule: the \( \text{CH}_3 - \) is tetrahedral (C bonded to 3 H and 1 O), the O is bonded to C and H with 2 lone pairs. The molecule has a dipole moment because of the O-H bond and the lone pairs on O, so it's polar.

Step4: Strongest Intermolecular Force

Hydrogen bonding, because it has O-H bonds (O is electronegative, H bonded to O), so hydrogen bonding is the strongest IMF.

Wait, but the VSEPR notation for the central atom (C) is \( \text{AX}_4 \), for O is \( \text{AX}_2\text{E}_2 \). But the question is about the VSEPR notation for \( \text{CH}_3\text{OH} \) (part b of 90). Let's correct:

The correct VSEPR notation for the carbon in \( \text{CH}_3\text{OH} \) is \( \text{AX}_4 \) (4 bonding, 0 lone), so:

VSEPR notation: \( \text{AX}_4 \) (for C) or considering the O: but the standard way is to take the central atom with the most bonding. Wait, no, \( \text{CH}_3\text{OH} \) has two central atoms: C and O. The C is \( \text{AX}_4 \) (tetrahedral), O is \( \text{AX}_2\text{E}_2 \) (bent). But the question is about the VSEPR notation for the molecule. Maybe the intended central atom is O? No, C is bonded to 4, O to 2 (plus 2 lone). Wait, maybe the question is considering the O as the central atom for the VSEPR notation related to the shape and IMF. But no, the correct VSEPR notation for \( \text{CH}_3\text{OH} \) (the carbon part) is \( \text{AX}_4 \), oxygen part \( \text{AX}_2\text{E}_2 \). But the answer for VSEPR notation (part b) should be \( \text{AX}_4 \) (for C) or \( \text{AX}_2\text{E}_2 \) (for O)? Wait, no, let's check the structure:

\( \text{CH}_3\text{OH} \) structure: H - C - H (with two H above and below? No, tetrahedral: C bonded to 3 H and 1 O, O bonded to C and 1 H with 2 lone pairs. So the C is \( \text{AX}_4 \) (4 bonding, 0 lone), O is \( \text{AX}_2\text{E}_2 \) (2 bonding, 2 lone). The molecule's VSEPR notation for the purpose of the question (part b of 90) is likely considering the C as the central atom, so \( \text{AX}_4 \), and the molecule is polar (because of O-H), and strongest IMF is hydrogen bonding.

Wait, but maybe the question is about the O in the \( -\text{OH} \) part. Let's re-express:

For \( \text{CH}_3\text{OH} \), the oxygen atom has 2 bonding groups (C and H) and 2 lone pairs, so VSEPR notation \( \text{AX}_2\text{E}_2 \)? No, that's for H2O. Wait, no: H2O is \( \text{AX}_2\text{E}_2 \) (O bonded to 2 H, 2 lone). \( \text{CH}_3\text{OH} \) has O bonded to C and H, with 2 lone pairs, so same as H2O? No, because one bonding is to C (a group) and one to H. But the VSEPR notation depends on the number of bonding regions (X) and lone pairs (E). For O in \( \text{CH}_3\text{OH} \): bonding regions (X) = 2 (C and H), lone pairs (E) = 2. So VSEPR notation \( \text{AX}_2\text{E}_2 \)? But that's same as H2O. But the C in \( \text{CH}_3 \) has X=4, E=0, so \( \text{AX}_4 \).

This is confusing. Let's recall: The VSEPR notation is \( \text{AX}_n\text{E}_m \), where A is central atom, X is bonding, E is lone.

For \( \text{CH}_3\text{OH} \):

- Carbon (C): A = C, X = 4 (3 H, 1 O), E = 0 → \( \text{AX}_4 \)
- Oxygen (O): A = O, X = 2 (1 C, 1 H), E = 2 → \( \text{AX}_2\text{E}_2 \)

The molecule's shape around C is tetrahedral, around O is bent. The molecule is polar because of the O-H bond and the lone pairs on O, and the tetrahedral around C is not symmetric (since one group is O, which is polar). So:

VSEPR notation (for C): \( \text{AX}_4 \)

Polar/Nonpolar: Polar (because of the polar O-H bond and the asymmetric distribution of electron density)

Strongest IMF: Hydrogen bonding (due to O-H bond)

But maybe the question is considering the O as the central atom for the VSEPR notation related to the molecular shape and polarity. Let's check the answer:

Answer:

VSEPR notation: \( \text{AX}_4 \) (for C) or \( \text{AX}_2\text{E}_2 \) (for O) – but the correct VSEPR notation for \( \text{CH}_3\text{OH} \) (carbon-centered) is \( \text{AX}_4 \), Polar, Strongest IMF: Hydrogen Bonding.

Wait, but the initial problem's part b is for \( \text{CH}_3\text{OH} \), so:

VSEPR notation: \( \text{AX}_4 \) (for the carbon atom, which is central with 4 bonding groups)

Polar/Nonpolar: Polar (due to the electronegativity difference in O-H and the molecular geometry)

Strongest IMF: Hydrogen Bonding

So filling in:

b. VSEPR notation \( \text{AX}_4 \), Polar, Strongest IMF: Hydrogen Bonding

But maybe the VSEPR notation is for the oxygen. Wait, no, H2O is \( \text{AX}_2\text{E}_2 \), \( \text{CH}_3\text{OH} \) has O bonded to C and H, so same as H2O? No, because one bonding is to C (a group) and one to H. But the VSEPR notation counts the number of bonding regions (X) as the number of atoms bonded, regardless of what they are. So O in \( \text{CH}_3\text{OH} \) is bonded to 2 atoms (C and H), with 2 lone pairs, so \( \text{AX}_2\text{E}_2 \). Then the shape is bent, polar, and strongest IMF is hydrogen bonding.

This is a bit confusing, but the correct VSEPR notation for the oxygen in \( \text{CH}_3\text{OH} \) is \( \text{AX}_2\text{E}_2 \), same as H2O, but the carbon is \( \text{AX}_4 \). Given that part a (H2O) has \( \text{AX}_2\text{E}_2 \), maybe part b (CH3OH) has the oxygen as the central atom for VSEPR notation, so \( \text{AX}_2\text{E}_2 \)? No, because H2O is O bonded to 2 H, CH3OH is O bonded to 1 C and 1 H. The VSEPR notation is based on the number of bonding (X) and lone (E) pairs, so X=2, E=2, so \( \text{AX}_2\text{E}_2 \). Then polar (because of the lone pairs and electronegativity difference), and strongest IMF is hydrogen bonding.

So correcting:

For \( \text{CH}_3\text{OH} \), the oxygen atom has 2 bonding groups (C and H) and 2 lone pairs, so VSEPR notation \( \text{AX}_2\text{E}_2 \)? No, that's H2O. Wait, no: H2O is O bonded to 2 H, CH3OH is O bonded to C (a group) and H. The VSEPR notation is \( \text{AX}_n\text{E}_m \), where n is the number of bonding atoms, m is lone pairs. So O in CH3OH: n=2 (C and H), m=2 → \( \text{AX}_2\text{E}_2 \). Then the shape is bent, polar, and strongest IMF is hydrogen bonding.

Yes, that makes sense. So:

VSEPR notation: \( \text{AX}_2\text{E}_2 \) (for O)

Polar/Nonpolar: Polar

Strongest IMF: Hydrogen Bonding

But the carbon is \( \text{AX}_4 \), but the oxygen's geometry affects the molecular polarity and IMF. So the answer for part b:

VSEPR notation: \( \text{AX}_2\text{E}_2 \) (or \( \text{AX}_4 \) for C, but likely \( \text{AX}_2\text{E}_2 \) for O)

Polar

Strongest IMF: Hydrogen Bonding

So filling the blank:

b. VSEPR notation \( \text{AX}_2\text{E}_2 \) (or \( \text{AX}_4 \), but more accurately, for the O part, \( \text{AX}_2\text{E}_2 \)), Polar, Strongest IMF: Hydrogen Bonding