use the following lewis diagram for methyl acetate to answer the questions: remember that geometry refers to the geometry defined by the atoms, not the electron pairs. the geometry about atom 1 is. the ideal value of the c - c - o angle at atom 2 is degrees. the geometry about atom 3 is.

Question

Accepted Answer

### For the geometry about atom 1:
# Explanation:
## Step 1: Analyze atom 1 (C)
Atom 1 is a carbon atom bonded to four atoms (four single bonds: three H and one C). The electron - domain geometry and molecular geometry for a central atom with four bonding domains (and no lone pairs) is tetrahedral.
## Step 2: Determine the geometry
Based on the VSEPR (Valence - Shell Electron - Pair Repulsion) theory, when a central atom has four bonding groups and no lone pairs, the geometry is tetrahedral.

# Answer: tetrahedral

### For the ideal value of the C - C - O angle at atom 2:
# Explanation:
## Step 1: Analyze atom 2 (C)
Atom 2 is a carbon atom involved in a double bond (C = O) and two single bonds (C - C and C - O). The electron - domain geometry around atom 2 is trigonal planar (three bonding domains: one double bond and two single bonds, and no lone pairs).
## Step 2: Determine the bond angle
In a trigonal planar geometry, the ideal bond angle between the atoms is $120^{\circ}$. So the C - C - O angle at atom 2 (where the central atom is atom 2 with trigonal planar geometry) has an ideal value of $120$ degrees.

# Answer: 120

### For the geometry about atom 3 (O):
# Explanation:
## Step 1: Analyze atom 3 (O)
Atom 3 is an oxygen atom bonded to two atoms (C - O single bond and C - O single bond) and has two lone pairs of electrons. But according to the problem statement, geometry refers to the geometry defined by the atoms (not electron pairs). So we consider the bonding atoms. The oxygen atom is bonded to two carbon atoms. But if we consider the electron - domain geometry (including lone pairs), it is tetrahedral, but for the atomic geometry (based on bonded atoms), since it is bonded to two atoms and has two lone pairs, but the problem says geometry is defined by atoms. Wait, no, the problem says "geometry refers to the geometry defined by the atoms, not the electron pairs". Wait, atom 3 (O) is bonded to two C atoms? Wait, no, looking at the Lewis structure (methyl acetate: $CH_3 - CO - O - CH_3$). Atom 3 is the O in the $ - O - CH_3$ part? Wait, no, let's re - examine. In methyl acetate ($CH_3COOCH_3$), the structure is $CH_3 - C( = O)-O - CH_3$. So atom 3 is the O between the $C = O$ and $CH_3$. Atom 3 (O) is bonded to two C atoms (one single bond to the carbonyl C and one single bond to the methyl C) and has two lone pairs. But the problem says geometry is defined by the atoms. Wait, no, the problem statement: "Remember that geometry refers to the geometry defined by the atoms, not the electron pairs." So for atom 3 (O), the atoms bonded to it are two C atoms? No, wait, in the structure, atom 3 is O, bonded to C (atom 2) and C (the methyl C). Wait, no, the Lewis structure: atom 1 is $CH_3$ C, atom 2 is $C = O$ C, atom 3 is O bonded to atom 2 and $CH_3$ C. So atom 3 is bonded to two C atoms. But the electron - domain geometry (including lone pairs) is tetrahedral, but the atomic geometry (based on bonded atoms) - but if we consider the atoms bonded to O (atom 3), it is two C atoms. But actually, the correct approach: the central atom (atom 3, O) has two bonding domains (two single bonds to C atoms) and two lone pairs. But the problem says geometry is defined by the atoms. Wait, maybe I made a mistake. Wait, no, the problem says "geometry refers to the geometry defined by the atoms, not the electron pairs". So we look at the number of atoms bonded to the central atom. Atom 3 (O) is bonded to two C atoms? No, in methyl acetate, the O in the ester group ($ - O - $) is bonded to two C atoms (the carbonyl C and the methyl C). But the geometry defined by the atoms: if we consider the two C atoms and the O, but the O has two lone pairs. But the problem says "geometry defined by the atoms". Wait, maybe the intended structure: atom 3 is the O in the $ - OCH_3$ group. So atom 3 (O) is bonded to one C (from $ - CO - $) and one C (from $ - CH_3$) and has two lone pairs. But according to the problem's definition (geometry by atoms), we consider the bonded atoms. But actually, the correct geometry for the atoms around O (when considering the bonded atoms) - no, maybe the problem is considering the electron - domain geometry but with the note. Wait, no, let's recall VSEPR. For a central atom with four electron domains (two bonding, two lone pairs), the electron - domain geometry is tetrahedral, but the molecular geometry (geometry of atoms) is bent. But the problem says "geometry refers to the geometry defined by the atoms, not the electron pairs". Wait, that is a bit confusing. Wait, maybe the problem has a typo, or maybe I misinterpret. Wait, atom 3: in the structure, atom 3 is O, bonded to two C atoms? No, in $CH_3COOCH_3$, the O is bonded to one C (the carbonyl C) and one C (the methyl C), so two bonding atoms. But with two lone pairs. But the problem says geometry is defined by the atoms. So if we consider the two C atoms bonded to O, the geometry of the atoms (the arrangement of the two C atoms around O) - but that would be linear? No, that can't be. Wait, no, the correct approach: the central atom (O) has two bonding groups and two lone pairs. The electron - domain geometry is tetrahedral, but the molecular geometry (based on atoms) is bent. But the problem says "geometry refers to the geometry defined by the atoms, not the electron pairs". So maybe the problem is considering the electron - domain geometry but the note is a bit misleading. Wait, no, let's check the structure again. Atom 3 is O, bonded to C (atom 2) and C (the $CH_3$ C). So the two C atoms and the O: the angle between the two C - O bonds. But if we consider the electron - domain geometry (including lone pairs) is tetrahedral, but the atomic geometry (of the bonded atoms) - but the problem says "geometry defined by the atoms". So maybe the answer is tetrahedral? No, wait, no. Wait, the problem says "geometry refers to the geometry defined by the atoms, not the electron pairs". So for atom 3 (O), the atoms bonded to it are two C atoms. So the geometry of the atoms (the two C atoms around O) - but that would be a bent geometry? No, if we have two atoms bonded to a central atom with two lone pairs, the atomic geometry (of the bonded atoms) is bent. But according to the VSEPR, when you have two bonding groups and two lone pairs, the molecular geometry (of the atoms) is bent. But the problem says "geometry defined by the atoms". So the geometry about atom 3 (O) is bent? Wait, no, maybe I made a mistake in the structure. Let's re - draw the Lewis structure of methyl acetate: $CH_3 - C( = O)-O - CH_3$. Atom 1: $CH_3$ C (tetrahedral), atom 2: $C = O$ C (trigonal planar), atom 3: O between $C = O$ and $CH_3$. Atom 3 (O) has two single bonds (to C of $C = O$ and to C of $CH_3$) and two lone pairs. The electron - domain geometry is tetrahedral (four electron domains: two bonding, two lone pairs), but the molecular geometry (based on atoms) is bent. But the problem says "geometry refers to the geometry defined by the atoms, not the electron pairs". So we look at the atoms bonded to O: two C atoms. The arrangement of these two C atoms around O: since there are two lone pairs, the angle between the two C - O bonds is less than $109.5^{\circ}$, but the geometry of the atoms (the two C atoms) around O is bent. Wait, but maybe the problem considers the electron - domain geometry but with the note. No, the problem says "geometry refers to the geometry defined by the atoms, not the electron pairs". So the answer should be bent? Wait, no, maybe I messed up. Wait, another approach: in the structure, atom 3 is O, bonded to two C atoms. The Lewis structure for the O: it has two bonding pairs (C - O) and two lone pairs. But the problem says "geometry defined by the atoms", so we ignore the lone pairs. So the O is bonded to two C atoms, so the geometry (of the atoms) is linear? No, that's not right. Wait, no, the VSEPR theory: the electron - domain geometry is determined by both bonding and lone pairs, but the molecular geometry (atomic geometry) is determined by the positions of the atoms. For a central atom with two bonding domains and two lone pairs, the molecular geometry is bent. So the geometry about atom 3 (O) is bent. But maybe the problem has a different interpretation. Wait, maybe atom 3 is the O in the carbonyl group? No, atom 2 is the C in the carbonyl group, atom 3 is the other O. Wait, I think I made a mistake earlier. Let's re - assign: in $CH_3COOCH_3$, the structure is:

- Atom 1: $CH_3$ (C bonded to 3 H and 1 C)
- Atom 2: $C = O$ (C bonded to 1 C (atom 1), 1 O (double bond), and 1 O (atom 3))
- Atom 3: O bonded to atom 2 (C) and $CH_3$ (C)

So atom 3 (O) is bonded to two C atoms (atom 2 and the $CH_3$ C) and has two lone pairs. The electron - domain geometry is tetrahedral (four electron domains: two bonding, two lone pairs), but the molecular geometry (of the atoms) is bent. So the geometry about atom 3 is bent. But maybe the problem considers the electron - domain geometry but the note is wrong? No, the problem says "geometry refers to the geometry defined by the atoms, not the electron pairs". So we consider only the bonded atoms. So O is bonded to two C atoms, so the geometry (of the atoms) is linear? No, that's incorrect. Wait, I think the correct answer for the geometry about atom 3 is tetrahedral? No, no. Wait, maybe the problem is using a different definition. Alternatively, maybe atom 3 is the O in the $ - O - CH_3$ group, and when considering the atoms (the two C atoms it is bonded to), but no. I think the correct answer for the geometry about atom 3 is bent. But I'm a bit confused. Wait, let's check standard geometries:

- Tetrahedral: 4 bonding domains, 0 lone pairs (e.g., $CH_4$)
- Trigonal planar: 3 bonding domains, 0 lone pairs (e.g., $BF_3$)
- Bent: 2 bonding domains, 2 lone pairs (e.g., $H_2O$)
- Linear: 2 bonding domains, 0 lone pairs (e.g., $CO_2$)

So for atom 3 (O), which has 2 bonding domains (two C - O bonds) and 2 lone pairs, the molecular geometry (of the atoms) is bent. So the geometry about atom 3 is bent.

# Answer: bent

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QUESTION IMAGE

Question

Explanation:

For the geometry about atom 1:

Step 1: Analyze atom 1 (C)

Step 2: Determine the geometry

Step 1: Analyze atom 2 (C)

Step 2: Determine the bond angle

Step 1: Analyze atom 3 (O)

Answer:

For the ideal value of the C - C - O angle at atom 2: