Question

select the correct answer.
in the covalently bonded chemical compound acetate ch₃coo⁻, the two carbon atoms have different oxidation states. what are its two oxidation states?
diagram of acetate structure: a bracket with two carbon atoms, one bonded to three h and one c, the other bonded to c, o, and o⁻
a. +1 and -2
b. +3 and -4
c. +3 and -3
d. +1 and -4

Explanation:

Step1: Analyze the first carbon (in $-CH_3$)

In the methyl group ($-CH_3$), each H has an oxidation state of +1. Let the oxidation state of C be $x$. The sum of oxidation states in a neutral group (here, the methyl group is part of the ion, but we consider the group's charge contribution). For $-CH_3$: $x + 3(+1)=0$ (since the methyl group is attached to the carbonyl carbon, and the overall charge of the ion is -1, but for the methyl C, we calculate its oxidation state considering its bonds). Wait, actually, in the acetate ion $[CH_3COO]^-$, let's label the two carbons: $C_1$ (in $CH_3$) and $C_2$ (in $COO$).

For $C_1$ (bonded to 3 H and 1 C): The oxidation state of H is +1, C - C bond is between two C atoms (same element, so oxidation state contribution 0). So for $C_1$: $x + 3(+1)+0 = 0$ (since the $CH_3$ group is neutral in terms of its internal charge before considering the rest? Wait, no, the overall ion is $-1$. Let's use the rule: oxidation state of O is -2 (in $COO$ part, except maybe in peroxides, but here it's a carboxylate), H is +1.

For $C_2$ (in $COO$): bonded to 1 C, 2 O (one double bond, one single bond). Let oxidation state of $C_2$ be $y$. The single-bonded O has -2, double-bonded O also has -2 (since in a double bond, each O contributes -2? Wait, oxidation state for O in most cases is -2. The charge of the ion is -1. Let's calculate the oxidation state of each C.

First, $C_1$ (methyl C): bonded to 3 H (+1 each) and 1 C (oxidation state 0 for the C - C bond). So the oxidation state of $C_1$: $x + 3(+1) + 0 = 0$ (since the $CH_3$ group is attached to $C_2$, and we can consider the oxidation state of $C_1$ by its bonds. Wait, actually, the formula is: sum of oxidation states in a compound/ion equals its charge. For the acetate ion $[CH_3COO]^-$, charge is -1.

Let's denote:

- Oxidation state of $C_1$ (CH₃ - C) as $x$
- Oxidation state of $C_2$ (C - COO⁻) as $y$
- H: +1 (3 H atoms)
- O: -2 (2 O atoms in COO⁻, one with a single bond, one with a double bond, but oxidation state of O is -2 in both cases)
- The ion has a charge of -1.

So the sum of oxidation states:

For $C_1$: bonded to 3 H (+1) and 1 C (y). So $x + 3(+1) + y =$? Wait, no, the entire ion:

Atoms: 2 C, 3 H, 2 O, charge -1.

Sum of oxidation states: $2x' + 3(+1) + 2(-2) = -1$, where $x'$ is the average, but we need individual. Wait, no, the two C atoms have different oxidation states, so we need to calculate each.

For $C_1$ (CH₃ - C):

Bonds: 3 H (each +1) and 1 C (C - C bond, so oxidation state contribution 0 from the C - C bond, since same element). So the oxidation state of $C_1$: $x + 3(+1) + 0 = 0$ (because the CH₃ group is a neutral fragment? No, better way: in organic compounds, for a C atom, oxidation state can be calculated by considering the electronegativity of bonded atoms. H is less electronegative than C, so C gets -1 for each H? Wait, no, standard rule: oxidation state of H is +1 (except in hydrides), O is -2 (except in peroxides, superoxides), F is -1, etc. For a C atom bonded to H and C:

- Each H bonded to C: C has -1 (since H is +1, so to balance, C is -1 per H? Wait, no, let's use the formula for oxidation state:

Oxidation state of an atom = (number of bonds to more electronegative atoms) - (number of bonds to less electronegative atoms)

For $C_1$ (CH₃ - C):

- Bonds to H (3): H is less electronegative than C, so each H gives C -1 (since C is more electronegative, so when C bonds to H, C "gains" an electron, so oxidation state -1 per H? Wait, no, oxidation state is the charge an atom would have if all bonds are ionic. So for a C - H bond, since H is less electronegative, the electron pair is closer to C, so C has -1, H has +1. For a C - C bond, since same electronegativity, each C has 0 from the C - C bond.

So for $C_1$: 3 C - H bonds (each gives C -1) and 1 C - C bond (gives 0). So oxidation state of $C_1$: $3(-1) + 0 = -3$? Wait, that can't be. Wait, no, let's use the standard method for oxidation state in ions.

The acetate ion is $CH_3COO^-$. Let's calculate oxidation state of each C:

First C (C1: CH₃):

- Bonds: 3 H (+1 each) and 1 C (C2). Let oxidation state of C1 be x.

Sum of oxidation states for CH₃ group: x + 3(+1) + oxidation state of C2 (but no, the CH₃ is attached to C2, so the entire ion's charge is -1. Let's consider the entire ion:

Atoms: C1, C2, 3 H, 2 O.

Charge: -1.

Oxidation state of H: +1 (3 H: 3*+1 = +3)

Oxidation state of O: -2 (2 O: 2*(-2) = -4)

Let oxidation state of C1 be x, C2 be y.

So x + y + 3 + (-4) = -1

x + y -1 = -1

x + y = 0 --> equation (1)

Now, for C1 (CH₃):

It is bonded to 3 H and 1 C (C2). The C - C bond: since same element, oxidation state contribution 0. So for C1:

x + 3(+1) + 0 (from C - C) = charge of CH₃ group. But CH₃ is part of the ion, so the charge of CH₃ group is neutral? No, the entire ion is -1, so the sum of all oxidation states is -1.

Wait, another approach: in the carboxylate group (COO⁻), the C (C2) is bonded to 2 O (one double, one single) and 1 C (C1). Let's calculate y (oxidation state of C2):

C2 is bonded to 1 C (C1, oxidation state x), 2 O (each -2). So the oxidation state of C2: y + x + 2*(-2) = charge of COO⁻ group. But the COO⁻ group has a charge of -1 (since the entire ion is CH₃COO⁻, so COO⁻ is -1). Wait, no, the entire ion is -1, so CH₃ (neutral) + COO⁻ (-1) = -1. So CH₃ is neutral, so oxidation state of C1 in CH₃:

In CH₃ (neutral), 3 H (+1 each) and 1 C (C1). So x + 3(+1) = 0 --> x = -3. Wait, that's different. Wait, CH₃ is a methyl group, which is neutral, so x + 3(+1) = 0 --> x = -3. Then from equation (1): x + y = 0 --> -3 + y = 0 --> y = +3. So C1 has oxidation state -3, C2 has +3. Wait, but the options are +3 and -3? Wait, option C is +3 and -3. Wait, let's check again.

Wait, maybe I made a mistake in the CH₃ group. Let's use the oxidation state rules for organic compounds. For a carbon atom:

- Each bond to H: C has -1 (since H is +1, so C is -1 per H)
- Each bond to C: 0 (same element)
- Each bond to O: +1 (since O is more electronegative, C loses an electron, so +1 per O bond)
- Each bond to N: +1 (N is more electronegative)
- Each bond to X (halogen): +1 (X is more electronegative)

So for C1 (CH₃ - C):

- Bonds: 3 H (each -1 for C) and 1 C (0). So oxidation state: 3*(-1) + 0 = -3.

For C2 (C - COO⁻):

- Bonds: 1 C (0), 1 double bond to O (2 bonds, each +1, so 2*+1 = +2), 1 single bond to O (+1). So total: 0 + 2 + 1 = +3.

So the two oxidation states are -3 and +3, which is option C.

Step2: Verify with options

The calculated oxidation states are -3 (for the methyl C) and +3 (for the carbonyl C). Looking at the options:

A. +1 and -2 → No

B. +3 and -4 → No

C. +3 and -3 → Yes

D. +1 and -4 → No

Answer:

C. +3 and -3