In the compound sodium hydrogen carbonate NaHCO3, what is the oxidation number of the carbon atom in the bicarbonate ion?

Introduction / Context:
Oxidation numbers are bookkeeping numbers used in chemistry to keep track of electron transfer in redox reactions and to understand the distribution of electrons within compounds. Sodium hydrogen carbonate, NaHCO3, also called baking soda, is a common compound that appears frequently in examples and exam questions. This problem asks for the oxidation number of carbon in NaHCO3, which requires careful application of standard oxidation number rules to each element in the formula.

Given Data / Assumptions:

• The compound is NaHCO3, written as sodium hydrogen carbonate.
• Sodium is a group 1 metal and usually has oxidation number +1 in its compounds.
• Hydrogen, when bonded to non metals, usually has oxidation number +1.
• Oxygen in most covalent compounds has oxidation number -2.
• The overall charge on NaHCO3 is zero because it is a neutral compound.

Concept / Approach:
To find the oxidation number of carbon, we write an equation based on the sum of oxidation numbers in a neutral compound being zero. Assign the usual oxidation numbers to sodium, hydrogen and oxygen, then let x be the oxidation number of carbon. The bicarbonate ion HCO3 minus carries a net charge of minus one, but we can work directly with the whole formula NaHCO3 by remembering that sodium contributes plus one and the total must be zero. Solving the algebraic equation gives the oxidation number of carbon, which turns out to be +4 in this compound.

Step-by-Step Solution:
Step 1: Write down the oxidation numbers that are known. Sodium Na in group 1 has oxidation number +1. Hydrogen H bonded to a non metal typically has +1. Oxygen O in oxides and most oxyanions has -2. Step 2: Let x represent the oxidation number of carbon C in NaHCO3. Step 3: Set up the sum of oxidation numbers for all atoms in one formula unit. There is 1 Na, 1 H, 1 C and 3 O atoms. The total must equal 0 for a neutral compound. Step 4: Write the equation: (+1 for Na) + (+1 for H) + (x for C) + 3 * (-2 for O) = 0. Step 5: Simplify the equation: 1 + 1 + x - 6 = 0 which becomes x - 4 = 0. Step 6: Solve for x to obtain x = +4. Therefore, the oxidation number of carbon in NaHCO3 is +4.

Verification / Alternative check:
Another way is to think of NaHCO3 as Na+ and HCO3 minus. Sodium is clearly +1, so the bicarbonate ion must have a total charge of minus one. In HCO3 minus, there is 1 H with +1, 3 O with -2 each and carbon with oxidation number x. The sum must equal -1, so 1 + x + 3 * (-2) = -1. This simplifies to 1 + x - 6 = -1, giving x - 5 = -1 and x = +4 again. Both approaches agree, confirming that carbon is in the +4 oxidation state in bicarbonate. This result is consistent with many other carbon oxygen compounds such as carbon dioxide, CO2, where carbon also has oxidation number +4.

Why Other Options Are Wrong:
- -4: This would require carbon to have gained so many electrons that the overall charge balance in NaHCO3 would not come out to zero, contradicting the neutrality of the compound.
- -3: A negative oxidation state for carbon is seen in some carbides, not in bicarbonate with multiple oxygens.
- +2: This would give an incorrect sum of oxidation numbers and would not match the known charge on the bicarbonate ion.
- 0: A zero oxidation number for carbon occurs in elemental carbon or in some symmetrical molecules, not in a strongly oxidised oxyanion like bicarbonate.

Common Pitfalls:
Students often forget to multiply the oxidation number of oxygen by three when there are three oxygen atoms, or they accidentally treat the entire HCO3 group as neutral when it actually has a minus one charge. Another frequent error is to mix up the rules for hydrogen in metal hydrides, where H is negative, with its behaviour in compounds with non metals, where H is positive. Carefully writing the oxidation number equation and double checking the arithmetic helps avoid these mistakes in exam situations.

Final Answer:
The oxidation number of carbon in sodium hydrogen carbonate NaHCO3 is +4.