GATE 2024 BT Q.33: solution and answer

GATE 2024 · BTQ.33 · 1 mark · NAT

The degree of reduction (reductance) for oxalic acid ( $C_{2} H_{2} O_{4}$ ) is ________.

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Answer and solution

ANSWER

The degree of reduction of oxalic acid is $γ = 1$ (available electrons per carbon atom).

Official answer: 1Source: Official answer key, GATE 2024 BT (IISc Bengaluru)

The degree of reduction $γ$ is the number of available electrons per carbon atom (a C-mol basis). Each element is given its valence: C $= + 4$ , H $= + 1$ , O $= - 2$ , and N $= - 3$ (nitrogen taken at the oxidation level of ammonia).
For a compound $C_{w} H_{x} O_{y} N_{z}$ : $γ = \frac{4 w + x - 2 y - 3 z}{w}$
Oxalic acid $C_{2} H_{2} O_{4}$ has $w = 2$ , $x = 2$ , $y = 4$ , $z = 0$ .
Available electrons per molecule: $4 (2) + 1 (2) - 2 (4) = 8 + 2 - 8 = 2$ .
Divide by the number of carbon atoms to put it on a C-mol basis: $γ = \frac{2}{2} = 1$
Check with the C-mol formula $C H O_{2}$ directly: $γ = 4 + 1 - 2 (2) = 1$ . This fits the picture of oxalic acid as a highly oxidised compound: $C O_{2}$ has $γ = 0$ , and glucose has $γ = 4$ .

KEY IDEA

Degree of reduction is counted per carbon atom (per C-mol), not per molecule. Add up the valences C $= 4$ , H $= 1$ , O $= - 2$ , N $= - 3$ and divide by the number of carbons: $γ = (4 w + x - 2 y - 3 z) / w$ .

WATCH OUT FOR

Stopping at the per-molecule count of 2 available electrons and forgetting to divide by the two carbon atoms. Using $+ 2$ for oxygen instead of $- 2$ is another slip, and it gives $(8 + 2 + 8) /2 = 9$ .

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GATE 2024 BT Q.33: solution and answer

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