Formula is better written as K^+[I3]^-; the underlined element is iodine in ce{I3^-.
Let oxidation number of each I in \(\ce{I3^-}\) be \(x\): \[ 3x = -1 \Rightarrow x = -\frac{1}{3} \]
Result: Oxidation number of I in \(\ce{I3^-}\) is \(-\dfrac{1}{3}\). This fractional value arises because \(\ce{I3^-}\) is a resonance-stabilised polyiodide ion with delocalised charge over three equivalent I atoms.
Underlined element: S.
Assign usual oxidation numbers: \[ 2(+1) + 4(x) + 6(-2) = 0 \] \[ 2 + 4x - 12 = 0 \Rightarrow 4x - 10 = 0 \Rightarrow x = +\frac{10}{4} = +2.5 \]
Average oxidation state of S in \(\ce{H2S4O6}\) is \(+2.5\).
Structurally, tetrathionic acid contains sulphur atoms in different environments (some at higher, some at lower oxidation states), and their average comes out as +2.5.
Underlined element: Fe.
\(\ce{Fe3O4}\) is a mixed oxide often written as \(\ce{FeO·Fe2O3}\), containing both Fe(II) and Fe(III).
Let the average oxidation state of Fe be \(x\): \[ 3x + 4(-2) = 0 \Rightarrow 3x - 8 = 0 \Rightarrow x = +\frac{8}{3} \]
Result: Average oxidation number of Fe is \(+\dfrac{8}{3}\), rationalised as 1 Fe in +2 state and 2 Fe in +3 state: \[ \frac{(+2) + 2(+3)}{3} = \frac{8}{3}. \]
There are two different carbons; usually the question underlines one specific C. Give oxidation numbers of both C atoms for clarity.
For C₁ in CH₃– (bonded to 3 H and 1 C): Count using whole molecule approach: Total for molecule: \[ x_1 + x_2 + 6(+1) + (-2) = 0 \Rightarrow x_1 + x_2 + 4 = 0 \]
For C₂ in –CH₂OH (one C, two H, one O, one H): Oxidation number around this carbon: \[ x_2 + 2(+1) + (-2) + (+1) = 0 \Rightarrow x_2 + 1 = 0 \Rightarrow x_2 = -1 \] Put in total equation: \[ x_1 + (-1) + 4 = 0 \Rightarrow x_1 + 3 = 0 \Rightarrow x_1 = -3 \]
Result: In \(\ce{CH3CH2OH}\): C in \(\ce{CH3-}\): –3; C in \(\ce{-CH2OH}\): –1. The carbon bonded to more electronegative O is less reduced (–1) than the terminal methyl carbon (–3).
Again, two distinct carbons.
For C₂ in –COOH group:
Consider the –COOH part: \(\ce{CO2H}\). Let oxidation number of that carbon be \(x_2\): \[ x_2 + 2(-2) + (+1) = 0 \Rightarrow x_2 - 4 + 1 = 0 \Rightarrow x_2 = +3 \]
For C₁ in CH₃– (same logic as in alkane methyl group): Treat the bond to C₂ as to an atom of similar electronegativity, so: \[ x_1 + 3(+1) = 0 \Rightarrow x_1 = -3 \]
Result: In \(\ce{CH3COOH}\): methyl carbon (C₁) = –3, carboxyl carbon (C₂) = +3. This illustrates that within one molecule, different atoms of the same element can have very different oxidation states, and an average value would not reflect their actual redox character.