What are the oxidation numbers on S atoms in $S_{4} O_{6}^{2 -}$

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$2.5, 2.5, 2.5, 2.5$

$7, - 2, - 2, 7$

$5, 0, 0, 5$

$6, - 1, - 1, 6$

answer is B.

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Detailed Solution

Oxidation Numbers of Sulfur Atoms in Tetrathionate Ion

Answer: The oxidation numbers of sulfur (S) atoms in the tetrathionate ion S₄O₆²⁻ are +5 and 0.

Detailed Explanation:

Structure of Tetrathionate Ion:

The tetrathionate ion S₄O₆²⁻ consists of a chain of four sulfur atoms with oxygen atoms bonded to the terminal sulfur atoms.

Terminal Sulfur Atoms: The two terminal sulfur atoms are each bonded to three oxygen atoms (similar to sulfate groups). Each terminal sulfur has an oxidation number of +5.
Central Sulfur Atoms: The two central (middle) sulfur atoms are bonded only to other sulfur atoms through S-S bonds. Each central sulfur has an oxidation number of 0.
Distribution: 2 sulfur atoms at oxidation state +5 and 2 sulfur atoms at oxidation state 0.

Verification:

Total oxidation numbers from sulfur atoms:

2 × (+5) + 2 × (0) = +10

Contribution from oxygen atoms: 6 × (−2) = −12

Net charge: +10 + (−12) = −2 ✓

This matches the charge on the tetrathionate ion.

Average Oxidation State:

While the average oxidation state of sulfur is +2.5, the actual oxidation states of individual sulfur atoms are +5 (terminal) and 0 (central) due to the unique structure of the ion.