Question

Which one of the the following compounds has the same oxidation state of all the S-atoms present in it?

• A. Na_{2}S_{2}O_{3}
• B. Na_{2}S_{2}O_{4}
• C. Na_{2}S_{3}O_{6}
• D. Na_{2}S_{5}O_{6}

Answer 1

Answer: (B)

Na_{2}S_{2}O_{4}

Answer 2

To determine which compound has all sulfur (S) atoms in the same oxidation state, we need to calculate the oxidation state of each S atom in the given compounds. We will use the following rules:

1. The oxidation state of Na is +1.
2. The oxidation state of O is -2 (except in peroxides, superoxides, etc., which are not the case here).
3. For a bond between two identical atoms (like S-S), the contribution to the oxidation state of each atom is 0.
4. For a bond between two different atoms, the electrons are assigned to the more electronegative atom.

Let's analyze each option:

A. Na₂S₂O₃ (Sodium Thiosulfate)

The thiosulfate ion (S₂O₃²⁻) has a structure analogous to the sulfate ion (SO₄²⁻) where one oxygen atom is replaced by a sulfur atom. The two sulfur atoms are not equivalent. One is a central sulfur (S_c) bonded to three oxygen atoms and one terminal sulfur atom, and the other is a terminal sulfur (S_t) bonded only to the central sulfur atom.

Let the oxidation state of S_c be $x$ and S_t be $y$.
The overall charge of Na₂S₂O₃ is 0.
$2(\text{Na}^{+}) + \text{S}\_c + \text{S}\_t + 3(\text{O}^{2-}) = 0$
$2(+1) + x + y + 3(-2) = 0$
$2 + x + y - 6 = 0$
$x + y = 4$

Based on the structure, the terminal sulfur atom (S_t) is often considered to be in a sulfide-like state, i.e., $y = -2$.
Substituting $y = -2$ into the equation:
$x + (-2) = 4$
$x = 6$

So, the oxidation states of the two sulfur atoms are +6 and -2. These are different.

B. Na₂S₂O₄ (Sodium Dithionite)

The dithionite ion (S₂O₄²⁻) contains an S-S bond. The structure is symmetric, with each sulfur atom bonded to two oxygen atoms and one other sulfur atom.

O⁻
|
S - S
|| ||
O O

Due to the symmetry, both sulfur atoms are equivalent. Therefore, they must have the same oxidation state.
Let the oxidation state of each S atom be $x$.
The overall charge of Na₂S₂O₄ is 0.
$2(\text{Na}^{+}) + 2(\text{S}) + 4(\text{O}^{2-}) = 0$
$2(+1) + 2x + 4(-2) = 0$
$2 + 2x - 8 = 0$
$2x = 6$
$x = +3$

Since both sulfur atoms are equivalent and have an oxidation state of +3, this compound fits the condition.

C. Na₂S₃O₆ (Sodium Trithionate)

The trithionate ion (S₃O₆²⁻) contains a chain of three sulfur atoms (S-S-S). The two terminal sulfur atoms (S_t) are equivalent, and the central sulfur atom (S_c) is different.

O₃S - S - SO₃²⁻

The central sulfur atom (S_c) is bonded only to other sulfur atoms. According to the rules for assigning oxidation states, bonds between identical atoms contribute 0 to the oxidation state. Therefore, the oxidation state of the central sulfur atom is 0.
Let the oxidation state of the terminal S atoms (S_t) be $x$.
The overall charge of Na₂S₃O₆ is 0.
$2(\text{Na}^{+}) + 2(\text{S}\_t) + \text{S}\_c + 6(\text{O}^{2-}) = 0$
$2(+1) + 2x + 0 + 6(-2) = 0$
$2 + 2x - 12 = 0$
$2x = 10$
$x = +5$

So, the oxidation states of the sulfur atoms are +5 (for the two terminal S atoms) and 0 (for the central S atom). These are different.

D. Na₂S₅O₆ (Sodium Pentathionate)

The pentathionate ion (S₅O₆²⁻) contains a chain of five sulfur atoms (S-S-S-S-S). The two terminal sulfur atoms (S_t) are equivalent, and the three central sulfur atoms (S_c) are equivalent among themselves.

O₃S - S - S - S - SO₃²⁻

The three central sulfur atoms (S_c) are bonded only to other sulfur atoms. Therefore, their oxidation state is 0.
Let the oxidation state of the terminal S atoms (S_t) be $x$.
The overall charge of Na₂S₅O₆ is 0.
$2(\text{Na}^{+}) + 2(\text{S}\_t) + 3(\text{S}\_c) + 6(\text{O}^{2-}) = 0$
$2(+1) + 2x + 3(0) + 6(-2) = 0$
$2 + 2x - 12 = 0$
$2x = 10$
$x = +5$

So, the oxidation states of the sulfur atoms are +5 (for the two terminal S atoms) and 0 (for the three central S atoms). These are different.

Based on the analysis, only Na₂S₂O₄ has all its sulfur atoms in the same oxidation state (+3).