Which among the following compounds is the most reducing compound?

The correct option is H₂S.

To determine which compound is the most reducing, we need to analyze the oxidation states of the central atoms in the given options and evaluate their ease of oxidation (acting as a reducing agent).
A reducing agent undergoes oxidation (loses electrons). Therefore, a compound where the central atom is in its lowest possible oxidation state will have the strongest tendency to lose electrons and act as the strongest (most) reducing agent.

Let's determine the oxidation states of the central atoms in each compound:

1. H₂S (Hydrogen sulfide):
The oxidation state of hydrogen is +1. Let x be the oxidation state of sulfur (S):
$2(+1)+x=0$
$x=-2$
Since sulfur belongs to Group 16, its valence shell electronic configuration is $s^2{p}^4$. The minimum possible oxidation state for sulfur is -2. In H₂S, sulfur is at its minimum oxidation state of -2. It cannot be reduced further; it can only be oxidized (e.g., to S, SO₂, or SO₄²⁻), making H₂S a strong reducing agent.

2. HNO₂ (Nitrous acid):
Let x be the oxidation state of nitrogen (N):
$(+1)+x+2(-2)=0$
$1+x-4=0$
$x=+3$
Nitrogen can have oxidation states ranging from -3 to +5. Since +3 is an intermediate oxidation state, HNO₂ can act as both an oxidizing agent and a reducing agent.

3. SnCl₂ (Tin(II) chloride):
Let x be the oxidation state of tin (Sn):
$x+2(-1)=0$
$x=+2$
Tin (Sn) belongs to Group 14 and has common oxidation states of +2 and +4. While Sn²⁺ can be oxidized to Sn⁴⁺, it is not as strong of a reducing agent as H₂S, where the sulfur atom is in its lowest possible oxidation state (-2) and has a high electron density.

4. H₂SO₃ (Sulfurous acid):
Let x be the oxidation state of sulfur (S):
$2(+1)+x+3(-2)=0$
$2+x-6=0$
$x=+4$
Here, sulfur is in the +4 oxidation state, which is an intermediate state (between -2 and +6). Thus, H₂SO₃ is a weaker reducing agent compared to H₂S.

Conclusion:
In H₂S, sulfur is present in its lowest possible oxidation state of -2. It readily loses electrons to be oxidized to higher oxidation states, making it the most powerful reducing agent among the given options.