Question

consider the following reaction, and identify the substance that is the reducing agent.
ncl₃(l) + 3 h₂o(l) → nh₃(aq) + 3 hocl(aq)
o n in nh₃
o n in ncl₃
o cl in hocl
o cl in ncl₃

Explanation:

Step1: Calculate oxidation number

Oxidation number rules: In \(NH_3\), \(H = + 1\). Let \(N=x\), then \(x+(+1)\times3 = 0\), \(x=-3\). In \(NCl_3\), \(Cl=-1\). Let \(N = y\), then \(y+(-1)\times3 = 0\), \(y = + 3\). In \(HOCl\), \(H = + 1\), \(O=-2\). Let \(Cl = z\), then \((+1)+(-2)+z = 0\), \(z = + 1\). In \(NCl_3\), \(Cl=-1\).

Step2: Determine reducing agent

A reducing agent is oxidized (loses electrons, oxidation number increases). In the reaction \(NCl_3(l)+3H_2O(l)\to NH_3(aq)+3HOCl(aq)\), the oxidation number of \(N\) changes from \(+3\) in \(NCl_3\) to \(-3\) in \(NH_3\) (reduction). The oxidation number of \(Cl\) changes from \(-1\) in \(NCl_3\) to \(+1\) in \(HOCl\) (oxidation). But we are looking for the reducing agent (species that gets oxidized). Wait, no, let's re - check.

Wait, another approach: The general formula for oxidation number of \(N\) in \(NCl_3\): Let \(N\) oxidation number be \(x\), \(x + 3\times(-1)=0\), \(x = + 3\). In \(NH_3\), \(N=-3\). The species that is oxidized (loses electrons) is the reducing agent. The \(Cl\) in \(NCl_3\) has an oxidation number of \(-1\) and in \(HOCl\) it is \(+1\). But the problem is about the atom in the compound. Wait, no, the reducing agent is the whole compound. Wait, no, the question is about the atom in the compound. Wait, no, let's calculate oxidation numbers properly.

For \(N\) in \(NCl_3\): Let \(N\) be \(x\), \(x+3\times(-1) = 0\), \(x = + 3\). For \(N\) in \(NH_3\): Let \(N\) be \(y\), \(y + 3\times(+1)=0\), \(y=-3\). The \(N\) in \(NCl_3\) is reduced (\(+3\to - 3\)). For \(Cl\) in \(NCl_3\): \(-1\), in \(HOCl\): \(+1\) (oxidized). But the options are about \(N\) or \(Cl\) in a compound.

Wait, the reducing agent is the species that donates electrons (gets oxidized). In the reaction, \(Cl\) in \(NCl_3\) (\(-1\)) is oxidized to \(+1\) in \(HOCl\). But the options:

• For \(N\) in \(NH_3\): \(N=-3\) (product, not reactant).
• For \(N\) in \(NCl_3\): \(N = + 3\) (reactant, but it is reduced).
• For \(Cl\) in \(HOCl\): \(Cl = + 1\) (product).
• For \(Cl\) in \(NCl_3\): \(Cl=-1\) (reactant) and is oxidized (\(-1\to+1\))

Answer:

Cl in \(NCl_3\)