The weight of chlorine gas undergoing oxidation when excess of chlorine is sent into hot concentrated caustic soda solution containing 24 grams of solute in it
$3 C l_{2} + 6 N a O H \to 5 N a C l + N a C l O_{3} + 3 H_{2} O$

Q. Consider the following reaction:

6 N a O H + 3 C l_{2} \to 5 N a C l + A + 3 H_{2} O

What is the oxidation number of chlorine in

A

(b) (i) What are the products formed when ammonia is oxidized with copper oxide?
(ii) Name one lead compound that can be used to oxidize hydrogen chloride to chlorine.

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When excess of ammonia is used: 3Cl2+8NH3⟶A+BWhen access of chlorine is used:3Cl2+4NH3⟶C+DOxidation state of nitrogen in compound B is:

The correct option is B 08NH3+3Cl2→6NH4Cl+N2$3Cl_2+4NH_3 \rightarrow 3NH_4Cl+NCl_3$ Compound B is N2 Here as the nitrogen is in elemental state so its oxidation state is 0 Hence option A is correct.

When excess of ammonia is used:
$3 C l_{2} + 8 N H_{3}$ $⟶$ $A + B$
When access of chlorine is used:
$3 C l_{2} + 4 N H_{3}$ $⟶$ $C + D$
Oxidation state of nitrogen in compound $B$ is:

The correct option is B $0$
$8 N H_{3} + 3 C l_{2} \to 6 N H_{4} C l + N_{2}$
$3Cl_2+4NH_3 \rightarrow 3NH_4Cl+NCl_3$

Compound B is $N_{2}$ Here as the nitrogen is in elemental state so its oxidation state is $0$

Hence option A is correct.