The reaction that takes place when Cl 2 gas is passed through conc- N aOH solution is-A- OxidationB- ReductionC- DisplacementD- Disproportionation

The correct option is D Disproportionation6NaOH + 3Cl_2 → 5NaCl + NaClO_3+3H_2O Disproportionation as the oxidation state of chlorine goes from 0 to 1 in NaC ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Common Oxidation States

The reaction ...

Question

The reaction that takes place when $C l_{2}$ gas is passed through conc. $N a O H$ solution is:

Oxidation

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Reduction

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Displacement

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Disproportionation

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D Disproportionation
$6 N a O H + 3 C l_{2} \to 5 N a C l + N a C l O_{3} + 3 H_{2} O$

Disproportionation as the oxidation state of chlorine goes from 0 to $- 1$ in $N a C l$ and to $+ 5$ in $N a C l O_{3}$ .

Suggest Corrections

Similar questions

Q. The reaction that takes place when

C l_{2}

gas is passed through conc.

N a O H

solution is :

Q. When

C O_{2}

gas is passed into

{M n O}_{4}^{2 -}

solution, a disproportionation reaction takes place.

3 M n O_{4}^{2 -} + 4 H^{+} \to M n O_{2} + 2 M n O_{4}^{-} + H_{2} O

If true enter 1, else enter 0.

$H_{2}$ $O_{2}$ acts as bleaching agent. In this reaction $H_{2}$ $O_{2}$ undergoes

When $C l_{2}$ gas is passed through hot and conc. solution of KOH, following compound is formed

Q. Assertion :Chlorine gas disproportionate in hot & conc. NaOH solution Reason: NaCl and NaOCl are formed in the above reaction.

Join BYJU'S Learning Program

The reaction that takes place when Cl2 gas is passed through conc. NaOH solution is:

The correct option is D Disproportionation6NaOH+3Cl2→5NaCl+NaClO3+3H2O Disproportionation as the oxidation state of chlorine goes from 0 to −1 in NaCl and to +5 in NaClO3.

The reaction that takes place when $C l_{2}$ gas is passed through conc. $N a O H$ solution is:

The correct option is D Disproportionation
$6 N a O H + 3 C l_{2} \to 5 N a C l + N a C l O_{3} + 3 H_{2} O$

Disproportionation as the oxidation state of chlorine goes from 0 to $- 1$ in $N a C l$ and to $+ 5$ in $N a C l O_{3}$ .