Which of the following reaction represent auto redox or disproportionation?

{C l}_{2} + O H^{-} \to {C l}^{-} + C l O_{3}^{-} + H_{2} O

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

P_{4} + N a O H \to P H_{3} + N a H_{2} P O_{2}

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

{N O}_{2}^{-} \to N O + N O_{3}^{-}

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

N H_{4} N O_{2} \to N_{2} + 2 H_{2} O

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

Open in App

Solution

The correct options are
A ${C l}_{2} + O H^{-} \to {C l}^{-} + C l O_{3}^{-} + H_{2} O$
C $P_{4} + N a O H \to P H_{3} + N a H_{2} P O_{2}$
D ${N O}_{2}^{-} \to N O + N O_{3}^{-}$
In option A, chlorine is oxidized as well as reduced. Hence, it is a disproportionation reaction.
In option B, phosphorus is oxidized as well as reduced. Hence, it is a disproportionation reaction.
In option C, nitrogen is oxidized as well as reduced. Hence, it is also a disproportionation reaction.
Option D is an example of comproportionation reaction. N atom is in two different oxidation states in reactants and is in same oxidation state in product.

Suggest Corrections

Similar questions

B + N a O H \to 2 N a_{3} B O_{3} + H_{2}

P_{4} + N a O H + H_{2} O \to N a H_{2} P O_{2} + P H_{3}

S + N a O H \to N a_{2} S_{2} O_{3} + N a_{2} S + H_{2} O

I . N H_{4} N O_{3} Δ \to N_{2} O + 2 H_{2} O

I I . N H_{4} N O_{2} Δ \to N_{2} + 2 H_{2} O

I I I . P C l_{5} Δ \to P C l_{3} + C l_{2}

P_{4} + N a O H \to P H_{3} + N a H_{2} P O_{2}

N a H_{2} P O_{2}

P H_{3}

P_{4}

P_{4} + N a O H \to N a H_{2} P O_{2} + P H_{3}

Join BYJU'S Learning Program