A process which involves no change in oxidation number is:

2 K C l O_{3} \to 2 K C l + 3 O_{2}

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N a O H + H C l \to N a C l + H_{2} O

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2 H_{2} S + S O_{2} \to 2 H_{2} O + 3 S

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C l_{2} + 2 N a O H \to N a C l + N a C l O + H_{2} O

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Solution

The correct option is B $N a O H + H C l \to N a C l + H_{2} O$
In acid-base (or) neutralization reactions, there is no change in the oxidation number.

In neutralization there is no transfer of electrons between species or atom didn't change its oxidation state.

Option B is correct.

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Similar questions

H C l + N a O H \to N a C l + H_{2} O + x c a l .

H_{2} S O_{4} + 2 N a O H \to N a_{2} S O_{4} + 2 H_{2} O + y c a l

2 {KClO}_{3} (s) \overset{∆}{\to} 2 KCl (s) + 3 O_{2} ↑

C a {CO}_{3} (s) \overset{∆}{\to} C a O (s) + {CO}_{2} ↑

H C l + N a O H \to N a C l + H_{2} O + x c a l

H_{2} S O_{4} + 2 N a O H \to N a_{2} S O_{4} + 2 H_{2} O + y c a l

x

y

H_{2} S

S O_{2}

2 H_{2} S + S O_{2} \to 2 H_{2} O + 3 S

C l_{2} + 2 N a O H \to N a C l + N a C l O + H_{2} O

3 N a C l O \to 2 N a C l + N a C l O_{3}

4 N a C l O_{3} \to 3 N a C l O_{4} + N a C l

C l_{2}

106.5

N a C l O_{3}

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