Choose the correct option regarding change in oxidation number of the underlined atom in the following conversion.

$C a O C l_{2} - -- - + H_{2} O \to C a (O H)_{2} + C l_{2}$

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

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

-1

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

None of these

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

Open in App

Solution

The correct option is A 0
In $C a O {C l}_{2}$ , $C l$ has two oxidation states: $+ 1$ and $- 1$ as shown in the figure.

$C a (O C l^{(+ 1)}) C l^{(- 1)} + H_{2} O ⟶ C a (O H)_{2} + C l^{(} 0)_{2}$
The net change in Oxidation state of $C l = (- 1) + (+ 1) = 0$
Hence, the A is the correct option.

Suggest Corrections

Similar questions

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

H_{2} S - - O_{5} + 2 K I \to K_{2} S O_{4} + H_{2} O + I_{2}

A s_{2} - -- - S_{3} + O H^{-} \to A s S_{4}^{3 -}

C a O C l_{2} + H_{2} O \to C l_{2} + C a (O H)_{2}

C a O C l_{2}

\frac{M}{x}

C a O C l_{2} + H_{2} O ⟶ C a (O H)_{2} + C l_{2}

2 K I + C l_{2} ⟶ 2 K C l + I_{2}

I_{2} + 2 N a_{2} S_{2} O_{3} ⟶ N a_{2} S_{4} O_{6} + 2 N a I

N a_{2} S_{2} O_{3}

I_{2}

127

C a O C l_{2}

50 %

Join BYJU'S Learning Program