Compare the structures of $H_2O$ and $H_2O_2$.
(N/A) In the gaseous phase, the water molecule $(H_2O)$ has a bent structure with a bond angle of $104.5^{\circ}$. The $O-H$ bond length is $95.7 \, pm$.
Hydrogen peroxide $(H_2O_2)$ has a non-planar structure in both the gas and solid phases. The dihedral angle in the gas phase is $111.5^{\circ}$, while in the solid phase, it is $90.2^{\circ}$.
Hydrogen peroxide $(H_2O_2)$ has a non-planar structure in both the gas and solid phases. The dihedral angle in the gas phase is $111.5^{\circ}$, while in the solid phase, it is $90.2^{\circ}$.
Explore More
Similar Questions
Match List $I$ with List $II$ and select the correct answer using the codes given below the lists:
$A$. Peroxide $1$. $C_3O_2$ $B$. Superoxide $2$. $PbO_2$ $C$. Dioxide $3$. $KO_2$ $D$. Suboxide $4$. $H_2O_2$
| $A$. Peroxide | $1$. $C_3O_2$ |
| $B$. Superoxide | $2$. $PbO_2$ |
| $C$. Dioxide | $3$. $KO_2$ |
| $D$. Suboxide | $4$. $H_2O_2$ |
Medium
View SolutionThe decomposition of hydrogen peroxide is an example of
Easy
View SolutionIn which of the following reactions does $H_2O_2$ act as a reducing agent?
$(I)$ $H_2O_2 + 2H^{+} + 2e^- \to 2H_2O$
$(II)$ $H_2O_2 \to O_2 + 2H^{+} + 2e^-$
$(III)$ $H_2O_2 + 2e^- \to 2OH^{-}$
$(IV)$ $H_2O_2 + 2OH^{-} \to O_2 + 2H_2O + 2e^-$
$(I)$ $H_2O_2 + 2H^{+} + 2e^- \to 2H_2O$
$(II)$ $H_2O_2 \to O_2 + 2H^{+} + 2e^-$
$(III)$ $H_2O_2 + 2e^- \to 2OH^{-}$
$(IV)$ $H_2O_2 + 2OH^{-} \to O_2 + 2H_2O + 2e^-$
Difficult
View SolutionIn which of the following reactions does hydrogen peroxide act as a reducing agent?
Which of the following statements is incorrect?
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo