Hydrogen peroxide Questions in English

(A) The mixture of $Na_2O_2$ and $HCl$ is commercially known as oxone.
It is primarily used for the bleaching of delicate fibres.

(D) $H_2O_2$ is prepared by the action of $H_2SO_4$ on hydrated $BaO_2$ $(BaO_2 \cdot 8H_2O)$.
Anhydrous $BaO_2$ is not used because it forms a protective layer of $BaSO_4$ on its surface,which prevents further reaction.
Therefore,the Assertion is incorrect because anhydrous $BaO_2$ is not used.
The Reason is also incorrect because hydrated $BaO_2$ is readily available and is the preferred reagent.
The reaction is: $BaO_2 \cdot 8H_2O + H_2SO_4 \to BaSO_4 + H_2O_2 + 8H_2O$.

(C) The decomposition of $H_{2}O_{2}$ is: $2H_{2}O_{2} \rightarrow 2H_{2}O + O_{2}$. This is correct.
$(b)$ Thermal decomposition of these salts also yields dioxygen:
$2KClO_{3} \xrightarrow{\Delta} 2KCl + 3O_{2}$
$2Pb(NO_{3})_{2} \xrightarrow{\Delta} 2PbO + 4NO_{2} + O_{2}$
$2NaNO_{3} \xrightarrow{\Delta} 2NaNO_{2} + O_{2}$. This is correct.
$(c)$ In the industrial process,$2$-ethylanthraquinol is oxidized by air to produce $H_{2}O_{2}$ and $2$-ethylanthraquinone,which is then recycled. This is correct.
$(d)$ $H_{2}O_{2}$ is used in the synthesis of sodium perborate from borax and sodium hydroxide. This is correct.
Therefore,all statements $(a)$,$(b)$,$(c)$,and $(d)$ are correct.

(A) $10$ volume solution of $H_{2}O_{2}$ means that $1 \, L$ of this $H_{2}O_{2}$ solution will give $10 \, L$ of oxygen at $STP$.
$\mathop{2H_{2}O_{2(l)}}\limits_{68 \, g}$ $\rightarrow \mathop{O_{2(g)} + 2H_{2}O_{(l)}}\limits_{22.7 \, L \, \text{at} \, STP}$
On the basis of the above equation,$22.7 \, L$ of $O_{2}$ is produced from $68 \, g \, H_{2}O_{2}$ at $STP$.
Therefore,$10 \, L$ of $O_{2}$ at $STP$ is produced from:
$\frac{68 \times 10}{22.7} \, g \approx 29.95 \, g \approx 30.3 \, g \, H_{2}O_{2}$.
Thus,the strength of $H_{2}O_{2}$ in $10$ volume $H_{2}O_{2}$ solution is approximately $30.3 \, g/L$.

(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 $90.2^{\circ}$.

(N/A) $i$. $PbS_{(s)} + 4H_2O_{2_{(aq)}} \rightarrow PbSO_{4_{(s)}} + 4H_2O_{(l)}$
$H_2O_2$ acts as an oxidizing agent. This is a redox reaction.
$ii$. $2MnO_{4_{(aq)}}^{-} + 5H_2O_{2_{(aq)}} + 6H_{(aq)}^{+} \rightarrow 2Mn_{(aq)}^{2+} + 8H_2O_{(l)} + 5O_{2_{(g)}}$
$H_2O_2$ acts as a reducing agent. This is a redox reaction.
$iii$. $CaO_{(s)} + H_2O_{(g)} \rightarrow Ca(OH)_{2_{(s)}}$
This is a hydration reaction.
$iv$. $AlCl_{3_{(s)}} + 3H_2O_{(l)} \rightarrow Al(OH)_{3_{(s)}} + 3HCl_{(aq)}$
This is a hydrolysis reaction.
$v$. $Ca_3N_{2_{(s)}} + 6H_2O_{(l)} \rightarrow 3Ca(OH)_{2_{(aq)}} + 2NH_{3_{(g)}}$
This is a hydrolysis reaction.

Hydrogen peroxide,$H_{2}O_{2}$,acts as an oxidizing as well as a reducing agent in both acidic and alkaline media.
Reactions involving oxidizing actions (where $H_{2}O_{2}$ is reduced to $H_{2}O$ or $OH^{-}$):
$1) \ 2Fe^{2+} + 2H^{+} + H_{2}O_{2} \longrightarrow 2Fe^{3+} + 2H_{2}O$
$2) \ PbS + 4H_{2}O_{2} \longrightarrow PbSO_{4} + 4H_{2}O$
Reactions involving reducing actions (where $H_{2}O_{2}$ is oxidized to $O_{2}$):
$1) \ 2MnO_{4}^{-} + 6H^{+} + 5H_{2}O_{2} \longrightarrow 2Mn^{2+} + 8H_{2}O + 5O_{2}$
$2) \ I_{2} + H_{2}O_{2} + 2OH^{-} \longrightarrow 2I^{-} + 2H_{2}O + O_{2}$

(N/A) $H_{2}O_{2}$ (hydrogen peroxide) acts as a strong oxidizing agent in both acidic and basic media.
It releases nascent oxygen $([O])$ which oxidizes the colored matter (chromophores) present in the substance.
This oxidation breaks the chemical bonds of the chromophores,rendering them colorless.
Consequently,the material appears white.

(N/A) $H_2O$ (Water): In the gas phase, $H_2O$ is a bent molecule with a bond angle of $104.5^{\circ}$ and an $O-H$ bond length of $95.7 \text{ pm}$. It is a highly polar molecule with $sp^3$ hybridization of the oxygen atom.
$H_2O_2$ (Hydrogen Peroxide): $H_2O_2$ has a non-planar 'open book' structure. In the gas phase, the dihedral angle is $111.5^{\circ}$, while in the solid phase, it is $90.2^{\circ}$. The $O-O$ bond length is $147.5 \text{ pm}$ and the $O-H$ bond length is $95.0 \text{ pm}$.

(N/A) $(i)$ Acidifying barium peroxide and removing excess water by evaporation under reduced pressure gives hydrogen peroxide.
$BaO_{2} \cdot 8H_{2}O_{(s)} + H_{2}SO_{4_{(aq)}} \longrightarrow BaSO_{4_{(s)}} + H_{2}O_{2_{(aq)}} + 8H_{2}O_{(l)}$
$(ii)$ Peroxodisulphate,obtained by electrolytic oxidation of acidified sulphate solutions at high current density,on hydrolysis yields hydrogen peroxide.
$2HSO_{4_{(aq)}}^{-}$ $\xrightarrow{\text{Electrolysis}} HO_{3}SOOSO_{3}H_{(aq)}$ $\xrightarrow{\text{Hydrolysis}} 2HSO_{4_{(aq)}}^{-} + 2H_{(aq)}^{+} + H_{2}O_{2_{(aq)}}$
This method is now used for the laboratory preparation of $D_{2}O_{2}$.
$K_{2}S_{2}O_{8_{(s)}} + 2D_{2}O_{(l)} \longrightarrow 2KDSO_{4_{(aq)}} + D_{2}O_{2_{(l)}}$
$(iii)$ Industrially,it is prepared by the auto-oxidation of $2$-alkylanthraquinols.
(2-ethyl anthraquinol) $\underset{\text { Oxidation }}{\stackrel{\text { Reduction }}{\rightleftharpoons}} H _2 O _2+$ (Oxidised product)
In this case,$1\%$ $H_{2}O_{2}$ is formed. It is extracted with water and concentrated to $\sim 30\%$ (by mass) by distillation under reduced pressure.
It can be further concentrated to $\sim 85\%$ by careful distillation under low pressure. The remaining water can be frozen out to obtain pure $H_{2}O_{2}$.
In the pure state,$H_{2}O_{2}$ is an almost colourless (very pale blue) liquid.
$H_{2}O_{2}$ is miscible with water in all proportions and forms a hydrate $H_{2}O_{2} \cdot H_{2}O$ ($mp$ $221 \ K$).
Physical Properties of Hydrogen Peroxide:

Property	Value
Melting point/$K$	$272.4$
Density (liquid at $298 \ K$)/$g \ cm^{-3}$	$1.44$
Boiling point (extrapolated)/$K$	$423$
Viscosity $(290 \ K)$/centipoise	$1.25$
Vapour pressure $(298 \ K)$/$mmHg$	$1.9$
Dielectric constant $(298 \ K)$	$70.7$
Density (solid at $268.5 \ K$)/$g \ cm^{-3}$	$1.64$
Electrical conductivity $(298 \ K)$/$\Omega^{-1} cm^{-1}$	$5.1 \times 10^{-8}$

Hydrogen peroxide has a non-planar structure.
$a$. In the gas phase, the $H_2O_2$ molecule has a dihedral angle of $111.5^{\circ}$. The $O-O$ bond length is $147.5 \text{ pm}$ and the $O-H$ bond length is $95.0 \text{ pm}$.
$b$. In the solid phase at $110 \text{ K}$, the $H_2O_2$ molecule has a dihedral angle of $90.2^{\circ}$. The $O-O$ bond length is $145.8 \text{ pm}$ and the $O-H$ bond length is $98.8 \text{ pm}$.

(N/A) Hydrogen peroxide acts as an oxidising as well as a reducing agent in both acidic and alkaline media. The reactions are as follows:
$(i)$ Oxidising effect in acidic medium:
$2 Fe^{2+}_{(aq)} + 2 H^{+}_{(aq)} + H_{2}O_{2(aq)} \longrightarrow 2 Fe^{3+}_{(aq)} + 2 H_{2}O_{(l)}$
$PbS_{(s)} + 4 H_{2}O_{2(aq)} \longrightarrow PbSO_{4(s)} + 4 H_{2}O_{(l)}$
$(ii)$ Reducing effect in acidic medium:
$2 MnO_{4}^{-} + 6 H^{+} + 5 H_{2}O_{2} \longrightarrow 2 Mn^{2+}_{(aq)} + 8 H_{2}O + 5 O_{2}$
$HOCl + H_{2}O_{2} \longrightarrow H_{3}O^{+} + Cl^{-} + O_{2}$
$(iii)$ Oxidising effect in basic medium:
$2 Fe^{2+} + H_{2}O_{2} \longrightarrow 2 Fe^{3+} + 2 OH^{-}$
$Mn^{2+} + H_{2}O_{2} \longrightarrow Mn^{4+} + 2 OH^{-}$
$(iv)$ Reducing effect in basic medium:
$I_{2} + H_{2}O_{2} + 2 OH^{-} \longrightarrow 2 I^{-} + 2 H_{2}O + O_{2}$
$2 MnO_{4}^{-} + 3 H_{2}O_{2} \longrightarrow 2 MnO_{2} + 3 O_{2} + 2 H_{2}O + 2 OH^{-}$

Hydrogen peroxide $(H_{2}O_{2})$ acts as both an oxidizing and a reducing agent in both acidic and alkaline media. Key reactions are described below:
$(i)$ Oxidizing effect in acidic medium:
$2 Fe_{(aq)}^{2+} + 2 H_{(aq)}^{+} + H_{2}O_{2_{(aq)}} \longrightarrow 2 Fe_{(aq)}^{3+} + 2 H_{2}O_{(l)}$
$PbS_{(s)} + 4 H_{2}O_{2_{(aq)}} \longrightarrow PbSO_{4_{(s)}} + 4 H_{2}O_{(l)}$
$(ii)$ Reducing effect in acidic medium:
$2 MnO_{4}^{-} + 6 H^{+} + 5 H_{2}O_{2} \longrightarrow 2 Mn_{(aq)}^{2+} + 8 H_{2}O + 5 O_{2}$
$HOCl + H_{2}O_{2} \longrightarrow H_{3}O^{+} + Cl^{-} + O_{2}$
$(iii)$ Oxidizing effect in basic medium:
$2 Fe^{2+} + H_{2}O_{2} \longrightarrow 2 Fe^{3+} + 2 OH^{-}$
$Mn^{2+} + H_{2}O_{2} \longrightarrow Mn^{4+} + 2 OH^{-}$
$(iv)$ Reducing effect in basic medium:
$I_{2} + H_{2}O_{2} + 2 OH^{-} \longrightarrow 2 I^{-} + 2 H_{2}O + O_{2}$
$2 MnO_{4}^{-} + 3 H_{2}O_{2} \longrightarrow 2 MnO_{2} + 3 O_{2} + 2 H_{2}O + 2 OH^{-}$

(B) $H_2O_2$ decomposes slowly on exposure to light.
$2 H_2O_2(l) \longrightarrow 2 H_2O(l) + O_2(g)$
In the presence of metal surfaces or traces of alkali (present in glass containers),the above reaction is catalysed.
It is,therefore,stored in wax-lined glass or plastic vessels in the dark.
Urea can be added as a stabiliser. It is kept away from dust because dust can induce explosive decomposition of the compound.

(N/A) $H_2O_2$ acts as a bleaching agent due to the release of nascent oxygen $([O])$ upon its decomposition.
$2H_2O_2 \rightarrow 2H_2O + 2[O]$
This nascent oxygen oxidizes the colored matter present in substances like textiles,paper pulp,leather,oils,and fats into colorless products,thereby acting as a bleaching agent.

(N/A) $(i)$ In daily life,it is used as a hair bleach and as a mild disinfectant. As an antiseptic,it is sold in the market as perhydrol.
$(ii)$ It is used to manufacture chemicals like sodium perborate and per-carbonate,which are used in high-quality detergents.
$(iii)$ It is used in the synthesis of hydroquinone,tartaric acid,and certain food products and pharmaceuticals (e.g.,cephalosporin).
$(iv)$ It is employed in industries as a bleaching agent for textiles,paper pulp,leather,oils,fats,etc.
$(v)$ Nowadays,it is also used in Environmental (Green) Chemistry. For example,in pollution control treatment of domestic and industrial effluents,oxidation of cyanides,and restoration of aerobic conditions to sewage wastes.

(B) The relationship between volume strength and molarity $(M)$ is given by: $Volume \ strength = 11.2 \times M$.
Given volume strength = $40$.
Therefore,$M = \frac{40}{11.2} \approx 3.57 \ M$.
Now,$\% \ W/V = \text{Molarity} \times \frac{\text{Molar mass of } H_2O_2}{10}$.
Molar mass of $H_2O_2 = (2 \times 1) + (2 \times 16) = 34 \ g/mol$.
$\% \ W/V = 3.57 \times \frac{34}{10} = 3.57 \times 3.4 = 12.138 \approx 12.14\%$.
Thus,the molarity is $3.57 \ M$ and the $\% \ W/V$ is $12.14\%$.

(C) The relationship between molarity $(M)$ and volume strength $(V)$ of $H_2O_2$ is given by the formula: $M = \frac{V}{11.2}$.
Given $M = 3.125 \ M$.
Substituting the value: $3.125 = \frac{V}{11.2}$.
$V = 3.125 \times 11.2 = 35$.
Therefore,the grade is $35$ volume.

(A) The strength of $H_2O_2$ in terms of $\% \ W/V$ is given by the formula: $\text{Strength} \ (\% \ W/V) = \frac{\text{Volume strength}}{5.6} = \frac{12}{5.6} \approx 2.14 \% \ W/V$.
Given mass of $H_2O_2 = 680 \ g$.
Since $\% \ W/V = \frac{\text{mass of solute in } g}{\text{volume of solution in } mL} \times 100$,we have $2.14 = \frac{680}{V} \times 100$.
$V = \frac{68000}{2.14} \approx 31775.7 \ mL \approx 31.78 \ L$.

(A) The strength of $H_2O_2$ is given as $8.32\% \ W/V$,which means $8.32 \ g$ of $H_2O_2$ is present in $100 \ mL$ of solution.
$1$. Calculation of Volume Strength:
Volume strength of $H_2O_2 = \text{Strength in } g/L \times \frac{11.2}{34}$.
Strength in $g/L = 8.32 \times 10 = 83.2 \ g/L$.
Volume strength $= 83.2 \times \frac{11.2}{34} \approx 27.4 \ V$.
$2$. Calculation of Molar Concentration:
Molarity $(M) = \frac{\text{Strength in } g/L}{\text{Molar mass of } H_2O_2}$.
Molar mass of $H_2O_2 = (2 \times 1) + (2 \times 16) = 34 \ g/mol$.
$M = \frac{83.2}{34} \approx 2.447 \ M$.
Thus,the volume strength is $27.4 \ V$ and molarity is $2.447 \ M$.

(N/A) In a basic medium,$H_2O_2$ acts as a reducing agent towards $I_2$. The balanced chemical equation is:
$I_{2(s)} + H_2O_{2(aq)} + 2OH^{-}_{(aq)} \rightarrow 2I^{-}_{(aq)} + 2H_2O_{(l)} + O_{2(g)}$

(A) The hydrolysis of peroxodisulphuric acid $(H_2S_2O_8)$ is a standard method for the preparation of hydrogen peroxide $(H_2O_2)$.
The reaction is given by: $H_2S_2O_8(aq) + 2H_2O(l) \rightarrow 2H_2SO_4(aq) + H_2O_2(aq)$.
Therefore,the product $x$ is $H_2O_2$.

(B) In the solid state,$H_2O_2$ adopts a non-planar structure due to the repulsion between the lone pairs of electrons on the oxygen atoms.
The dihedral angle is $90.2^{\circ}$ and the $O-O-H$ bond angle is $101.9^{\circ}$.

(N/A) $(i)$ Perhydrol
$(ii)$ $-SO_3H$ (Sulphonic acid)
$(iii)$ $1.5 \times 10^{-3} \ pm, 50-200 \ pm$
$(iv)$ Alkali metals (Group $1$) and Alkaline earth metals (Group $2$)

(N/A) $(i)$ When $PbS$ reacts with hydrogen peroxide,it forms $PbSO_4$ and water.
$PbS_{(s)} + 4H_2O_{2_{(aq)}} \to PbSO_{4_{(s)}} + 4H_2O_{(l)}$
$(ii)$ Methanol is formed by the reaction of carbon monoxide with $H_2$ gas in the presence of a cobalt catalyst.
$CO_{(g)} + 2H_{2_{(g)}} \xrightarrow{\text{cobalt catalyst}} CH_3OH_{(l)}$

(N/A) The preparation of $D_2O_2$ can be carried out by the action of $D_2SO_4$ on barium peroxide $(BaO_2)$.
$BaO_2 + D_2SO_4 \rightarrow BaSO_4 + D_2O_2$

(D) By definition,$5$ volumes $H_2O_2$ solution means that $1 \ L$ of this solution produces $5 \ L$ of $O_2$ gas at $STP$ upon decomposition.
The decomposition reaction is: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$.
According to the stoichiometry,$2 \times 34 \ g$ $(68 \ g)$ of $H_2O_2$ produces $22.7 \ L$ of $O_2$ at $STP$.
Therefore,$22.7 \ L$ of $O_2$ at $STP$ is produced by $68 \ g$ of $H_2O_2$.
So,$5 \ L$ of $O_2$ at $STP$ will be produced by: $\frac{68 \times 5}{22.7} \ g \approx 14.98 \ g$.
The strength of the solution is approximately $15 \ g/L$.

(N/A) $(i)$ $H_2O_2$ has a non-planar structure. The gas phase structure has a dihedral angle of $111.5^{\circ}$ and the solid phase structure has a dihedral angle of $90.2^{\circ}$.
$(ii)$ $H_2O_2$ is a better oxidizing agent than water because the $O-O$ bond in $H_2O_2$ is weak and can easily break to release oxygen atoms or accept electrons to form water.
Examples:
$1.$ $H_2O_2$ oxidizes an acidified solution of $KI$ to $I_2$,which gives a blue color with starch solution:
$2KI + H_2SO_4 + H_2O_2 \longrightarrow K_2SO_4 + 2H_2O + I_2$
$2.$ $H_2O_2$ oxidizes black $PbS$ to white $PbSO_4$:
$PbS + 4H_2O_2 \longrightarrow PbSO_4 + 4H_2O$

$O_2^{2-}$ is the peroxide ion.
$O_2^-$ is the superoxide ion.
$(i)$ Peroxide ion reacts with water to form hydrogen peroxide $(H_2O_2)$ and hydroxide ions $(OH^-)$:
$O_2^{2-} + 2H_2O \to 2OH^- + H_2O_2$
$(ii)$ Superoxide ion reacts with water to produce hydrogen peroxide $(H_2O_2)$,oxygen $(O_2)$,and hydroxide ions $(OH^-)$:
$2O_2^- + 2H_2O \to 2OH^- + H_2O_2 + O_2$

(N/A) $(i)$ The name of the compound is hydrogen peroxide,$H_2O_2$. It acts as an oxidising agent as well as a reducing agent in both acidic and basic media.
$(ii)$ $H_2O_2$ is unstable and decomposes in the presence of light and dust particles. Urea is added as a stabilizer (negative catalyst) to inhibit its decomposition.
The decomposition reaction is: $2H_2O_{2(aq)} \xrightarrow{h\nu} 2H_2O_{(l)} + O_{2(g)}$
Due to its strong oxidising properties,$H_2O_2$ is used in pollution control to oxidise harmful cyanides and foul-smelling sulphides present in domestic and industrial effluents. It also aids in sewage treatment by supplying oxygen for the oxidation of organic matter.

The Lewis structure of $H_2O_2$ is represented as $H-\ddot{O}-\ddot{O}-H$. In this structure,each oxygen atom is bonded to one hydrogen atom and the other oxygen atom,with two lone pairs of electrons on each oxygen atom.

(N/A) The following chemical equations indicate the oxidising and reducing nature of $H_{2}O_{2}$.
$(i)$ $H_{2}O_{2}$ acts as an oxidising agent by oxidising acidified $KI$ to $I_{2}$:
$2KI + H_{2}O_{2} + H_{2}SO_{4} \longrightarrow I_{2} + K_{2}SO_{4} + 2H_{2}O$
$(ii)$ $H_{2}O_{2}$ acts as a reducing agent by reducing acidified $KMnO_{4}$ to $Mn^{2+}$ ions:
$2KMnO_{4} + 3H_{2}SO_{4} + 5H_{2}O_{2} \longrightarrow K_{2}SO_{4} + 2MnSO_{4} + 8H_{2}O + 5O_{2}$

(N/A) The bleaching action of $H_{2}O_{2}$ is due to the release of nascent oxygen upon its decomposition.
$H_{2}O_{2} \longrightarrow H_{2}O + [O]$
The nascent oxygen $[O]$ is a strong oxidizing agent. It reacts with the colouring matter present in substances and oxidizes it to a colourless product.
$\text{Colouring matter} + [O] \longrightarrow \text{Colourless matter}$
Due to this oxidation property,$H_{2}O_{2}$ is used for bleaching delicate materials such as silk,wool,feathers,and ivory.

(N/A) $H_{2}O_{2}$ cannot be concentrated by heating because it decomposes much below its boiling point to give $H_{2}O$ and $O_{2}$.
$2H_{2}O_{2} \longrightarrow 2H_{2}O + O_{2}$
Therefore,concentration of $H_{2}O_{2}$ is carried out in a number of stages:
$(i)$ The $H_{2}O_{2}$ formed (about $1\%$) is extracted with water.
$(ii)$ The aqueous solution is concentrated by distillation under reduced pressure to give $30\% H_{2}O_{2}$ solution.

(N/A) Hydrogen peroxide $(H_{2}O_{2})$ is decomposed by the rough surfaces of glass,alkali oxides present in glass,and light to form water $(H_{2}O)$ and oxygen $(O_{2})$.
The decomposition reaction is: $2 H_{2}O_{2} \rightarrow 2 H_{2}O + O_{2}$.
To prevent this catalytic decomposition,$H_{2}O_{2}$ is usually stored in coloured paraffin wax-coated plastic or teflon bottles.

(N/A) $H_{2}SO_{4}$ acts as a catalyst for the decomposition of $H_{2}O_{2}$. Therefore,weaker acids such as $H_{3}PO_{4}$ are preferred over $H_{2}SO_{4}$ for preparing $H_{2}O_{2}$ from peroxides.
$3BaO_{2} + 2H_{3}PO_{4} \rightarrow Ba_{3}(PO_{4})_{2} + 3H_{2}O_{2}$

(N/A) Concentration of $H_2O_2$: The $H_2O_2$ solution obtained industrially is about $30\%$ by mass. It can be concentrated to $\sim 85\%$ by careful distillation under reduced pressure. The remaining water can be frozen out to obtain pure $H_2O_2$.
Structural Differences:
- $H_2O$ has a bent (angular) structure with a bond angle of $104.5^\circ$ and $O-H$ bond length of $95.7 \text{ pm}$.
- $H_2O_2$ has a non-planar 'open book' structure. In the gas phase, the dihedral angle is $111.5^\circ$, and in the solid phase, it is $90.2^\circ$.
Uses of $H_2O_2$:
$1$. It is used as an antiseptic and mouthwash (sold as perhydrol).
$2$. It is used in the manufacture of chemicals like sodium perborate and sodium carbonate.
$3$. It is used as a bleaching agent for textiles, paper pulp, and oils.

Industrially,$H_2O_2$ is prepared by the auto-oxidation of $2-alkylanthraquinols$.
$2-ethylanthraquinol \underset{H_2/Pd}{\overset{O_2 \text{ (air)}}{\rightleftarrows}} H_2O_2 + \text{oxidised product}$
In this process,$1\% H_2O_2$ is formed. It is extracted with water and concentrated to $\sim 30\%$ (by mass) by distillation under reduced pressure.
It can be further concentrated to $\sim 85\%$ by careful distillation under low pressure. The remaining water can be frozen out to obtain pure $H_2O_2$.
$(i)$ Oxidising effect in acidic medium:
$2Fe^{2+}_{(aq)} + 2H^+_{(aq)} + H_2O_{2(aq)} \longrightarrow 2Fe^{3+}_{(aq)} + 2H_2O_{(l)}$
$PbS_{(s)} + 4H_2O_{2(aq)} \longrightarrow PbSO_{4(s)} + 4H_2O_{(l)}$
$(ii)$ Reducing effect in acidic medium:
$2MnO_4^- + 6H^+ + 5H_2O_2 \longrightarrow 2Mn^{2+}_{(aq)} + 8H_2O + 5O_2$
$HOCl + H_2O_2 \longrightarrow H_3O^+ + Cl^- + O_2$
$(iii)$ Oxidising effect in basic medium:
$2Fe^{2+} + H_2O_2 \longrightarrow 2Fe^{3+} + 2OH^-$
$Mn^{2+} + H_2O_2 \longrightarrow Mn^{4+} + 2OH^-$
$(iv)$ Reducing effect in basic medium:
$I_2 + H_2O_2 + 2OH^- \longrightarrow 2I^- + 2H_2O + O_2$
$2MnO_4^- + 3H_2O_2 \longrightarrow 2MnO_2 + 3O_2 + 2H_2O + 2OH^-$

(N/A) $(i)$ Molar mass of $H_2O_2 = 34 \, g \, mol^{-1}$.
$1 \, L$ of $5 \, M$ solution of $H_2O_2$ contains $34 \times 5 = 170 \, g$ of $H_2O_2$.
Therefore,$2 \, L$ of $5 \, M$ solution of $H_2O_2$ contains $170 \times 2 = 340 \, g$ of $H_2O_2$.
$(ii)$ $200 \, mL$ $(0.2 \, L)$ of $5 \, M$ solution contains $H_2O_2 = 5 \, mol \, L^{-1} \times 0.2 \, L = 1 \, mol$.
Mass of $1 \, mol$ of $H_2O_2 = 34 \, g$.
The decomposition reaction is: $2H_2O_2 \rightarrow 2H_2O + O_2$.
From the stoichiometry,$2 \, mol$ of $H_2O_2$ $(68 \, g)$ produces $1 \, mol$ of $O_2$ $(32 \, g)$.
Thus,$1 \, mol$ of $H_2O_2$ $(34 \, g)$ will produce $\frac{32}{2} = 16 \, g$ of $O_2$.

(N/A) $(i)$ Since the colourless liquid $A$ contains only hydrogen and oxygen and decomposes slowly on exposure to light but is stabilised by the addition of urea,liquid $A$ is hydrogen peroxide $(H_2O_2)$. The structure of $H_2O_2$ is an open book structure.
$(ii)$ The decomposition reaction is:
$2 H_2O_2 \xrightarrow{h\nu} 2 H_2O + O_2$

(B) The relationship between volume strength and molarity is given by: $\text{Volume strength} = 11.2 \times \text{Molarity}$.
$\text{Molarity} = \frac{5.6}{11.2} = 0.5 \ M$.
Assuming $1 \ L$ $(1000 \ mL)$ of solution:
$\text{Mass of solution} = 1000 \ mL \times 1 \ g/mL = 1000 \ g$.
$\text{Moles of } H_2O_2 = \text{Molarity} \times \text{Volume in Liters} = 0.5 \ mol/L \times 1 \ L = 0.5 \ mol$.
$\text{Mass of solute } (H_2O_2) = 0.5 \ mol \times 34 \ g/mol = 17 \ g$.
$\text{Mass percentage} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100 = \frac{17}{1000} \times 100 = 1.7 \%$.

(A) The decomposition of $H_{2}O_{2}$ is given by: $2H_{2}O_{2}(aq) \rightarrow 2H_{2}O(l) + O_{2}(g)$.
At $STP$ ($273 \, K$ and $1 \, atm$),$1 \, mole$ of $O_{2}$ gas occupies $22.4 \, L$.
From the stoichiometry,$2 \, moles$ of $H_{2}O_{2}$ produce $1 \, mole$ of $O_{2}$.
Therefore,$1 \, mole$ of $H_{2}O_{2}$ produces $11.2 \, L$ of $O_{2}$ at $STP$.
Volume strength $= Molarity \times 11.2 = 8.9 \times 11.2 = 99.68$.
Rounding off to the nearest integer,we get $100$.

(A) $H_2O_2$ has a non-planar (open book) structure due to the repulsion between the lone pairs of electrons on the oxygen atoms.
In its pure state,it is a nearly colorless (very pale blue) liquid.

(A) $H_2O_2$ is prepared by the action of dilute sulfuric acid on hydrated barium peroxide $(BaO_2 \cdot 8H_2O)$.
The chemical reaction is:
$BaO_2 \cdot 8H_2O_{(s)} + H_2SO_{4(aq)} \rightarrow BaSO_{4(s)} + H_2O_{2(aq)} + 8H_2O_{(l)}$.

(B) Statement $I$ is true: $H_{2}O_{2}$ acts as an oxidising agent in basic medium,e.g.,$2Fe^{2+} + H_{2}O_{2} \rightarrow 2Fe^{3+} + 2OH^{-}$. It also acts as a reducing agent in basic medium,e.g.,$2MnO_{4}^{-} + 3H_{2}O_{2} \rightarrow 2MnO_{2} + 3O_{2} + 2H_{2}O + 2OH^{-}$.
Statement $II$ is true: The basic principle of the hydrogen economy is the transportation and storage of energy in the form of liquid or gaseous dihydrogen. Energy is transmitted as dihydrogen rather than as electric power.

(B) $H_2O_2$ is used in the treatment of effluents (industrial waste).
$H_2O_2$ acts as both an oxidising agent $(O.A.)$ and a reducing agent $(R.A.)$ in both acidic and basic media.
$H_2O_2$ has a non-planar,open-book structure. Therefore,the two hydroxyl groups do not lie in the same plane.
$H_2O_2$ is miscible with water in all proportions due to hydrogen bonding.
Thus,statements $A, B,$ and $D$ are correct.

(D) $H_{2}O_{2}$ acts as an oxidizing agent in basic medium:
$Mn^{2+} + H_{2}O_{2} \rightarrow Mn^{4+} + 2OH^{-}$
$H_{2}O_{2}$ acts as a reducing agent in basic medium:
$I_{2} + H_{2}O_{2} + 2OH^{-} \rightarrow 2I^{-} + 2H_{2}O + O_{2}$
$H_{2}O_{2}$ acts as an oxidizing agent in acidic medium:
$PbS_{(s)} + 4H_{2}O_{2(aq)} \rightarrow PbSO_{4(s)} + 4H_{2}O_{(\ell)}$
Thus,reactions $A$ and $B$ occur in basic medium.

(B) In the reaction $2I^{-} + H_{2}O_{2} + 2H^{+} \rightarrow I_{2} + 2H_{2}O$,the oxidation state of iodine increases from $-1$ to $0$,which indicates that $I^{-}$ is oxidized to $I_{2}$ by $H_{2}O_{2}$.
Thus,$H_{2}O_{2}$ acts as an oxidizing agent.
In the other reactions,$H_{2}O_{2}$ acts as a reducing agent.
Hence,the correct answer is $(B)$.

(B) $1.$ $PbO_{2}$ does not react with water to give $H_{2}O_{2}$.
$2.$ $Na_{2}O_{2} + 2 H_{2}O \rightarrow 2 NaOH + H_{2}O_{2}$. This reaction occurs readily at room temperature.
$3.$ $SnO_{2}$ is an acidic oxide and does not yield $H_{2}O_{2}$ with water.
$4.$ $BaO_{2} \cdot 8 H_{2}O$ requires the addition of a dilute acid (like $H_{2}SO_{4}$) to produce $H_{2}O_{2}$,not just water.

(C) In a basic medium,hydrogen peroxide $(H_{2}O_{2})$ acts as a reducing agent towards iodine $(I_{2})$.
The balanced chemical equation for this redox reaction is:
$I_{2} + H_{2}O_{2} + 2 OH^{-} \longrightarrow 2 I^{-} + 2 H_{2}O + O_{2}$
Thus,the product formed is the iodide ion $(I^{-})$.