Unit 09 Hydrogen
Hydrogen is the first element in the periodic table. Hydrogen has electronic configuration 1s’. The position is anomalous since in some properties it resembles alkali metals and in some it resembles halogens.
There are three isotopes of hydrogen : protium, deuterium and tritium.
Dihydrogen has two nuclear spin isomers called ortho and para-dihydrogen.
Hydrogen combines with a large number of metals as well as non-metals to form compounds which are collectively called hydrides
Resemblance with alkali metals
1. Electronic configuration - $\mathrm{H}$ contains one electron in its valence shell $\mathrm{Is}^{1}$ same as $n \mathrm{~s}^{1}$ in alkali metals.
2. Electropositive character $-\mathrm{H}$ can lose one electron to form $\mathrm{H}^{+}$
$\qquad \mathrm{H} \longrightarrow \mathrm{H}^{+}+\mathrm{e}^{-}$
3. Oxidation state $-\mathrm{H}$ shows +1 oxidation state
$\qquad \mathrm{H}^{+} \mathrm{Cl}^{-}$
$\qquad \mathrm{Na}^{+} \mathrm{Cl}^{-}$
4. Liberation at cathode - When aqueous solution of $\mathrm{HCl}$ is electrolyzed, $\mathrm{H} _{2}$ is liberated at cathode in the same way as alkali metals are liberated at cathode during electrolysis of their fused halides
5. Reducing character-Hydrogen acts as strong reducing agent.
$\mathrm{Fe} _{3} \mathrm{O} _{4}(\mathrm{~s})+4 \mathrm{H} _{2}(\mathrm{~g}) \xrightarrow{\Delta} 3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H} _{2} \mathrm{O}(\mathrm{g})$
Resemblance with halogens
1. Electronic configuration - Like halogens, $\mathrm{H}$ has one electron less than the nearest inert gas.
2. Electronegative character $-\mathrm{H}$ can gain $1 \mathrm{e}^{-}$to form hydride
$\mathrm{H}+\mathrm{e}^{-} \longrightarrow \mathrm{H}^{-}$
3. Oxidation state-H shows -1 oxidation state $\mathrm{Na}^{+} \mathrm{H}^{-}$.
4. Liberation at anode - When fused alkali metal hydrides are subjected to electrolysis, hydrogen is liberated at anode.
$$ \begin{aligned} & \ & 2 \mathrm{NaH}(\mathrm{I}) \xrightarrow{\text { electrolysis }} \begin{array}{l} \text { Cathode } \\ 2 \mathrm{Na}(\mathrm{l}) \end{array}+\begin{array}{l} \text { anode } \\ \mathrm{H} _{2}(\mathrm{~g}) \end{array} \end{aligned} $$
Isotopes of hydrogen
$\qquad { } _{1}^{1} \mathrm{H} \qquad \qquad { } _{1}^{2} \mathrm{H} \qquad\qquad { } _{1}^{3} \mathrm{H}$
protium $\qquad $ deuterium $\qquad $ tritium
They differ from each other only in the number of neutrons in the nucleus. Since they have same electronic configuration $1 s^{1}$, they have similar chemical properties.
Dihydrogen is the most abundant element in the universe $(70 %$ of the total man of the universe)
Preparation
1. By action of water on active metals
$$ \begin{aligned} & 2 \mathrm{Na}(\mathrm{s})+2 \mathrm{H} _{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{H} _{2}(\mathrm{~g}) \\ & \mathrm{Ca}(\mathrm{s})+2 \mathrm{H} _{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Ca}(\mathrm{OH}) _{2}(\mathrm{aq})+\mathrm{H} _{2}(\mathrm{~g}) \\ & 3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H} _{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{Fe} _{3} \mathrm{O} _{4}(\mathrm{~s})+4 \mathrm{H} _{2}(\mathrm{~g}) \end{aligned} $$
2. By electrolysis of water
$2 \mathrm{H} _{2} \mathrm{O}(\mathrm{I}) \xrightarrow{\text { electrolysis }} 2 \mathrm{H} _{2}(\mathrm{~g})+\mathrm{O} _{2}(\mathrm{~g})$
3. By reaction of metals with alkalies
$$ \begin{aligned} \mathrm{Zn}(\mathrm{s})+2 \mathrm{NaOH}(\mathrm{aq}) \xrightarrow{\Delta} & \mathrm{Na} _{2} \mathrm{ZnO} _{2}(\mathrm{aq})+\mathrm{H} _{2}(\mathrm{~g}) \\ & \text { Sod. zincate } \\ \mathrm{Be}(\mathrm{s})+2 \mathrm{NaOH}(\mathrm{aq}) \xrightarrow{\Delta} & \mathrm{Na} _{2} \mathrm{BeO} _{2}(\mathrm{aq})+\mathrm{H} _{2}(\mathrm{~g}) \\ & \text { Sod. beryllate } \end{aligned} $$
4. By reaction of metals with acids
$$ \begin{aligned} & \mathrm{Zn}(\mathrm{s})+\mathrm{H} _{2} \mathrm{SO} _{4}(\mathrm{aq}) \longrightarrow \mathrm{ZnSO} _{4}(\mathrm{aq})+\mathrm{H} _{2}(\mathrm{~g}) \\ & \mathrm{Mg}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{MgCl} _{2}(\mathrm{aq})+\mathrm{H} _{2}(\mathrm{~g}) \end{aligned} $$
Coal Gasification - Process of producing syn gas from coke or coal.
$$ \begin{aligned} & \mathrm{C}(\mathrm{s})+\mathrm{H} _{2} \mathrm{O}(\mathrm{g}) \xrightarrow[\mathrm{Ni}]{\Delta} \mathrm{CO}(\mathrm{g})+\mathrm{H} _{2}(\mathrm{~g}) \\ & \text { coke \quad steam \qquad \qquad water gas } \end{aligned} $$
Mixture of $\mathrm{CO}$ and $\mathrm{H} _{2}$ is used for synthesis of methane and other hydrocarbons so it is known as syn gas.
Water gas shift reaction - Conversion of water gas into syn gas
$$ \underset{\text { Water gas }}{\mathrm{CO}(\mathrm{g})+\mathrm{H} _{2} \mathrm{O}(\mathrm{g})}+\mathrm{H} _{2} \mathrm{O}(\mathrm{g}) \xrightarrow[\mathrm{FeCrO} _{4}]{\text { 673K }} \underset{\text { syn gas }}{\mathrm{CO} _{2}(\mathrm{~g})+2 \mathrm{H} _{2}(\mathrm{~g})} $$
Properties of dihydrogen
Stable and relatively inert at room temperature due to high $\mathrm{H}-\mathrm{H}$ bond dissociation enthalpy.
Highly combustible gas burns with a pale blue flame to form water, not a supporter of combustion.
Reactions of dihydrogen
$$ \begin{aligned} & \mathrm{H} _{2}(\mathrm{~g})+\mathrm{S}(\mathrm{I}) \xrightarrow{700 \mathrm{~K}} \mathrm{H} _{2} \mathrm{~S}(\mathrm{~g}) \\ & \mathrm{C}(\mathrm{s})+2 \mathrm{H} _{2}(\mathrm{~g}) \xrightarrow{\Delta} \mathrm{CH} _{4}(\mathrm{~g}) \\ & \mathrm{H} _{2}(\mathrm{~g})+\mathrm{F} _{2}(\mathrm{~g}) \xrightarrow{\Delta} 2 \mathrm{HF}(\mathrm{g}) \\ & \mathrm{H} _{2}(\mathrm{~g})+\mathrm{Cl} _{2}(\mathrm{~g}) \xrightarrow[\mathrm{hv}]{\Delta} 2 \mathrm{HCl}(\mathrm{g}) \\ & \mathrm{N} _{2}(\mathrm{~g})+3 \mathrm{H} _{2}(\mathrm{~g}) \xrightarrow[\mathrm{Fe}, \mathrm{Mo}]{673 \mathrm{~K}, 20 \mathrm{~atm}} 2 \mathrm{NH} _{3}(\mathrm{~g}) \text { (Haber’s process) }+\Delta \mathrm{H} \\ & \mathrm{CuO}(\mathrm{s})+\mathrm{H} _{2}(\mathrm{~g}) \xrightarrow{\Delta} \mathrm{Cu}(\mathrm{s})+\mathrm{H} _{2} \mathrm{O}(\mathrm{g}) \text { (hydrogen as reducing agent) } \\ & \mathrm{CH} _{2}=\mathrm{CH} _{2}(\mathrm{~g})+\mathrm{H} _{2}(\mathrm{~g}) \xrightarrow[\Delta]{\mathrm{NiorPt}} \mathrm{CH} _{3}-\mathrm{CH} _{3}(\mathrm{~g}) \text { (hydrogenation) } \end{aligned} $$
Uses:
1. Liquid hydrogen (mixed with liquid oxygen) is used as rocket fuel.
2. Atomic hydrogen and oxy - hydrogen torch for cutting and welding.
3. In synthesis of ammonia
4. In manufacturing of vanaspati fat by hydrogenation of poly unsaturated vegetable oil.
5. In manufacturing of bulk organic chemicals
6. In manufacturing of metal hydrides
Hydrides
1. Ionic hydrides : formed by s block elements LiH, NaH, KH. They are stoichiometric.
2. Metallic or interstitial hydrides : formed by $d$ and $f$ block elements. They are non stoichiometric eg. $\mathrm{LaH} _{287} \mathrm{Yb} _{255} \mathrm{H}$
3. Molecular / covalent hydrides: formed by $p$ block elements
a) Electron deficient hydrides : formed by group 13 elements. They do not have sufficient number of electrons to form normal covalent bonds.
They exist in polymeric forms eg. $\mathrm{BH} _{3}, \mathrm{AlH} _{3}$
b) Electron precise hydrides - formed by group 14 elements. They can neither act as lewis acid or lewis base eg. $\mathrm{CH} _{4}, \mathrm{SiH} _{4}$
c) Electron rich hydrides - formed by group 15, 16, 17 elements. They can act as Lewis base due to presence of lone pair of electrons eg. $\mathrm{NH} _{3}, \mathrm{HF}, \mathrm{H} _{2} \mathrm{O}$.
Water
Properties
It has high boiling point due to presence of intermolecular hydrogen bonding.
Due to its high dielectric constant (78.39) it is a universal solvent.
Amphoteric character - It can act as acid as well as base
$\mathrm{H} _{2} \mathrm{O}(\mathrm{I})+\mathrm{NH} _{3}(\mathrm{~g}) \longrightarrow \mathrm{NH} _{4}^{+}+\mathrm{OH}^{-}$
acid $\quad $ base
$\mathrm{H} _{2} \mathrm{O}(\mathrm{I})+\mathrm{H} _{2} \mathrm{~S}(\mathrm{~g}) \longrightarrow \mathrm{H} _{3} \mathrm{O}^{+}+\mathrm{HS}^{-}$
base $\quad $ acid
Oxidising agent
$ 2 \mathrm{Na}(\mathrm{s})+2 \mathrm{H} _{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{H} _{2}(\mathrm{~g})$
Reducing agent
$2 \mathrm{~F} _{2}(\mathrm{~g})+2 \mathrm{H} _{2} \mathrm{O}(\mathrm{I}) \longrightarrow \mathrm{O} _{2}(\mathrm{~g})+4 \mathrm{H}^{+}+4 \mathrm{~F}^{-}$
Water has a higher specific heat, thermal conductivity, surface tension, dipole moment & dielectric constant.
It is an excellent solvent for transportation of ions and molecules required for plant and animal metabolism.
Due to hydrogen bonding with polar molecules, even covalent compounds like alcohols and carbohydrates dissolve in water.
Structure of Water
- Water is regarded as a universal solvent.
- Depending upon the behavior of water towards soap, water may be classified as soft or hard water.
- Degree of hardness is expressed in ppm.
Types of hardness of water
1. Temporary hardness - due to bicarbonates of $\mathrm{Ca}$ and $\mathrm{Mg}$.
2. Permanent hardness - due to chlorides and sulphates of $\mathrm{Ca}$ and $\mathrm{Mg}$.
Methods of removing hardness
Temporary hardness
1. By boiling
$$ \begin{aligned} & \mathrm{Ca}\left(\mathrm{HCO} _{3}\right) _{2} \xrightarrow{\Delta} \mathrm{CaCO} _{3}(\mathrm{~s})+\mathrm{CO} _{2}(\mathrm{~g})+\mathrm{H} _{2} \mathrm{O} \\ & \mathrm{Mg}\left(\mathrm{HCO} _{3}\right) _{2} \xrightarrow{\Delta} \mathrm{MgCO} _{3}(\mathrm{~s})+\mathrm{CO} _{2}(\mathrm{~g})+\mathrm{H} _{2} \mathrm{O} \end{aligned} $$
2. By Clark’s process - calculated amount of quick lime is added
$$ \begin{aligned} & \mathrm{CaO}(\mathrm{s})+\mathrm{H} _{2} \mathrm{O} \longrightarrow \mathrm{Ca}(\mathrm{OH}) _{2}(\mathrm{~s}) \\ & \text { quick lime } \quad \text { slaked line } \\ & \mathrm{Ca}\left(\mathrm{HCO} _{3}\right) _{2}+\mathrm{Ca}(\mathrm{OH}) _{2} \longrightarrow 2 \mathrm{CaCO} _{3}(\mathrm{~s})+2 \mathrm{H} _{2} \mathrm{O}(\mathrm{I}) \\ & \mathrm{Mg}\left(\mathrm{HCO} _{3}\right) _{2}+\mathrm{Ca}(\mathrm{OH}) _{2} \longrightarrow \mathrm{CaCO} _{3}(\mathrm{~s})+\mathrm{MgCO} _{3}(\mathrm{~s}) \downarrow+2 \mathrm{H} _{2} \mathrm{O}(\mathrm{I}) \end{aligned} $$
Permanent hardness
1. Ion exchange method $-\mathrm{Ca}^{2+}$ and $\mathrm{Mg}^{2+}$ present in hard water are exchanged by ions present in ion exchangers. (sodium aluminium silicate) $2 \mathrm{NaZ}+\mathrm{M}^{2+} \longrightarrow \mathrm{MZ} _{2}+2 \mathrm{Na}^{+}$
Cation exchange resin - They can exchange $\mathrm{H}^{+}$ions with cations such as $\mathrm{Ca}^{2+}$ and $\mathrm{Mg}^{2+}$ ions present in hard water.
$\underset{\text { cation exchange resin }}{2 \mathrm{RCOOH}}+\underset{\text { (from hard water) }}{\mathrm{CaCl} _{2}} \longrightarrow \underset{\text { (exhausted resin) }} {(RCOO) _{2}} \mathrm{Ca}+2 \mathrm{Ca}^{+}+2 \mathrm{Cl}^{-}$
Anion exchange resin - They can exchange $\mathrm{OH}^{-}$ions with anions such as $\mathrm{Cl}^{-}$and $\mathrm{SO} _{4}{ }^{2-}$ present in water.
$\mathrm{R}-\mathrm{N}^{+} \mathrm{H} _{3} \mathrm{OH} \qquad \qquad + \qquad \qquad \mathrm{Cl}^{-} \longrightarrow \mathrm{R}-\mathrm{N}^{+} \mathrm{H} _{3} \mathrm{Cl}^{-}+\mathrm{OH}^{-}$
Anion exchange resin $\qquad \qquad \qquad $ from hard water
$\mathrm{H}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{H} _{2} \mathrm{O}$
Water obtained is free from all cations and anions. Water obtained by this process is called de-iosised water.
2. Calgon Process
$2 \mathrm{CaCl} _{2} \qquad \qquad + \qquad \qquad \mathrm{Na} _{2}\left[\mathrm{Na} _{4}\left(\mathrm{PO} _{3}\right) _{6}\right] \longrightarrow \mathrm{Na} _{2}\left[\mathrm{Ca} _{2}\left(\mathrm{PO} _{3}\right) _{6}\right]+4 \mathrm{NaCl}$
from hard water $\qquad \qquad \qquad $ sod. hexametaphosphate (calgon)
Reactions of heavy water $\left(\mathrm{D} _{2} \mathrm{O}\right)$
1. $2 \mathrm{D} _{2} \mathrm{O} \xrightarrow{\text { Electroysis }} 2 \mathrm{D} _{2}(\mathrm{~g})+\mathrm{O} _{2}(\mathrm{~g})$
2. $\mathrm{Ca}(\mathrm{s})+2 \mathrm{D} _{2} \mathrm{O} \longrightarrow \mathrm{Ca}(\mathrm{OD}) _{2}+\mathrm{D} _{2}(\mathrm{~g})$
3. $\mathrm{CaO}(\mathrm{s})+\mathrm{D} _{2} \mathrm{O} \longrightarrow \mathrm{Ca}(\mathrm{OD}) _{2}$
4. $\mathrm{P} _{4} \mathrm{O} _{10}(\mathrm{~s})+6 \mathrm{D} _{2} \mathrm{O}(\mathrm{I}) \longrightarrow 4 \mathrm{D} _{3} \mathrm{PO} _{4}(\mathrm{l})$ deuterophosphoric acid
5. $\mathrm{SO} _{3}+\mathrm{D} _{2} \mathrm{O}(\mathrm{I}) \longrightarrow \mathrm{D} _{2} \mathrm{SO} _{4}$ deuterosulphuric acid
6. $\mathrm{Ca} _{3} \mathrm{P} _{2}(\mathrm{~s})+6 \mathrm{D} _{2} \mathrm{O} \longrightarrow 3 \mathrm{Ca}(\mathrm{OD}) _{2}+2 \mathrm{PD} _{3}(\mathrm{~g})$ deuterophosphine
Uses:
1. Tracer compound for studying mechanism of reactions.
2. As a moderator in nuclear reactions to slow down fast moving neutrons.
Hydrogen Peroxide
Preparation
1. From sodium peroxide
$\mathrm{Na} _{2} \mathrm{O} _{2}(\mathrm{~s})+\mathrm{H} _{2} \mathrm{SO} _{4}(\mathrm{aq}) \longrightarrow \mathrm{Na} _{2} \mathrm{SO} _{4}(\mathrm{aq})+\mathrm{H} _{2} \mathrm{O} _{2}$
2. From barium peroxide
$\mathrm{BaO} _{2} \cdot 8 \mathrm{H} _{2} \mathrm{O}+\mathrm{H} _{2} \mathrm{SO} _{4} \longrightarrow \mathrm{BaSO} _{4}+\mathrm{H} _{2} \mathrm{O} _{2}+2 \mathrm{H} _{2} \mathrm{O}$
Strength of $\mathrm{H} _{2} \mathrm{O} _{2}$ solution
1. Volume strength $=5.6 x$ Normality
$$ \begin{aligned} & =5.6 \times \frac{\% \text { strength }}{\text { Eq wt of } \mathrm{H} _{2} \mathrm{O} _{2}(17)} \\ & =5.6 \times \frac{\text { strength in gL }}{\text { Eq wt of } \mathrm{H} _{2} \mathrm{O} _{2}(17)} \end{aligned} $$
2. Volume strength $=11.2 \times$ Molarity
$$ \begin{aligned} & =11.2 \times \frac{\% \text { strength }}{\text { Mol. Wt of } \mathrm{H} _{2} \mathrm{O} _{2}(34)} \\ & =5.6 \times \frac{\text { strength in gL }}{\text { Eq wt of } \mathrm{H} _{2} \mathrm{O} _{2}(17)} \end{aligned} $$
Properties
More viscous than water since molecules are more highly associated through $\mathrm{H}$ bonds than $\mathrm{H} _{2} \mathrm{O}$ molecules.
Behaves as weak acid
$\mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{I}) \longrightarrow \mathrm{H}^{+}+\mathrm{HO} _{2}^{-}$(hydroperoxide iorn)
$\mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{I}) \longrightarrow 2 \mathrm{H}^{+}+\mathrm{O} _{2}^{2-}$ (peroxide iron)
Oxidising character
In acidic medium
$\mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{I})+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H} _{2} \mathrm{O}(\mathrm{I})$
$2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{I})+2 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{H} _{2} \mathrm{O}(\mathrm{l})$
$2 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{I})+2 \mathrm{H}^{+} \longrightarrow \mathrm{I} _{2}+2 \mathrm{H} _{2} \mathrm{O}(\mathrm{I})$
$\mathrm{PbS}(\mathrm{s})+4 \mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{I}) \longrightarrow \mathrm{PbSO} _{4}(\mathrm{~s})+\mathrm{H} _{2} \mathrm{O}(\mathrm{l})$
black white
Restores the white colour of lead paintings which have blackened due to formation of lead sulphide.
In alkaline medium
$\mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{OH}^{-}$
$\mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{aq})+10 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{CrO} _{4}^{2-}(\mathrm{aq})+8 \mathrm{H} _{2} \mathrm{O}(\mathrm{l})$
$\mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{aq})+2 \mathrm{OH}^{-} \longrightarrow \mathrm{MnO} _{2}(\mathrm{~s})+2 \mathrm{H} _{2} \mathrm{O}(\mathrm{l})$
Reducing character
In acidic medium
$\mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{aq}) \longrightarrow 2 \mathrm{H}^{+}+\mathrm{O} _{2}(\mathrm{~g})+2 \mathrm{e}^{-}$
$\mathrm{Cr} _{2} \mathrm{O} _{7}^{2-}(\mathrm{aq})+8 \mathrm{H}^{+}+3 \mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+7 \mathrm{H} _{2} \mathrm{O}(\mathrm{I})+3 \mathrm{O} _{2}(\mathrm{~g})$
$2 \mathrm{MnO} _{4}^{-}(\mathrm{aq})+6 \mathrm{H}^{+}+5 \mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{aq}) \longrightarrow 2 \mathrm{Mn}^{2+}(\mathrm{aq})+8 \mathrm{H} _{2} \mathrm{O}+5 \mathrm{O} _{2}(\mathrm{~g})$
In alkaline medium
$\mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{aq})+2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H} _{2} \mathrm{O}+\mathrm{O} _{2}(\mathrm{~g})+2 \mathrm{e}^{-}$
$\mathrm{I} _{2}+\mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{aq})+2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{I}^{-}+2 \mathrm{H} _{2} \mathrm{O}+\mathrm{O} _{2}$
$2 \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{aq})+2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{O} _{2}(\mathrm{~g})+2 \mathrm{H} _{2} \mathrm{O}$
Bleaching action-due to liberation of nascent oxygen produced on decomposilion.
$\mathrm{H} _{2} \mathrm{O} _{2}(\mathrm{aq}) \longrightarrow \mathrm{H} _{2} \mathrm{O}+[0]$
colouring matter $+[0] \longrightarrow$ colourless matter
It liberates $\mathrm{I} _{2}$ with $\mathrm{KI}$ which gives blue colour with starch.
Structure of $\mathrm{H} _{2} \mathrm{O} _{2}$
Structure of $\mathrm{H} _{2} \mathrm{O} _{2}\left(\mathrm{a}\right.$ ) in the gas phase, dihedral angle $=111.5^{\circ}$ and (b) in the solid phase at $100 \mathrm{~K}$, dihedral angle $=90.2^{\circ}$.
Hydrogen peroxide acts as an oxidizing agent, reducing agent and bleaching agent.
Strength of $\mathrm{H} _{2} \mathrm{O}$ is expressed in terms of volume of $\mathrm{O} _{2}$ liberated on heating its unit volume.
The search for an alternative source of energy has given rise to the hydrogen economy.
Liquid hydrogen is an important fuel.
Hydrogen as a Fuel
Dihydrogen releases large quantities of heat on combustion.
Pollutants in combustion of dihydrogen will be less than petrol. The only pollutant will be the oxide of dinitrogen which is present as an impurity. Thus hydrogen economy is an alternative the bane principle is the transportation and storage of energy in the form of liquid or gaseous dihydrogen. Advantage is that energy is transmitted in the form of dihydrogen and not as electric power.
Solved Problems
1. $\mathrm{HClO} _{4} \mathrm{H} _{2} \mathrm{O}$ is
a) A covalent solid containing hydrogen bonded $\mathrm{H} _{2} \mathrm{O}$
b) An ionic solid containing $\mathrm{H}^{+}$ions
c) An ionic solid containing $\mathrm{ClO} _{4}^{-}$ions
d) Aliquid
Show Answer
Solution: $\mathrm{HClO} _{4} \mathrm{H} _{2} \mathrm{O}$ is $\left[\mathrm{H} _{3} \mathrm{O}^{+}\right]\left[\mathrm{ClO} _{4}^{-}\right]$in the same way as $\mathrm{NH} _{4} \mathrm{Cl}$ is $\left[\mathrm{NH} _{4}^{+}\right][\mathrm{Cl}]$ after the proton shift. Therefore answer is $\mathrm{c}$.2. $\mathrm{ZnH} _{2}$ is a/an
a) ionichydride
b) covalent hydride
c) interstitial hydride
d) intermediate hydride
Show Answer
Solution : Like $\mathrm{ZnH} _{2}, \mathrm{CuH} _{2}, \mathrm{CdH} _{2}, \mathrm{MgH} _{2}$ and $\mathrm{HgH} _{2}$ have properties intermediate between ionic and covalent hydrides. Therefore answer is d.3. The bond dissociation energy of $\mathrm{H} _{2}, \mathrm{D} _{2}$ and $\mathrm{T} _{2}$ follow the order (D-Deuterium, $\mathrm{T}$-Tritium)
a) $\mathrm{H} _{2}>\mathrm{D} _{2}>\mathrm{T} _{2}$
b) $\mathrm{T} _{2}>\mathrm{D} _{2}>\mathrm{H} _{2}$
c) $\mathrm{D} _{2}>\mathrm{T} _{2}>\mathrm{H} _{2}$
d) $\mathrm{T} _{2}>\mathrm{H} _{2}>\mathrm{D} _{2}$
Show Answer
Solution: Bond dissociation energy of
$\mathrm{T} _{2}$ is $446.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{D} _{2}$ is $443.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and
$\mathrm{H} _{2}$ is $435.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Therefore the answer is b.
4. When NaH reacts with water, the gas liberated is
a) oxygen
b) hydrogen
c) steam
d) ozone
Show Answer
Solution: $\mathrm{NaH}+\mathrm{H} _{2} \mathrm{O} \rightarrow \mathrm{NaOH}+\mathrm{H} _{2}$ Therefore the answer is b.5. $\mathrm{H} _{2} \mathrm{O} _{2}$ on standing decomposes to produce
a) $\mathrm{H} _{2}$
b) $\mathrm{H} _{2}$ and $\mathrm{O} _{2}$
c) $\mathrm{H} _{2} \mathrm{O}$ and $\mathrm{O} _{2}$
d) $\mathrm{OH}^{-}$
Show Answer
Solution: $2 \mathrm{H} _{2} \mathrm{O} _{2} \rightarrow 2 \mathrm{H} _{2} \mathrm{O}+\mathrm{O} _{2}$ Therefore the answer is $\mathrm{c}$.6. Dipole moment of $\mathrm{H} _{2} \mathrm{O} _{2}$ is
a) Equal to that of water
b) Greater than that of water
c) Less than that of water
d) Unpredictable
Show Answer
Solution: Due to presence of two polar $0-\mathrm{H}$ bonds and non-planar structure, dipole moment of $\mathrm{H} _{2} \mathrm{O} _{2}$ becomes quite high.
Therefore the answer is $b$.
7. A hair dye available in market generally contains two bottles, one containing dye and other contains hydrogen peroxide. Two bottles are mixed before applying the dye. Function of $\mathrm{H} _{2} \mathrm{O} _{2}$ is to
a) Reduce the dye
b) Oxidise the dye
c) Dilute the dye
d) Acidify the dye solution
Show Answer
Solution : Bleaching action of $\mathrm{H} _{2} \mathrm{O} _{2}$ is due to its oxidative property.
Therefore the answer is $b$.
8. What percentage strength of $\mathrm{H} _{2} \mathrm{O} _{2}$ will be marked as ’ 20 vol'?
a) $3 \%$
b) $6 \%$
c) $9 \%$
d) $12 \%$
Hint: $2 \mathrm{H} _{2} \mathrm{O} _{2} \longrightarrow 2 \mathrm{H} _{2} \mathrm{O}+\mathrm{O} _{2}$
Show Answer
Solution: $2 \mathrm{H} _{2} \mathrm{O} _{2} \longrightarrow 2 \mathrm{H} _{2} \mathrm{O}+\mathrm{O} _{2}$
$22400 \mathrm{~cm}^{3}=68 \mathrm{~g}$ (Since, mass of $2 \mathrm{H} _{2} \mathrm{O} _{2}=68 \mathrm{~g}$ )
$1 \mathrm{~cm}^{3}=\frac{68}{22400} \mathrm{~g}$
$20 \mathrm{~cm}^{3}=\frac{68 \mathrm{~g}}{22400 \mathrm{~cm} 3} \times 20 \mathrm{~cm}^{3}$
Therefore, percentage strength
$\frac{68 \times 20}{22400} \times 1000$
$=6.06 \% \approx 6 \%$
Therefore answer is $b$.
Practice Questions
Question 1. In which of the following reactions, $\mathrm{H} _{2} \mathrm{O} _{2}$ is acting as a reducing agent?
a) $\mathrm{SO} _{2}+\mathrm{H} _{2} \mathrm{O} _{2} \longrightarrow \mathrm{H} _{2} \mathrm{SO} _{4}$
b) $2 \mathrm{KI}+\mathrm{H} _{2} \mathrm{O} _{2} \longrightarrow 2 \mathrm{KOH}+\mathrm{I} _{2}$
c) $\mathrm{PbS}+4 \mathrm{H} _{2} \mathrm{O} _{2} \longrightarrow \mathrm{PbSO} _{4}+4 \mathrm{H} _{2} \mathrm{O}$
d) $\mathrm{Ag} _{2} \mathrm{O}+\mathrm{H} _{2} \mathrm{O} _{2} \longrightarrow 2 \mathrm{Ag}+\mathrm{H} _{2} \mathrm{O}+\mathrm{O} _{2}$
Show Answer
Answer: (d)Question 2. $10 \mathrm{~mL}$ of $\mathrm{H} _{2} \mathrm{O} _{2}$ solution is treated with $\mathrm{Kl}$ and titration of liberated $\mathrm{I} _{2}$, required $10 \mathrm{~mL}$ of $1 \mathrm{~N}$ hypo. Thus $\mathrm{H} _{2} \mathrm{O} _{2}$ is
a) $1 \mathrm{~N}$
b) 5-6 volume
c) $17 \mathrm{gL}^{-1}$
d) all are correct
Show Answer
Answer: (d)Question 3. 30 volumes $\mathrm{H} _{2} \mathrm{O} _{2}$ means
a) $30 \% \mathrm{H} _{2} \mathrm{O} _{2}$
b) $30 \mathrm{~cm}^{3}$ of the solution contains $1 \mathrm{~g}$ of $\mathrm{H} _{2} \mathrm{O} _{2}$
c) $1 \mathrm{~cm}^{3}$ of the solution liberates $30 \mathrm{~cm}^{3}$ of $\mathrm{O} _{2}$ at STP
d) $30 \mathrm{~cm}^{3}$ of the solution contains one mole of $\mathrm{H} _{2} \mathrm{O} _{2}$
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Answer: (c)Question 4. In alkaline medium, $\mathrm{H} _{2} \mathrm{O} _{2}$ reacts with $\mathrm{Fe}^{3+}$ and $\mathrm{Mn}^{2+}$ respectively to give
a) $\mathrm{Fe}^{4+}$ and $\mathrm{Mn}^{4+}$
b) $\mathrm{Fe}^{2+}$ and $\mathrm{Mn}^{2+}$
c) $\mathrm{Fe}^{2+}$ and $\mathrm{Mn}^{4+}$
d) $\mathrm{Fe}^{4+}$ and $\mathrm{Mn}^{2+}$
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Answer: (c)Question 5. In transforming 0.01 mole of $\mathrm{PbS}$ to $\mathrm{PbSO} _{4}$, the volume of ’ 10 volume’ $\mathrm{H} _{2} \mathrm{O} _{2}$ required will be
a) $11.2 \mathrm{~mL}$
b) $22.4 \mathrm{~mL}$
c) $33.6 \mathrm{~mL}$
d) $44.8 \mathrm{~mL}$
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Answer: (d)Question 6. Blackened oil painting can be restored into original form by the action of
a) Chlorine
b) $\mathrm{BaO} _{2}$
c) $\mathrm{H} _{2} \mathrm{O} _{2}$
d) $\mathrm{MnO} _{2}$
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Answer: (c)Question 7. In the calgon process of softening of water, which of the following is used?
a) Sodium hexametaphosphate
b) Hydrated sodium aluminium silicate
c) Cation exchange resins
d) Anion exchange resins
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Answer: (a)Question 8. The reagent commonly used to determine hardness of water quantitatively is
a) Oxalic acid
b) Disodium salt of EDTA
c) Sodium citrate
d) Sodium thiosulphate
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Answer: (b)Question 9. The hardness of water sample containing 0.002 mole of magnesium sulphate dissolved in a litre of water is expressed as:
a) $20 \mathrm{ppm}$
b) $240 \mathrm{ppm}$
c) $2000 \mathrm{ppm}$
d) $120 \mathrm{ppm}$
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Answer: (b)Question 10. In the following reaction using isotope ${ }^{18} \mathrm{O}^{2} \mathrm{H} _{2} \mathrm{O} _{2}$
$2 \mathrm{MnO} _{4}^{-}+3 \mathrm{H} _{2} \mathrm{O} _{2}^{18} \longrightarrow 2 \mathrm{MnO} _{2}+3 \mathrm{O} _{2}+2 \mathrm{H} _{2} \mathrm{O}+2 \mathrm{OH}^{-}$
Isotopic oxygen goes:
a) both in $\mathrm{O} _{2}$
b) both in $\mathrm{MnO} _{2}$
c) both in $\mathrm{OH}^{-}$
d) one in $\mathrm{O} _{2}$ and one in $\mathrm{MnO} _{2}$