Hydrogen Questions in English

(N/A) Atomic hydrogen atoms are produced by the dissociation of dihydrogen $(H_2)$ with the help of an electric arc.
This process releases a huge amount of energy $(435.88 \ kJ \ mol^{-1})$.
This energy is utilized to generate a temperature of approximately $4000 \ K$,which is ideal for welding and cutting metals.
Atomic hydrogen is allowed to recombine on the surface to be welded,releasing the stored energy to generate the required high temperature.

(N/A) Saline hydrides (e.g.,$NaH, LiH$) react with water to form a base and hydrogen gas.
The chemical equation for the reaction is:
$MH_{(s)} + H_2O_{(l)} \to MOH_{(aq)} + H_{2(g)}$
This reaction is violent and produces fire.
$CO_2$ is a well-known fire extinguisher because it is denser than air and acts as a blanket,cutting off the supply of dioxygen $(O_2)$ to the fire.
However,$CO_2$ cannot be used to extinguish fires caused by saline hydrides. This is because saline hydrides are strong reducing agents and can reduce $CO_2$ to carbon (soot) and metal carbonates/oxides,which can further fuel the reaction or cause an explosion.
$2MH + CO_2 \to M_2CO_3 + H_2$ (or similar reduction products).
Therefore,$CO_2$ should not be used.

(N/A) $(i)$ The electrical conductance of a hydride depends upon its ionic or covalent nature. $BeH_{2}$ is a covalent polymeric hydride and does not conduct. $CaH_{2}$ is an ionic hydride,which conducts electricity in the molten state. $TiH_{2}$ is a metallic (interstitial) hydride and conducts electricity at room temperature. Thus,the increasing order is: $BeH_{2} < CaH_{2} < TiH_{2}$.
$(ii)$ The ionic character depends on the electronegativity difference. As electronegativity decreases from $Li$ to $Cs$,the ionic character increases: $LiH < NaH < CsH$.
$(iii)$ Bond dissociation enthalpy depends on bond strength. $D-D$ has a higher bond dissociation enthalpy than $H-H$ due to higher reduced mass. $F-F$ has the lowest bond dissociation enthalpy due to strong inter-electronic repulsions between lone pairs. Thus,the increasing order is: $F-F < H-H < D-D$.
$(iv)$ Ionic hydrides are strong reducing agents. $NaH$ is ionic and a strong reducing agent. $MgH_{2}$ is covalent,and $H_{2}O$ is a covalent hydride with high stability. Thus,the increasing order of reducing property is: $H_{2}O < MgH_{2} < NaH$.

(N/A) The elements of atomic numbers $15, 19, 23,$ and $44$ are phosphorus $(P)$,potassium $(K)$,vanadium $(V)$,and ruthenium $(Ru)$ respectively.
$1)$ Hydride of phosphorus $(PH_3)$:
Phosphorus forms a covalent molecular hydride. It is an electron-rich hydride due to the presence of a lone pair on the phosphorus atom.
$2)$ Hydride of potassium $(KH)$:
Potassium is a highly electropositive alkali metal. It forms an ionic (saline) hydride with dihydrogen. It is crystalline and non-volatile.
$3)$ Hydrides of vanadium $(V)$ and ruthenium $(Ru)$:
Both vanadium and ruthenium are $d$-block transition metals. They form metallic (interstitial) hydrides,which are often non-stoichiometric and conduct electricity.
$4)$ Behaviour towards water:
- Potassium hydride $(KH)$ reacts violently with water: $KH_{(s)} + H_2O_{(l)} \to KOH_{(aq)} + H_{2_{(g)}}$.
- Phosphine $(PH_3)$ is slightly soluble in water and acts as a very weak Lewis base.
- Metallic hydrides of vanadium and ruthenium are generally inert towards water.
Thus,the reactivity towards water follows the order: $KH > PH_3 > (V, Ru)H$.

(N/A) $(i)$ Hydrogen economy: It is a technique of using dihydrogen in an efficient way,involving the transportation and storage of dihydrogen in the form of liquid or gas. It is eco-friendly and releases more energy than petrol,making it suitable for use in fuel cells.
$(ii)$ Hydrogenation: It is the addition of dihydrogen to another reactant,typically in the presence of a catalyst,to reduce a compound. For example,the hydrogenation of vegetable oil using nickel as a catalyst produces edible fats.
$(iii)$ Syngas: It is a mixture of carbon monoxide $(CO)$ and dihydrogen $(H_2)$. It is also known as synthesis gas or water gas and is used for the synthesis of methanol. It is produced by the reaction of steam with hydrocarbons or coke at high temperatures: $C_nH_{2n+2} + nH_2O \xrightarrow[Ni]{1270\ K} nCO + (3n+1)H_2$.
$(iv)$ Water-gas shift reaction: It is the reaction of carbon monoxide in syngas with steam in the presence of a catalyst to increase the yield of dihydrogen: $CO_{(g)} + H_2O_{(g)} \xrightarrow[Catalyst]{673\ K} CO_{2_{(g)}} + H_{2_{(g)}}$.
$(v)$ Fuel-cell: These are devices that produce electricity directly from fuels like dihydrogen in the presence of an electrolyte. They are preferred due to their eco-friendly nature and high energy output per unit mass.

(N/A) Hydrogen has an atomic number of $1$ and an electronic configuration of $1s^1$.
Because of this configuration,it shows similarities with both alkali metals (Group $1$) and halogens (Group $17$).
Like alkali metals,it can lose one electron to form $H^+$,and like halogens,it can gain one electron to form $H^-$.
Due to these dual properties,its position in the periodic table is considered anomalous,and it is often placed separately.

(N/A) Hydrogen is the first element in the periodic table. Its placement has been a subject of discussion due to its unique properties.
$1$. Resemblance to alkali metals: Hydrogen has an electronic configuration of $1s^1$, which is similar to the outer electronic configuration $(ns^1)$ of alkali metals (Group $1$). Like alkali metals, it forms unipositive ions $(H^+)$ and forms oxides, halides, and sulfides.
$2$. Resemblance to halogens: Like halogens (Group $17$), hydrogen is short by one electron to achieve the nearest noble gas configuration. It forms a diatomic molecule $(H_2)$ and forms covalent compounds.
$3$. Differences: Unlike alkali metals, hydrogen has a very high ionization enthalpy $(1312 \text{ kJ mol}^{-1})$ and does not possess metallic characteristics. Unlike halogens, its reactivity is much lower.
$4$. Unique nature: The loss of an electron from a hydrogen atom results in a nucleus $(H^+)$ of size $\sim 1.5 \times 10^{-3} \text{ pm}$, which is extremely small compared to normal atomic/ionic sizes ($50$ to $200 \text{ pm}$). Consequently, $H^+$ does not exist freely.
Conclusion: Due to these unique properties, hydrogen is best placed separately in the periodic table.

(N/A) hydrogen atom has only one electron,which is one electron less than the stable inert gas configuration of helium $(1s^2)$.
To attain the stable electronic configuration of helium,each hydrogen atom shares its single electron with another hydrogen atom.
This sharing of electrons results in the formation of a stable diatomic molecule $(H_2)$.

(N/A) Occurrence of dihydrogen:
$(i)$ Dihydrogen is the most abundant element in the universe and is the principal element in the solar atmosphere.
(ii) However,due to its light nature,it is much less abundant ($0.15 \%$ by mass) in the earth's atmosphere.
(iii) In the combined form,it constitutes $15.4 \%$ of the earth's crust and the oceans.

(N/A) Hydrogen has three isotopes: Protium $(^1_1H)$,Deuterium ($^2_1H$ or $D$),and Tritium ($^3_1H$ or $T$).
These isotopes differ from one another in the number of neutrons present in the nucleus.
$1$. Protium $(^1_1H)$: Contains $0$ neutrons.
$2$. Deuterium $(^2_1H)$: Contains $1$ neutron.
$3$. Tritium $(^3_1H)$: Contains $2$ neutrons.
Among these,only Tritium is radioactive. The mass ratio of these isotopes (Protium : Deuterium : Tritium) is $1 : 2 : 3$.

(N/A) Hydrogen has three isotopes: protium,$^1_1H$,deuterium,$^2_1H$ (or $D$),and tritium,$^3_1H$ (or $T$). These isotopes differ from one another in the number of neutrons present in the nucleus.
Ordinary hydrogen (protium) has no neutrons,deuterium (heavy hydrogen) has one neutron,and tritium has two neutrons.
In $1934$,Harold $C$. Urey received the Nobel Prize for separating the hydrogen isotope of mass number $2$ using physical methods.
Protium is the most abundant form. Terrestrial hydrogen contains $0.0156 \%$ of deuterium,primarily as $HD$.
Tritium is rare,with a concentration of about one atom per $10^{18}$ atoms of protium. Among these,only tritium is radioactive,emitting low-energy $\beta^-$ particles with a half-life of $t_{1/2} = 12.33 \text{ years}$.
Since these isotopes share the same electronic configuration,they exhibit nearly identical chemical properties. The primary difference lies in their reaction rates,which is attributed to their different bond dissociation enthalpies.

(N/A) In the laboratory,dihydrogen is prepared by the reaction of granulated zinc with dilute hydrochloric acid:
$Zn(s) + 2H^{+}(aq) \longrightarrow Zn^{2+}(aq) + H_{2}(g)$
It can also be prepared by the reaction of zinc with aqueous alkali,such as sodium hydroxide:
$Zn(s) + 2NaOH(aq) \longrightarrow Na_{2}ZnO_{2}(aq) + H_{2}(g)$
In this reaction,sodium zincate $(Na_{2}ZnO_{2})$ is formed as a byproduct.

(N/A) The production of dihydrogen from 'coal gasification' (syngas) can be increased by reacting the carbon monoxide present in the syngas mixture with steam in the presence of iron chromate $(FeCrO_4)$ as a catalyst. This process is known as the 'water-gas shift reaction'.
The chemical equation for this reaction is:
$CO_{(g)} + H_2O_{(g)} \xrightarrow{673 \ K, \text{FeCrO}_4} CO_{2_{(g)}} + H_{2_{(g)}}$

(N/A) The electrolytic method for the bulk preparation of dihydrogen involves the following processes:
$(i)$ Electrolysis of acidified water: Pure water is a poor conductor of electricity. Therefore,traces of an acid or base are added to make it conducting. Electrolysis of this acidified water using platinum electrodes yields hydrogen at the cathode.
$2 H_2O_{(l)} \stackrel{\text{Electrolysis}}{\text{Traces of acid/base}} 2 H_{2_{(g)}} + O_{2_{(g)}}$
$(ii)$ Electrolysis of aqueous barium hydroxide: High purity $(> 99.95 \%)$ dihydrogen is obtained by electrolysing a warm aqueous solution of barium hydroxide between nickel electrodes.
$(iii)$ Electrolysis of brine: Dihydrogen is obtained as a byproduct in the manufacture of sodium hydroxide and chlorine by the electrolysis of brine solution ($NaCl$ solution).
At anode: $2 Cl_{(aq)}^{-} \longrightarrow Cl_{2_{(g)}} + 2 e^{-}$
At cathode: $2 H_2O_{(l)} + 2 e^{-} \longrightarrow H_{2_{(g)}} + 2 OH_{(aq)}^{-}$
Role of electrolyte: Pure water is a poor conductor of electricity due to very low ionization. The electrolyte (acid,base,or salt) increases the concentration of ions in the solution,thereby increasing its electrical conductivity and facilitating the flow of current required for the electrolysis process.

(N/A) $(i)$ Electrolysis of acidified water using platinum electrodes gives hydrogen.
$2 H_{2}O_{(l)} \stackrel{\text{Electrolysis}}{\text{Traces of acid/base}} 2 H_{2(g)} + O_{2(g)}$
$(ii)$ High purity $(> 99.95 \%)$ dihydrogen is obtained by electrolysing warm aqueous barium hydroxide solution between nickel electrodes.
$(iii)$ It is obtained as a byproduct in the manufacture of sodium hydroxide and chlorine by the electrolysis of brine solution. During electrolysis,the reactions that take place are:
At anode: $2 Cl_{(aq)}^{-} \longrightarrow Cl_{2(g)} + 2 e^{-}$
At cathode: $2 H_{2}O_{(l)} + 2 e^{-} \longrightarrow H_{2(g)} + 2 OH_{(aq)}^{-}$
The overall reaction is:
$2 Na_{(aq)}^{+} + 2 Cl_{(aq)}^{-} + 2 H_{2}O_{(l)}$ $\longrightarrow Cl_{2(g)} + H_{2(g)} + 2 Na_{(aq)}^{+} + 2 OH_{(aq)}^{-}$
$(iv)$ Reaction of steam on hydrocarbons or coke at high temperatures in the presence of a catalyst yields hydrogen.
$C_{n}H_{2n+2} + n H_{2}O \stackrel{1270 \ K, Ni}{\longrightarrow} nCO + (2n+1) H_{2}$
e.g.,$CH_{4(g)} + H_{2}O_{(g)} \stackrel{1270 \ K, Ni}{\longrightarrow} \underbrace{CO_{(g)} + 3 H_{2(g)}}_{\text{water gas}}$
The mixture of $CO$ and $H_{2}$ is called water gas. As this mixture is used for the synthesis of methanol and a number of hydrocarbons,it is also called synthesis gas or 'syngas'. Nowadays,'syngas' is produced from sewage,saw-dust,scrap wood,newspapers,etc. The process of producing 'syngas' from coal is called 'coal gasification'.
$C_{(s)} + H_{2}O_{(g)} \stackrel{1270 \ K}{\longrightarrow} CO_{(g)} + H_{2(g)}$

$(i) \, H_{2(g)} + M_m O_{o(s)} \xrightarrow{\Delta } m M_{(s)} + o H_2 O_{(g)}$
$(ii) \, CO_{(g)} + 2 H_{2(g)} \xrightarrow[catalyst]{\Delta } CH_3 OH_{(l)}$
$(iii) \, C_3 H_{8(g)} + 3 H_2 O_{(g)} \xrightarrow[Ni, 1270 \ K]{\Delta } 3 CO_{(g)} + 7 H_{2(g)}$
$(iv) \, Zn_{(s)} + 2 NaOH_{(aq)} \xrightarrow{heat} Na_2 ZnO_{2(aq)} + H_{2(g)}$

(N/A) $(i)$ $H_2$ is a colourless,odourless,tasteless,and combustible gas.
$(ii)$ It is lighter than air and insoluble in water.

(N/A) The chemical behaviour of dihydrogen is determined,to a large extent,by its bond dissociation enthalpy.
The $H-H$ bond dissociation enthalpy is the highest for a single bond between two atoms of any element.
The dissociation of dihydrogen into its atoms is only $0.081 \%$ at $2000 \ K$,which increases to $95.5 \%$ at $5000 \ K$.
It is relatively inert at room temperature due to the high $H-H$ bond enthalpy. Thus,atomic hydrogen is typically produced at high temperatures in an electric arc or under ultraviolet radiations.
Since its $1s$ orbital is incomplete with $1s^{1}$ electronic configuration,it combines with almost all elements.
It accomplishes reactions by $(i)$ loss of the only electron to give $H^{+}$,$(ii)$ gain of an electron to form $H^{-}$,and $(iii)$ sharing electrons to form a single covalent bond.
Reaction with halogens: It reacts with halogens $(X_{2})$ to give hydrogen halides $(HX)$: $H_{2(g)} + X_{2(g)} \longrightarrow 2HX_{(g)}$ $(X = F, Cl, Br, I)$. While the reaction with fluorine occurs even in the dark,the reaction with iodine requires a catalyst.
Reaction with dioxygen: It reacts with dioxygen to form water. The reaction is highly exothermic: $2H_{2(g)} + O_{2(g)} \xrightarrow{\text{Catalyst or heat}} 2H_{2}O_{(l)}$,$\Delta H^{\ominus} = -285.9 \ kJ \ mol^{-1}$.
Reaction with dinitrogen: With dinitrogen,it forms ammonia: $3H_{2(g)} + N_{2(g)} \xrightarrow{673 \ K, 200 \ atm, Fe} 2NH_{3(g)}$,$\Delta H^{\ominus} = -92.6 \ kJ \ mol^{-1}$. This is the method for the manufacture of ammonia by the Haber process.

(N/A) The chemical behaviour of dihydrogen is determined,to a large extent,by its bond dissociation enthalpy.
The $H-H$ bond dissociation enthalpy is the highest for a single bond between two atoms of any element.
The dissociation of dihydrogen into its atoms is only $0.081 \%$ at $2000 \ K$,which increases to $95.5 \%$ at $5000 \ K$.
It is relatively inert at room temperature due to the high $H-H$ bond enthalpy. Thus,atomic hydrogen is produced at high temperatures in an electric arc or under ultraviolet radiations.
Since its orbital is incomplete with $1s^1$ electronic configuration,it combines with almost all elements.
It accomplishes reactions by $(i)$ loss of the only electron to give $H^+$,$(ii)$ gain of an electron to form $H^-$,and $(iii)$ sharing electrons to form a single covalent bond.
Reaction with halogens: It reacts with halogens $(X_2)$ to give hydrogen halides $(HX)$: $H_{2(g)} + X_{2(g)} \longrightarrow 2HX_{(g)}$ (where $X = F, Cl, Br, I$). While the reaction with fluorine occurs even in the dark,with iodine it requires a catalyst.
Reaction with dioxygen: It reacts with dioxygen to form water. The reaction is highly exothermic: $2H_{2(g)} + O_{2(g)} \xrightarrow{\text{Catalyst or Heat}} 2H_2O_{(l)}$,$\Delta H^{\ominus} = -285.9 \ kJ \ mol^{-1}$.
Reaction with dinitrogen: With dinitrogen,it forms ammonia: $3H_{2(g)} + N_{2(g)} \xrightarrow{673 \ K, 200 \ atm, Fe} 2NH_{3(g)}$,$\Delta H^{\ominus} = -92.6 \ kJ \ mol^{-1}$. This is the method for the manufacture of ammonia by the Haber process.

(N/A) The largest single use of $Dihydrogen$ is in the synthesis of ammonia,which is used in the manufacture of nitric acid and nitrogenous fertilizers.
$Dihydrogen$ is used in the manufacture of vanaspati fat by the hydrogenation of polyunsaturated vegetable oils like soyabean,cotton seeds,etc.
It is used in the manufacture of bulk organic chemicals,particularly methanol:
$CO_{(g)} + 2H_{2(g)} \xrightarrow{\text{cobalt catalyst}} CH_3OH_{(l)}$
It is widely used for the manufacture of metal hydrides and for the preparation of hydrogen chloride,a highly useful chemical. In metallurgical processes,it is used to reduce heavy metal oxides to metals.
$Dihydrogen$ is used in fuel cells for generating electrical energy. It has many advantages over conventional fossil fuels and electric power,as it does not produce pollution and releases greater energy per unit mass of fuel compared to gasoline.
Atomic hydrogen and oxy-hydrogen torches are used for cutting and welding. Atomic hydrogen atoms (produced by dissociation of $Dihydrogen$ using an electric arc) are allowed to recombine on the surface to be welded,generating temperatures up to $4000 \ K$.

(N/A) Saline hydrides react violently with water to produce $H_2$ gas,which is highly combustible.
$NaH_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(aq)} + H_{2_{(g)}}$
$CO_2$ cannot be used as a fire extinguisher in this case because it reacts with saline hydrides. For example,$CO_2$ is reduced by $NaH$ to form sodium formate:
$2NaH_{(s)} + CO_{2(g)} \longrightarrow HCOONa_{(s)}$
Since this reaction is exothermic and can provide oxygen or react further,$CO_2$ is not suitable for extinguishing fires involving saline hydrides.

(N/A) Dihydrogen,under certain reaction conditions,combines with almost all elements,except noble gases,to form binary compounds,called hydrides.
If $E$ is the symbol of an element,then the hydride can be expressed as $EH_{x}$ (e.g.,$MgH_{2}$) or $E_{m}H_{n}$ (e.g.,$B_{2}H_{6}$).
The hydrides are classified into three categories:
$(i)$ Ionic or saline or salt-like hydrides: These are stoichiometric compounds of dihydrogen formed with most of the $s$-block elements which are highly electropositive in character.
$(ii)$ Covalent or molecular hydrides: These are compounds of dihydrogen formed with most of the $p$-block elements.
$(iii)$ Metallic or non-stoichiometric hydrides: These are formed by many $d$-block and $f$-block elements. These hydrides are often non-stoichiometric,being deficient in hydrogen.

(N/A) Dihydrogen,under certain reaction conditions,combines with almost all elements,except noble gases,to form binary compounds called hydrides. If '$E$' is the symbol of an element,then the hydride can be expressed as $EH_x$ (e.g.,$MgH_2$) or $E_mH_n$ (e.g.,$B_2H_6$).
The hydrides are classified into three categories:
$1.$ Ionic or saline or salt-like hydrides: These are stoichiometric compounds of dihydrogen formed with most of the $s$-block elements which are highly electropositive in character. Example: $NaH$,$CaH_2$.
$2.$ Covalent or molecular hydrides: These are compounds of dihydrogen formed with most of the $p$-block elements. Example: $CH_4$,$NH_3$,$H_2O$,$HF$.
$3.$ Metallic or non-stoichiometric hydrides: These are formed by many $d$-block and $f$-block elements. They are often non-stoichiometric,being deficient in hydrogen. Example: $LaH_{2.87}$,$YbH_{2.55}$.

(N/A) These are stoichiometric compounds of dihydrogen formed with most of the $s$-block elements which are highly electropositive in character. Examples include $LiH$,$BeH_{2}$,and $MgH_{2}$.
In fact,$BeH_{2}$ and $MgH_{2}$ are polymeric in structure.
The ionic hydrides are crystalline,non-volatile,and non-conducting in the solid state. However,their melts conduct electricity,and on electrolysis,they liberate dihydrogen gas at the anode,which confirms the existence of the $H^{-}$ ion.
$2 H^{-}(\text{melt}) \xrightarrow{\text{anode}} H_{2}(g) + 2 e^{-}$
Saline hydrides react violently with water,producing dihydrogen gas.
$NaH(s) + H_{2}O(l) \longrightarrow NaOH(aq) + H_{2}(g)$
Lithium hydride is relatively unreactive at moderate temperatures with $O_{2}$ or $Cl_{2}$. It is,therefore,used in the synthesis of other useful hydrides,for example:
$8 LiH + Al_{2}Cl_{6} \longrightarrow 2 LiAlH_{4} + 6 LiCl$
$2 LiH + B_{2}H_{6} \longrightarrow 2 LiBH_{4}$

(N/A) $(i)$ $BeH_2$ is covalent (polymeric),$CaH_2$ is ionic,and $TiH_2$ is metallic (interstitial). Metallic hydrides conduct electricity,while ionic hydrides conduct only in the molten state. The order is $BeH_2 < CaH_2 < TiH_2$.
$(ii)$ Ionic character increases as the electronegativity difference between the metal and hydrogen increases. Down the group,electronegativity of the alkali metal decreases,so $LiH < NaH < CsH$.
$(iii)$ Bond dissociation enthalpy depends on bond strength. $F-F$ has a weak bond due to lone pair repulsions. $D-D$ is stronger than $H-H$ due to lower zero-point energy. The order is $F-F < H-H < D-D$.
$(iv)$ Reducing property depends on the ease of releasing hydride ions or hydrogen. $NaH$ is a strong ionic hydride,$MgH_2$ is covalent,and $H_2O$ is not a reducing agent. The order is $H_2O < MgH_2 < NaH$.

(N/A) $(i)$ The element $(Z=15)$ is a $p$-block element,so it forms a covalent hydride,e.g.,$PH_{3}$.
$(ii)$ The element $(Z=19)$ is an $s$-block element,so it forms an ionic hydride,e.g.,$KH$.
$(iii)$ The element $(Z=23)$ is a $d$-block element,so it forms a non-stoichiometric (interstitial) hydride,e.g.,$VH_{1.6}$.
$(iv)$ The element $(Z=44)$ belongs to the $d$-block (Group $8$). Elements of Group $7, 8, 9$ do not form hydrides (this is known as the hydride gap).
Regarding their behaviour towards water: Only the ionic hydride $(KH)$ reacts vigorously with water to produce dihydrogen gas: $KH + H_{2}O \longrightarrow KOH + H_{2}$.

(N/A) Molecular hydrides are classified based on the number of electrons and bonds in their Lewis structures:
$(i)$ Electron-deficient hydrides: These have fewer electrons than required for writing a conventional Lewis structure. Example: Diborane $(B_{2}H_{6})$. Elements of group $13$ form such compounds and act as Lewis acids (electron acceptors).
$(ii)$ Electron-precise hydrides: These have the exact number of electrons required to write their conventional Lewis structures. Example: $CH_{4}$. Elements of group $14$ form such compounds,which typically exhibit tetrahedral geometry.
$(iii)$ Electron-rich hydrides: These have excess electrons present as lone pairs. Example: Elements of group $15-17$ form such compounds (e.g.,$NH_{3}$ has $1$ lone pair,$H_{2}O$ has $2$,and $HF$ has $3$ lone pairs). They act as Lewis bases (electron donors) and can form hydrogen bonds.

(N/A) An electron-deficient hydride is a type of molecular hydride that has too few electrons to write its conventional Lewis structure.
Regarding structure,these hydrides often exhibit multi-center bonding (e.g.,$3$-center-$2$-electron bonds) to compensate for the lack of electrons,as seen in diborane $(B_{2}H_{6})$.
Regarding chemical reactions,because they have an incomplete octet or lack sufficient electrons to form standard covalent bonds,they act as $Lewis$ acids,meaning they are electron-pair acceptors.
They readily react with $Lewis$ bases to complete their octets or stabilize their structures.

(N/A) Dihydrogen forms molecular compounds with most of the $p$-block elements. Most familiar examples are $CH_{4}$,$NH_{3}$,$H_{2}O$,and $HF$.
For convenience,hydrogen compounds of non-metals are considered as hydrides.
Being covalent,they are volatile compounds.
Molecular hydrides are further classified according to the relative numbers of electrons and bonds in their Lewis structure into: $(i)$ electron-deficient,$(ii)$ electron-precise,and $(iii)$ electron-rich hydrides.
An electron-deficient hydride has too few electrons for writing its conventional Lewis structure. Example: Diborane $(B_{2}H_{6})$. All elements of group $13$ form electron-deficient compounds and act as Lewis acids (electron acceptors).
Electron-precise compounds have the required number of electrons to write their conventional Lewis structures. All elements of group $14$ form such compounds (e.g.,$CH_{4}$),which are tetrahedral in geometry.
Electron-rich hydrides have excess electrons present as lone pairs. Elements of group $15-17$ form such compounds. For example,$NH_{3}$ has $1$ lone pair,$H_{2}O$ has $2$ lone pairs,and $HF$ has $3$ lone pairs. They behave as Lewis bases (electron donors). The presence of lone pairs on highly electronegative atoms like $N$,$O$,and $F$ results in hydrogen bond formation,leading to the association of molecules.

(N/A) Non-stoichiometric hydrides are metallic hydrides where the ratio of hydrogen to metal is not a simple integer,and they do not follow the law of constant composition. These are often hydrogen-deficient. Examples include $LaH_{2.87}$ and $PdH_{0.6-0.8}$.
Alkali metals (Group $1$) form ionic or saline hydrides (e.g.,$LiH, NaH$),which are stoichiometric. They do not form non-stoichiometric hydrides because they have a strong tendency to form ionic bonds by transferring electrons to hydrogen,rather than forming interstitial metallic structures.

(N/A) Metallic hydrides,also known as interstitial hydrides,are formed by many $d$-block and $f$-block elements.
These hydrides are often non-stoichiometric,meaning they are deficient in hydrogen (e.g.,$LaH_{2.87}, TiH_{1.5-1.8}, PdH_{0.6-0.8}$).
In these structures,hydrogen atoms occupy the interstitial sites within the metal lattice.
Certain transition metals,particularly $Pd$ and $Pt$,possess the unique ability to absorb and accommodate a very large volume of hydrogen gas.
Because of this high absorption capacity,these metallic hydrides can serve as efficient storage media for hydrogen,acting as a potential source of energy for future applications.

(N/A) Metallic hydrides are formed by many $d$-block and $f$-block elements.
However,the metals of group $7$,$8$,and $9$ do not form hydrides. Even from group $6$,only chromium forms $CrH$. These hydrides conduct heat and electricity,though not as efficiently as their parent metals.
Unlike saline hydrides,they are almost always non-stoichiometric,being deficient in hydrogen.
For example: $LaH_{2.87}$,$YbH_{2.55}$,$TiH_{1.5-1.8}$,$ZrH_{1.3-1.75}$,$VH_{0.56}$,$NiH_{0.6-0.7}$,$PdH_{0.6-0.8}$,etc. In such hydrides,the law of constant composition does not hold.
In these hydrides,hydrogen occupies interstitial space in the metal lattice,producing distortion without any change in its type. Consequently,they are termed as interstitial hydrides.
Except for the hydrides of $Ni$,$Pd$,$Ce$,and $Ac$,other hydrides of this class have a lattice different from that of the parent metal.
The property of absorption of hydrogen on transition metals is widely used in catalytic reduction/hydrogenation reactions for the preparation of a large number of compounds. Some metals (e.g.,$Pd$,$Pt$) can accommodate a very large volume of hydrogen and,therefore,can be used as storage media.
This property has high potential for hydrogen storage and as a source of energy.

$(i) \ PbS_{(s)} + 4 H_2O_{2_{(aq)}} \longrightarrow PbSO_{4_{(s)}} + 4 H_2O_{(l)}$ (Redox reaction)
$(ii) \ 2 MnO_{4_{(aq)}}^{-} + 5 H_2O_{2_{(aq)}} + 6 H_{(aq)}^{+}$ $\longrightarrow 2 Mn_{(aq)}^{2+} + 8 H_2O_{(l)} + 5 O_{2_{(g)}}$ (Redox reaction)
$(iii) \ CaO_{(s)} + H_2O_{(g)} \longrightarrow Ca(OH)_{2_{(s)}}$ (Hydration reaction)
$(iv) \ AlCl_{3_{(g)}} + 6 H_2O_{(l)} \longrightarrow [Al(H_2O)_6]_{(aq)}^{3+} + 3 Cl_{(aq)}^{-}$ (Hydration reaction)
$(v) \ Ca_3N_{2_{(s)}} + 6 H_2O_{(l)} \longrightarrow 3 Ca(OH)_{2_{(aq)}} + 2 NH_{3_{(aq)}}$ (Hydrolysis reaction)

(N/A) Dihydrogen releases large quantities of heat on combustion. Dihydrogen can release more energy than petrol (about $3$ times).
The only pollutants will be the oxides of dinitrogen (due to the presence of dinitrogen as impurity with dihydrogen).
$A$ cylinder of compressed dihydrogen weighs about $30$ times as much as a tank of petrol containing the same amount of energy.
Tanks of metal alloy like $NaNi_{5}$,$Ti-TiH_{2}$,$Mg-MgH_{2}$ etc. are in use for storage of dihydrogen in small quantities.
The basic principle of hydrogen economy is the transportation and storage of energy in the form of liquid or gaseous dihydrogen.
Advantage of hydrogen economy is that energy is transmitted in the form of dihydrogen and not as electric power.

(N/A) $(i)$ $\text{Hydrogen economy}$: The basic principle of hydrogen economy is the storage and transportation of energy in the form of liquid or gaseous dihydrogen.
$(ii)$ $\text{Hydrogenation}$: The process of adding dihydrogen to unsaturated organic compounds in the presence of catalysts to form saturated compounds.
$(iii)$ $\text{Syngas}$: $A$ mixture of $CO$ and $H_2$ is called synthesis gas or syngas.
$(iv)$ $\text{Water-gas shift reaction}$: The reaction of carbon monoxide of syngas with steam in the presence of a catalyst $(FeCrO_4)$ to increase the production of dihydrogen: $CO_{(g)} + H_2O_{(g)} \xrightarrow[673 \ K]{\text{catalyst}} CO_{2(g)} + H_{2(g)}$.
$(v)$ $\text{Fuel-cell}$: $A$ device that converts the chemical energy of a fuel (like hydrogen,methane,methanol) directly into electrical energy.

(C) The $d$-block and $f$-block elements (transition and inner transition metals) form interstitial hydrides. In these compounds,hydrogen atoms occupy the interstitial sites (voids) in the metal lattice.

(N/A) Hydrogen has three isotopes: $(i)$ Protium $({ }_{1}^{1} H)$,$(ii)$ Deuterium (${ }_{1}^{2} H$ or $D$),and $(iii)$ Tritium (${ }_{1}^{3} H$ or $T$).

(B) Hydrogen exists as a diatomic molecule $(H_2)$ because it requires one electron to complete its valence shell,similar to the halogen group $(Group \ 17)$ elements like $F_2$,$Cl_2$,$Br_2$,and $I_2$.

(A) The reaction between lithium hydride $(LiH)$ and aluminium chloride $(Al_2Cl_6)$ is a standard method for the preparation of lithium aluminium hydride $(LiAlH_4)$.
The balanced chemical equation is:
$8 LiH + Al_2Cl_6 \to 2 LiAlH_4 + 6 LiCl$

(B) The maximum amount of industrial hydrogen is obtained from petrochemical sources (such as steam reforming of hydrocarbons like methane). Approximately $77\%$ of the global hydrogen production is derived from petrochemicals.

(C) Among the isotopes of hydrogen,tritium ($^3_1H$ or $T$) is radioactive in nature. It undergoes $\beta-$decay to form helium-$3$ $(^3_2He)$ by emitting a $\beta-$particle (electron,$^0_{-1}e$). The reaction is: $^3_1H \rightarrow ^3_2He + ^0_{-1}e$.

(B) Hydrogen $(H)$ has an electronic configuration of $1s^1$. By accepting one electron,it achieves the configuration $1s^2$,which is the stable electronic configuration of the noble gas Helium $(He)$.

(A) Water gas is a mixture of $CO$ and $H_{2}$. It is primarily used in the industrial synthesis of methanol $(CH_{3}OH)$ by the reaction: $CO(g) + 2H_{2}(g) \xrightarrow{ZnO/Cr_{2}O_{3}} CH_{3}OH(g)$.

(N/A) $(1)$ The oxygen atom in the crystal structure of ice is surrounded by four hydrogen atoms at a distance of $276 \ pm$.
$(2)$ Tritium is the rarest isotope of hydrogen, and its relative abundance compared to protium is $1: 10^{18}$.

(D) $1$. False: Metals of group $7$,$8$ and $9$ do not form hydrides at all. This region of the periodic table is known as the hydride gap.
$2$. False: This process produces syngas $(CO + H_2)$,not pure dihydrogen. Pure dihydrogen is obtained by the electrolysis of acidified water or other specific methods.
$3$. True: Hydrogen has a very high ionization enthalpy $(1312 \ kJ \ mol^{-1})$ compared to alkali metals due to its small size and the proximity of the electron to the nucleus.
$4$. True: Hydrogen shows metallic character by forming $H^+$ ions and non-metallic character by forming $H^-$ ions or covalent bonds.

(A) $(1)$ False. $d$-block elements form interstitial (or metallic) hydrides,not molecular hydrides. Molecular hydrides are typically formed by $p$-block elements.
$(2)$ False. Dihydrogen does not react with dioxygen at normal temperature because the $H-H$ bond dissociation energy is very high. The reaction requires a catalyst,heating,or an electric spark to proceed.

(A) The correct matches are as follows:
$(1)$ Ionic hydride: These are formed by $s$-block elements. However,$BeH_2$ and $MgH_2$ are polymeric and covalent. The question implies a standard classification where ionic hydrides are typically $s$-block (except $Be, Mg$). Given the options,let's re-evaluate:
$(1)$ Ionic hydride: $NaH$ (not listed).
$(2)$ Interstitial hydride: $TiH$ (metallic/interstitial).
$(3)$ Electron-precise hydride: $CH_4$ (Group $14$ elements).
$(4)$ Electron-rich hydride: $H_2O$ (Group $15-17$ elements).
Note: $BeH_2$ is polymeric covalent,$B_2H_6$ is electron-deficient. Based on standard classification: $(1)$ Ionic hydride: None of the options perfectly fit,but often $s$-block hydrides are categorized here. $(2)$ Interstitial: $TiH$. $(3)$ Electron-precise: $CH_4$. $(4)$ Electron-rich: $H_2O$. The provided solution $1-p$ is technically incorrect as $BeH_2$ is covalent. Correct matching is $(1-None, 2-q, 3-r, 4-s)$.

(A) $1$. $VH$ is a metallic (interstitial) hydride,so $1-b$.
$2$. $HF$ is a molecular (covalent) hydride,so $2-d$.
$3$. $MgH_2$ is an ionic (saline) hydride,so $3-a$.
$4$. $B_2H_6$ is an electron-deficient (molecular) hydride,but in the context of classification types provided,it is often categorized as electron-deficient or electron-precise depending on the specific textbook context; however,based on standard classification,$B_2H_6$ is electron-deficient. Given the options,$4-c$ is the intended match for electron-precise/deficient classification.
Therefore,the correct match is $1-b, 2-d, 3-a, 4-c$.

Water gas is produced when superheated steam is passed over red-hot coke or coal at $1270 \ K$ in the presence of a nickel catalyst:
$C_{(s)} + H_2O_{(g)} \xrightarrow[Ni]{1270 \ K} CO_{(g)} + H_{2(g)}$
It is difficult to obtain pure $H_2$ from water gas because $CO$ is hard to remove. To increase the production of $H_2$,$CO$ is oxidized to $CO_2$ by mixing it with more steam and passing the mixture over an $FeCrO_4$ catalyst at $673 \ K$:
$CO_{(g)} + H_{2(g)} + H_2O_{(g)} \xrightarrow[FeCrO_4]{673 \ K} CO_{2(g)} + 2H_{2(g)}$
This process is known as the water gas shift reaction. The $CO_2$ produced can be removed by scrubbing the mixture with sodium arsenite solution or by passing it through water under $30 \ atm$ pressure,leaving behind pure $H_2$.

(N/A) Metallic or interstitial hydrides are formed by many $d$-block and $f$-block metals. These hydrides are conductors of heat and electricity.
Except for saline hydrides,almost all hydrides are primarily non-stoichiometric and are deficient in $H_2$.
For example,$LaH_{2.87}, YbH_{2.55}, TiH_{1.5-1.8}, VH_{0.56}, NiH_{0.6-0.7}, PdH_{0.6-0.8}$,etc. The law of constant composition does not apply to these hydrides.

Molecular Hydrides	Metallic Hydrides
These are primarily formed by $p$-block elements and some $s$-block elements $(Be, Mg)$.	These are primarily formed by elements of groups $3, 4, 5, 10, 11, 12$ and $f$-block elements.
They form volatile compounds with low melting and boiling points.	They form hard compounds that exhibit some metallic character.
They are insulators (non-conductors) of electricity.	They are conductors of electricity.