Hydrogen Questions in English

(A) Copper $(Cu)$ is less reactive than hydrogen in the electrochemical series,so it cannot displace hydrogen from dilute acids like $HCl$. Therefore,it will not produce $H_2$ gas.

(A) Heavy water is formed by the combination of the heavy isotope of hydrogen,$_1^2D$ (deuterium),and oxygen.
$2D_2 + O_2 \to 2D_2O$
Here,$D_2$ is deuterium and $D_2O$ is heavy water.

(C) The industrial production of hydrogen from water gas is known as the water-gas shift reaction.
$CO(g) + H_2O(g) \xrightarrow[673 \ K]{Fe_2O_3, Cr_2O_3} CO_2(g) + H_2(g)$
In this process,carbon monoxide is oxidized to carbon dioxide using steam.
The resulting $CO_2$ is then removed by scrubbing the mixture with an alkali solution (like $NaOH$ or $K_2CO_3$),which absorbs the $CO_2$ and leaves pure $H_2$ gas.

(D) The reaction of saline hydrides (like $NaH$) with water produces very high purity hydrogen $(99.9 \%)$.
The chemical equation is: $NaH(s) + H_2O(l) \to NaOH(aq) + H_2(g)$.
This method is specifically used to obtain ultra-pure hydrogen.

(D) Saline hydrides,also known as ionic or salt-like hydrides,are formed by the reaction of $s$-block elements with dihydrogen at high temperatures.
They are stoichiometric compounds consisting of a metal cation and a hydride anion $\left( H^- \right)$.
Because they are ionic,they possess high melting points and conduct electricity in the molten state.
They react vigorously with water to produce hydrogen gas and are strong reducing agents.
Therefore,the statement that they are covalent in nature is incorrect.

(D) The reaction between $KF$ and $HF$ is given by:
$KF + HF \to K^+ + HF_2^-$
The formation of the $HF_2^-$ ion occurs due to strong hydrogen bonding between the fluoride ion $(F^-)$ and the hydrogen fluoride $(HF)$ molecule.

(A) $2NaNO_3 \xrightarrow{\Delta} 2NaNO_2 + O_2$
$2Pb_3O_4 \xrightarrow{\Delta} 6PbO + O_2$
$2KClO_3 \xrightarrow{\Delta} 2KCl + 3O_2$
$MgO$ is a highly stable oxide with a very high melting point. It does not decompose to release $O_2$ upon heating in a Bunsen flame and is often used as a drying agent (dehydrating agent) for gases.

(B) In a laboratory,anhydrous $CaCl_2$ is commonly used as a drying agent in a desiccator because it effectively absorbs moisture from the surrounding environment.

(A) $CaO$ (Quicklime) acts as a drying agent because it reacts with moisture $(H_2O)$ to form $Ca(OH)_2$.
It also reacts with acidic gases like $CO_2$ to form $CaCO_3$.
The reactions are:
$CaO + H_2O \to Ca(OH)_2$
$CaO + CO_2 \to CaCO_3$

(A) Alkali and alkaline earth metals react with water to produce hydrogen gas and metal hydroxides due to their high electropositive character.
For example:
$Mg + 2H_2O \to Mg(OH)_2 + H_2$
In contrast,the other options involve the formation of hydrogen peroxide $(H_2O_2)$ or sulfuric acid,not hydrogen gas $(H_2)$.

(D) Alkali metals are highly reactive and act as strong reducing agents. When they react with acids,they typically displace hydrogen to form $H_2$ gas. Hydrazoic acid $(HN_3)$,perxenic acid $(H_4XeO_6)$,and boric acid $(H_3BO_3)$ all contain acidic hydrogen atoms that can be displaced by alkali metals to release $H_2$ gas. Therefore,all the given acids produce $H_2$ gas upon reaction with alkali metals.

(B) In fused $CaH_2$,the dissociation occurs as: $CaH_2(l) \rightarrow Ca^{2+} + 2H^-$.
At the cathode,the reduction of $Ca^{2+}$ ions occurs: $Ca^{2+} + 2e^- \rightarrow Ca$.
At the anode,the oxidation of hydride ions $(H^-)$ occurs: $2H^- \rightarrow H_2(g) + 2e^-$.
Therefore,hydrogen gas is produced at the anode.

(C) The Bosch process involves the reaction of water gas $(CO + H_2)$ with steam to produce hydrogen gas and carbon dioxide.
The chemical reaction is: $CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)$.
In this process,$Fe_2O_3$ (ferric oxide) is used as the primary catalyst,while $Cr_2O_3$ (chromic oxide) is used as a promoter to increase the efficiency of the catalyst.
Therefore,the mixture $Fe_2O_3 + Cr_2O_3$ is used.

(C) Water gas,also known as syngas,is produced by passing steam over red-hot coke. The chemical reaction is: $C(s) + H_2O(g) \xrightarrow{1273 \ K} CO(g) + H_2(g)$. Thus,it is a mixture of carbon monoxide $(CO)$ and dihydrogen $(H_2)$.

(B) mixture of $O_2$ and $He$ is used for asthma patients because helium is less soluble in blood than nitrogen,which helps in easier breathing under pressure.

(B) Liquid ammonia is used in refrigeration systems primarily because it possesses a high heat of vaporisation.
When liquid ammonia evaporates,it absorbs a large amount of heat from its surroundings,making it an efficient refrigerant.

(C) $TiH_{1.73}$ is a non-stoichiometric hydride formed by transition metals.
These are known as metallic or interstitial hydrides.

(B) Hydrogen cannot be prepared by the action of dil. $H_2SO_4$ on copper or mercury as these metals are less reactive than hydrogen and cannot displace it from acids.
In the case of lead,the reaction stops after some time due to the formation of an insoluble layer of $PbSO_4$ on the metal surface.
Iron $(Fe)$ reacts readily with dil. $H_2SO_4$ to produce hydrogen gas: $Fe(s) + H_2SO_4(aq) \rightarrow FeSO_4(aq) + H_2(g)$.

(C) . $Zn + 2HCl(dil.) \to ZnCl_2 + H_2$ (Produces $H_2$)
$B$. $NaH + H_2O \to NaOH + H_2$ (Produces $H_2$)
$C$. $4Zn + 10HNO_3(dil.) \to 4Zn(NO_3)_2 + N_2O + 5H_2O$ (Produces $N_2O$,not $H_2$)
$D$. $2HCOONa \xrightarrow{\Delta } Na_2C_2O_4 + H_2$ (Produces $H_2$)
Therefore,$Zn + HNO_3(dil.)$ cannot be used for the preparation of $H_2$ because $HNO_3$ is a strong oxidizing agent and oxidizes $H_2$ to $H_2O$.

(D) The radioisotope of hydrogen is Tritium,represented as $_1H^3$ or $_1^3H$.
It consists of $1$ proton and $2$ neutrons.
The sum of the number of neutrons and protons is $1 + 2 = 3$.

(B) The reduction of metal oxides by hydrogen depends on the position of the metal in the electrochemical series.
Metals that are more electropositive than hydrogen (i.e.,have a more negative standard reduction potential) cannot be reduced by hydrogen.
Potassium $(K)$ is a highly reactive metal with a very negative reduction potential,so $K_2O$ cannot be reduced by hydrogen.
Conversely,oxides of metals like $Ag$ and $Fe$ can be reduced by hydrogen because they are less electropositive than hydrogen.

(D) The statement in option $(a)$ is incorrect because $H_3O^{+}$ does not exist freely in solution; it exists as $H_9O_4^{+}$ or similar hydrated forms.
However,the question asks for incorrect statements among the provided options.
Option $(b)$ is incorrect because dihydrogen acts as a strong reducing agent in many reactions (e.g.,reduction of metal oxides).
Option $(c)$ is incorrect because protium $(_1^1H)$ is the most common isotope,not tritium.
Since both $(b)$ and $(c)$ are incorrect statements,option $(d)$ is the correct choice.

(D) On the industrial scale,hydrogen is prepared from water gas using the water-gas shift reaction:
$CO + H_2O(g) \xrightarrow{\text{catalyst}} CO_2 + H_2$
The resulting $CO_2$ is removed by scrubbing the mixture with an alkali solution (e.g.,$NaOH$ or $K_2CO_3$):
$CO_2 + 2NaOH \longrightarrow Na_2CO_3 + H_2O$
Thus,$CO$ is oxidized to $CO_2$ using steam in the presence of a catalyst,and the $CO_2$ is subsequently removed by absorption in an alkali.

(D) Very pure hydrogen $(99.9 \%)$ is obtained by the reaction of water with metal hydrides such as sodium hydride $(NaH)$.
The chemical equation is:
$NaH(s) + H_2O(l) \rightarrow NaOH(aq) + H_2(g)$

(B) Water gas is a mixture of carbon monoxide $(CO)$ and dihydrogen $(H_2)$.
It is also known as synthesis gas or syngas.
It is produced by the action of steam on red-hot coke: $C(s) + H_2O(g) \xrightarrow{1273 \ K} CO(g) + H_2(g)$.

(B) In the Bosch process,water gas $(CO + H_2)$ is reacted with steam in the presence of an iron chromate catalyst to produce hydrogen gas:
$CO + H_2 + H_2O \xrightarrow{\text{Catalyst}} CO_2 + 2H_2$

(B) An oxidising agent itself undergoes reduction,meaning its oxidation number decreases. When $H_2$ is converted to the hydride ion $H^-$,it is reduced and acts as an oxidising agent.
In the reaction with lithium,hydrogen acts as an oxidising agent:
$H_2 + 2Li \rightarrow 2LiH$
Here,the oxidation state of $H$ changes from $0$ to $-1$.
In the other reactions,hydrogen acts as a reducing agent:
$H_2 + I_2 \rightarrow 2HI$ ($H$ goes from $0$ to $+1$)
$3H_2 + N_2 \rightarrow 2NH_3$ ($H$ goes from $0$ to $+1$)
$H_2 + S \rightarrow H_2S$ ($H$ goes from $0$ to $+1$)

(A) The number of protons in $H$,$D$,and $T$ is the same ($1$ proton each). Therefore,the number of protons in $H_2$,$D_2$,and $T_2$ is equal ($2$ protons each). The order $H_2 < D_2 < T_2$ for the number of protons is incorrect.
Bond energy increases with the mass of the isotope due to the decrease in zero-point energy,so $H_2 < D_2 < T_2$ is correct.
Boiling point increases with molecular mass and van der Waals forces,so $H_2 < D_2 < T_2$ is correct.
The number of neutrons in $H$,$D$,and $T$ is $0$,$1$,and $2$ respectively. Thus,the number of neutrons in $H_2$,$D_2$,and $T_2$ is $0$,$2$,and $4$ respectively,making $H_2 < D_2 < T_2$ correct.

(NONE) $1$. The combustion of diborane $(B_2H_6)$ is highly exothermic: $B_2H_6 + 3O_2 \rightarrow B_2O_3 + 3H_2O + \text{energy}$. This statement is correct.
$2$. $B_2H_6$ reacts with $CO$ to form the adduct $BH_3 \cdot CO$. This statement is correct.
$3$. The reaction of $K_4[Fe(CN)_6]$ with conc. $H_2SO_4$ produces $CO$ gas: $K_4[Fe(CN)_6] + 6H_2SO_4 + 6H_2O \rightarrow 2K_2SO_4 + FeSO_4 + 3(NH_4)_2SO_4 + 6CO$. This statement is correct.
$4$. Syn gas (synthesis gas) is a mixture of $CO$ and $H_2$. This statement is also correct.
Note: All given statements are chemically correct. However,in standard competitive chemistry contexts,if a question asks for an incorrect statement and all are correct,there may be a typo in the source. Based on standard chemical facts,all options provided are true.

(C) Hydrides are classified based on the number of electrons and bonds present in the molecule.
$1$. Electron-deficient hydrides have fewer electrons than required for bonding (e.g.,$B_2H_6$).
$2$. Electron-precise hydrides have the exact number of electrons required to form covalent bonds,typically found in group $14$ elements (e.g.,$CH_4, SiH_4, GeH_4$).
$3$. Electron-rich hydrides have excess electrons as lone pairs,typically found in groups $15, 16,$ and $17$ (e.g.,$NH_3, H_2O, HF$).
Therefore,$SiH_4$ is an electron-precise hydride.

(C) The ionisation energy of $H$ is low.
Due to this,$H$ readily loses its one electron to produce $H^{+}$ ion.
This $H^{+}$ ion is commonly known as a proton because it contains only $1$ proton and no neutron.

(B) Soda-acid fire extinguishers contain a solution of sodium hydrogen carbonate $(NaHCO_3)$ and sulphuric acid $(H_2SO_4)$.
When the extinguisher is inverted,these two solutions come into contact and react to produce carbon dioxide gas $(CO_2)$.
The reaction is: $2NaHCO_3 + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O + 2CO_2$.
The released $CO_2$ gas helps in extinguishing the fire.

(D) Commonly used drying agents in laboratories include anhydrous calcium chloride $(CaCl_2)$,anhydrous phosphorus pentaoxide $(P_2O_5)$,sodium sulfate $(Na_2SO_4)$,and magnesium sulfate $(MgSO_4)$.
Both anhydrous $P_2O_5$ and anhydrous $CaCl_2$ are effective dehydrating agents used to remove moisture from substances.
Therefore,both options $(A)$ and $(C)$ are correct.

(B) The reaction between beryllium chloride $(BeCl_2)$ and lithium aluminium hydride $(LiAlH_4)$ is a standard method for the preparation of beryllium hydride $(BeH_2)$.
The balanced chemical equation is:
$2BeCl_2 + LiAlH_4 \rightarrow 2BeH_2 + LiCl + AlCl_3$
Comparing this with the given equation,$X$ corresponds to $BeH_2$.

(D) Diborane $(B_2H_6)$ can undergo substitution reactions with methyl groups to form various methyl diboranes.
The known methyl diboranes are: monomethyl diborane $(B_2H_5(CH_3))$,dimethyl diborane $(B_2H_4(CH_3)_2)$,trimethyl diborane $(B_2H_3(CH_3)_3)$,and tetramethyl diborane $(B_2H_2(CH_3)_4)$.
The compound $B_2H(CH_3)_5$ (pentamethyl diborane) does not exist because it would require the substitution of a bridging hydrogen atom,which is not possible under standard conditions.
Therefore,option $D$ is the correct answer.

(B) Carbongene is a mixture of $95\% \ O_2$ and $5\% \ CO_2.$
It is used as an antidote for $CO$ poisoning because the $5\% \ CO_2$ stimulates the respiratory center in the brain to increase the rate of breathing,which helps in eliminating $CO$ from the blood.

(A) The boiling point of substances depends on the magnitude of intermolecular forces,which are primarily van der Waals forces in these non-polar molecules.
These forces increase with an increase in the molecular mass of the species.
The molecular masses are:
$H_2 = 2 \ g/mol$
$HD = 3 \ g/mol$
$D_2 = 4 \ g/mol$
$T_2 = 6 \ g/mol$
Since the molecular mass increases in the order $H_2 < HD < D_2 < T_2$,the boiling point follows the same order.
Therefore,the correct order is $H_2 < HD < D_2 < T_2$.

(A) Non-stoichiometric hydrides,also known as interstitial hydrides,are formed by many $d$-block and $f$-block elements.
Specifically,metals of group $7$,$8$,and $9$ do not form hydrides (this is known as the hydride gap).
$Zr$ and $Ti$ (group $4$ elements) form non-stoichiometric hydrides.
Alkali metals $(Li, Na)$ form stoichiometric ionic hydrides.

(D) $LiH$ is an ionic hydride.
$1$. Molten $LiH$ conducts electricity,which is a characteristic property of ionic compounds.
$2$. During the electrolysis of molten $LiH$,the hydride ion $(H^-)$ migrates to the anode and is oxidized to $H_2$ gas $(2H^- \rightarrow H_2 + 2e^-)$,confirming the presence of $H^-$ ions.
$3$. The reaction with water $(LiH + H_2O \rightarrow LiOH + H_2)$ is a characteristic reaction of ionic hydrides.
Therefore,all the given facts support the ionic nature of $LiH$.

(A) Hydrolysis of $CaC_2$ produces ethyne $(C_2H_2)$:
$CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2$
In $C_2H_2$:
Number of carbon atoms $(n)$ = $2$
Degree of Unsaturation $(DU)$ = $C + 1 - \frac{H}{2} = 2 + 1 - \frac{2}{2} = 2$
Since $DU = n = 2$,the condition is satisfied.

(D) The reaction is: $2BeCl_2 + LiAlH_4 \to 2BeH_2 + LiCl + AlCl_3$.
Thus,the compound $(X)$ is $BeH_2$.
In the monomer form,$BeH_2$ is a linear molecule with $sp$ hybridization.
It contains two $Be-H$ covalent bonds,which are $2c-2e^\Theta$ bonds.
There are no $3c-4e^\Theta$ bonds in the monomeric $BeH_2$ molecule.
Therefore,the total number of $3c-4e^\Theta$ bonds is $0$.

(B) covalent electron-deficient hydride is one where the number of valence electrons is less than that required for bonding.
$Al$ forms $AlH_3$,which exists as a polymeric structure $(AlH_3)_n$ and is electron-deficient.
$SiH_4$ and $HCl$ are covalent but are electron-precise.
$CaH_2$ is an ionic (saline) hydride.

(B) The reaction of magnesium with hot water produces hydrogen gas:
$Mg_{(s)} + 2H_2O_{(l)} \rightarrow Mg(OH)_{2(s)} + H_{2(g)}$
In contrast,$Na_2O_2 + 2HCl \rightarrow 2NaCl + H_2O_2$,$BaO_2 + 2HCl \rightarrow BaCl_2 + H_2O_2$,and $H_2S_2O_8 + 2H_2O \rightarrow 2H_2SO_4 + H_2O_2$ produce hydrogen peroxide $(H_2O_2)$.

(A) The calorific value is defined as the amount of heat produced by the complete combustion of $1 \, g$ of a fuel.
Hydrogen $(H_2)$ has the highest calorific value among common fuels,which is approximately $150 \, kJ/g$.
In contrast,hydrocarbons like $CH_4$ or $LPG$ have significantly lower calorific values (around $50 \, kJ/g$ to $55 \, kJ/g$).

(C) Interstitial hydrides are formed by the absorption of hydrogen into the metal lattice.
These hydrides are non-stoichiometric,meaning their composition varies with temperature and pressure.
They are generally less dense than the parent metal because the crystal lattice expands upon the inclusion of hydrogen atoms.
Not all $d-$ block metals form these hydrides; metals of groups $7, 8,$ and $9$ do not form them,which is known as the hydride gap.

(D) The isotopes of hydrogen are Protium $(^1H)$,Deuterium $(^2H)$,and Tritium $(^3H)$.
Protium has $1$ proton and $0$ neutrons (sum = $1$).
Deuterium has $1$ proton and $1$ neutron (sum = $2$).
Tritium has $1$ proton and $2$ neutrons (sum = $3$).
Thus,the maximum sum of the number of neutrons and protons in an isotope of hydrogen is $3$.

(A) Electron-precise hydrides are those in which the central atom has exactly the number of electrons required to complete its octet through covalent bonding,with no lone pairs remaining.
In $GeH_4$,the central atom $Ge$ (Group $14$) has $4$ valence electrons.
It forms $4$ covalent bonds with $4$ hydrogen atoms,resulting in $8$ electrons in the valence shell of $Ge$ (a complete octet) with no lone pairs.
Therefore,$GeH_4$ is an electron-precise hydride.

(A) The compositions of the given fuels are as follows:
$A$. Water gas: It is a mixture of $CO$ and $H_2$ $(iii)$.
$B$. Producer gas: It is a mixture of $CO$ and $N_2$ $(i)$.
$C$. Coal gas: It is a mixture of $CO$,$H_2$,$CH_4$,and $CO_2$ $(iv)$.
$D$. Natural gas: It consists mainly of methane $(ii)$.
Therefore,the correct matching is $A-iii, B-i, C-iv, D-ii$.

(D) $A.$ $Zn + 2NaOH \text{ (dil.)} \longrightarrow Na_2ZnO_2 + H_2(g)$ (Produces $H_2$ gas).
$B.$ $NaH + H_2O \longrightarrow NaOH + H_2(g)$ (Produces $H_2$ gas).
$C.$ $Na + (x+y)NH_3 \longrightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$ (The ammoniated electron solution is unstable and slowly decomposes to produce $H_2$ gas).
$D.$ $Cu + \text{dil. } H_2SO_4 \longrightarrow \text{No reaction}$. Since $Cu$ is below $H$ in the electrochemical series,it cannot displace $H_2$ from dilute acids.

(B) The reaction between diborane $(B_2H_6)$ and sodium hydride $(NaH)$ in an ether solvent yields sodium borohydride $(NaBH_4)$:
$B_2H_6 + 2NaH \to 2NaBH_4$.
In the borohydride ion $([BH_4]^-)$,the Boron atom is bonded to four Hydrogen atoms via single covalent bonds.
Since the Boron atom has a steric number of $4$ (four sigma bonds and zero lone pairs),its hybridisation is $sp^3$.