Hydrogen Questions in English

(C) For Protium $(P)$: $e^- = 1, p^+ = 1, n^0 = 0 \Rightarrow$ Sum = $2$.
For Deuterium $(D)$: $e^- = 1, p^+ = 1, n^0 = 1 \Rightarrow$ Sum = $3$.
For Tritium $(T)$: $e^- = 1, p^+ = 1, n^0 = 2 \Rightarrow$ Sum = $4$.
Therefore,Tritium $(T)$ has the maximum sum.

(D) Carbogen is a mixture of $95\% \, O_2$ and $5\% \, CO_2$ (or sometimes up to $10\% \, CO_2$).
It is used in medical practice for patients suffering from pneumonia or other respiratory issues to stimulate breathing.
It is also effectively used to treat victims of carbon monoxide $(CO)$ poisoning,as the $CO_2$ content stimulates the respiratory center in the brain to increase the rate of breathing,helping to displace $CO$ from hemoglobin.

(A) $1$. Hydrolysis of calcium imide $(CaNH)$: $CaNH + 2H_2O \rightarrow Ca(OH)_2 + NH_3(g) (B)$.
$2$. Oxidation of ammonia $(NH_3)$ by bleaching powder $(CaOCl_2)$: $2NH_3 + 3CaOCl_2 \rightarrow N_2(g) (C) + 3CaCl_2 + 3H_2O$.
$3$. Reaction of nitrogen $(N_2)$ with magnesium $(Mg)$: $3Mg + N_2 \rightarrow Mg_3N_2(s) (D)$.
$4$. Hydrolysis of magnesium nitride $(Mg_3N_2)$: $Mg_3N_2 + 6H_2O \rightarrow 3Mg(OH)_2 + 2NH_3(g) (B)$.
Thus,$(B) = NH_3$,$(C) = N_2$,and $(D) = Mg_3N_2$.

(C) Generally,metals react with dilute acids to produce hydrogen gas. However,$HNO_3$ is a strong oxidizing agent and usually oxidizes $H_2$ to $H_2O$.
Extremely dilute nitric acid $(2\%)$ is an exception,as it is not a strong enough oxidizing agent to oxidize $H_2$.
Therefore,the reaction is: $Mn + 2HNO_3 (\text{extremely dil.}) \to Mn(NO_3)_2 + H_2 \uparrow$.

(C) Among the isotopes of hydrogen,$Protium$ $(^1H)$ and $Deuterium$ ($^2H$ or $D$) are stable isotopes.
$Tritium$ ($^3H$ or $T$) is a radioactive isotope of hydrogen that emits low-energy $\beta$-particles.

(C) Ionic hydrides,also known as saline or salt-like hydrides,are formed by the reaction of dihydrogen with highly electropositive $s$-block elements (alkali and alkaline earth metals).
$NaH$ is an ionic hydride because it is formed by the reaction of sodium (an alkali metal) with hydrogen.
$AlH_3$ and $BH_3$ are covalent polymeric hydrides,and $NH_3$ is a covalent molecular hydride.

(D) The thermal decomposition of oxalic acid $(H_2C_2O_4)$ is given by:
$H_2C_2O_4 \xrightarrow{\Delta} CO (A) + CO_2 (B) + H_2O (C)$
Gas $(A)$ is $CO$,which burns with a blue flame and is oxidized to $CO_2$ (gas $B$).
Gas $(A)$ reacts with $Cl_2$ in the presence of sunlight or charcoal to form phosgene $(COCl_2)$:
$CO + Cl_2 \to COCl_2 (D)$
Thus,$A = CO$,$B = CO_2$,and $D = COCl_2$.

(D) The correct match for Hydrolith is $CaH_2$,which is an ionic (saline) hydride,not an electron-deficient hydride. Electron-deficient hydrides are typically formed by Group $13$ elements (e.g.,$B_2H_6$). Therefore,the match in option $D$ is incorrect.

(D) Very pure hydrogen $(99.9\%)$ is obtained by the electrolysis of warm aqueous barium hydroxide solution between nickel electrodes.
This process is preferred because it yields high-purity hydrogen compared to other industrial methods like steam reforming of hydrocarbons,which often contain impurities like $CO$ and $CO_2$.

(B) Hydrogen resembles halogens because both have a similar tendency to gain a single electron to attain a stable noble gas configuration.
The electronic configuration of hydrogen is $1s^{1}$. It requires one more electron to achieve the stable electronic configuration of helium,which is a noble gas.
Similarly,the valence shell electronic configuration of halogens is $ns^{2}np^{5}$. They also require one more electron to attain the stable electronic configuration of the nearest noble gas.

(C) Metals like $Mg$ and $Mn$ are highly reactive and can reduce dilute $HNO_3$ to produce hydrogen gas $(H_2)$.
Specifically,the reaction is: $Mn(s) + 2HNO_3(dil) \rightarrow Mn(NO_3)_2(aq) + H_2(g)$.
Other metals like $Cu$ are less reactive than hydrogen and do not displace it from dilute acids.
$Zn$ with concentrated $H_2SO_4$ produces $SO_2$ gas instead of $H_2$.

(B) Molecular hydrogen $(H_2)$ is stable due to its complete duplet configuration.
Atomic hydrogen is highly reactive because it has a single unpaired electron.
Nascent hydrogen is also highly reactive,but slightly less so than free atomic hydrogen due to its immediate generation in a reaction medium.
Occluded hydrogen (hydrogen adsorbed on metal surfaces like $Pd$,$Pt$,or $Ni$) is the least reactive among these three forms.
Therefore,the order of reactivity is: $\text{Atomic} > \text{Nascent} > \text{Occluded } H$.

(B) The reaction between lithium aluminium hydride $(LiAlH_4)$ and silicon tetrachloride $(SiCl_4)$ is a reduction reaction.
The balanced chemical equation is:
$LiAlH_4 + SiCl_4 \to LiCl + AlCl_3 + SiH_4$
Thus,the products formed are lithium chloride $(LiCl)$,aluminium chloride $(AlCl_3)$,and silane $(SiH_4)$.

(A) In the reaction of lithium hydride $(LiH)$ with diborane $(B_2H_6)$,lithium borohydride is formed without the evolution of hydrogen gas:
$2LiH + B_2H_6 \longrightarrow 2Li[BH_4]$
In contrast,the other reactions listed liberate hydrogen gas:
$1$. Electrolysis of acidified water: $2H_2O \longrightarrow 2H_2 + O_2$
$2$. Zinc with aqueous alkali: $Zn + 2NaOH + 2H_2O \longrightarrow Na_2[Zn(OH)_4] + H_2$
$3$. Sodium in liquid ammonia: $2Na + 2NH_3 \longrightarrow 2NaNH_2 + H_2$

(D) $H_2$ is a highly inflammable gas. Therefore,the statement that it is a non-inflammable gas is incorrect.

(C) Soda-acid fire extinguishers work on the principle of the reaction between an acid and a carbonate or bicarbonate salt.
They contain a solution of $NaHCO_3$ (sodium bicarbonate) and a separate bottle of $H_2SO_4$ (sulfuric acid).
When the extinguisher is activated,the acid reacts with the bicarbonate to release $CO_2$ gas,which helps in extinguishing the fire.
Therefore,the correct component is $NaHCO_3$.

(C) Hydrogen has three naturally occurring isotopes:
$1$. Protium $({}_1^1H)$
$2$. Deuterium (${}_1^2H$ or $D$)
$3$. Tritium (${}_1^3H$ or $T$)
Therefore,the isotopes of hydrogen are protium,deuterium,and tritium.

(A) The total number of isotopes of hydrogen is $3$.
These are protium $({}_{1}^{1}H)$,deuterium (${}_{1}^{2}H$ or $D$),and tritium (${}_{1}^{3}H$ or $T$).
Among these,only tritium $({}_{1}^{3}H)$ is a radioactive isotope.
Therefore,the total number of isotopes is $3$ and the number of radioactive isotopes is $1$.
Hence,the correct option is $A$.

(A) The reactivity of halogens towards hydrogen decreases down the group as the oxidizing power of halogens decreases.
$F_2$ reacts even in the dark,$Cl_2$ reacts in the presence of sunlight,and $Br_2$ requires heat.
However,the reaction between $H_2$ and $I_2$ is slow and reversible,requiring a catalyst (like finely divided platinum or nickel) and high temperature to proceed effectively.

(C) $NaH$ (Sodium hydride) is formed by the transfer of an electron from sodium to hydrogen,resulting in an ionic lattice structure. Therefore,it is classified as a saline or ionic hydride.

(D) $H_2$ produces fewer pollutants compared to petrol because its combustion does not produce $CO$ or $CO_2$.
$(b)$ $A$ cylinder of compressed dihydrogen weighs $\sim 30$ times more than a petrol tank for the same energy output,making storage difficult.
$(c)$ Dihydrogen is stored in tanks of metal alloys like $NaNi_5$ to safely contain the gas.
$(d)$ The energy released per gram of liquid dihydrogen is $142 \ kJ$,while for $LPG$ it is approximately $50 \ kJ$. Thus,statement $(d)$ is incorrect as the values are swapped.
Therefore,statements $(a), (b),$ and $(c)$ are correct.

(A) An electron-deficient hydride is one that does not have sufficient electrons to form normal covalent bonds,typically having fewer than $8$ electrons in the valence shell of the central atom.
$B_2H_6$,$GaH_3$,and $AlH_3$ are electron-deficient hydrides because the central atoms ($B$,$Ga$,$Al$) have incomplete octets.
$SiH_4$ is a group $14$ hydride where the central silicon atom has a complete octet ($8$ electrons),making it an electron-precise hydride,not an electron-deficient one.

(C) Saline (ionic) hydrides react with water to produce $H_2$ gas: $MH + H_2O \rightarrow MOH + H_2$. This is correct.
$(b)$ The reaction $3LiAlH_4 + 4BF_3 \rightarrow 2B_2H_6 + 3LiF + 3AlF_3$ is a standard method for the preparation of diborane. This is correct.
$(c)$ $PH_3$ has a lone pair on phosphorus,making it an electron-rich hydride. $CH_4$ has exactly the number of electrons required for bonding,making it an electron-precise hydride. This is correct.
$(d)$ Molecular hydrides are formed by $p$-block elements and are covalent in nature. Both $HF$ and $CH_4$ are covalent molecular hydrides. This is correct.
Therefore,all statements $(a), (b), (c),$ and $(d)$ are correct.

(C) An oxidising agent is a substance that gains electrons or causes another substance to be oxidised. In the reaction $Ca + H_2 \rightarrow CaH_2$,the oxidation state of $Ca$ changes from $0$ to $+2$ (oxidation),and the oxidation state of $H$ changes from $0$ to $-1$ (reduction). Since hydrogen is reduced,it acts as an oxidising agent. In all other options $(N_2, Br_2, S)$,hydrogen is oxidised from $0$ to $+1$,acting as a reducing agent.

(A) The liquid with the lowest boiling point is distilled first. Boiling point is generally related to molecular mass and intermolecular forces.
The boiling points of the given substances are:
$H_2$: $20.3 \ K$
$N_2$: $77.4 \ K$
$O_2$: $90.2 \ K$
$CO_2$: $194.7 \ K$ (sublimation point at $1 \ atm$)
Since liquid $H_2$ has the lowest boiling point,it will be distilled first.

(D) The enthalpy of fusion of a substance,also known as latent heat of fusion,is the change in its enthalpy resulting from providing energy to a specific quantity of the substance to change its state from a solid to a liquid at constant pressure.
The enthalpy of fusion is almost always a positive quantity. $Helium$ is the only known exception,as it has a negative enthalpy of fusion at certain conditions.

(A) $CaNH + 2H_2O \to Ca(OH)_2 + \mathop{NH_{3(g)}}\limits_{(B)}$
$2NH_3(B) + 3CaOCl_2 \to \mathop{N_{2(g)}}\limits_{(C)} + 3CaCl_2 + 3H_2O$
$N_2(C) + 3Mg \to \mathop{Mg_3N_2}\limits_{(D)}$
$Mg_3N_2(D) + 6H_2O \to 3Mg(OH)_2 + 2NH_3(B)$
Thus,$(B) = NH_3$,$(C) = N_2$,and $(D) = Mg_3N_2$.

(C) The reaction between sodium borohydride $(Na[BH_4])$ and boron trifluoride $(BF_3)$ under anhydrous conditions (typically in an ether solvent like diglyme) is a standard laboratory method for the preparation of diborane.
The balanced chemical equation is: $3 Na[BH_4] + 4 BF_3 \longrightarrow 2 B_2H_6(g) + 3 Na[BF_4]$.
In this reaction,$B_2H_6$ (diborane) is the primary gaseous product formed at room temperature.

(A) The reaction sequence is as follows:
$1$. $Na + NH_{3(g)} \xrightarrow{\Delta} NaNH_2 + \frac{1}{2} H_2$. Here,$[X]$ is $NaNH_2$ (sodamide).
$2$. $NaNH_2 + N_2O \to NaN_3 + H_2O$. Here,$[Y]$ is $NaN_3$ (sodium azide).
$3$. $2NaN_3 \xrightarrow{\Delta} 2Na + 3N_2(g)$. Here,$[Z]$ is $N_2$ (pure nitrogen gas).
Thus,the correct gas $[Z]$ is $N_2$.

(A) Calcium imide $(CaNH)$ on hydrolysis gives gas $NH_3$ and $Ca(OH)_2$. So $(B)$ is $NH_3$.
$CaNH + 2H_2O \rightarrow NH_3 + Ca(OH)_2$
$NH_3$ on oxidation by bleaching powder $(CaOCl_2)$ gives gas $N_2$. So $(C)$ is $N_2$.
$3CaOCl_2 + 2NH_3 \rightarrow 3CaCl_2 + 3H_2O + N_2$
$N_2$ on reaction with magnesium gives compound $Mg_3N_2$. So $(D)$ is $Mg_3N_2$.
$N_2 + 3Mg \rightarrow Mg_3N_2$
$Mg_3N_2$ on hydrolysis gives $NH_3$,which is $(B)$.
$Mg_3N_2 + 6H_2O \rightarrow 3Mg(OH)_2 + 2NH_3$
Therefore,$(B), (C)$ and $(D)$ are $NH_3, N_2, Mg_3N_2$.

(B) When steam is passed over red hot iron,the chemical reaction is as follows:
$3Fe(s) + 4H_2O(g) \rightarrow Fe_3O_4(s) + 4H_2(g)$
In this reaction,iron reacts with steam to form magnetic iron oxide $(Fe_3O_4)$ and hydrogen gas $(H_2)$.

(C) When steam is passed over red hot iron,the chemical reaction is as follows:
$3Fe(s) + 4H_2O(g) \rightarrow Fe_3O_4(s) + 4H_2(g)$
In this reaction,iron $(Fe)$ reacts with steam $(H_2O)$ to form magnetic iron oxide $(Fe_3O_4)$ and hydrogen gas $(H_2)$.

(B) $NH_3$ is a basic gas. It cannot be dried using acidic drying agents like $P_4O_{10}$,$H_2SO_4$,or $CaCl_2$ (which forms an adduct $CaCl_2 \cdot 8NH_3$).
Therefore,it is dried using quicklime,$CaO$,which is a basic drying agent.

(A) An alkaline solution of pyrogallol is a well-known absorbent used to remove oxygen from gas mixtures rapidly.

(A) Concentrated $H_2SO_4$ acts as a dehydrating agent. When formic acid $(HCOOH)$ is heated with concentrated $H_2SO_4$,it undergoes dehydration to produce carbon monoxide $(CO)$ and water $(H_2O)$.
$HCOOH \xrightarrow{conc. H_2SO_4} H_2O + CO$

(C) Sulfuric acid $(H_2SO_4)$ has a strong affinity for water because it forms stable hydrates with water,such as $H_2SO_4 \cdot H_2O$,$H_2SO_4 \cdot 2H_2O$,etc.,by releasing a large amount of energy.

(B) In the laboratory,hydrogen sulfide $(H_2S)$ is prepared by the action of dilute sulfuric acid $(H_2SO_4)$ on ferrous sulfide $(FeS)$.
The chemical reaction is: $FeS(s) + H_2SO_4(aq) \to FeSO_4(aq) + H_2S(g)$.

(D) Hydrogenation is the addition of $H_2$ gas to unsaturated compounds.
In the presence of a $Ni$ catalyst,hydrogen gas is added to vegetable oils to convert them into saturated fats (vanaspati).
This process is known as the hydrogenation of oils.
It helps in improving the texture,shelf life,and stability of the oils.

(A) The reaction for the complete hydrolysis of diborane $(B_2H_6)$ at room temperature is:
$B_2H_6 + 6H_2O \longrightarrow 2H_3BO_3 + 6H_2 \uparrow$
Here,the product containing the underlined atom $(B)$ is $H_3BO_3$,which is orthoboric acid.
The suffix used for $H_3BO_3$ is $-ic$ (boric acid).
Therefore,the product is an oxy acid with an $-ic$ suffix.

(B) Heavy hydrogen is known as deuterium ($^2_1H$ or $D$).
In deuterium,the atomic number $(Z)$ is $1$ and the mass number $(A)$ is $2$.
The number of neutrons $(n)$ is calculated as $n = A - Z$.
Therefore,$n = 2 - 1 = 1$.

(A) The reaction between coke $(C)$ and steam $(H_2O)$ at high temperature $(1273 \ K)$ in the presence of $Fe_2O_3$ (catalyst) and $Cr_2O_3$ (promoter) is known as the Bosch process.
The chemical reaction is: $C(s) + H_2O(g) \xrightarrow{Fe_2O_3/Cr_2O_3} CO(g) + H_2(g)$.
This mixture of $CO$ and $H_2$ is called water gas or syngas.

(D) 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 passing steam over red-hot coke at high temperatures:
$C(s) + H_2O(g) \xrightarrow{1273 \ K} CO(g) + H_2(g)$

(D) Hydrides are classified as electron-deficient,electron-precise,or electron-rich.
$CH_4$ is an electron-precise hydride because it has the exact number of electrons required to form its covalent bonds ($8$ electrons in the valence shell of carbon).
Electron-precise hydrides are generally neutral in nature.
$PH_3$ is an electron-rich hydride (contains lone pairs),$NaBH_4$ is an ionic/complex hydride,and $HBr$ is an acidic hydride.

(A) Hydrogen gas $(H_2)$ does not cause atmospheric pollution.
When hydrogen is burnt in air,the primary product is water $(H_2O)$,with only trace amounts of nitrogen oxides.
In hydrogen fuel cells,the only emission is water vapour.
Conversely,$SO_2$ contributes to acid rain and particulate matter.
$CO_2$ is a major greenhouse gas.
$CO$ is a toxic gas that can lead to the formation of secondary pollutants like tropospheric ozone.
Therefore,$H_2$ is the correct answer.

(C) In molten ionic hydrides like $NaH$,the hydride ion $(H^-)$ is present.
During electrolysis,the following reactions occur:
At the anode: $2H^- \rightarrow H_2(g) + 2e^-$.
At the cathode: $Na^+ + e^- \rightarrow Na(s)$.
However,for $MgH_2$ (molten) or $NaH$ (molten),the electrolysis of the hydride ion produces $H_2$ at the anode.
Specifically,for molten $NaH$,$H_2$ is produced at the anode.
If we consider the electrolysis of $NaH$ (molten),the reaction at the cathode is $Na^+ + e^- \rightarrow Na$.
Actually,the question refers to the electrolysis of ionic hydrides where $H^-$ is oxidized to $H_2$ at the anode.
Among the given options,$NaH$ (molten) is the standard example where $H_2$ is evolved at the anode.
Note: In some contexts,if $H_2$ is produced at both,it implies a specific setup,but for standard ionic hydrides,$H_2$ is produced at the anode.

(B) Ionic hydrides react with water to produce a basic solution and hydrogen gas.
For example: $LiH + H_2O \longrightarrow LiOH + H_2 \uparrow$
Here,$LiOH$ is a strong base,making the resulting solution basic.

(C) The reaction of methane with steam in the presence of a nickel catalyst at $725\,^oC$ is known as steam reforming of methane.
The chemical equation for this reaction is:
$CH_4(g) + H_2O(g) \xrightarrow{Ni/Al_2O_3, 725\,^oC} CO(g) + 3H_2(g)$
This process produces synthesis gas (syngas),which is a mixture of carbon monoxide $(CO)$ and hydrogen $(H_2)$.

(D) $B_2H_6$ reacts with $NH_3$,primary $(I^o)$,and secondary $(II^o)$ amines to form an ionic compound of the type $[BH_2(x)_2]^+ [BH_4]^-$.
However,with tertiary $(III^o)$ amines like $(CH_3)_3N$,$B_2H_6$ undergoes symmetric cleavage to form a simple adduct: $B_2H_6 + 2(CH_3)_3N \longrightarrow 2(CH_3)_3N \cdot BH_3$.

(C) Concentrated $H_2SO_4$ acts as a strong dehydrating agent. When it reacts with sugar $(C_{12}H_{22}O_{11})$,it removes water molecules from the sugar,leaving behind carbon. The reaction is: $C_{12}H_{22}O_{11} \xrightarrow{conc. H_2SO_4} 12C + 11H_2O$. This process is known as dehydration.