Hydrogen Questions in English

(D) The Assertion $(A)$ is correct because the hydrogen atom consists of one proton and one electron. When the electron is lost,only the nucleus (a proton) remains,which has a size of approximately $1.5 \times 10^{-3} \ pm$.
The Reason $(R)$ is also correct because a proton $(H^+)$ is highly reactive and cannot exist independently in the gas phase or aqueous solution; it always associates with other molecules or ions (e.g.,$H_3O^+$).
However,the Reason $(R)$ does not explain why the size of the proton is $1.5 \times 10^{-3} \ pm$. Therefore,both statements are correct,but $R$ is not the correct explanation of $A$.

(C) In an ice crystal,the structure is highly ordered and three-dimensional.
Each water molecule is hydrogen bonded to $4$ neighbouring water molecules in a tetrahedral arrangement.

(D) Water gas is a mixture of $CO$ and $H_2$. When water gas reacts with $H_2$ in the presence of a cobalt catalyst,it produces methanol.
$CO + 2H_2 \xrightarrow{Co} CH_3OH$

(A) The first reaction is the water-gas shift reaction (or coal gasification),which requires a high temperature of approximately $1270 \ K$ to proceed.
The second reaction is the water-gas shift reaction,which is carried out at a lower temperature of approximately $673 \ K$ in the presence of an iron chromate catalyst to increase the yield of $H_2$.
Thus,$T_1 \approx 1270 \ K$ and $T_2 \approx 673 \ K$.
Therefore,$T_1 > T_2$.

(D) The isotopes of hydrogen are protium $(^1H)$,deuterium $(^2H)$,and tritium $(^3H)$.
Assertion $A$ is true because the physical properties (such as boiling point,melting point,density) of these isotopes differ significantly due to the large relative mass difference between them.
Reason $R$ is also true because the mass of deuterium is approximately double that of protium,and tritium is triple,which is a very large percentage difference compared to isotopes of other elements.
Therefore,the mass difference is the direct cause of the difference in physical properties,making $R$ the correct explanation of $A$.

(A) Statement-$I$ is correct: The reaction of methane with steam over a heated $Ni$ catalyst is known as steam reforming of hydrocarbons,which produces syngas $(CO + H_2)$. The reaction is: $CH_{4(g)} + H_2O_{(g)} \xrightarrow[1270 \ K]{Ni} CO_{(g)} + 3H_{2(g)}$.
Statement-$II$ is correct: Sodium nitrite reacts with ammonium chloride to produce nitrogen gas,sodium chloride,and water. The reaction is: $NaNO_{2(aq)} + NH_4Cl_{(aq)} \rightarrow N_{2(g)} + NaCl_{(aq)} + 2H_2O_{(\ell)}$.

(A) $BeH_2$ is an electron-deficient (hypovalent) hydride with a polymeric structure,making it less stable compared to the other given hydrides which have complete octets or are more ionic in nature.

(A) $B_2H_6$ is an electron-deficient hydride because it has fewer electrons than required for normal covalent bonding.
$HF$ is an electron-rich hydride because it has lone pairs of electrons on the central atom.
$CH_4$ is an electron-precise hydride because it has the exact number of electrons required for covalent bonding.
$MgH_2$ is a saline (ionic) hydride formed by the reaction of $Mg$ with $H_2$.
Therefore,the correct matching is $A-III, B-II, C-IV, D-I$.

(B) Isotopes of hydrogen (protium,deuterium,and tritium) have the same electronic configuration,which leads to almost identical chemical properties.
However,they differ in their rates of reaction because of the difference in their bond dissociation enthalpies,which arises due to the difference in their isotopic masses.
Therefore,both $A$ and $R$ are correct,and $R$ is the correct explanation of $A$.

(A) The water-gas shift reaction is represented by the equation: $CO(g) + H_2O(g) \xrightarrow{\text{Iron chromate}} CO_2(g) + H_2(g)$.
In this reaction,the oxidation state of carbon in $CO$ is $+2$,and in $CO_2$ it is $+4$.
Since the oxidation state of carbon increases from $+2$ to $+4$,$CO$ is oxidized to $CO_2$.

(D) Statement $A$ is correct: Hydrogen acts as a reducing agent for heavy metal oxides.
Statement $B$ is correct: Heavy water $(D_2O)$ is used as a tracer to study reaction mechanisms.
Statement $C$ is correct: Hydrogenation of oils produces saturated fats (vanaspati ghee).
Statement $D$ is incorrect: The $H-H$ bond dissociation enthalpy $(435.88 \ kJ \ mol^{-1})$ is actually the highest for a single bond between two atoms of any element.
Statement $E$ is incorrect: Hydrogen can only reduce oxides of metals that are less active than iron (e.g.,$Cu, Pb, Ag$). It cannot reduce oxides of highly active metals like $Na, K, Ca$.
Therefore,statements $D$ and $E$ are incorrect.

(A) Small amines such as $NH_3$,$CH_3NH_2$,and $(CH_3)_2NH$ undergo unsymmetrical cleavage of diborane to form ionic products of the type $[BH_2(X)_2]^+ [BH_4]^-$.
The reaction is: $B_2H_6 + 2X \rightarrow [BH_2(X)_2]^+ [BH_4]^-$.
Large amines,such as $(CH_3)_3N$,undergo symmetrical cleavage of diborane to form borane-amine adducts: $B_2H_6 + 2N(CH_3)_3 \rightarrow 2H_3B \leftarrow N(CH_3)_3$.
Therefore,$X$ can be $NH_3$,$CH_3NH_2$,or $(CH_3)_2NH$.

(D) High purity dihydrogen $(> 99.95\%)$ is obtained by the electrolysis of warm aqueous barium hydroxide solution between $Ni$ electrodes.
In this process,the electrolyte is $Ba(OH)_2$ and the electrodes used are $Ni$ electrodes.
This method is preferred for producing high-purity hydrogen gas.

(A) Water gas is a mixture of carbon monoxide $(CO)$ and hydrogen $(H_2)$.
It is produced by the reaction of methane with steam at high temperature in the presence of a nickel catalyst:
$CH_{4(g)} + H_2O_{(g)} \xrightarrow{1270 \ K, \ Ni} CO_{(g)} + 3H_{2(g)}$
Thus,the mixture of $CO_{(g)}$ and $H_{2(g)}$ is known as water gas or syngas.

(D) The electrolysis of warm aqueous $Ba(OH)_2$ solution between nickel electrodes is a standard laboratory method used to produce high-purity dihydrogen $(H_2)$.
This process yields dihydrogen with a purity greater than $99.95 \%$.

(A) $1$. Gasification of coal: $C(s) + H_2O(g) \xrightarrow{1270 K} CO(g) + H_2(g)$. This process produces $H_2$ (syngas),but the question asks for processes involving the use of dihydrogen.
$2$. Formation of vanaspati ghee: This is the hydrogenation of vegetable oils using $H_2$ gas in the presence of a nickel catalyst.
$3$. Preparation of $HCl$: $H_2(g) + Cl_2(g) \xrightarrow{h\nu} 2HCl(g)$. This process uses $H_2$.
$4$. Preparation of metal hydride: $2M(s) + nH_2(g) \rightarrow 2MH_n(s)$. This process uses $H_2$.
$5$. Gasification of coal (specifically the water-gas shift reaction) is often confused,but the primary gasification step produces $H_2$ rather than consuming it as a reactant in the same way hydrogenation or synthesis does. However,in the context of industrial processes,gasification is a method of production,not a consumption process for $H_2$.

(A) Hydrogen resembles halogens in several properties:
$1$. Both exist as diatomic molecules $(H_2, F_2, Cl_2, Br_2, I_2)$.
$2$. Both have high ionisation enthalpy values.
$3$. Both form uninegative ions $(H^-, F^-, Cl^-, Br^-, I^-)$ by gaining one electron.
$4$. Both form covalent compounds with non-metals.
Among the given options,the ionisation enthalpy of hydrogen is comparable to that of halogens,making it the most appropriate similarity in terms of periodic trends.

(D) In the Bosch process,water gas $(CO + H_{2})$ is reacted with steam to produce $CO_{2}$ and $H_{2}$.
$Fe_{2}O_{3}$ acts as a catalyst,while $Cr_{2}O_{3}$ is used as a promoter to increase the efficiency of the catalyst.

(C) The hydride $NiH$ is an example of a metallic or non-stoichiometric hydride.
These hydrides are often electron-deficient,meaning they do not have a sufficient number of electrons to form normal covalent bonds,which is why they are also referred to as non-stoichiometric hydrides.

(C) $HF$ is a molecular (covalent) hydride because it is formed by the sharing of electrons between hydrogen and a non-metal (fluorine).
$KH$,$NaH$,and $LiH$ are examples of saline or ionic hydrides,which are formed by the transfer of electrons from an alkali metal to hydrogen.

(B) The formula of hydrolith is $CaH_{2}$.
Calcium hydride $(CaH_{2})$ is a salt-like binary hydride that is commonly used as a reducing agent and as a convenient source of hydrogen gas.

(A) The reaction $CO + H_2O \rightleftharpoons CO_2 + H_2$ is known as the water-gas shift reaction.
In industrial processes,this reaction is carried out at approximately $500^{\circ} C$ using iron chromite $(Fe_2O_3-Cr_2O_3)$ as a catalyst to increase the yield of hydrogen.

(D) The reaction $CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)$ is known as the water-gas shift reaction.
This reaction is carried out at about $500^{\circ} C$ in the presence of an iron-chromium $(Fe-Cr)$ catalyst to produce dihydrogen from water gas.

(D) Water gas is a fuel gas that consists of a mixture of carbon monoxide $(CO)$ and hydrogen $(H_2)$.
It is produced by passing steam over red-hot coke:
$C_{(s)} + H_2O_{(g)} \xrightarrow{1273 \ K} CO_{(g)} + H_{2_{(g)}}$
Therefore,the correct composition is $CO_{(g)} + H_{2_{(g)}}$.

(C) Hydrogen exhibits oxidation states of $-1$ (in metal hydrides) and $+1$ (in most compounds).
Oxygen does not have an oxidation state of $-2$ in all compounds; for example,in $OF_2$,it is $+2$.
Hydrogen has a significantly higher ionisation enthalpy compared to alkali metals due to its small size and the proximity of its electron to the nucleus.

(D) When an ionic hydride of $s$-block elements (e.g.,$NaH$) is electrolyzed in its molten state,the following reactions occur:
At the cathode: $Na^+ + e^- \rightarrow Na$
At the anode: $2H^- \rightarrow H_2 + 2e^-$
Thus,dihydrogen gas $(H_2)$ is liberated at the anode.

(A) The given reaction is the water-gas shift reaction.
In this industrial process,$CO$ reacts with steam at $623 \ K$ in the presence of an iron chromate $(Fe_2O_3 \cdot Cr_2O_3)$ catalyst to produce $CO_2$ and $H_2$.

(A) In the manufacture of hydrogen from water gas $(CO + H_2)$,the process is known as the water-gas shift reaction.
The reaction is: $CO(g) + H_2O(g) \xrightarrow{\text{catalyst}} CO_2(g) + H_2(g)$.
$CO$ is oxidized to $CO_2$ with steam in the presence of an iron chromate catalyst.
The resulting $CO_2$ is then removed by scrubbing the mixture with a solution of sodium arsenite or alkali (like $KOH$ or $NaOH$) to obtain pure hydrogen.

(D) Water gas is primarily composed of carbon monoxide and hydrogen gas,i.e.,$CO_{(g)} + H_{2(g)}$.
It is a valuable industrial feedstock and is used in the production of chemicals,fuels,and as a reducing agent in metallurgical applications.

(B)
Hydrogen is highly inflammable and hence storage is difficult.
Hydrogen has a low critical temperature and is therefore not easily liquefiable.

(C) An electron-deficient hydride is one that does not have sufficient electrons to form the required number of covalent bonds for its structure.
In $B_2H_6$ (diborane),there are $8$ covalent bonds,which would normally require $16$ electrons.
However,$B_2H_6$ has only $12$ valence electrons available for bonding.
Therefore,it is an electron-deficient hydride.

(D) The reaction between sodium hydride $(NaH)$ and diborane $(B_{2}H_{6})$ occurs in an ether solvent,typically diethyl ether,$(C_{2}H_{5})_{2}O$,to produce sodium borohydride $(NaBH_{4})$.
The chemical equation is: $2NaH + B_{2}H_{6} \xrightarrow{(C_{2}H_{5})_{2}O} 2NaBH_{4}$.
Here,'$A$' is $(C_{2}H_{5})_{2}O$ and '$B$' is $NaBH_{4}$,which is a well-known reducing agent.

(D) Ionic hydrides are formed by the transfer of electrons from electropositive metals to hydrogen.
They are typically crystalline solids with high melting and boiling points,making them non-volatile.
They do not conduct electricity in the solid state but conduct in the molten state or in solution.
Therefore,being volatile is not a property of ionic hydrides.

(A) The catalytic hydrogenation of carbon monoxide $(CO)$ in the presence of $ZnO-Cr_2O_3$ catalyst at high temperature $(673 \ K)$ and high pressure $(300 \ atm)$ produces methanol $(CH_3OH)$.
The chemical equation is:
$CO + 2H_2 \xrightarrow[\text{High pressure } (673 \ K, 300 \ atm)]{ZnO-Cr_2O_3} CH_3OH$

(A) Concentrated $H_2SO_4$ is a strong dehydrating agent.
It removes water from sugar $(C_{12}H_{22}O_{11})$,leaving behind a black mass of carbon.
This process is known as charring.
$C_{12}H_{22}O_{11} \xrightarrow{\text{conc. } H_2SO_4} 12C + 11H_2O$

(D) Mercury is the only metal that exists as a liquid at room temperature. This is a correct statement.
$(b)$ Among non-metals,carbon (in the form of diamond) has the highest melting point. This is a correct statement.
$(c)$ Hydrogen is the most abundant element in the universe,accounting for approximately $75 \%$ of the elemental mass. This is a correct statement.
$(d)$ Oxygen is the most abundant element in the earth's crust,comprising about $46.6 \%$ by mass. This is a correct statement.
Therefore,all four statements are correct.

(A) The reaction of carbon monoxide with hydrogen yields different products depending on the catalyst used:
$1$. $CO + 3 H_2 \xrightarrow{Ni} CH_4 + H_2 O$ (Catalyst $A = Ni$)
$2$. $CO + 2 H_2 \xrightarrow{ZnO-Cr_2 O_3} CH_3 OH$ (Catalyst $B = ZnO-Cr_2 O_3$)
$3$. $O_2 + 2 H_2 \xrightarrow{Pt} 2 H_2 O$ (Catalyst $C = Pt$)
Therefore,the catalysts $A, B, C$ are $Ni, ZnO-Cr_2 O_3, Pt$ respectively.

(A) The hydride gap refers to the region in the periodic table where elements do not form stable hydrides under normal conditions.
These transition metals exhibit a very low affinity for hydrogen in their standard oxidation states.
Specifically,the metals belonging to group $7$,$8$,and $9$ of the periodic table do not form hydrides.
Therefore,the group $7, 8, 9$ series is known as the hydride gap.

(D) The reactions provided are industrial processes for the production of hydrogen and ammonia:
$I$. This is the Haber process for the synthesis of ammonia,where $X$ is $Fe$ (Iron) with $Mo$ as a promoter.
$II$. This is the water-gas shift reaction,where $Y$ is $FeCrO_4$ (Iron chromate).
$III$. This is the steam reforming of methane,where $Z$ is $Ni$ (Nickel).
Therefore,the catalysts $X$,$Y$,and $Z$ are Iron,iron chromate,and nickel respectively.

(D) The reactions are as follows:
$(a)$ $2NaBH_4 + I_2 \rightarrow B_2H_6 + 2NaI + H_2 \uparrow$
$(b)$ $B_2H_6 + 6H_2O \rightarrow 2B(OH)_3 + 6H_2 \uparrow$
$(c)$ $3B_2H_6 + 6NH_3$ $\rightarrow 3[BH_2(NH_3)_2]^+[BH_4]^-$ $\xrightarrow{\Delta} 2B_3N_3H_6 + 12H_2 \uparrow$
$(d)$ $B_2H_6 + 3O_2 \rightarrow B_2O_3 + 3H_2O$
In reaction $(d)$,water is formed instead of dihydrogen gas. Therefore,dihydrogen is not evolved in the burning of diborane in oxygen.

(B) Tritium is radioactive and emits low-energy $\beta$-particles,not $\gamma$-rays.
$H^{+}$ does not exist freely in aqueous solution; it exists as the hydronium ion,$H_3O^{+}$.
Tritium is found in the ratio of $1$ atom per $10^{18}$ atoms of protium.
Hydrogen is less reactive than halogens,which are strong oxidizing agents.
Therefore,the correct statement is that $H^{+}$ does not exist freely.

(A) $H-H$ bond enthalpy $= 435.88 \ kJ \ mol^{-1}$.
$D-D$ bond enthalpy $= 443.35 \ kJ \ mol^{-1}$.
Due to the difference in their bond dissociation enthalpies,the rates of reactions involving protium and deuterium differ significantly.
Thus,both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.

(B) Hydrogen has three isotopes: $H, D, T$.
Possible molecules are formed by combinations of these isotopes taken two at a time (including identical pairs).
The possible combinations are: $H-H, D-D, T-T, H-D, H-T, D-T$.
Thus,there are $6$ possible hydrogen molecules.

(C) Bond length depends upon the electrostatic force of attraction between the nuclei and the shared pair of electrons.
Since isotopes have the same number of protons and electrons,the electronic environment remains identical.
Therefore,the bond length is the same for both $H_2$ and $D_2$.

(C) An oxidising agent is a substance that gains electrons or causes another substance to be oxidised. In this process,the oxidising agent itself gets reduced.
In the reaction $2Na_{(s)} + H_{2(g)} \longrightarrow 2NaH_{(s)}$,the oxidation state of $H$ changes from $0$ in $H_2$ to $-1$ in $NaH$.
Since the oxidation state of hydrogen decreases,it is being reduced.
Because $H_2$ is being reduced,it acts as an oxidising agent for $Na$ (which is oxidised from $0$ to $+1$).
In the other options,$H_2$ acts as a reducing agent because it is being oxidised (oxidation state increases from $0$ to $+1$).

(C) $1$. Ionic (or saline) hydrides are stoichiometric compounds formed by $s$-block elements. They are crystalline,non-volatile,and non-conducting solids in their solid state. They react violently with water to produce $H_2$ gas.
$2$. Group $14$ elements (like $CH_4$,$SiH_4$) have the required number of electrons to form normal covalent bonds,hence they are called electron-precise hydrides.
$3$. Covalent (or molecular) hydrides are generally volatile compounds with low boiling points and melting points,as they are held together by weak van der Waals forces. Therefore,the statement that covalent hydrides are non-volatile is incorrect.

(D) Hydrides are classified based on their electronic structure:
$1$. Saline (ionic) hydrides: Formed by $s$-block elements,e.g.,$(NaH)$.
$2$. Electron-deficient hydrides: Have fewer electrons than required for bonding,e.g.,$(B_2H_6)$.
$3$. Electron-precise hydrides: Have the exact number of electrons for bonding,e.g.,$(CH_4)$.
$4$. Electron-rich hydrides: Have lone pairs of electrons,e.g.,$(H_2O, NH_3, HF)$.
In option $(D)$,$(HF)$ is an electron-rich hydride because it has lone pairs on the fluorine atom,not an electron-precise hydride. Therefore,$(D)$ is incorrectly matched.

(A) Electron-precise hydrides are those that have the exact number of electrons required to form the necessary covalent bonds in their structure.
These hydrides do not have any lone pairs or electron deficiency.
Group $14$ elements (like $C, Si, Ge, Sn, Pb$) form hydrides of the type $EH_4$,which are electron-precise.
Examples include $CH_4, SiH_4, GeH_4, SnH_4,$ and $PbH_4$.

(D) Hydrides of group $15$ elements (like $NH_3$,$PH_3$) possess a lone pair of electrons on the central atom,which gives them the ability to act as electron donors. Therefore,they behave as Lewis bases,not Lewis acids.
Option $(D)$ is incorrect.