Hydrogen Questions in English

(A) $B_2H_6$ (Diborane) is an electron-deficient hydride because it has fewer electrons than required for conventional bonding.
$NH_3$ (Ammonia) has one lone pair of electrons on the nitrogen atom,making it an electron-rich hydride,not electron-precise.
$H_2O$ (Water) has two lone pairs of electrons on the oxygen atom,making it an electron-rich hydride.
Therefore,only statements $(i)$ and $(iii)$ are correct.

(D) $NH_3$ is an electron-rich hydride because the $N$-atom belongs to group $15$ and uses three of its valence electrons to form three $N-H$ bonds,while the remaining two electrons form a lone pair.
$SiH_4$ is an electron-precise hydride.
$CrH$ is a metallic (interstitial) hydride.
$HF$ is an electron-rich hydride due to the presence of lone pairs on the $F$-atom.

(C) Statement $I$ is correct: $LiH, BeH_2$,and $MgH_2$ are saline (ionic) hydrides that exhibit significant covalent character due to the small size and high polarizing power of the cations.
Statement $II$ is incorrect: Saline hydrides are non-volatile,high-melting crystalline solids due to strong electrostatic forces of attraction between ions.
Statement $III$ is incorrect: Electron-precise hydrides (e.g.,$CH_4, SiH_4$) have the required number of electrons to form normal covalent bonds and do not act as Lewis bases.
Statement $IV$ is correct: Chromium forms a non-stoichiometric interstitial hydride with the formula $CrH$ (or $CrH_{1.7}$ depending on conditions).
Therefore,statements $I$ and $IV$ are correct.

(A) $NaH$ is an ionic (saline) hydride,which is non-volatile and crystalline.
Thus,statement $(A)$ is correct.
$MgH_2$ is a polymeric hydride,as shown by its structure containing bridging hydrogen atoms.
Thus,statement $(B)$ is correct.
$NH_3$ has a lone pair on the $N$-atom,making it an electron-rich hydride,not electron-precise.
Thus,statement $(C)$ is incorrect.
$H_2O$ has two lone pairs on the $O$-atom,making it an electron-rich hydride.
Thus,statement $(D)$ is correct.
Therefore,statements $(A)$,$(B)$,and $(D)$ are correct.

(B) The correct matching is as follows:
$(A)$ Electron precise hydrides have the exact number of electrons needed to form covalent bonds. These are formed by group $14$ elements such as $CH_4$ and $SiH_4$. Thus,$A-(i)$.
$(B)$ Saline hydrides (also known as ionic hydrides) are compounds formed between hydrogen and the most active metals,especially alkali and alkaline-earth metals such as $MgH_2$. Thus,$B-(iii)$.
$(C)$ Electron-deficient hydrides are compounds in which the central atom has an incomplete octet,such as boron in $B_2H_6$. Thus,$C-(iv)$.
$(D)$ Electron-rich hydrides have one or more lone pairs of electrons around the central more electronegative element,such as $H_2O$. Thus,$D-(ii)$.
Therefore,the correct match is $A-(i), B-(iii), C-(iv), D-(ii)$.

(A) The region of the periodic table from group $7$ to $9$ is referred to as the hydride gap because these elements do not form hydrides.
Examples of such elements are $Mn, Fe, Co, Ru$,etc.
These elements do not form hydrides due to their low affinity for hydrogen in their normal oxidation states.
Hence,the correct option is $(A)$.

(C) $(I)$ $B_2H_6$ (diborane) is an electron deficient hydride. This is due to the presence of $3C-2e^-$ bridge bonds.
$(II)$ $NH_3$ is an electron rich hydride due to the presence of a lone pair of electrons on the nitrogen atom.
$(III)$ $NaH$ is an ionic (saline) hydride; hydrides of alkali metals are ionic in nature. Thus,statement $III$ is incorrect.
$(IV)$ $YbH_{2.55}$ is an interstitial (non-stoichiometric) hydride. Here,hydrogen atoms occupy the interstitial spaces in the metal lattice. Thus,statement $IV$ is correct.
Therefore,statements $I$,$II$,and $IV$ are correct. The correct option is $(C)$.

(D) Statement $(i)$: $NaH_{(s)} + H_2O_{(l)} \rightarrow NaOH_{(aq)} + H_{2(g)}$. This reaction is highly exothermic and violent. Thus,statement $(i)$ is correct.
Statement $(ii)$: $NH_3$ has a lone pair of electrons on the $N$ atom,making it an electron-rich hydride. Thus,statement $(ii)$ is correct.
Statement $(iii)$: Saline (or ionic) hydrides are formed by $s$-block elements. Nickel $(Ni)$ is a $d$-block element and forms metallic (or interstitial) hydrides,not saline hydrides. Thus,statement $(iii)$ is incorrect.
Therefore,statements $(i)$ and $(ii)$ are correct.

(C) $(i)$ $Zn$ is an amphoteric metal. It reacts with both acids and bases to liberate $H_2$ gas:
$Zn_{(s)} + 2HCl_{(aq)} \longrightarrow ZnCl_{2(aq)} + H_{2(g)}$
$Zn_{(s)} + 2NaOH_{(aq)} + 2H_2O_{(l)} \longrightarrow Na_2[Zn(OH)_4]_{(aq)} + H_{2(g)}$
Thus,statement $(i)$ is correct.
$(ii)$ $Ti$ and $Zr$ are transition metals that form non-stoichiometric interstitial hydrides. Thus,statement $(ii)$ is correct.
$(iii)$ The viscosity of $D_2 O$ is higher than that of $H_2 O$ due to stronger hydrogen bonding in $D_2 O$. Thus,statement $(iii)$ is incorrect.

(A) Statement $(i)$ is correct: In $CuSO_4 \cdot 5H_2O$, four water molecules are coordinated to the $Cu^{2+}$ ion, and one water molecule is hydrogen-bonded to the sulphate ion.
Statement $(ii)$ is correct: Lanthanum and zirconium form non-stoichiometric (interstitial) hydrides, such as $LaH_{2.87}$ and $ZrH_{1.9}$.
Statement $(iii)$ is incorrect: In the solid form of $H_2O$ (ice), each oxygen atom is surrounded by four other oxygen atoms in a tetrahedral arrangement at a distance of $276 \ pm$, not octahedral.
Therefore, statements $(i)$ and $(ii)$ are correct.

(D) The correct classification of hydrides is as follows:
$A$. Electron-deficient hydride: These have fewer electrons than required for bonding,e.g.,$B_2H_6$ $(II)$.
$B$. Electron-precise hydride: These have the exact number of electrons required for bonding,e.g.,$CH_4$ $(I)$.
$C$. Electron-rich hydride: These have excess electrons (lone pairs),e.g.,$PH_3$ $(V)$.
$D$. Saline (ionic) hydride: These are formed by $s$-block elements,e.g.,$CaH_2$ $(III)$.
$IV$. $NiH_{0.6}$ is a metallic or non-stoichiometric hydride.
Therefore,the correct match is $A-II, B-I, C-V, D-III$.

(B) Statement $(I)$ is correct: The chemical formula of Calgon is $Na_6(PO_3)_6$ (sodium hexametaphosphate).
Statement $(II)$ is incorrect: Exhausted permutit (hydrated sodium aluminium silicate) is regenerated by $10 \% NaCl$ solution,not $KCl$.
Statement $(III)$ is incorrect: Calcium stearate is a soap scum and is insoluble in water.
Statement $(IV)$ is correct: In the structure of ice,each oxygen atom is tetrahedrally surrounded by four other oxygen atoms through hydrogen bonding.
Therefore,statements $(I)$ and $(IV)$ are correct.

(C) Let us analyze each reaction:
$I) 2NaBH_4 + I_2 \longrightarrow 2NaI + B_2H_6 + H_2$. This reaction produces $H_2$.
$II) B_2H_6 + 2N(CH_3)_3 \longrightarrow 2BH_3 \cdot N(CH_3)_3$. This is an adduct formation reaction and does not produce $H_2$.
$III) 2Al + 2NaOH + 6H_2O \longrightarrow 2Na[Al(OH)_4] + 3H_2$. This reaction produces $H_2$.
$IV) 2BF_3 + 6NaH \longrightarrow B_2H_6 + 6NaF$. This reaction does not produce $H_2$.
Therefore,only reactions $I$ and $III$ produce $H_2$ as a product.

(C) The reaction of diborane $(B_2H_6)$ with metal hydrides $(MH)$ in the presence of diethyl ether yields metal borohydrides $(MBH_4)$.
For example: $B_2H_6 + 2LiH \longrightarrow 2LiBH_4$.
Therefore,$X$ is a metal hydride.

(C) $NaBH_4$ is a mild reducing agent. When sodium borohydride reacts with iodine,it produces diborane,sodium iodide,and hydrogen gas. This reaction involves the oxidation of sodium borohydride with iodine in diglyme to yield diborane. This approach is common in the industrial production of $B_2H_6$.
The balanced chemical equation is:
$2NaBH_4 + I_2 \xrightarrow{\text{Diglyme}} B_2H_6 + 2NaI + H_2$

(A) Sodium hydride reacts with diborane in the presence of diethyl ether to produce sodium tetrahydroborate$(III)$.
The balanced chemical equation is:
$2NaH + B_2H_6 \xrightarrow{\text{Diethyl ether}} 2Na[BH_4]$
Therefore,$X$ is $Na[BH_4]$.

(C) The reaction of $BF_3$ with $NaH$ at $450 \ K$ produces diborane $(B_2 H_6)$ as $X$:
$2 BF_3 + 6 NaH \xrightarrow{450 \ K} 6 NaF + B_2 H_6 (X)$
When diborane $(X)$ reacts with $LiH$ in the presence of diethyl ether,it forms lithium borohydride $(LiBH_4)$ as $Y$:
$B_2 H_6 + 2 LiH \xrightarrow{\text{Diethyl ether}} 2 LiBH_4 (Y)$
Therefore,$Y$ is $LiBH_4$.

(C) The preparation of water gas involves passing steam over red-hot coke at a temperature of $1273 \ K$. The resulting mixture of $CO$ and $H_2$ is known as water gas or synthesis gas.
$C_{(s)} + H_2O_{(g)} \xrightarrow{1273 \ K} CO_{(g)} + H_{2(g)}$

(B) Coal gasification is the process of producing coal gas,which is a mixture of carbon monoxide $(CO)$ and hydrogen $(H_2)$,also known as synthesis gas or syn gas,from coal.
The chemical equation for this process is:
$C_{(s)} + H_2O_{(g)} \xrightarrow{1273 \ K} CO_{(g)} + H_{2(g)}$

(C) Dehydration of formic acid $(HCOOH)$ with conc. $H_2SO_4$ is a laboratory method,not a commercial one.
$(B)$ Direct oxidation of carbon in limited supply of oxygen is difficult to control on a commercial scale.
$(C)$ Passing steam over hot coke at high temperatures produces water gas $(CO + H_2)$,which is a standard commercial process:
$C(s) + H_2O(g) \xrightarrow{1273 \ K} CO(g) + H_2(g)$
$(D)$ Heating limestone $(CaCO_3)$ produces $CO_2$,not $CO$.
Therefore,option $(C)$ is the correct answer.

(C) The mixture of $CO$ and $H_2$ is known as water gas or synthesis gas $(syn \ gas)$.
$C_{(s)} + H_2O_{(g)} \rightarrow CO_{(g)} + H_{2(g)}$ is the coal gasification reaction.
The production of dihydrogen can be increased by reacting carbon monoxide with steam in the presence of iron chromate as a catalyst at $673 \ K$.
$CO_{(g)} + H_2O_{(g)} \xrightarrow[\text{iron chromate}]{673 \ K} CO_{2(g)} + H_{2(g)}$
This specific reaction is known as the water-gas shift reaction.

(D) The given reaction represents the industrial preparation of methanol from a mixture of carbon monoxide and hydrogen,also known as synthesis gas or syngas.
In this process,the gaseous mixture is subjected to a pressure of $200-300 \ atm$ and passed over a heated catalyst mixture of $ZnO$ and $Cr_2 O_3$ maintained at $573-673 \ K$.
This reaction results in the formation of methanol vapors,which are subsequently condensed to a liquid state.
$CO + 2 H_2 \xrightarrow[\substack{200-300 \ atm, 573-673 \ K}]{ZnO-Cr_2 O_3(X)} CH_3 OH$

(A) $i$. The reaction $H_{2(g)} + F_{2(g)} \longrightarrow 2HF_{(g)}$ is highly spontaneous and occurs even in the dark due to the high reactivity of $F_2$.
$ii$. The Haber process,$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,is an exothermic reaction $(\Delta H = -92.4 \ kJ/mol)$,not endothermic.
$iii$. $HF$ is an electron-rich hydride because the fluorine atom has three lone pairs of electrons.
Therefore,statements $i$ and $iii$ are correct.

(B) Ammonium nitrate $(NH_4NO_3)$ on thermal decomposition produces nitrous oxide $(N_2O)$ and water $(H_2O)$.
$NH_4NO_3 \longrightarrow N_2O + 2H_2O$
$N_2O$ is a colourless,neutral gas (laughing gas).
The structure of $N_2O$ is $N \equiv N^+ - O^-$,which is linear in shape.

(A) Producer gas is a mixture of $CO$ and $N_2$.
Its calorific value is low primarily due to the high percentage of non-combustible nitrogen $(N_2)$ gas present in the mixture.

(B) The reaction of calcium metal with water is: $Ca + 2H_2O \longrightarrow Ca(OH)_2 + H_2(g)$.
The reaction of calcium hydride with water is: $CaH_2 + 2H_2O \longrightarrow Ca(OH)_2 + 2H_2(g)$.
Both reactions produce $H_2$ gas as a product upon hydrolysis.

(B) $HCl$ gas is dried by passing it through concentrated $H_2SO_4$ because concentrated $H_2SO_4$ is a strong dehydrating agent and absorbs moisture from the gas.
$HCl$ gas reacts with $NH_3$ to form white fumes of $NH_4Cl$ $(NH_3 + HCl \rightarrow NH_4Cl)$.
Both statements are chemically correct,but the reaction of $HCl$ with $NH_3$ is not the reason why $H_2SO_4$ is used to dry $HCl$ gas. Therefore,$(R)$ is not the correct explanation of $(A)$.

(A) $HCOOH \xrightarrow{\text{Conc. } H_2SO_4} \underset{(B)}{CO} + \underset{(A)}{H_2O}$
$3 \underset{(B)}{CO} + Fe_2O_3 \xrightarrow{\Delta} \underset{(D)}{2Fe} + \underset{(C)}{3CO_2}$
$CaCO_3 + 2HCl \text{ (dil.)} \longrightarrow \underset{(C)}{CO_2} + H_2O + CaCl_2$
From the reactions,$B$ is $CO$ and $C$ is $CO_2$.
The reduction of $Fe_2O_3$ by $CO$ yields $Fe$ $(D)$ and $CO_2$ $(C)$.
Therefore,$D$ is $Fe$.

(B) The industrial preparation of methanol involves the catalytic hydrogenation of carbon monoxide.
The chemical reaction is: $CO(g) + 2H_2(g) \xrightarrow[ZnO-Cr_2O_3]{573-673 \ K, 200-300 \ atm} CH_3OH(g)$.
Comparing this with the given ratio of $1:2$ for gases $X$ and $Y$,we identify $X = CO$ and $Y = H_2$.

(C) Hydrogen exists as a diatomic molecule $(H_2)$ and has one electron in its outermost shell $(1s^1)$.
When a hydrogen atom loses an electron,it forms a proton $(H^+)$.
Due to its extremely small size (radius $\approx 10^{-15} \ m$),the bare proton $(H^+)$ cannot exist freely in solution; it is always associated with other molecules or ions (e.g.,forming $H_3O^+$ in water).
Therefore,statement $3$ is false because the cation $(H^+)$ cannot exist freely.
Additionally,hydrogen can form ionic compounds with highly electropositive metals (e.g.,$NaH$,$CaH_2$),making statement $4$ also technically false in a broad context,but statement $3$ is the most fundamentally incorrect regarding the stability of the hydrogen cation.

(C) In reaction $I$,the catalytic hydrogenation of carbon monoxide to formaldehyde $(HCHO)$ is carried out using copper $(Cu)$ as a catalyst at high temperature and pressure.
In reaction $II$,the catalytic hydrogenation of carbon monoxide to methane $(CH_4)$ is carried out using nickel $(Ni)$ as a catalyst at approximately $573 \ K$.
Therefore,the catalysts $X$ and $Y$ are $Cu$ and $Ni$ respectively.

(A) When electric current is passed through an ionic hydride in the molten state,hydrogen gas is liberated at the anode.
In hydrides of $s-$block elements,hydrogen exists in the $-1$ oxidation state.
It is oxidized to hydrogen gas $(H_2)$ at the anode because oxidation always occurs at the anode.
The reaction is:
$2H^{-} \rightarrow H_2 + 2e^{-}$ (at anode)

(C) $I$. $Zn + 2HCl \ (dil.) \longrightarrow ZnCl_2 + H_2 \uparrow$
$II$. $Zn + 2NaOH \ (aq.) \longrightarrow Na_2ZnO_2 + H_2 \uparrow$
$III$. Heating calcium hydrogen carbonate $\left[Ca(HCO_3)_2\right]$ results in:
$Ca(HCO_3)_2 \stackrel{\Delta}{\longrightarrow} CaCO_3 + CO_2 + H_2O$
Since reaction $(III)$ does not produce $H_2$,only reactions $(I)$ and $(II)$ produce dihydrogen.

(B) Hydrogen has three naturally occurring isotopes: Protium $(^1_1H)$,Deuterium $(^2_1D)$,and Tritium $(^3_1T)$.
Protium is the most abundant isotope with a natural abundance of $99.985 \%$.
Deuterium has a natural abundance of $0.015 \%$.
Tritium is radioactive and exists in trace amounts with a natural abundance of approximately $10^{-16} \%$.

(A) $(i)$ $Zn + 2NaOH_{(aq)} \longrightarrow Na_2ZnO_2 + H_{2(g)}$ (Liberates $H_2$)
$(ii)$ $HCOOH \xrightarrow[\text{Conc. } H_2SO_4]{373 \ K} H_2O + CO$ (Does not liberate $H_2$)
$(iii)$ $CH_{4(g)} + H_2O_{(g)} \xrightarrow[Ni]{1270 \ K} CO_{(g)} + 3H_{2(g)}$ (Liberates $H_2$)
$(iv)$ $Zn + 2H^+_{(aq)} \longrightarrow Zn^{2+} + H_{2(g)}$ (Liberates $H_2$)
$(v)$ $C_{(s)} + H_2O_{(g)} \xrightarrow{1270 \ K} CO_{(g)} + H_{2(g)}$ (Liberates $H_2$)
Therefore,reactions $(i)$,$(iii)$,$(iv)$,and $(v)$ liberate $H_2$ gas.

(D) Both hydrogen $(H_2)$ and deuterium $(D_2)$ molecules have the same bond length of $74 \ pm$.
However, they differ in their physical and chemical properties such as bond energy, melting point, and boiling point due to the difference in their isotopic masses.
Based on the data:
- Bond length: $74 \ pm$ (Same)
- Bond energy: $436 \ kJ \ mol^{-1}$ for $H_2$ vs $443.3 \ kJ \ mol^{-1}$ for $D_2$
- Melting point: $0.00 \ ^\circ C$ for $H_2$ vs $3.81 \ ^\circ C$ for $D_2$
- Boiling point: $100.0 \ ^\circ C$ for $H_2$ vs $101.42 \ ^\circ C$ for $D_2$
Therefore, the bond length is the same.

(C) Electron-rich hydrides are those which have one or more lone pairs of electrons on the central atom.
These are typically formed by elements of groups $15, 16,$ and $17$.
Examples include $NH_3$ (one lone pair),$H_2O$ (two lone pairs),and $HF$ (three lone pairs).
In the given options,$NH_3$ and $H_2O$ are electron-rich hydrides.
$B_2H_6$ and $AlH_3$ are electron-deficient hydrides.
$NaH$ and $MgH_2$ are saline or ionic hydrides.
$CH_4$ and $SiH_4$ are electron-precise hydrides.

(C) Statement $(I)$ is incorrect because the reaction between $NaH$ and $H_2O$ releases hydrogen gas,not oxygen gas: $NaH + H_2O \rightarrow NaOH + H_2 \uparrow$.
Statement $(II)$ is correct; $CH_4$ and $SiH_4$ have the required number of electrons to form normal covalent bonds,making them electron-precise hydrides.
Statement $(III)$ is incorrect; $NH_3$ and $H_2O$ are electron-rich hydrides because they possess lone pairs of electrons on the central atom.
Statement $(IV)$ is correct; hydrides of $Be$ and $Mg$ ($BeH_2$ and $MgH_2$) are electron-deficient and exist in polymeric structures.
Thus,statements $(II)$ and $(IV)$ are correct.

(D) Statement $(I)$ is incorrect because in saline (ionic) hydrides,hydrogen exists as the hydride ion $(H^-)$,meaning its oxidation state is $-1$.
Statement $(II)$ is correct because lanthanum hydride $(LaH_{2.87})$ is a non-stoichiometric interstitial hydride.
Statement $(III)$ is correct because $NH_3$ and $H_2O$ contain highly electronegative atoms ($N$ and $O$) bonded to hydrogen,allowing for hydrogen bonding.
Statement $(IV)$ is correct because during the electrolysis of molten $NaH$,the $H^-$ ions migrate to the anode and are oxidized to $H_2$ gas $(2H^- \rightarrow H_2 + 2e^-)$.
Therefore,statements $(II)$ and $(III)$ are correct.

(C) Electron-precise hydrides are formed by group $14$ elements.
These elements have exactly the number of electrons required to form normal covalent bonds,such as in $CH_4$,$SiH_4$,$GeH_4$,and $SnH_4$.

(A) Beryllium hydride $(BeH_2)$ is prepared by the reaction of beryllium chloride $(BeCl_2)$ with lithium aluminium hydride $(LiAlH_4)$ in an ethereal solution.
The chemical reaction is: $2BeCl_2 + LiAlH_4 \rightarrow 2BeH_2 + LiCl + AlCl_3$.

(C) $ZnH_2$ is an example of an interstitial (metallic) hydride.
$NH_3$,$CH_4$,and $H_2O$ are examples of covalent (molecular) hydrides.

(A) Permutit,also known as zeolite,is a hydrated sodium aluminosilicate with the general formula $Na_2Al_2Si_2O_8 \cdot xH_2O$.
It is used to remove the hardness of water by ion exchange.
During the process,the $Na^{+}$ ions in the zeolite are replaced by $Ca^{2+}$ and $Mg^{2+}$ ions present in hard water,making the water soft.
When all the $Na^{+}$ ions are replaced by $Ca^{2+}$ or $Mg^{2+}$ ions,the permutit becomes exhausted.
Therefore,exhausted permutit does not contain $Na^{+}$ ions.

(A) $I$. $2NaBH_4 + I_2 \rightarrow B_2H_6 + 2NaI + H_2 \uparrow$ (Hydrogen is evolved)
$II$. $B_2H_6 + 3O_2 \rightarrow B_2O_3 + 3H_2O$ (No hydrogen is evolved)
$III$. $8BF_3 + 6NaH \rightarrow B_2H_6 + 6NaBF_4$ (No hydrogen is evolved)
$IV$. $B_2H_6 + 6H_2O \rightarrow 2H_3BO_3 + 6H_2 \uparrow$ (Hydrogen is evolved)
Thus,in reactions $I$ and $IV$,hydrogen gas is evolved.

(A) The primary components of synthesis gas are hydrogen $(H_2)$ and carbon monoxide $(CO)$.
It is also known as syn gas.
It is used as a fuel gas.
Hence,option $(A)$ is the correct option.

(A) Producer gas is a mixture of $CO$ and $N_2$.
Its calorific value is low due to the high percentage of non-combustible nitrogen gas $(N_2)$ present in the mixture.