Hydrogen Questions in English

(C) Heavy water is chemically known as deuterium oxide,represented by the formula $D_2O$.
It is composed of deuterium,an isotope of hydrogen with a mass number of $2$,instead of the common protium isotope found in ordinary water $(H_2O)$.

(C) $I.$ Atomic hydrogen is produced by passing an electric arc through hydrogen gas at low pressure. This statement is correct.
$II.$ Hydrogen gas is a reducing agent,but it does not reduce aluminum oxide $(Al_2O_3)$. However,it does not reduce metallic aluminum itself because aluminum is more reactive than hydrogen. This statement is correct.
$III.$ Finely divided metals like palladium,platinum,and nickel have the property of adsorbing large volumes of hydrogen gas on their surface. This is known as occlusion. This statement is correct.
$IV.$ The reaction of $Na$ with $C_2H_5OH$ produces sodium ethoxide and hydrogen gas,but it does not produce 'nascent' hydrogen in the chemical sense (which refers to hydrogen at the moment of its liberation). This statement is incorrect.
Therefore,statements $I, II,$ and $III$ are correct.

(C) An oxidizing agent is a substance that gains electrons or causes the oxidation of another substance. In the reaction with metals,hydrogen acts as an oxidizing agent because it gains electrons to form hydride ions $(H^-)$.
For example,the reaction with sodium is: $2Na + H_2 \rightarrow 2NaH$.
In this reaction,sodium $(Na)$ is oxidized to $Na^+$,and hydrogen $(H_2)$ is reduced to $H^-$,thus acting as an oxidizing agent.

(D) Hydrogen reacts directly with highly electropositive metals like alkali metals and alkaline earth metals to form metal hydrides.
Among the given options,$Ca$ (Calcium) is an alkaline earth metal.
It reacts with hydrogen at high temperatures to form calcium hydride: $Ca(s) + H_2(g) \rightarrow CaH_2(s)$.
$Au$,$Cu$,and $Ni$ do not react directly with hydrogen under normal conditions to form stable hydrides.

(A) Hydrogen has the electronic configuration $1s^1$. By accepting one electron,it achieves the configuration $1s^2$,which is the stable electronic configuration of the inert gas Helium $(He)$.
Halogens also have a tendency to accept one electron to achieve the nearest inert gas configuration (e.g.,$Cl + e^- \rightarrow Cl^-$). Therefore,in this behavior,hydrogen resembles halogens.

(C) proton $(H^+)$ is essentially a bare nucleus with no electrons. Due to its extremely small size (radius $\approx 10^{-15} \ m$),it has a very high charge density. Consequently,it has a very high hydration energy compared to other ions because it interacts strongly with water molecules.

(D) In $H_2$ molecules,there are two forms based on the nuclear spin of the two hydrogen atoms: ortho-hydrogen and para-hydrogen.
In ortho-hydrogen,the nuclear spins of the two protons are in the same direction (parallel).
In para-hydrogen,the nuclear spins of the two protons are in the opposite direction (anti-parallel).
Therefore,the nuclear spins are in the opposite direction to each other.

(C) In molten lithium hydride $(LiH)$,the compound dissociates into $Li^+$ and $H^-$ ions.
During electrolysis,the $H^-$ ions migrate towards the anode (positive electrode).
At the anode,the $H^-$ ions undergo oxidation to release hydrogen gas: $2H^- \rightarrow H_2(g) + 2e^-$.
Therefore,hydrogen gas is liberated at the anode.

(C) Ortho and para hydrogen are nuclear spin isomers of the hydrogen molecule $(H_2)$.
In ortho-hydrogen,the nuclear spins of the two protons are parallel (in the same direction).
In para-hydrogen,the nuclear spins of the two protons are anti-parallel (in opposite directions).
Therefore,they differ in their nuclear spin states.

(C) Hydrogen resembles alkali metals in some properties such as electropositive character and forming $+1$ oxidation state compounds. However,hydrogen is a non-metal and its ionization enthalpy is much higher than that of alkali metals. Therefore,it does not act as a strong reducing agent like alkali metals. In covalent hydrides,hydrogen typically exhibits a $+1$ oxidation state (e.g.,$CH_4$,$NH_3$),whereas it shows a $-1$ oxidation state only in ionic (saline) hydrides (e.g.,$NaH$). Thus,statement $D$ is also incorrect in the context of covalent hydrides,but statement $C$ is the most fundamentally incorrect property compared to alkali metals.

(C) Hydrolith is the common name for calcium hydride $(CaH_2)$.
It is a saline or ionic hydride formed by the reaction of calcium metal with hydrogen gas at high temperatures.
$Ca + H_2 \rightarrow CaH_2$.

(D) The correct statement is that $d-$ and $f-$block elements form metallic (or interstitial) hydrides,not ionic hydrides. Ionic hydrides are primarily formed by $s-$block elements (alkali and alkaline earth metals,except $Be$ and $Mg$). Therefore,the statement in option $D$ is incorrect.

(A) Elements that form stable,covalent,and hydrogen-bonded hydrides are those with high electronegativity and small atomic size.
These elements are $N$,$O$,and $F$.
They form hydrides like $NH_3$,$H_2O$,and $HF$,which exhibit strong intermolecular hydrogen bonding due to the high electronegativity difference between hydrogen and the central atom.

(C) Hydrogen has three naturally occurring isotopes: Protium $(^1H)$,Deuterium ($^2H$ or $D$),and Tritium ($^3H$ or $T$).
Ortho-hydrogen and Para-hydrogen are not isotopes; they are allotropic forms of molecular hydrogen $(H_2)$ based on the relative orientation of the nuclear spins of the two hydrogen atoms in the molecule.
Therefore,Ortho-hydrogen is not an isotope of hydrogen.

(D) Hydrogen is a non-metal,whereas alkali metals are metals. This non-metallic nature is the fundamental property that distinguishes hydrogen from alkali metals,even though both have a valence shell configuration of $ns^1$.

(D) Hydrogen acts as a reducing agent for metal oxides of metals that are less reactive than hydrogen in the reactivity series.
$CuO + H_2 \rightarrow Cu + H_2O$
$Fe_2O_3 + 3H_2 \rightarrow 2Fe + 3H_2O$
$SnO_2 + 2H_2 \rightarrow Sn + 2H_2O$
Aluminium is more reactive than hydrogen,so $Al_2O_3$ cannot be reduced by $H_2$.

(D) The isotopes of hydrogen are protium $(^1H)$,deuterium $(^2H)$,and tritium $(^3H)$.
The heaviest isotope is tritium ($^3H$ or $^3_1H$).
In tritium,the number of protons $(Z)$ is $1$.
The mass number $(A)$ is $3$.
The number of neutrons $(N)$ is calculated as $A - Z = 3 - 1 = 2$.
The sum of the number of neutrons and protons is $N + Z = 2 + 1 = 3$.

(A) Heavy water is deuterium oxide,represented as $D_2O$.
It is formed by the combination of deuterium (a heavy isotope of hydrogen) and oxygen.
Option $A$ correctly describes its composition.
Option $B$ is incorrect because $D_2O$ has a higher density than $H_2O$,but the property of maximum density at $4\,^oC$ is a characteristic of ordinary water $(H_2O)$.
Option $C$ is incorrect because $D_2O$ is heavier than $H_2O$ due to the higher atomic mass of deuterium.

(A) The industrial production of hydrogen from water gas (also known as the Bosch process) involves the 'water-gas shift reaction'.
$1$. The mixture of $CO$ and $H_2$ is mixed with excess steam and passed over an iron chromate catalyst at $673 \ K$.
$2$. The reaction is: $CO(g) + H_2O(g) \xrightarrow{FeCrO_4} CO_2(g) + H_2(g)$.
$3$. The resulting $CO_2$ is then removed by scrubbing the mixture with a solution of sodium arsenite or potassium hydroxide (alkali),leaving behind pure $H_2$.

(A) Heavy water $(D_2O)$ is typically obtained by the prolonged electrolysis of ordinary water. However,in the context of laboratory preparation or specific chemical methods,the fractional distillation or exchange reactions are used. Among the given options,none of the chemical additions ($MgCO_3$,$CaSO_4$,$Na_2SO_4$,$CaCO_3$) directly produce heavy water. This question appears to be based on a specific textbook context where the preparation of heavy water is discussed via electrolysis. Given the standard curriculum,if this is a multiple-choice question from a specific source,it is often noted that heavy water is obtained as a byproduct of the electrolysis of water. Since the provided options do not list electrolysis,this question might be flawed or referring to a specific industrial process not listed. However,based on standard chemistry,none of these options are correct. Assuming a correction is needed for a standard exam context,the question is often associated with the electrolysis of water.

(C) The electrolysis of water,especially using a dilute solution of an acid or alkali with platinum electrodes,produces hydrogen of very high purity $(99.9\%)$.
Other methods like the reaction of methane with steam or the reaction of saline hydrides with water often contain impurities from the starting materials or side reactions.

(C) The hydrogen molecule $(H_2)$ consists of a strong $H-H$ covalent bond.
Due to the small size of hydrogen atoms and the high electronegativity difference being zero,the bond dissociation energy is very high $(435.88 \ kJ \ mol^{-1})$.
Because of this high bond dissociation energy,the $H_2$ molecule is chemically inert under normal conditions.

(C) Synthesis gas or syngas is a mixture of carbon monoxide $(CO)$ and dihydrogen $(H_2)$.
It is also known as water gas.
The reaction for its production is: $C(s) + H_2O(g) \xrightarrow{1270 K} CO(g) + H_2(g)$.

(D) Carbogen is a medical gas mixture consisting of $95\%$ oxygen $(O_2)$ and $5\%$ carbon dioxide $(CO_2)$. It is used in the treatment of various medical conditions,such as carbon monoxide poisoning and certain respiratory issues.

(D) Water gas $(CO + H_2)$,Producer gas $(CO + N_2)$,and Coal gas $(CO + H_2 + CH_4 + CO_2)$ are all widely used as industrial fuels due to their high calorific value and combustible nature. Therefore,the correct option is $D$.

(A) Semi-water gas is produced by passing a mixture of air and steam over red-hot coke.
It is essentially a mixture of water gas $(CO + H_2)$ and producer gas $(CO + N_2)$.
Therefore,it is a mixture of water gas and producer gas.

(D) Ammonia $(NH_3)$ is widely used as a refrigerant in industrial refrigeration systems because of its high latent heat of vaporization and excellent thermodynamic properties. In its liquid form,it absorbs heat during evaporation,effectively cooling the surroundings.

(A) The reaction of hydrogen sulfide $(H_2S)$ with oxygen $(O_2)$ depends on the amount of oxygen supplied.
When $H_2S$ is burned in a limited supply of oxygen,it produces water and sulfur:
$2H_2S(g) + O_2(g) \rightarrow 2H_2O(l) + 2S(s)$
In excess oxygen,it produces sulfur dioxide:
$2H_2S(g) + 3O_2(g) \rightarrow 2H_2O(l) + 2SO_2(g)$
Given the options provided,the reaction yielding water and sulfur is the standard representation for the combustion of $H_2S$ in limited oxygen.

(D) Concentrated $H_2SO_4$ is a strong dehydrating agent.
When sugar $(C_{12}H_{22}O_{11})$ reacts with concentrated $H_2SO_4$,the acid 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,which causes the sugar to turn black.

(A) Hydrogen fluoride $(HF)$ exists as a polymer $(HF)_n$ or $H_2F_2$ due to strong intermolecular hydrogen bonding.
Because of this,it does not ionize easily to provide free ions for electrical conduction.
In contrast,$HCl$,$HBr$,and $HI$ are strong acids that ionize readily in solution to conduct electricity.

(D) The mixture of $KF$ and $HF$ is used for the preparation of $F_{2}$ gas by electrolysis.
In the anhydrous state,$KF$ dissociates into $K^{+}$ and $F^{-}$ ions.
The $F^{-}$ ion then reacts with $HF$ to form the hydrogen difluoride ion,$HF_{2}^{-}$.
Therefore,the mixture contains $K^{+}$ and $HF_{2}^{-}$ ions.

(B) In a soda-acid fire extinguisher,the two main components are a solution of sodium bicarbonate $(NaHCO_3)$ and sulfuric acid $(H_2SO_4)$.
When the extinguisher is activated,the acid reacts with the bicarbonate to produce carbon dioxide gas $(CO_2)$,which helps in extinguishing the fire.
The chemical reaction is: $2NaHCO_3 + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O + 2CO_2 \uparrow$.

(A) $1$. $MgO$ (Magnesium oxide) is a basic oxide that does not decompose to release $O_2$ upon heating.
$2$. It is highly hygroscopic and is often used as a desiccant to remove water vapor.
$3$. $NaNO_3$,$Pb_2O_4$ (or $Pb_3O_4$),and $KClO_3$ all release $O_2$ upon thermal decomposition.
$4$. Therefore,the correct option is $A$.

(A) Calcium oxide $(CaO)$,also known as quicklime,is a basic oxide.
It reacts with water vapor to form calcium hydroxide: $CaO + H_2O \rightarrow Ca(OH)_2$.
It also reacts with acidic gases like $CO_2$ to form calcium carbonate: $CaO + CO_2 \rightarrow CaCO_3$.
Therefore,it acts as a drying agent that can remove both water vapor and carbon dioxide.

(D) The acid $HNO_2$ (nitrous acid) contains nitrogen in the $+3$ oxidation state.
Since the oxidation state of nitrogen can increase (up to $+5$ in $HNO_3$) or decrease (down to $-3$ in $NH_3$),it acts as both an oxidizing and a reducing agent.
Additionally,it can act as a ligand to form complexes with transition metals.
Therefore,$HNO_2$ exhibits all three properties.

(C) Statement $1$ is correct: Atomic hydrogen is produced by passing $H_2$ gas through an electric arc at high temperature.
Statement $2$ is correct: Hydrogen gas is not a strong enough reducing agent to reduce $Al_2O_3$ to $Al$.
Statement $3$ is correct: Finely divided palladium (or platinum) can adsorb a large volume of $H_2$ gas,a property known as occlusion.
Statement $4$ is incorrect: The reaction $C_2H_5OH + Na \to C_2H_5ONa + \frac{1}{2} H_2$ produces molecular hydrogen,not nascent hydrogen. Nascent hydrogen is typically generated in situ during chemical reactions (e.g.,$Zn + HCl$).
Therefore,statements $1, 2,$ and $3$ are correct.

(C) Ionic hydrides are stoichiometric compounds of dihydrogen formed with most of the $s$-block elements. They are crystalline,non-volatile,and non-conducting in the solid state. However,when in a molten state,they conduct electricity and liberate dihydrogen gas at the $anode$,not the $cathode$. Therefore,statement $(C)$ is incorrect.

(A) Hydrogen has $1s^1$ electronic configuration. It can lose one electron to form $H^+$,gain one electron to form $H^-$,or share one electron to form covalent bonds (e.g.,$H_2$,$CH_4$).
Thus,the Assertion is correct.
Hydrogen forms electrovalent bonds (ionic hydrides like $NaH$,$CaH_2$) and covalent bonds (like $H_2O$,$NH_3$).
Since the ability to form these bonds is a direct consequence of its ability to lose,gain,or share electrons,the Reason is the correct explanation of the Assertion.

(B) Spin isomerism is shown by $H_2$ (hydrogen) molecules.
It arises due to the different relative orientations of the nuclear spins of the two hydrogen atoms in the molecule.
In $ortho-hydrogen$,the spins of the nuclei of the two atoms are in the same direction.
In $para-hydrogen$,the spins of the nuclei of the two atoms are in opposite directions.

(B) The assertion is true because $H_2SO_4$ molecules are associated through a large number of intermolecular hydrogen bonds,which increases its viscosity compared to water.
The reason is also true because concentrated $H_2SO_4$ is highly hygroscopic and has a strong affinity for water.
However,the high viscosity is due to hydrogen bonding,not the affinity for water. Thus,the reason is not the correct explanation of the assertion.

(B) Hydrogen has three isotopes:
$1$. Protium $(_{1}^{1}H)$: Number of neutrons $(x) = 1 - 1 = 0$
$2$. Deuterium $(_{1}^{2}H)$: Number of neutrons $(y) = 2 - 1 = 1$
$3$. Tritium $(_{1}^{3}H)$: Number of neutrons $(z) = 3 - 1 = 2$
The sum of neutrons is $(x + y + z) = 0 + 1 + 2 = 3$.

(N/A) $(i)$ Dihydrogen reacts with chlorine to form hydrogen chloride $(HCl)$. The reaction is: $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$. Here,dihydrogen is oxidised to $H^+$ and chlorine is reduced to $Cl^-$,forming a covalent molecule.
$(ii)$ Dihydrogen reacts with sodium to form sodium hydride $(NaH)$. The reaction is: $2Na(s) + H_2(g) \rightarrow 2NaH(s)$. Here,an electron is transferred from $Na$ to $H$,forming an ionic compound,$Na^+H^-$.
$(iii)$ Dihydrogen reduces copper $(II)$ oxide to copper metal. The reaction is: $CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(l)$. Dihydrogen acts as a reducing agent,getting oxidised to water $(H_2O)$.

(N/A) Hydrogen has three naturally occurring isotopes:
$1.$ Protium,$\_1^1H$
$2.$ Deuterium,$\_1^2H$ or $D$
$3.$ Tritium,$\_1^3H$ or $T$
The mass numbers of these isotopes are $1$,$2$,and $3$ respectively.
Therefore,the mass ratio of protium,deuterium,and tritium is $1: 2: 3$.

(B) The ionization enthalpy of the hydrogen atom is very high $\left(1312 \, kJ \, mol^{-1}\right)$.
Due to this high value,it is very difficult to remove its only electron to form a monoatomic ion.
Consequently,hydrogen prefers to share its electron with another hydrogen atom to form a stable covalent bond,existing as a diatomic $(H_2)$ molecule.

(N/A) $(i)$ The balanced reaction is: $o H_{2(g)} + M_mO_{o(s)} \xrightarrow{\Delta} m M_{(s)} + o H_2O_{(l)}$
$(ii)$ The reaction is: $CO_{(g)} + 2H_{2(g)} \xrightarrow{\Delta, \text{ catalyst}} CH_3OH_{(l)}$
$(iii)$ The reaction is: $C_3H_{8(g)} + 3H_2O_{(g)} \xrightarrow{\Delta, \text{ catalyst}} 3CO_{(g)} + 7H_{2(g)}$
$(iv)$ The reaction is: $Zn_{(s)} + 2NaOH_{(aq)} \xrightarrow{\text{heat}} Na_2ZnO_{2(aq)} \text{ (Sodium Zincate)} + H_{2(g)}$

(N/A) The bond dissociation enthalpy of the $H-H$ bond is very high $(435.88 \, kJ \, mol^{-1})$.
This high value indicates that the $H-H$ bond is very stable and strong.
Consequently,dihydrogen is relatively inert at room temperature.
It requires high temperatures or the presence of a catalyst to react with other elements.
Additionally,the high ionization enthalpy $(1312 \, kJ \, mol^{-1})$ of hydrogen makes it difficult to lose an electron to form $H^+$ ions,which is why it primarily forms covalent bonds rather than ionic ones.

(N/A) Molecular hydrides are classified on the basis of the total number of electrons and bonds in their Lewis structures as:
$1.$ Electron-deficient hydrides: These have fewer electrons than required for writing their conventional Lewis structures. Example: Diborane $(B_2H_6)$. In $B_2H_6$,there are six bonds in total,out of which only four are regular two-centered-two-electron bonds. The remaining two are three-centered-two-electron bonds,meaning two electrons are shared by three atoms.
$2.$ Electron-precise hydrides: These have a sufficient number of electrons to be represented by their conventional Lewis structures. Example: Methane $(CH_4)$. It forms four regular bonds where two electrons are shared by two atoms.
$3.$ Electron-rich hydrides: These contain excess electrons as lone pairs. Example: Ammonia $(NH_3)$. It has three regular bonds and one lone pair of electrons on the nitrogen atom.

(N/A) Non-stoichiometric hydrides are hydrogen-deficient compounds formed by the reaction of dihydrogen with $d$-block and $f$-block elements. These hydrides do not follow the law of constant composition. For example: $LaH_{2.87}$, $YbH_{2.55}$, $TiH_{1.5-1.8}$ etc.
Alkali metals form stoichiometric hydrides. These hydrides are ionic in nature. Hydride ions have comparable sizes $(208 \text{ pm})$ with alkali metal ions. Hence, strong binding forces exist between the constituting metal and hydride ion. As a result, stoichiometric hydrides are formed.
Alkali metals will not form non-stoichiometric hydrides.