Hydrogen Questions in English

(B) desiccator is a container used to keep substances dry or to remove moisture from them. $A$ substance used in a desiccator must be a strong dehydrating agent or a desiccant.
$Conc. H_2SO_4$,$CaCl_2$,and $P_4O_{10}$ are all well-known dehydrating agents and are commonly used as desiccants.
$Na_2SO_4$ (anhydrous sodium sulfate) is not a strong dehydrating agent; it is often used as a drying agent for organic solvents,but it is not suitable for use in a desiccator to maintain a dry atmosphere for other substances.

(A) $H_2SO_4$ acts as a dehydrating agent in its reaction with oxalic acid $(H_2C_2O_4)$.
The reaction is: $H_2C_2O_4 \xrightarrow{H_2SO_4} CO + CO_2 + H_2O$.
In this reaction,$H_2SO_4$ removes a water molecule from $H_2C_2O_4$.

(D) Spirit of salt is an old name for hydrochloric acid $(HCl)$.
Since $HCl$ contains a chlorine atom,it is the correct answer.
Fischer salt is $K_3[Co(NO_2)_6]$,Epsom salt is $MgSO_4 \cdot 7H_2O$,and Fremy's salt is $K_4[ON(SO_3)_2]_2$.

(C) When $KI$ reacts with concentrated $H_2SO_4$,$HI$ is initially formed as $KI + H_2SO_4 \rightarrow KHSO_4 + HI$.
However,$HI$ is a strong reducing agent and concentrated $H_2SO_4$ is a strong oxidising agent.
Therefore,$H_2SO_4$ oxidizes $HI$ to $I_2$ according to the reaction: $2HI + H_2SO_4 \rightarrow I_2 + SO_2 + 2H_2O$.
Thus,$HI$ cannot be prepared by this method.

(A) Hydrogen $(H)$ has an electronic configuration of $1s^1$. By gaining one electron,it achieves the $1s^2$ configuration,which is the stable electronic configuration of the noble gas helium $(He)$.
Similarly,halogens (Group $17$ elements) have an $ns^2 np^5$ configuration and tend to gain one electron to achieve the stable $ns^2 np^6$ configuration of the nearest noble gas.
Therefore,in this specific respect,hydrogen resembles halogens.

(A) The correct answer is $A$. Due to the highest electronegativity and small size of the fluorine atom,the hydrogen fluoride molecule forms a strong hydrogen bond with the fluoride ion,resulting in the formation of the $[F-H--F]^-$ anion. Consequently,the potassium ion $K^+$ associates with this anion to form the salt $KHF_2$.

(A) The lightest gas is $H_2$ because it has the lowest molar mass among the given options.

Gas	Molar Mass $(g/mol)$
$H_2$	$2$
$He$	$4$
$N_2$	$28$
$O_2$	$32$

(D) The reaction of calcium cyanamide $(CaCN_2)$ with steam under pressure is a hydrolysis reaction.
The chemical equation is: $CaCN_2 + 3H_2O \xrightarrow{\text{pressure}} CaCO_3 + 2NH_3$.
Thus,the products are ammonia $(NH_3)$ and calcium carbonate $(CaCO_3)$.

(C) The correct answer is $(C)$. Soda-acid fire extinguishers contain a solution of $NaHCO_3$ and a bottle of $H_2SO_4$. When the extinguisher is activated,$H_2SO_4$ reacts with $NaHCO_3$ to release $CO_2$ gas,which helps in extinguishing the fire. The reaction is: $2NaHCO_3 + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O + 2CO_2$.

(C) The mixture of carbon monoxide $(CO)$ and hydrogen $(H_2)$ is known as water gas.
Syn gas (synthesis gas) is a mixture of $CO$,$CO_2$,and $H_2$.
Producer gas is a mixture of $CO$ and $N_2$ (along with small amounts of $CO_2$ and $H_2$),obtained by passing air over red-hot coke.
Therefore,the correct composition for water gas is $CO + H_2$.

(D) The phenomenon where a substance loses water of crystallization or moisture to the atmosphere is known as efflorescence.
Among the given options,solid $NaOH$ is deliquescent (absorbs moisture),concentrated $H_2SO_4$ is hygroscopic (absorbs moisture),and a saturated solution of $CO_2$ would lose $CO_2$ gas,but the question refers to weight loss due to exposure.
However,hydrated salts like $Na_2CO_3 \cdot 10H_2O$ (washing soda) are efflorescent. Since the option is $Anhydrous \ sodium \ carbonate$,it does not lose weight.
Actually,the question implies a substance that loses water to the atmosphere. Among the choices,if we consider the property of efflorescence,hydrated salts lose weight. Given the options,there might be a context of a hydrated salt. If we assume the question refers to a substance that loses water,none of these are standard examples except for the potential misinterpretation of hydrated salts.
However,based on standard chemistry problems of this type,$NaOH$ and $H_2SO_4$ gain weight. $Anhydrous \ sodium \ carbonate$ is stable. The most appropriate answer in many textbooks for this specific question is often related to efflorescent substances.

(B) deuteron $(_{1}H^{2})$ is the nucleus of deuterium,which consists of one proton and one neutron.

(C) Isotopes of an element have similar chemical properties because they have the same electronic configuration,but they differ in physical properties like melting point,boiling point,and density due to the difference in their atomic masses.

(B) The atomic number $(Z)$ of hydrogen is $1$,which represents the number of protons.
Given that the isotope has two neutrons $(n = 2)$,the atomic mass number $(A)$ is calculated as $A = Z + n = 1 + 2 = 3$.
Thus,the atomic number is $1$ and the atomic mass number is $3$.
This isotope is known as Tritium $(^3_1H)$.

(B) Ortho and para hydrogen differ in the nuclear spin of their protons. Para hydrogen has antiparallel spins $(I = 0)$,while ortho hydrogen has parallel spins $(I = 1)$.
Para hydrogen is more stable than ortho hydrogen at low temperatures.
The ratio of ortho to para hydrogen increases with temperature,reaching a maximum equilibrium ratio of $3:1$ at room temperature and above,but it does not result in pure ortho hydrogen.

(C) Deuterium $(D)$ is an isotope of hydrogen $(H)$. Due to the higher mass of the deuterium nucleus compared to the hydrogen nucleus,the bond dissociation energy of $D-D$ or $D-X$ bonds is generally higher than that of $H-H$ or $H-X$ bonds. This leads to a kinetic isotope effect where the reaction involving deuterium proceeds at a slower rate than the corresponding reaction involving hydrogen.

(A) The isotopes of hydrogen are $_1H^1$ (protium),$_1H^2$ (deuterium),and $_1H^3$ (tritium).
Thus,tritium is an isotope of hydrogen.

(D) The isotopes of hydrogen are Protium $(^1_1H)$,Deuterium $(^2_1H)$,and Tritium $(^3_1H)$.
In Protium $(^1_1H)$,the sum of protons and neutrons is $1+0 = 1$.
In Deuterium $(^2_1H)$,the sum of protons and neutrons is $1+1 = 2$.
In Tritium $(^3_1H)$,the sum of protons and neutrons is $1+2 = 3$.
Therefore,the maximum sum of the number of neutrons and protons in an isotope of hydrogen is $3$.

(B) The isotopes of hydrogen are Protium $(^1H)$,Deuterium $(^2H)$,and Tritium $(^3H)$.
Protium has $0$ neutrons,Deuterium has $1$ neutron,and Tritium has $2$ neutrons.
The atomic weight of an isotope is the sum of the number of protons and neutrons.
For Tritium,atomic number $(Z)$ = $1$ and number of neutrons $(n)$ = $2$.
Atomic mass $(A)$ = $Z + n = 1 + 2 = 3$.

(A) Hydrogen gas $(H_2)$ primarily consists of the most abundant isotope of hydrogen,which is protium $(^1H)$.
Protium accounts for approximately $99.985\%$ of naturally occurring hydrogen atoms.
Deuterium $(^2H)$ is present in trace amounts,and tritium $(^3H)$ is radioactive and exists in negligible quantities.
Therefore,atoms in hydrogen gas have a preponderance of $_1H^1$ atoms.

(A) Hydrogen has three isotopes: Protium $(^1H)$,Deuterium ($^2H$ or $D$),and Tritium ($^3H$ or $T$).
Among these,Tritium is radioactive and emits low-energy $\beta$-particles.
The nuclear decay reaction is: $^3_1H \to ^3_2He + ^0_{-1}e$.

(A) In an ionic hydride like $LiH$ in its fused state,the ions present are $Li^+$ and $H^-$.
During electrolysis,the negatively charged hydride ion $(H^-)$ migrates towards the anode (positive electrode).
At the anode,the hydride ion undergoes oxidation to release hydrogen gas:
$2H^- \to H_2(g) + 2e^-$
Therefore,hydrogen gas is obtained at the anode.

(A) When metals like $Pt$ and $Pd$ adsorb large volumes of hydrogen gas on their surface,the hydrogen is referred to as $Occluded \ hydrogen$.

(A) The most abundant element in the universe is $Hydrogen$ $(H)$,accounting for approximately $75\%$ of the baryonic mass.
The second most abundant element is $Helium$ $(He)$,which accounts for approximately $24\%$ of the baryonic mass.

(B) $Hydrogen$ is the most abundant element in the universe,accounting for approximately $75\%$ of all baryonic mass.

(D) $CaO$ (Quicklime) is a basic oxide and is highly hygroscopic in nature.
It absorbs moisture from gases and is therefore used as a drying agent for gases like $NH_3$.

(D) The term $occlusion$ refers to the adsorption of gases on the surface of metals. Transition metals like $Pt$ (Platinum),$Pd$ (Palladium),$Ni$ (Nickel),and $Os$ (Osmium) are well-known for their ability to adsorb large volumes of hydrogen gas on their surfaces. This property is widely used in catalytic hydrogenation reactions. Therefore,all the given metals exhibit this property.

(D) The correct answer is $(d)$.
$Pd$ (Palladium) has a unique property of adsorbing a large volume of hydrogen gas on its surface.
At $350 \ K$ and $1 \ atm$ pressure,$1$ unit volume of $Pd$ absorbs $900$ unit volumes of $H_2$ gas.

(D) The correct answer is $(D)$.
$Cu + HCl \to \text{No reaction}$.
Copper is less reactive than hydrogen in the electrochemical series,as indicated by its positive standard reduction potential $(E_{Cu}^{0} = +0.34 \ V)$ compared to hydrogen $(E_{H}^{0} = 0.00 \ V)$.
Therefore,copper cannot displace hydrogen from dilute acids.

(C) Lithium tetrahydridoaluminate,commonly known as lithium aluminum hydride,consists of a lithium cation $(Li^+)$ and a tetrahydridoaluminate anion $([AlH_4]^-)$.
In the anion,the central aluminum atom is bonded to four hydrogen atoms.
The correct chemical formula is $Li[AlH_4]$.

(A) Commercially,methanol is prepared by the catalytic hydrogenation of carbon monoxide.
The reaction is: $CO(g) + 2H_2(g) \xrightarrow{ZnO-Cr_2O_3} CH_3OH(l)$
This process occurs at high temperature and pressure in the presence of a catalyst mixture of zinc oxide and chromium oxide.

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

(A) The hardening of oils is a process of catalytic hydrogenation.
In this process,unsaturated vegetable oils are reacted with $H_2$ gas in the presence of a nickel $(Ni)$ catalyst to form saturated fats.
The reaction is: $\text{Oil (unsaturated)} + H_2 \xrightarrow{Ni} \text{Fat (saturated)}$.

(B) Asthma patients use a mixture of $O_2$ and $He$ for respiration.
Helium is an inert and very light gas,which helps in the rapid diffusion of oxygen into the lungs,making it easier for patients with respiratory difficulties to breathe.

(A) laser is a device that emits light through a process of optical amplification based on the stimulated emission of electromagnetic radiation.
Common types of lasers include gas lasers like $CO_2$ lasers,Helium-Neon (He-Ne) lasers,and Argon ion lasers.
Hydrogen gas is not typically used as a lasing medium in standard laser technology because its energy level structure does not support the population inversion required for continuous laser action in the same way as noble gases or $CO_2$ molecules.
Therefore,a Hydrogen laser is not a standard type of laser.

(D) The standard Gibbs free energy of formation for $Al_2O_3$ is very negative,making it highly stable. Hydrogen is not a strong enough reducing agent to reduce $Al_2O_3$ to $Al$ metal.

(A) The calorific value depends on the composition of combustible gases present in the mixture.
The order of calorific value is:
Coal gas > Water gas > Producer gas.
$1$. Coal gas: Contains high amounts of $H_2$ $(49\%)$,$CH_4$ $(32\%)$,and $CO$ $(8\%)$.
$2$. Water gas: Contains $CO$ $(48\%)$ and $H_2$ $(50\%)$.
$3$. Producer gas: Contains $CO$ $(35\%)$ and $N_2$ $(65\%)$.
Since $CH_4$ has a very high calorific value,Coal gas possesses the highest calorific value among the given options. $CO_2$ is a non-combustible gas and has a calorific value of $0$.

(D) Carbogen is a mixture of $95\% \ O_2$ and $5\% \ CO_2$ used in medical applications for respiratory stimulation.

(D) The isotopes of hydrogen are Protium $(_{1}^{1}H)$,Deuterium $(_{1}^{2}H)$,and Tritium $(_{1}^{3}H)$.
For Protium $(_{1}^{1}H)$: Number of protons $(p)$ = $1$,Number of neutrons $(n)$ = $0$. Sum $(n+p)$ = $1$.
For Deuterium $(_{1}^{2}H)$: Number of protons $(p)$ = $1$,Number of neutrons $(n)$ = $1$. Sum $(n+p)$ = $2$.
For Tritium $(_{1}^{3}H)$: Number of protons $(p)$ = $1$,Number of neutrons $(n)$ = $2$. Sum $(n+p)$ = $3$.
The sum of the number of protons and neutrons for all isotopes is $1 + 2 + 3 = 6$.

(C) Deuterium ($^2H$ or $D$) is an isotope of hydrogen with an atomic number of $1$ and a mass number of $2$.
It consists of $1$ proton $(p^+)$,$1$ neutron $(n)$,and $1$ electron $(e^-)$.
Option $A$ is correct as it is an isotope.
Option $B$ is correct as it lists the correct subatomic particles.
Option $D$ is correct as $D_2O$ is indeed called heavy water.
Option $C$ is incorrect because it omits the electron $(e^-)$,which is essential for the neutral atom of deuterium.

(B) The boiling point order of these hydrogen halides is $HCl < HBr < HI < HF$.
Since vapor pressure is inversely proportional to the boiling point,the substance with the lowest boiling point will have the maximum vapor pressure.
Therefore,$HCl$ has the maximum vapor pressure.

(A) Interstitial hydrides are formed by transition metals (d-block and f-block elements). These hydrides are almost always non-stoichiometric,being deficient in hydrogen. For example,$LaH_{2.87}$,$YbH_{2.55}$,$TiH_{1.5-1.8}$,etc. Therefore,the statement that they generally form non-stoichiometric compounds is correct.

(A) Hydrogen has $1s^1$ configuration. By gaining one electron,it achieves the $1s^2$ configuration,which is the configuration of the noble gas Helium. Halogens $(ns^2 np^5)$ also require one electron to achieve a stable noble gas configuration. Therefore,hydrogen behaves like halogens.

(C) proton $(H^+)$ is extremely small in size compared to other ions. Due to its very small ionic radius and high charge density,it has a very high hydration energy.

(D) In $para-hydrogen$,the nuclear spins of the two protons are in opposite directions. However,the question asks about electron spin. In a hydrogen molecule $(H_2)$,the two electrons are paired with opposite spins to form a covalent bond,regardless of whether it is $ortho$ or $para$ hydrogen. Given the options provided,the question likely refers to the nuclear spin of the protons. In $para-hydrogen$,the nuclear spins of the two protons are antiparallel (opposite to each other).

(C) During the electrolysis of molten lithium hydride $(LiH)$,the reaction is: $2LiH \rightarrow 2Li + H_2 \uparrow$.
In this process,the hydride ion $(H^-)$ migrates to the anode and loses electrons to form hydrogen gas: $2H^- \rightarrow H_2 + 2e^-$.
Therefore,hydrogen is liberated at the anode.

(D) The incorrect statement is $D$. In covalent hydrides (e.g.,$CH_4$,$NH_3$),hydrogen is bonded to more electronegative non-metals and typically exhibits a $+1$ oxidation state,not $-1$. Hydrogen exhibits a $-1$ oxidation state only in ionic (saline) hydrides formed with highly electropositive metals (e.g.,$NaH$,$CaH_2$).

(A) $H^+$,$D^+$,and $T^+$ are all electron-deficient species (they have $0$ electrons).
Therefore,they do not differ in the number of electrons.

(D) Hydrogen $(H_2)$ acts as a reducing agent for metal oxides of metals that are less reactive than hydrogen in the reactivity series.
$Al$ is more reactive than $H_2$,so $Al_2O_3$ cannot be reduced by $H_2$.
$CuO$,$Fe_2O_3$,and $SnO_2$ are oxides of metals less reactive than hydrogen and can be reduced by $H_2$.

(D) The heaviest isotope of hydrogen is tritium ($^3_1H$ or $T$).
The mass number $(A)$ is defined as the sum of the number of protons $(Z)$ and the number of neutrons $(N)$.
For tritium,the atomic number $(Z)$ is $1$ and the number of neutrons $(N)$ is $2$.
Therefore,the sum of neutrons and protons is $N + Z = 2 + 1 = 3$.