Hydrogen Questions in English

(C) Protium $(^1H)$,deuterium ($^2H$ or $D$),and tritium ($^3H$ or $T$) are isotopes of hydrogen with mass numbers $1, 2,$ and $3$ respectively.
They have the same electronic configuration $(1s^1)$.
However,their natural abundance is not in the ratio $1 : 2 : 3$; protium is the most abundant $(>99.9\%)$,while tritium is radioactive and found in trace amounts.

(B) The three isotopes of hydrogen are $H^1$ (Protium),$H^2$ (Deuterium),and $H^3$ (Tritium).
For a diatomic molecule $XY$,the number of possible combinations is given by the formula $\frac{n(n+1)}{2}$,where $n$ is the number of isotopes.
Here,$n = 3$.
Number of molecules = $\frac{3(3+1)}{2} = \frac{3 \times 4}{2} = 6$.
The possible molecules are: $H^1H^1, H^1H^2, H^1H^3, H^2H^2, H^2H^3, H^3H^3$.

(C) . Pure hydrogen is obtained by the electrolysis of $Ba(OH)_2$ solution in a $U$-tube using nickel electrodes. The gas is liberated at the cathode and is passed over heated platinum gauze to remove oxygen if present as an impurity.

(B) In the Bosch process,water gas is used to produce hydrogen gas. The reaction is as follows:
$CO + H_2 + H_2O \xrightarrow{\text{catalyst}} CO_2 + 2H_2$
Here,the mixture of $CO$ and $H_2$ is known as water gas.

(B) Deuterium $(\,_1^2H)$ and hydrogen $(\,_1^1H)$ have the same atomic number but different mass numbers.
Due to the same electronic configuration,they exhibit similar chemical properties.
However,due to the difference in mass,they exhibit different physical properties.

(B) Tritium $(^3_1H)$ is a radioactive isotope of hydrogen.
It undergoes radioactive decay by emitting a $\beta$-particle (electron) to form Helium-$3$ $(^3_2He)$.
The nuclear reaction is: $_1^3H \to \,_2^3He + \,_{-1}^{\,\,0}e$.

(A) The conversion of atomic hydrogen into ordinary hydrogen is represented by the reaction: $2H(g) \rightarrow H_2(g)$.
This process involves the formation of a chemical bond,which releases energy.
The enthalpy change for this reaction is $\Delta H = -104.5 \ kcal$.
Since the value of $\Delta H$ is negative,the reaction is an exothermic change.

(B) . $Lavoisier$ gave the name hydrogen,which is derived from the Greek words 'hydro' (water) and 'genes' (forming),meaning water maker.

(B) The triatomic form of hydrogen,$H_3$,is known as Hyzone.
Therefore,the correct option is $(B)$.

(B) The chemical reaction for the preparation of lithium aluminium hydride is given by:
$4LiH + AlCl_3 \xrightarrow{\text{Ether}} LiAlH_4 + 3LiCl$
From the balanced chemical equation,it is clear that $4$ moles of $LiH$ react with $1$ mole of $AlCl_3$. Thus,$LiH$ is used in excess.

(C) Ionic hydrides,such as $LiH$ or $CaH_2$,contain the hydride ion $(H^-)$.
In these compounds,the hydrogen atom is in the $-1$ oxidation state.
Since hydrogen can easily lose an electron to form $H_2$ gas,these compounds act as strong reducing agents.

(A) Sodium borohydride $(NaBH_4)$ is soluble in water,but it is unstable in the presence of water.
When $NaBH_4$ is dissolved in water,it undergoes hydrolysis to produce sodium hydroxide $(NaOH)$,boric acid or borates,and hydrogen gas $(H_2)$:
$NaBH_4 + 2H_2O \rightarrow NaBO_2 + 4H_2$.
However,in basic conditions,the reaction is slower. The formation of $OH^{-}$ ions (from $NaOH$) makes the solution alkaline,which helps in stabilizing the remaining $BH_4^-$ ions and prevents further rapid decomposition.

(D) $H_2$ does not react directly with $Au$,$Cu$,or $Ni$ under normal conditions.
With $Ca$,it reacts to form calcium hydride $(CaH_2)$.
The chemical equation is: $Ca + H_2 \to CaH_2$.

(C) The correct answer is $(C)$.
Anhydrous $CaCl_2$ is a common drying agent used in the laboratory for the fast drying of neutral gases.

(A) Hydrogen $(H)$ has the electronic configuration $1s^1$. By accepting one electron,it achieves the stable configuration of Helium $(He)$,which is $1s^2$.
$H + e^- \to H^- (1s^2)$
Similarly,halogens (Group $17$ elements) have a valence shell configuration of $ns^2 np^5$. By accepting one electron,they achieve the stable noble gas configuration $(ns^2 np^6)$.
$F + e^- \to F^- ([He] 2s^2 2p^6)$
Therefore,hydrogen resembles halogens in its ability to gain an electron to complete its shell.

(A) Hydrogen exhibits a $+1$ oxidation state by losing its electron (e.g.,in $HCl$) and a $-1$ oxidation state by gaining an electron (e.g.,in $NaH$). Thus,it can form bonds in both $+1$ and $-1$ oxidation states. Therefore,the correct option is $A$.

(C) $SO_2$,$NH_3$,and $CO_2$ are polar or acidic gases that show significant solubility in water due to dipole-dipole interactions or chemical reactions.
$H_2$ is a non-polar diatomic molecule with very weak van der Waals forces,making it essentially insoluble in water.

(B) The correct answer is $(B)$.
Although carbon is well-known for its ability to form a vast number of organic compounds due to catenation,hydrogen forms the maximum number of compounds in chemistry when considering both organic and inorganic compounds,as it bonds with almost every other element in the periodic table.

(C) Zinc does not react with cold water to produce hydrogen gas.
Zinc reacts with hot $NaOH$ solution to produce sodium zincate and hydrogen gas: $Zn + 2NaOH \to Na_2ZnO_2 + H_2$.
Zinc reacts with dilute $HCl$ to produce zinc chloride and hydrogen gas: $Zn + 2HCl \to ZnCl_2 + H_2$.
Zinc reacts with concentrated sulphuric acid to produce zinc sulphate,sulphur dioxide,and water,not hydrogen gas: $Zn + 2H_2SO_4 \to ZnSO_4 + SO_2 + 2H_2O$.
Therefore,hydrogen is not obtained when zinc reacts with concentrated sulphuric acid.

(C) Heavy water is chemically known as deuterium oxide,represented as $D_2O$.
In this compound,$D$ represents deuterium,which is an isotope of hydrogen with the symbol $_1H^2$ or $D$.

(C) The correct option is $C$.
$D_2O$ (heavy water) has a higher freezing point $(3.8 \ ^\circ C)$ compared to ordinary water $H_2O$ $(0 \ ^\circ C)$.
Therefore,the statement that $D_2O$ freezes at a lower temperature is false.

(C) The reaction of active metals with water or dilute mineral acids like $HCl$ or $H_2SO_4$ results in the evolution of $H_2$ gas,which is known as dihydrogen.
For example: $2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)$ and $Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)$.

(D) Hydrogen gas can be produced by the reaction of water with active metals and metal hydrides.
$1$. Reaction with metals: $Mg + 2 H_2O \to Mg(OH)_2 + H_2 \uparrow$
$2$. Reaction with metal hydrides: $LiH + H_2O \to LiOH + H_2 \uparrow$
Therefore,both $(a)$ and $(c)$ are correct.

(B) The correct answer is $(B)$.
$NaBH_4$ is a stable reducing agent that reacts very slowly with cold water.
Statement $(B)$ is incorrect because the reaction is not violent.
Statements $(A)$,$(C)$,and $(D)$ are factually correct properties of $NaBH_4$.

(D) The reaction of $CaH_2$ with water is: $CaH_2 + 2H_2O \to Ca(OH)_2 + 2H_2 \uparrow$
The reaction of $Ca$ with water is: $Ca + 2H_2O \to Ca(OH)_2 + H_2 \uparrow$
Both $CaH_2$ and $Ca$ react with water to produce hydrogen gas. Therefore,the correct option is $(d)$.

(D) The reactivity series of metals indicates that hydrogen can reduce oxides of metals that are less reactive than hydrogen.
$Cu$,$Fe$,and $Sn$ are less reactive than hydrogen,so their oxides ($CuO$,$Fe_2O_3$,$SnO_2$) can be reduced by $H_2$.
However,$Al$ is more reactive than hydrogen,so $H_2$ cannot reduce $Al_2O_3$ to $Al$.
Therefore,the correct option is $(D)$.

(A) The reactivity series of metals determines whether a metal can displace hydrogen from an acid or water.
$Cu$ (Copper) is placed below hydrogen in the electrochemical series,meaning it has a lower reduction potential than hydrogen.
Therefore,$Cu$ cannot displace hydrogen from dilute $HCl$.
In contrast,$Fe$,$Mg$,and $Na$ are more reactive than hydrogen and will react to produce $H_2$ gas.
Thus,the pair $Cu + HCl$ (dil.) will not produce dihydrogen gas.

(C) Hydrogen is a non-metal under normal conditions. However,at very high pressure,the hydrogen molecules are compressed to such an extent that the electron clouds overlap,allowing it to exhibit metallic properties. Therefore,the correct option is $(C)$.

(C) The two allotropic forms of dihydrogen $(H_2)$ are ortho-hydrogen and para-hydrogen.
In ortho-hydrogen,the nuclear spins are parallel,whereas in para-hydrogen,the nuclear spins are antiparallel.
Due to the difference in nuclear spin alignment,the para-hydrogen form has lower energy than the ortho-hydrogen form at low temperatures.

(C) The correct option is $(C)$.
Saline hydrides (ionic hydrides) react violently with water to produce hydrogen gas,which is highly flammable.
These fires cannot be extinguished using water or $CO_2$ because saline hydrides can reduce $CO_2$ at high temperatures,potentially releasing oxygen or causing further reactions.
Therefore,sand is used to smother the fire and cut off the oxygen supply.

(A) Metals of groups $7$,$8$,and $9$ do not form metallic hydrides. This phenomenon is known as the hydride gap.

(C) Hydrogen is the most abundant element in the universe,accounting for approximately $3/4$ of its total mass.
Helium is the second most abundant element,comprising almost all of the remaining $25 \%$.
Oxygen is the third most abundant element in the universe.
On Earth,however,oxygen is the most common element,making up about $47 \%$ of the Earth's mass.

(A) The equilibrium between ortho and para hydrogen is temperature-dependent. $Ortho-hydrogen$ is more stable at higher temperatures,while $para-hydrogen$ is more stable at lower temperatures. At $0 \ K$,hydrogen exists almost entirely as $para-hydrogen$. Therefore,by decreasing the temperature,the equilibrium shifts towards the formation of $para-hydrogen$,allowing pure $para-hydrogen$ to be obtained.

(B) When sodium formate $(HCOONa)$ is heated,it undergoes thermal decomposition to produce hydrogen gas and sodium oxalate.
The chemical reaction is: $2HCOONa_{(s)} \xrightarrow{\Delta} H_{2(g)} \uparrow + (COONa)_{2(s)}$
Therefore,the correct option is $B$.

(B) The reaction between alkali metals like $K$ and acids like $HCl$ is highly exothermic and explosive in nature.
$2K(s) + 2HCl(aq) \to 2KCl(aq) + H_2(g)$ (violent reaction).
Therefore,this is the most dangerous method.

(D) Hydrogen is a colourless,odourless,and tasteless gas.

(D) Hydrogen is a non-metal,whereas all other members of group $1$ (alkali metals) are metals. This fundamental difference in physical state and bonding behavior distinguishes hydrogen from the alkali metals.

(A) The correct answer is $(A)$.
$CaCl_2$ (Calcium chloride) is used as a dehydrating agent in the laboratory because it is highly hygroscopic,meaning it readily absorbs moisture from the surroundings.

(D) $Anhydrous \ CaCl_2$ (Anhydrous calcium chloride) is a common drying agent used in the laboratory for neutral gases.
It absorbs moisture to form a hydrate,$CaCl_2 \cdot 6H_2O$.
It is effective for drying neutral gases like $N_2$,$O_2$,and $CO_2$.

(D) Aluminium oxide $(Al_2O_3)$ cannot be reduced by hydrogen even under very hot conditions because $Al$ is more reactive than $H$.
Hydrogen can only reduce metal oxides of metals that are less reactive than hydrogen in the metal reactivity series,such as $Cu$,$Fe$,and $Sn$.

(A) Producer gas is obtained by passing air over red-hot coke. The reaction is $2C(s) + O_2(g) + 3.76N_2(g) \rightarrow 2CO(g) + 3.76N_2(g)$. Therefore,it is a mixture of carbon monoxide $(CO)$ and nitrogen $(N_2)$.

(B) . The thermal decomposition of ammonium nitrate is given by the reaction: $NH_4NO_{3(s)} \xrightarrow{\Delta} N_2O_{(g)} + 2H_2O_{(g)}$.
Thus,it decomposes into nitrous oxide $(N_2O)$ and water $(H_2O)$.

(B) The reaction between $NH_4Cl$ and $KNO_2$ is a double displacement reaction followed by the thermal decomposition of the unstable intermediate $NH_4NO_2$.
$NH_4Cl + KNO_2 \to NH_4NO_2 + KCl$
$NH_4NO_2 \xrightarrow{\Delta} N_2 + 2H_2O$
Thus,the final products are nitrogen gas $(N_2)$ and water.

(D) $HNO_2$ (nitrous acid) exhibits oxidising,reducing,and complex-forming properties.
$1$. It acts as an oxidising agent by reducing to $NO$ $(2HNO_2 \to 2NO + H_2O + [O])$.
$2$. It acts as a reducing agent by oxidising to $HNO_3$ $(HNO_2 + [O] \to HNO_3)$.
$3$. It can form complexes with transition metals,such as the formation of nitrosonium complexes or metal-nitrito complexes.

(A) The reaction of calcium cyanamide with steam under pressure is as follows:
$CaCN_2 + 3H_2O \to CaCO_3 + 2NH_3$
Thus,the products are ammonia $(NH_3)$ and calcium carbonate $(CaCO_3)$.

(B) Liquid ammonia is used in refrigeration because it has a high heat of vaporisation.

(A) $H_2SO_4$ acts as a strong oxidizing agent and a dehydrating agent. It is also used as a sulphonating agent in organic synthesis and is known for being highly viscous due to extensive hydrogen bonding. However,it is not a reducing agent; rather,it oxidizes other substances while being reduced itself.

(D) The sugar,a carbohydrate such as $C_{12}H_{22}O_{11}$,is mixed with concentrated sulfuric acid,$H_2SO_4$.
The sulfuric acid acts as a strong dehydrating agent and removes water molecules from the carbohydrate,leaving behind elemental carbon.
The reaction is: $C_{12}H_{22}O_{11} \xrightarrow{H_2SO_4} 12C + 11H_2O$.
This process turns the sugar black due to the formation of carbon.

(A) When $H_2S$ is reacted with a limited supply of $O_2$,it undergoes oxidation to form water and elemental sulfur.
The balanced chemical equation is: $2H_2S + O_2 \to 2H_2O + 2S$.

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