Hydrogen Questions in English

(D) $H^+$ (proton) will have a very large hydration energy due to its extremely small ionic size.
Hydration energy $\propto \frac{1}{\text{Size}}$.

(C) The hydride ion $({H^{-}})$ has $1 + 1 = 2$ electrons,which corresponds to the electronic configuration $1s^2$.
$He$ (Helium) has an atomic number of $2$,so it also has $2$ electrons with the configuration $1s^2$.
Since both species have the same number of electrons,they are isoelectronic.

(D) Tritium is an isotope of hydrogen represented as $_1^3H$.
The atomic number $(Z)$ is $1$,which represents the number of protons.
The mass number $(A)$ is $3$.
The number of neutrons $(n)$ is calculated as $n = A - Z = 3 - 1 = 2$.
Therefore,the nucleus of tritium contains $1$ proton and $2$ neutrons.

(B) Tritium is an isotope of hydrogen represented as $^3_1H$.
The atomic number $(Z)$ is $1$,which represents the number of protons.
The mass number $(A)$ is $3$.
The number of neutrons $(n)$ is calculated as $n = A - Z = 3 - 1 = 2$.
Therefore,the correct option is $(B)$.

(A) The correct answer is $Hydrogen$.
Tritium ($^3_1H$ or $T$) is a radioactive isotope of the element $Hydrogen$. It contains one proton and two neutrons in its nucleus.

(B) Heavy hydrogen is deuterium,represented as $_1^2H$ or $D$.
The atomic number $(Z)$ is $1$ and the mass number $(A)$ is $2$.
The number of neutrons is calculated as $N = A - Z = 2 - 1 = 1$.
Therefore,the correct option is $(B)$.

(A) $NaH$ and $CaH_2$ are ionic (saline) hydrides because they are formed by the transfer of electrons from electropositive metals to hydrogen.
$NH_3$ and $B_2H_6$ are covalent (molecular) hydrides because they are formed by the sharing of electrons between non-metals and hydrogen.
Therefore,the correct option is $A$.

(C) The ionization of a hydrogen atom $(H)$ involves the loss of its single electron to form a hydrogen ion $(H^+)$.
The reaction is represented as: $H \rightarrow H^+ + e^-$.
Since a neutral hydrogen atom consists of $1$ proton and $1$ electron (with no neutrons),the removal of the electron leaves behind only the nucleus,which is a single proton.
Therefore,the $H^+$ ion is commonly referred to as a proton.

(A) The amount of energy required to break the $H-H$ bond of $1 \, mole$ of gaseous hydrogen is $436 \, kJ \, mol^{-1}$.
This is known as the bond dissociation enthalpy.
The reaction is represented as: $H_{2(g)} + 436 \, kJ \, mol^{-1} \rightarrow H_{(g)} + H_{(g)}$.

(A) The correct statement is that hydrogen exhibits oxidation states of $+1$ (in most compounds) and $-1$ (in metal hydrides like $NaH$).
Therefore,option $A$ is correct.

(B) In $MH_2$,$M$ represents an alkaline earth metal (Group $2$ element).
Since the metal $M$ is in its $+2$ oxidation state,the sum of oxidation numbers must be zero.
Let the oxidation number of $H$ be $x$.
$1 \times (+2) + 2 \times (x) = 0$
$2 + 2x = 0$
$2x = -2$
$x = -1$.
Therefore,hydrogen in metal hydrides like $MH_2$ exhibits a $-1$ oxidation state.

(D) The correct answer is $(d)$.
Hydrogen exhibits properties similar to both alkali metals (Group $1$) and halogens (Group $17$).
Because it shares electronic configuration characteristics with alkali metals,it can indeed be placed in the first group of the periodic table.
Therefore,the statement that it cannot be placed in the first group is incorrect.

(B) Hydrogen exhibits properties similar to halogens because it can gain one electron to achieve a stable noble gas configuration (like $He$) and forms hydrides $(H^-)$ which are analogous to halides ($Cl^-$,$Br^-$,etc.). Therefore,it forms hydrides like chlorides.

(A) The ionization energy of hydrogen is $1312 \ kJ \ mol^{-1}$,which is significantly higher than that of alkali metals (e.g.,$Li = 520 \ kJ \ mol^{-1}$) and significantly lower than that of halogens (e.g.,$F = 1681 \ kJ \ mol^{-1}$).
Therefore,hydrogen does not fit perfectly into either group based on its ionization energy.

(B) Hydrogen has only one orbital which is strongly attracted by the nucleus.
Thus,it has a very small size compared to alkali metals,and hence it has a high ionization enthalpy.

(A) Sodium hydride $(NaH)$ reacts with water to produce hydrogen gas,making it useful in hydrogen generators.
The chemical reaction is: $NaH + H_2O \to NaOH + H_2 \uparrow$

(D) Metal hydrides $(MH)$ react with water to produce metal hydroxide and hydrogen gas.
For example: $KH + H_2O \to KOH + H_2 \uparrow$.
Therefore,the correct option is $(D)$.

(A) Hydrogen burns in air with a light bluish flame.

(A) Ortho and para hydrogen are forms of molecular hydrogen $(H_2)$ based on the relative orientation of nuclear spins,not isotopes. Isotopes are atoms of the same element with different numbers of neutrons. Protium $(^1H)$,deuterium $(^2H)$,and tritium $(^3H)$ are the isotopes of hydrogen.

(D) The boiling point of liquid $H_2$ is approximately $-252.87 \ ^\circ C$ $(20.28 \ K)$,which is the lowest among the given substances. Since the substance with the lowest boiling point distills first,liquid $H_2$ is distilled first.

(D) Very dilute nitric acid $(HNO_3)$ acts as a typical acid when reacting with metals like $Mg$ or $Mn$ because it is not a strong enough oxidizing agent at this concentration to oxidize the hydrogen gas produced.
The chemical reaction is:
$Mg(s) + 2HNO_3(aq) \to Mg(NO_3)_2(aq) + H_2(g) \uparrow $
Therefore,the correct option is $(D)$.

(A) The reaction of magnesium $(Mg)$ with water $(H_2O)$ produces magnesium hydroxide $(Mg(OH)_2)$ and hydrogen gas $(H_2)$:
$Mg + 2H_2O \to Mg(OH)_2 + H_2 \uparrow$
The other reactions listed produce hydrogen peroxide $(H_2O_2)$ instead of hydrogen gas.

(C) Hydrogen exhibits properties similar to both alkali metals and halogens.
$1$. Like alkali metals $(ns^1)$,hydrogen has one electron in its valence shell $(1s^1)$ and can lose it to form $H^+$.
$2$. Like halogens $(ns^2 np^5)$,hydrogen is one electron short of the nearest noble gas configuration and can gain one electron to form $H^-$.
Therefore,hydrogen shows resemblance to both groups.

(A) The difference between ortho and para hydrogen arises from the relative orientation of the nuclear spins of the two hydrogen atoms in the $H_2$ molecule.
In ortho-hydrogen,the spins of the two nuclei are parallel (same direction).
In para-hydrogen,the spins of the two nuclei are anti-parallel (opposite direction).
Therefore,they differ in their proton spin orientation.

(C) Metals react with water or dilute mineral acids to produce hydrogen gas in the form of $H_2$ molecules,which is known as dihydrogen.
For example: $Mg(s) + 2H_2O(l) \to Mg(OH)_2(aq) + H_2(g) \uparrow$
Similarly,with dilute acids: $Zn(s) + 2HCl(aq) \to ZnCl_2(aq) + H_2(g) \uparrow$

(A) Metals that are more reactive than hydrogen can displace it from dilute acids.
$Mg$ is more reactive than hydrogen,so it reacts with $HCl$ to produce hydrogen gas.
$Mg + 2HCl \to MgCl_2 + H_2 \uparrow$
$Cu$,$P$,and $Pt$ are less reactive than hydrogen and do not displace it from $HCl$.

(A) At room temperature,ordinary hydrogen exists as an equilibrium mixture of $75\%$ ortho-hydrogen and $25\%$ para-hydrogen.
In ortho-hydrogen,the nuclear spins of the two protons are parallel (same direction).
In para-hydrogen,the nuclear spins of the two protons are antiparallel (opposite direction).

(D) . Hydrogen cannot reduce $Al_2O_3$ because $Al$ has a much higher affinity for oxygen than hydrogen does. The standard Gibbs free energy of formation for $Al_2O_3$ is significantly more negative than that of water,making the reduction reaction thermodynamically unfavorable.

(D) Helium is a noble gas and does not combine with hydrogen.

(B) The phenomenon where metals like $Pt$ and $Pd$ absorb large volumes of hydrogen gas is called occlusion.
Therefore,the hydrogen absorbed by the metal is known as $Occluded \ hydrogen$.

(C) The reducing power of hydrogen species depends on their reactivity.
$1$. Nascent hydrogen is highly reactive because it is generated in situ.
$2$. Atomic hydrogen is also highly reactive due to the presence of an unpaired electron.
$3$. Dihydrogen $(H_2)$ is a stable molecule with a strong $H-H$ bond $(436 \ kJ \ mol^{-1})$,making it the least reactive and thus the poorest reducing agent among the given options.

(C) The heaviest isotope of hydrogen is tritium,represented as $_1H^3$ or $^3_1H$.
It contains $1$ proton,$1$ electron,and $2$ neutrons.
The sum of protons,electrons,and neutrons is $1 + 1 + 2 = 4$.

(D) The $D_2$ molecule consists of two deuterium atoms ($D$ or $_1^2H$).
Each deuterium atom has $1$ proton and $1$ neutron,totaling $2$ nucleons per atom.
Therefore,the total number of nucleons in a $D_2$ molecule is $2 \times 2 = 4$ nucleons.

(C) $Ortho-hydrogen$ and $para-hydrogen$ are allotropic forms of molecular hydrogen that differ in the relative orientation of their nuclear spins.
Because they have the same electronic configuration,they exhibit identical chemical properties.
However,they differ in physical properties such as thermal conductivity,heat capacity,and magnetic properties due to the difference in their nuclear spin states.

(A) The heat of adsorption for ortho-hydrogen and para-hydrogen is slightly different due to the difference in their nuclear spin states and rotational energy levels.
Experimental data shows that the difference in the heat of adsorption between ortho-hydrogen and para-hydrogen is approximately $0.4 \ kJ \ mol^{-1}$.

(B) Isoelectronic species are those that have the same number of electrons.
For the hydride ion $H^{-}$,the number of electrons is $1 + 1 = 2$.
For Helium $(He)$,the atomic number is $2$,so it has $2$ electrons.
Thus,$H^{-}$ and $He$ are isoelectronic,both having the configuration $1s^2$.

(C) The reaction between water gas $(CO + H_2)$ and steam $(H_2O)$ at $500 \, ^oC$ in the presence of $Fe_3O_4$ (catalyst) and $Cr_2O_3$ (promoter) is known as the Bosch process.
The chemical equation is: $CO(g) + H_2O(g) \xrightarrow{Fe_3O_4, Cr_2O_3, 500 \, ^oC} CO_2(g) + H_2(g)$.

(D) Hydrogen can lose one electron to form $H^+$ (e.g.,in $HCl$).
It can gain one electron to form $H^-$ (e.g.,in $NaH$).
It can also share one electron to form covalent bonds (e.g.,in $H_2$ or $CH_4$).
Therefore,hydrogen exhibits all three modes of combination.

(C) The position of hydrogen in the periodic table is unique because it resembles both alkali metals ($ns^1$ configuration) and halogens ($ns^2np^5$ configuration). However,hydrogen is a much lighter element than the alkali metals or halogens,and its properties differ significantly from these groups,making its placement ambiguous.

(D) The statement in option $(D)$ is incorrect. Heavy water $(D_2O)$ is a stable compound and is not radioactive or unstable. Hydrogen exists as a diatomic molecule $(H_2)$,can form both $H^+$ and $H^-$ ions,and the common isotope of hydrogen $(^1H)$ is the only nucleus known to contain no neutrons.

(A) Ionic hydrides consist of $M^+H^-$ ions in the molten state.
When electric current is passed,the hydride ion $(H^-)$ migrates towards the anode.
At the anode,the hydride ion undergoes oxidation to release hydrogen gas:
$H^- \to \frac{1}{2}H_2 + e^-$
Therefore,hydrogen is liberated at the anode.

(C) Hydrogen has an oxidation state of $0$ in its elemental form $(H_2)$.
It can lose electrons to form $H^+$ (oxidation state $+1$),acting as a reducing agent (e.g.,in the reduction of metal oxides).
It can also gain electrons to form $H^-$ (oxidation state $-1$),acting as an oxidising agent (e.g.,in the formation of metal hydrides like $NaH$).
Therefore,it can act as both an oxidising and a reducing agent.

(C) Hydrogen exhibits dual behavior because it can lose an electron to form a cation $(H^+)$ or gain an electron to form an anion $(H^-)$.
$1. \text{Electropositive nature: } H \to H^+ + e^-$
$2. \text{Electronegative nature: } H + e^- \to H^-$

(D) . Alkali metals are good reducing agents because of their low ionization energy. Hydrogen also exhibits similar reducing character by losing an electron to form $H^+$ ion.

(D) The position of hydrogen in the periodic table is anomalous because it resembles both alkali metals ($ns^1$ configuration) and halogens ($ns^2 np^5$ configuration). Due to its unique properties,no single position in the periodic table can fully explain all its characteristics. Therefore,the correct option is $D$.

(B) The correct answer is $(b)$.
Hydrogen readily combines with non-metals (like $Cl_2$,$O_2$,etc.) to form covalent compounds where hydrogen exhibits a positive oxidation state.
For example: $H_2 + Cl_2 \to 2HCl$.
In $HCl$,hydrogen has a $+1$ oxidation state,which indicates its electropositive character.

(C) Hydrogen exhibits an oxidation state of $+1$ in compounds with non-metals (e.g.,$HF$).
Hydrogen exhibits an oxidation state of $-1$ in compounds with electropositive metals (e.g.,$NaH$).
Hydrogen exhibits an oxidation state of $0$ in its elemental form (e.g.,$H_2$).
Therefore,the oxidation states shown by hydrogen are $+1, -1, 0$.

(B) When hydrogen reacts with highly electropositive metals (like alkali metals),it forms metal hydrides where hydrogen acts as an anion $(H^-)$.
$2Na + H_2 \to 2Na^+H^-$
In these compounds,hydrogen exhibits an oxidation state of $-1$,which indicates its electronegative character relative to the metal.

(A) The dissociation of fused sodium hydride is given by: $NaH \rightleftharpoons Na^{+} + H^{-}$.
At the anode,the hydride ion $(H^{-})$ undergoes oxidation to release hydrogen gas:
$H^{-} \to H + e^{-}$
$H + H \to H_2$ (gas).
Therefore,hydrogen is liberated at the anode.

(C) Hydrogen exhibits properties similar to both alkali metals and halogens.
Like alkali metals,it has a $1s^1$ configuration and can lose one electron to form $H^+$.
Like halogens,it can gain one electron to form $H^-$ or share an electron to form covalent bonds (e.g.,$HCl$,$H_2$).
Therefore,it resembles both.