Hydrogen bonding Questions in English

(B) $I.$ False: In the gaseous phase,$CH_3COOH$ exists as a dimer due to hydrogen bonding,which increases its molecular weight and decreases its vapour pressure compared to the monomeric state,but the statement implies a deviation from expected behavior that is actually characteristic of the liquid phase or specific conditions. However,standard textbooks note that $CH_3COOH$ vapor pressure is lower than expected due to dimerization.
$II.$ True: Acetic acid forms hydrogen-bonded dimers in non-polar solvents like benzene.
$III.$ False: $H_2O$ has more extensive hydrogen bonding than $CH_3OH$,making it more viscous.
$IV.$ False: $CH_3OCH_3$ (dimethyl ether) is polar and soluble in water due to hydrogen bonding with water molecules.
Thus,the sequence is $F, T, F, F$.

(C) $HF$ forms a linear polymer due to intermolecular hydrogen bonding.
In $HF$,the highly electronegative fluorine atom of one molecule forms a hydrogen bond with the hydrogen atom of another molecule,resulting in a zig-zag linear polymeric chain structure represented as $-(H-F)n-$.

(A) $o-$Nitrophenol exhibits intramolecular $H$-bonding,which restricts the association between molecules.
$p-$Nitrophenol exhibits intermolecular $H$-bonding,leading to the association of molecules and the formation of a larger molecular aggregate.
Due to stronger intermolecular forces in $p-$nitrophenol,it has a higher boiling point and lower vapour pressure compared to $o-$nitrophenol.
Therefore,the vapour pressure of $o-$nitrophenol is higher than that of $p-$nitrophenol.

(B) Hydrogen bonding requires a hydrogen atom to be covalently bonded to a highly electronegative atom like $F$,$O$,or $N$.
In $K_2HPO_4$,the $HPO_4^{2-}$ ion contains an $OH$ group,allowing for hydrogen bonding.
In $K_2HPO_3$,the structure is $[HPO_3]^{2-}$,where the hydrogen atom is directly bonded to the phosphorus atom ($P-H$ bond). Since $P$ is not sufficiently electronegative,it does not form hydrogen bonds.
Formic acid $(HCOOH)$ and Chloral hydrate $(CCl_3CH(OH)_2)$ both contain $OH$ groups capable of hydrogen bonding.

(D) Option $A$: $HCl_2^-$ is a known species that can be stabilized by large cations.
Option $B$: $NH_4F$ contains $NH_4^+$ and $F^-$ ions,both of which can form hydrogen bonds with water.
Option $C$: $H_2O_{(s)}$ (ice) has an open cage-like structure with large voids,which increases its volume and decreases its density compared to $H_2O_{(l)}$,causing it to float.
Option $D$: In a single $H_2O$ molecule,there is only one $O$ atom; therefore,the concept of distance between two adjacent $O$ atoms within a single molecule is physically meaningless. Thus,this statement is incorrect.

(D) The strength of a hydrogen bond depends on the electronegativity difference between the atoms involved and the charge density.
Hydrogen bonds involving ions (like $F^{-}...H-F$) are significantly stronger than neutral hydrogen bonds.
The interaction $F^{-}H...F$ involves a fluoride ion $(F^{-})$ and a hydrogen fluoride molecule $(HF)$.
This is a very strong hydrogen bond because the fluoride ion has a high negative charge density and the $H-F$ bond is highly polarized,leading to a very strong electrostatic attraction.
Therefore,the strength follows the order $F^{-}H...F > F^{-}H...O > O^{-}H...F > O^{-}H...O$.

(D) The ion $H_5O_2^+$ is a hydrated proton, $[H_3O^+ \cdot H_2O]$, which exists in acidic solutions.
$H_3O_2^-$ is a hydrated hydroxide ion, $[OH^- \cdot H_2O]$, which exists in basic solutions.
$HF_2^-$ is a well-known bifluoride ion formed by strong hydrogen bonding between $F^-$ and $HF$.
Since all these ions are stable under appropriate conditions, the correct answer is $D$.

(A) Hydrogen bonding strength depends on the electronegativity difference between the atoms involved.
$F$ is the most electronegative element in the periodic table.
Therefore,the $H$-bond formed between $F$ and $H$ (i.e.,$F - H - - - F$) is the strongest because the electrostatic attraction is maximized due to the high partial charges on the atoms.

(B) The strength of intermolecular $H$-bonding depends on the electronegativity of the atom attached to the hydrogen atom and the polarity of the bond.
In $CH_3NH_2$ (methylamine),$CH_3OH$ (methanol),and $C_6H_5OH$ (phenol),$H$-bonding exists.
Phenol $(C_6H_5OH)$ exhibits the strongest intermolecular $H$-bonding because the oxygen atom is attached to a phenyl ring,which increases the acidity of the phenolic hydrogen compared to aliphatic alcohols or amines,leading to a higher partial positive charge $(\delta )$ on the hydrogen atom.
Therefore,the correct option is $(B)$.

(C) Viscosity is primarily determined by the extent of intermolecular hydrogen bonding.
Dimethyl ether $(CH_3OCH_3)$ has no hydrogen bonding,so it has the lowest viscosity.
Methyl alcohol $(CH_3OH)$ has one $-OH$ group,allowing for some hydrogen bonding.
Water $(H_2O)$ has two $-OH$ groups per molecule,leading to more extensive hydrogen bonding than methyl alcohol.
Glycerol $(CH_2(OH)CH(OH)CH_2OH)$ has three $-OH$ groups,resulting in the most extensive hydrogen bonding network,making it the most viscous.
Therefore,the correct order is: $\text{dimethyl ether} < \text{methyl alcohol} < \text{water} < \text{glycerol}$.

(C) $HF$ exhibits strong intermolecular hydrogen bonding.
Due to the high electronegativity of the $F$ atom compared to $H$,a significant dipole is created,leading to strong hydrogen bonds.
These strong intermolecular forces require more energy to overcome,resulting in the highest boiling point among hydrogen halides ($HCl$,$HBr$,$HI$).

(A) $o-$nitrophenol exhibits intramolecular hydrogen bonding,while $p-$nitrophenol exhibits intermolecular hydrogen bonding.
Due to intermolecular hydrogen bonding,$p-$nitrophenol molecules associate,leading to a higher boiling point compared to $o-$nitrophenol.
Therefore,the statement that the $B.P.$ of $o-$nitrophenol is greater than $p-$nitrophenol is incorrect.
In $SO_3$,sulfur uses $d-$orbitals to form $p\pi-d\pi$ bonds with oxygen.
In the cyclic trimer $S_3O_9$,there are $3$ bridging oxygen atoms connecting the sulfur atoms.

(A) The boiling point of a substance depends on the strength of intermolecular forces.
$HF$ exhibits strong intermolecular hydrogen bonding,which leads to a significantly higher boiling point $(293 \ K)$.
$NH_3$ also exhibits hydrogen bonding,but it is weaker than in $HF$ because $N$ is less electronegative than $F$,resulting in a lower boiling point $(240 \ K)$.
$H_2S$ does not exhibit significant hydrogen bonding and relies on weaker dipole-dipole interactions,resulting in the lowest boiling point among the three $(213 \ K)$.
Therefore,the correct order is $HF > NH_3 > H_2S$.

(A) The strength of hydrogen bonding depends on the electronegativity of the atom bonded to hydrogen and the partial negative charge density on the electronegative atom.
$HF$ has the strongest hydrogen bonding due to the highest electronegativity of fluorine.
In $H_2O$,the oxygen atom is bonded to two hydrogen atoms,resulting in a higher partial negative charge density compared to $H_2O_2$,where the oxygen atoms are bonded to each other and one hydrogen atom,leading to a lower partial negative charge on each oxygen.
$H_2S$ does not form significant hydrogen bonds because sulfur is not sufficiently electronegative.
Therefore,the correct order of strength is $HF > H_2O > H_2O_2 > H_2S$.

(D) $1$. The $m.p.$ of $H_2O$ and $NH_3$ are indeed the highest in their respective groups ($Group$ $16$ and $Group$ $15$) due to the presence of strong intermolecular $H$-bonding.
$2$. The $b.p.$ of hydrides of $Group$ $14$ $(CH_4, SiH_4, GeH_4, SnH_4)$ increases with increasing molecular mass due to stronger van der Waals forces. Thus,$CH_4$ has the lowest $b.p.$
$3$. Formic acid $(HCOOH)$ exists as a dimer in the vapor state and in non-polar solvents due to intermolecular $H$-bonding.
$4$. Since all statements are correct,the correct option is $D$.

(C) Intermolecular $H$-bonding occurs between different molecules of the same or different substances.
$(I)$ Acetic acid forms a dimer via intermolecular $H$-bonding.
$(II)$ $o$-nitrophenol exhibits intramolecular $H$-bonding (chelation) due to the proximity of the $-OH$ and $-NO_2$ groups.
$(III)$ $m$-nitrophenol exhibits intermolecular $H$-bonding because the $-OH$ and $-NO_2$ groups are not close enough for intramolecular bonding.
$(IV)$ Boric acid $(H_3BO_3)$ forms a layered structure in the solid state held together by intermolecular $H$-bonding.
Thus,$(I)$,$(III)$,and $(IV)$ exhibit intermolecular $H$-bonding.

(A) Hydrogen bonding occurs when a hydrogen atom is covalently bonded to a highly electronegative atom $(F, O, N)$ and is attracted to another electronegative atom.
In the pair $CH_3COOH$ and $H_2O$,both molecules contain $O-H$ bonds,allowing for strong intermolecular $H$-bonding.
In other options,either one of the components lacks the necessary $H$ atom attached to a highly electronegative atom (like $CH_4$ or $PH_3$) or the interaction is not sufficient to be classified as $H$-bonding.

(B) The boiling point of hydrogen halides depends on the intermolecular forces present.
$HF$ exhibits strong intermolecular hydrogen bonding,which results in a significantly higher boiling point.
For the remaining hydrogen halides ($HCl$,$HBr$,$HI$),the boiling point increases with increasing molecular mass due to the increase in the strength of Van der Waals forces.
Since $HCl$ has the lowest molecular mass among these,it has the weakest Van der Waals forces and therefore the lowest boiling point.
The order of boiling points is: $HCl < HBr < HI < HF$.

(D) The correct answer is $(D)$.
In option $(A)$,$H_2O$ has more extensive hydrogen bonding than $CH_3OH$,so $H_2O > CH_3OH$ is correct.
In option $(B)$,$NH(CH_3)_2$ has intermolecular hydrogen bonding while $N(CH_3)_3$ does not,so the order $N(CH_3)_3 > NH(CH_3)_2$ is actually incorrect as well,but usually,in such multiple-choice questions,we look for the most prominent discrepancy. However,$CH_3N_3$ (methyl azide) has only dipole-dipole interactions,whereas $HN_3$ (hydrazoic acid) exhibits intermolecular hydrogen bonding,making $HN_3 > CH_3N_3$. Thus,the order $CH_3N_3 > HN_3$ is incorrect.

(D) In the crystal lattice of ice,each water molecule is linked to four other water molecules by hydrogen bonds.
Within each water molecule,the $O-H$ bonds are covalent.
Between the water molecules,there exist intermolecular hydrogen bonds.
Therefore,the structure of ice is maintained by both covalent bonds and intermolecular hydrogen bonds.

(B) The boiling point of a substance is significantly influenced by intermolecular forces.
Water $(H_2O)$ exhibits extensive intermolecular $H$-bonding due to the presence of two $H$ atoms attached to a highly electronegative oxygen atom,resulting in a high boiling point of $100\,^{\circ}C$.
Methanol $(CH_3OH)$ also exhibits $H$-bonding,but to a lesser extent than water because it has only one $H$ atom capable of $H$-bonding and the electron-donating $+I$ effect of the $CH_3$ group reduces the polarity of the $O-H$ bond,resulting in a lower boiling point of $65\,^{\circ}C$.
Dimethyl ether $(CH_3OCH_3)$ lacks $H$-bonding as there is no $H$ atom directly bonded to an oxygen atom; it only experiences weak dipole-dipole interactions and London dispersion forces,leading to a very low boiling point.

(B) The correct option is $B$. In ice,the $H$-bond length is approximately $2.76 \ \mathring{A}$,whereas the $O-H$ covalent bond length is approximately $0.96 \ \mathring{A}$.
$H$-bonds are intermolecular forces of attraction,which are significantly weaker and longer than the strong intramolecular covalent bonds that hold the atoms together within a single water molecule.

(A) The strength of the $H$ bond depends on the electronegativity of the atom bonded to the hydrogen atom.
As the electronegativity of the atom increases,the polarity of the $H-X$ bond increases,leading to a stronger $H$ bond.
The electronegativity order of the atoms is $F > O > N$.
Therefore,the strength of the $H$ bond follows the order $H \dots F > H \dots O > H \dots N$.

(B) $o-$nitrophenol and $p-$nitrophenol can be separated by steam distillation.
In $o-$nitrophenol,the $-OH$ and $-NO_2$ groups are adjacent,allowing for the formation of stable intramolecular hydrogen bonding. This reduces the intermolecular forces between molecules,making it more volatile and easily steam distilled.
In $p-$nitrophenol,the $-OH$ and $-NO_2$ groups are at opposite positions,preventing intramolecular hydrogen bonding. Instead,it forms strong intermolecular hydrogen bonding,which leads to molecular association,higher boiling point,and lower volatility,making it difficult to steam distill.

(B) Ice has a highly ordered,open cage-like structure due to extensive hydrogen bonding.
When ice melts at $0\,^{\circ}C$,some of these hydrogen bonds break,causing the rigid structure to collapse.
This results in the water molecules coming closer together,which leads to a contraction in volume.

(C) hydrogen bond is formed when a hydrogen atom is covalently bonded to a highly electronegative atom (like $F, O, N$) and is attracted to another electronegative atom with a lone pair.
In option $(A)$,the $H$ atom of $H_2O$ (which is more acidic) is attracted to the lone pair on the $N$ atom of $NH_3$. This is a correct representation.
In option $(B)$,the $H$ atom of the $-OH$ group in $o$-chlorophenol forms an intramolecular hydrogen bond with the $Cl$ atom. While $Cl$ is not as electronegative as $F, O, N$,it can participate in weak hydrogen bonding in specific geometries.
In option $(C)$,the $H$ atom of $NH_3$ is shown bonded to the $O$ atom of $H_2O$. Since $N$ is less electronegative than $O$,the $N-H$ bond is less polar than the $O-H$ bond. Therefore,the $H$ atom in $NH_3$ is less acidic and less capable of forming a strong hydrogen bond compared to the $H$ atom in $H_2O$. Thus,$(C)$ is the least favorable representation among the given choices.

(B) In solid $HF$,each fluorine atom is covalently bonded to one hydrogen atom and hydrogen-bonded to another hydrogen atom. The arrangement of bonding pairs $(bp)$ and lone pairs $(lp)$ around the fluorine atom is tetrahedral. This results in a zig-zag chain structure where the $H-F \dots H$ bond angle is approximately $120^{\circ}$. The correct representation is a zig-zag chain where $F$ is covalently bonded to $H$ and hydrogen-bonded to the next $H$ in the chain,which corresponds to the structure where $H$ atoms are positioned between $F$ atoms in a zig-zag manner.

(D) $(a) \, H_2 \dots H_2O$: Dipole-induced dipole interaction (van der Waals force) energy is $< 8 \, kJ/mol$.
$(b) \, HCl \dots HCl$: Dipole-dipole interaction (van der Waals force) energy is $< 8 \, kJ/mol$.
$(c) \, F^{-} \dots HF$: Ion-dipole $H$-bond (very strong $H$-bond) energy is $> 42 \, kJ/mol$.
$(d) \, HCN \dots NH_3$: $H$-bond energy typically ranges from $8 - 42 \, kJ/mol$.

(C) $NH_3$ molecules exhibit intermolecular hydrogen bonding.
Due to this,it shows a much higher boiling point compared to $PH_3$.

(D) The enthalpy of vaporization generally increases with an increase in molecular mass due to stronger van der Waals forces. However,$NH_3$ exhibits an exceptionally high enthalpy of vaporization due to the presence of intermolecular $H$-bonding. Therefore,the correct order of increasing enthalpy of vaporization is $PH_3 < AsH_3 < NH_3$.

(D) The $H-S$ bond is weaker than the $H-O$ bond due to the larger size of the sulfur atom compared to the oxygen atom.
Consequently,$H_2S$ has a greater tendency to release a proton $(H^+)$ compared to $H_2O$,making it more acidic.

(A) $H_2O$ has the highest boiling point among the hydrides of group $16$ elements because it exhibits extensive intermolecular hydrogen bonding.

(D) $HF$ molecules are associated due to strong intermolecular $H$-bonding,which is not present in other hydrogen halides. This association leads to a higher boiling point,causing $HF$ to exist as a liquid at room temperature.

(C) Hydrogen bonding occurs when hydrogen is covalently bonded to highly electronegative atoms like $F$,$O$,or $N$.
In $NH_3$,$H_2O$,and $C_2H_5OH$,hydrogen bonding is present.
Iodine is not sufficiently electronegative to form hydrogen bonds.
Therefore,hydrogen bonding does not play a role in the boiling point of $HI$.

(C) The strength of a hydrogen bond is directly proportional to the electronegativity difference between the atoms involved. Since $F$ is the most electronegative element,the $F-H \dots F$ hydrogen bond is the strongest among the given options.

(A) $HF$ exists as a liquid due to the presence of strong intermolecular hydrogen bonding. In contrast,all other hydrogen halides ($HCl$,$HBr$,$HI$) exist as gases at room temperature.

(C) Hydrogen bonding strength depends on the electronegativity of the atom bonded to hydrogen.
Since fluorine $(F)$ is the most electronegative element,the $H-F$ bond exhibits the strongest hydrogen bonding.

(C) $KF$ reacts with $HF$ to form potassium hydrogen difluoride $(KHF_2)$.
In the solid state,$KHF_2$ is an ionic compound consisting of potassium ions $(K^{+})$ and hydrogen difluoride ions $([HF_2]^{-})$.
The $[HF_2]^{-}$ ion is formed due to strong hydrogen bonding between the fluoride ion $(F^{-})$ and the hydrogen fluoride molecule $(HF)$: $F^{-} \cdots H-F \rightarrow [F-H-F]^{-}$.
Therefore,the compound contains $K^{+}$ and $[HF_2]^{-}$ ions.

(B) Volatility is inversely proportional to the strength of intermolecular forces.
$1$. In $p-\text{nitrophenol}$,strong intermolecular $H$-bonding exists,leading to higher boiling point and lower volatility.
$2$. In $o-\text{nitrophenol}$,intramolecular $H$-bonding occurs,which reduces intermolecular attraction,making it more volatile.
Therefore,the correct order of volatility is $p-\text{nitrophenol} < o-\text{nitrophenol}$.

(B) $HF$ forms a linear polymeric structure due to intermolecular $H$-bonding.
In this structure,each $F$ atom is bonded to one $H$ atom covalently and to another $H$ atom via a hydrogen bond,resulting in a zigzag chain structure.
$H_2O$ forms a three-dimensional network structure due to $H$-bonding.
$NH_3$ forms a pyramidal structure and does not form a linear polymer.
$HCl$ does not exhibit significant $H$-bonding.

(A) The boiling points of hydrides of group $16$ elements depend on both molecular mass and intermolecular forces.
As we move down the group from $O$ to $Te$,the molecular mass increases,which leads to an increase in van der Waals forces,generally increasing the boiling point.
However,$H_2O$ molecules are held together by strong intermolecular hydrogen bonding,which is absent in the other hydrides ($H_2S$,$H_2Se$,$H_2Te$).
Due to this strong hydrogen bonding,$H_2O$ has an anomalously high boiling point compared to the other hydrides in the series.

(A) The strength of intermolecular forces depends on the nature of the interaction.
$HF^{-} \dots H_2O$ represents an ion-dipole interaction,which is significantly stronger than dipole-induced dipole or London dispersion forces.
$HF^{-} \dots Cl_2$ and $H_2O \dots Cl_2$ are dipole-induced dipole interactions.
$Cl_2 \dots Cl_2$ is a London dispersion force (induced dipole-induced dipole),which is the weakest among the given options.
Therefore,the ion-dipole interaction $HF^{-} \dots H_2O$ is the strongest.

(D) In $KHF_2$,the anion is $[F-H-F]^-$,which contains a strong symmetric hydrogen bond between the two fluoride ions.
In $KH_2PO_4$,the phosphate units are linked by intermolecular hydrogen bonds,forming a polymeric structure.
In $KH_2PO_2$,the phosphorus atom is bonded to two hydrogen atoms directly,and there are no acidic protons available to form hydrogen bonds.
Therefore,both $(A)$ and $(B)$ exhibit hydrogen bonding.

(A) The strength of intermolecular forces depends on the nature of the interaction.
$HF \dots H_2O$ involves hydrogen bonding,which is the strongest type of dipole-dipole interaction among the given options.
$HF \dots Cl_2$ involves dipole-induced dipole interaction.
$H_2O \dots Cl_2$ involves dipole-induced dipole interaction.
$Cl_2 \dots Cl_2$ involves London dispersion forces,which are the weakest.
Therefore,the hydrogen bond between $HF$ and $H_2O$ is the strongest.

(A) Intramolecular hydrogen bonding occurs within the same molecule,typically forming a stable ring structure (chelation).
$1$. $o$-Fluorophenol exhibits intramolecular hydrogen bonding between the $-OH$ group and the $F$ atom.
$2$. Salicylaldehyde exhibits intramolecular hydrogen bonding between the $-OH$ group and the $-CHO$ group.
$3$. Chloral hydrate $(CCl_3CH(OH)_2)$ exhibits intramolecular hydrogen bonding between the two hydroxyl groups.
$4$. Urea $(NH_2CONH_2)$ exhibits intermolecular hydrogen bonding,where hydrogen bonds form between different molecules,leading to a high melting point and solid state structure. It does not exhibit intramolecular hydrogen bonding.

(C) The boiling point of these compounds is primarily determined by the strength of intermolecular hydrogen bonding.
$H_2O$ has the highest boiling point because each molecule can form up to $4$ hydrogen bonds,creating a strong three-dimensional network.
$HF$ has the next highest boiling point due to strong hydrogen bonding,though it forms fewer bonds per molecule than $H_2O$.
$NH_3$ has a lower boiling point than $HF$ because the electronegativity difference between $N$ and $H$ is smaller,leading to weaker hydrogen bonds.
$HCl$ does not exhibit significant hydrogen bonding and relies on dipole-dipole interactions,resulting in the lowest boiling point among the given compounds.
Therefore,the decreasing order is $II > III > IV > I$.

(B) Hydrogen bonding occurs in compounds where $H$ atoms are covalently bonded to highly electronegative atoms like $F$,$O$,or $N$.
In $HCl$,the chlorine atom is not sufficiently electronegative to facilitate hydrogen bonding.
Therefore,$HCl$ does not exhibit hydrogen bonding.

(A) Acetic acid $(CH_3COOH)$ exists as a dimer in the vapor phase and in non-polar solvents due to intermolecular hydrogen bonding.
Two molecules of acetic acid are held together by two hydrogen bonds to form a cyclic dimer structure.
$o$-nitrophenol exhibits intramolecular hydrogen bonding,which prevents dimerization.
$AlCl_3$ exists as a dimer $(Al_2Cl_6)$ due to coordinate covalent bonding,not hydrogen bonding.

(B) Intermolecular hydrogen bonding occurs between different molecules of the same or different substances.
$H_2O$,$HF$,and $CH_3COOH$ exhibit intermolecular hydrogen bonding.
$o$-nitrophenol exhibits intramolecular hydrogen bonding,where the hydrogen atom is bonded to an oxygen atom within the same molecule,forming a stable six-membered ring.

(A) Volatility is inversely proportional to intermolecular hydrogen bonding and boiling point.
$o$-Hydroxybenzoic acid (salicylic acid) exhibits intramolecular hydrogen bonding,which reduces its intermolecular association compared to the other options.
Due to weaker intermolecular forces,$o$-hydroxybenzoic acid has the lowest boiling point and therefore the maximum volatility among the given compounds.