Hydrogen bonding Questions in English

(A) Orthophosphoric acid $(H_3PO_4)$ exhibits high viscosity due to extensive intermolecular hydrogen bonding. Each molecule of $H_3PO_4$ can form multiple hydrogen bonds with neighboring molecules,leading to a highly associated structure.

(C) $H_2O$ molecules are associated through intermolecular hydrogen bonding due to the high electronegativity and small size of the oxygen atom.
In contrast,$H_2S$ does not exhibit hydrogen bonding because sulfur has low electronegativity and a larger atomic size,resulting in weaker intermolecular forces (van der Waals forces) and making it a gas at room temperature.

(A) The boiling points of hydrides of group $16$ elements follow the order: $H_2O > H_2Te > H_2Se > H_2S$.
$H_2O$ has an anomalously high boiling point due to the presence of strong intermolecular hydrogen bonding,which is absent in the other hydrides of this group.
Therefore,the correct option is $A$.

(A) $HF$ is a liquid at room temperature due to the presence of strong intermolecular hydrogen bonding,whereas other hydrogen halides exist as gases.

(C) $HF$ is a liquid at room temperature while $HCl$ is a gas. This is due to extensive $H$-bonding in $HF$,which increases its boiling point.

(C) . Hydrogen bonding is absent in $HI$ because iodine is not sufficiently electronegative to form a hydrogen bond.
In contrast,hydrogen bonding is present in $NH_3$,$H_2O$,and $C_2H_5OH$ due to the presence of highly electronegative atoms like $N$,$O$,or $F$ bonded to hydrogen.

(D) The molar heat of vaporisation depends on the strength of intermolecular forces.
In $HF$,hydrogen bonding is present.
In $HCl$,$HBr$,and $HI$,only London dispersion forces (a type of Van der Waals force) are present.
As the size of the halogen atom increases from $Cl$ to $I$,the polarisability and the strength of Van der Waals forces increase.
Therefore,$HI$ has the strongest intermolecular forces among the hydrogen halides ($HCl$,$HBr$,$HI$),leading to the highest molar heat of vaporisation.
Note: While $HF$ has hydrogen bonding,the magnitude of the Van der Waals forces in $HI$ due to its large size and high polarisability results in a higher molar heat of vaporisation compared to $HF$.

(A) The boiling point of hydrogen halides depends on the strength of intermolecular forces.
In $HF$,the molecules are associated due to strong intermolecular hydrogen bonding.
This results in a significantly higher boiling point compared to $HCl$,$HBr$,and $HI$,where only weak van der Waals forces exist.
Therefore,$HF$ has the highest boiling point.

(A) Hydrogen bonding occurs when hydrogen is covalently bonded to a highly electronegative atom such as $F$,$O$,or $N$.
In the given options,$F$ is the most electronegative element.
Therefore,$HF$ exhibits strong intermolecular hydrogen bonding,whereas $HCl$,$HBr$,and $HI$ do not show significant hydrogen bonding due to the lower electronegativity of $Cl$,$Br$,and $I$.

(A) Hydrogen bonding occurs when a hydrogen atom is covalently bonded to a highly electronegative atom like $F$,$O$,or $N$.
Among the given options,$C_2H_5OH$ (ethanol) contains an $-OH$ group where $H$ is directly bonded to an oxygen atom,which is highly electronegative.
In $CH_3-O-CH_3$ (dimethyl ether),$(CH_3)_2C=O$ (acetone),and $CH_3CHO$ (acetaldehyde),there is no $H$ atom directly bonded to an $O$ or $N$ atom,so they cannot form intermolecular hydrogen bonds.
Therefore,hydrogen bonding is maximum in $C_2H_5OH$.

(C) The vapour pressure of a substance is inversely proportional to its boiling point.
Lower boiling point implies higher vapour pressure.
The boiling point order of the given hydrogen halides is $HCl < HBr < HI < HF$.
$HF$ has the highest boiling point due to strong intermolecular hydrogen bonding.
$HCl$ has the lowest boiling point among the given options,therefore it has the maximum vapour pressure.

(A) $Ortho$-nitrophenol and $para$-nitrophenol can be separated by steam distillation.
$o$-nitrophenol is steam volatile due to the presence of intramolecular hydrogen bonding,which reduces the intermolecular forces of attraction.
In contrast,$p$-nitrophenol exhibits intermolecular hydrogen bonding,leading to the association of molecules,which increases its boiling point and makes it less volatile.
Intramolecular hydrogen bonding occurs in $ortho$-nitrophenol because the $-NO_2$ and $-OH$ groups are in close proximity to each other.

(B) Glycerol undergoes extensive hydrogen bonding due to the presence of $3$ $-OH$ groups.
As a result,the glycerol molecules are highly associated,and thus it has high viscosity.

(A) Glycerol $(CH_2OH-CHOH-CH_2OH)$ contains three $-OH$ groups,whereas propanol $(CH_3CH_2CH_2OH)$ contains only one $-OH$ group.
Due to the presence of more $-OH$ groups in glycerol,it forms a more extensive network of intermolecular hydrogen bonding compared to propanol.
As a result,more energy is required to break these intermolecular forces,leading to a significantly higher boiling point for glycerol.

(B) The compound $CH_3OH$ has the highest boiling point among the given options.
This is due to the presence of intermolecular hydrogen bonding in $CH_3OH$,which is a strong attractive force compared to the dipole-dipole interactions or London dispersion forces present in the other compounds.

(B) Methanol $(CH_3OH)$ has a higher boiling point than methyl thiol $(CH_3SH)$ because the oxygen atom in methanol is highly electronegative,allowing for the formation of intermolecular hydrogen bonding.
In contrast,sulfur in $CH_3SH$ is less electronegative and does not form significant hydrogen bonds.

(D) In non-polar solvents like benzene,carboxylic acids such as acetic acid $(CH_3COOH)$ exist as dimers due to intermolecular hydrogen bonding.
This association is represented as:
$CH_3-C(=O)OH \cdots O=C(OH)CH_3$
This leads to the formation of a stable eight-membered ring structure held together by two hydrogen bonds.

(A) When two ice cubes are pressed together,the pressure causes local melting at the interface. As the pressure is released,the water molecules re-orient and form $H$-bonds across the interface,causing the two cubes to fuse into one.

(D) single water molecule $(H_2O)$ can participate in a maximum of $4$ hydrogen bonds.
This is because each water molecule has two hydrogen atoms that can form hydrogen bonds with the lone pairs of oxygen atoms of neighboring water molecules,and the oxygen atom has two lone pairs that can accept hydrogen bonds from the hydrogen atoms of neighboring water molecules.
Thus,the total number of hydrogen bonds per water molecule is $2 + 2 = 4$.

(B) Hydrogen bonding occurs when a hydrogen atom is covalently bonded to a highly electronegative atom such as $F$,$O$,or $N$.
In $H_2O$,the oxygen atom is highly electronegative and small in size,which allows it to form strong hydrogen bonds.

(B) In ice,water molecules are held together by hydrogen bonds in a three-dimensional open cage-like structure.
This structure occupies more volume per unit mass compared to liquid water,which results in a lower density for ice.
Therefore,liquid water has a higher density than ice due to the nature of its hydrogen bonding network.

(B) Due to strong hydrogen bonding,$HF$ molecules associate to form a linear polymeric structure.
The structure can be represented as: $\cdots H-F \cdots H-F \cdots H-F \cdots H-F \cdots$

(A) $p$-nitrophenol molecules are associated through intermolecular hydrogen bonding,which leads to a higher boiling point.
In contrast,$o$-nitrophenol molecules exhibit intramolecular hydrogen bonding,which restricts association between molecules,resulting in a lower boiling point.

(C) Hydrogen bonding is not present in $H_2S$ because sulfur has a large atomic size and low electronegativity compared to oxygen or fluorine.

(A) When two ice balls are pressed together,the pressure causes local melting at the contact point. Upon releasing the pressure,the water refreezes,and the formation of $H$-bonds between the water molecules across the interface holds the two pieces together.

(A) Both $H_2O$ and $HF$ form hydrogen bonds.
In $H_2O$,each oxygen atom has two lone pairs and two hydrogen atoms,allowing it to form a maximum of $4$ hydrogen bonds per molecule.
In $HF$,although fluorine is highly electronegative,it has only one hydrogen atom available for bonding,limiting its ability to form a network compared to $H_2O$.

(D) Water molecules $(H_2O)$ are associated through intermolecular $H$-bonding,which leads to its liquid state at room temperature. In contrast,$H_2S$ does not exhibit $H$-bonding due to the lower electronegativity of sulfur compared to oxygen.

(D) In the structure of ice,each oxygen atom is tetrahedrally surrounded by four other water molecules through hydrogen bonding. Due to this tetrahedral geometry,the $H - O - H$ bond angle is approximately $109^o 28'$.

(A) single water molecule $(H_2O)$ has two hydrogen atoms and two lone pairs on the oxygen atom.
Each hydrogen atom can form one $H$-bond with an electronegative atom of another molecule,and each lone pair on the oxygen atom can accept one $H$-bond from a hydrogen atom of another molecule.
Therefore,a single water molecule can form a maximum of $4$ $H$-bonds.

(C) Among the given compounds,$C_2H_5OH$ (Ethyl alcohol) exhibits the strongest hydrogen bonding due to the high electronegativity of the oxygen atom bonded to a hydrogen atom,which is more effective than the $N-H$ bonding in amines or ammonia,and the absence of $O-H$ bonds in diethyl ether.

(B) $p$-Nitrophenol exhibits intermolecular hydrogen bonding,which leads to the association of molecules and higher boiling point.
In contrast,$o$-nitrophenol exhibits intramolecular hydrogen bonding,which prevents the association of molecules.
Due to the lack of intermolecular association,$o$-nitrophenol has a lower boiling point and is more volatile than $p$-nitrophenol.

(C) Hydrogen bonding occurs when a hydrogen atom is covalently bonded to a highly electronegative atom such as $F, O,$ or $N$.
In $H_2O$ (water),$HF$ (hydrogen fluoride),and glycerine $(C_3H_8O_3)$,hydrogen is bonded to $O$ or $F$,allowing for hydrogen bonding.
In $H_2S$ (hydrogen sulphide),the electronegativity of sulfur is not high enough to facilitate significant hydrogen bonding.
Therefore,the correct option is $C$.

(C) The correct decreasing order of boiling point is $H_2O > H_2Te > H_2Se > H_2S$.
$H_2O$ has the highest boiling point due to strong intermolecular hydrogen bonding.
For the remaining hydrides $(H_2S, H_2Se, H_2Te)$,the boiling point increases with increasing molecular mass due to the increase in van der Waals forces of attraction.

(C) Intramolecular hydrogen bonding occurs within a single molecule,whereas intermolecular hydrogen bonding occurs between different molecules.
$H_2O_2$,$HCN$,and concentrated $CH_3COOH$ exhibit intermolecular hydrogen bonding.
Cellulose contains multiple hydroxyl $(-OH)$ groups within its polymeric structure,which allow for the formation of intramolecular hydrogen bonds between the chains and within the glucose units,stabilizing its structure.

(D) When an $H$ atom is directly bonded to a highly electronegative atom like $N$,$O$,or $F$,a strong dipole is created,leading to the formation of intermolecular hydrogen bonding.
In liquid $CH_3OH$ (methanol),the $H$ atom is bonded to an $O$ atom,which allows for the formation of intermolecular hydrogen bonds between adjacent molecules.
To convert liquid $CH_3OH$ to a gas,these intermolecular hydrogen bonds must be overcome,as they are the dominant attractive forces holding the liquid molecules together.

(A) The boiling point of hydrogen halides depends on the intermolecular forces of attraction.
In $HF$ molecules,the high electronegativity and small size of the fluorine atom lead to the formation of strong intermolecular hydrogen bonding.
This results in a significantly higher boiling point compared to other hydrogen halides $(HI, HBr, HCl)$,where only weaker van der Waals forces are present.
Thus,the order is $HF > HI > HBr > HCl$.

(C) The strength of a hydrogen bond depends on the electronegativity difference between the atoms involved in the bond.
Greater the electronegativity difference,the stronger the hydrogen bond.
Fluorine $(F)$ is the most electronegative element.
Therefore,the $F-H \cdots F$ hydrogen bond is the strongest among the given options.

(B) The interaction energy between two dipoles (dipole-dipole interaction) is proportional to $1/r^3$ for stationary dipoles,where $r$ is the distance between the molecules.
Hydrogen bonding is a specific,strong type of dipole-dipole interaction.
London dispersion forces depend on $1/r^6$.
Ion-ion interaction depends on $1/r$.
Ion-dipole interaction depends on $1/r^2$.

(D) Fluorobenzene $(C_6H_5F)$ has a higher boiling point than benzene $(C_6H_6)$ primarily due to its higher molecular mass and dipole-dipole interactions.
Hydrogen bonding is not present in fluorobenzene because the fluorine atom is bonded to a carbon atom,not a hydrogen atom,and there are no $N-H$,$O-H$,or $F-H$ bonds to act as donors.
In contrast,options $(a)$,$(b)$,and $(c)$ are all consequences of hydrogen bonding.

(A) Ortho-nitrophenol exhibits intramolecular hydrogen bonding,where the hydrogen atom of the hydroxyl group is bonded to the oxygen atom of the adjacent nitro group within the same molecule.
Boric acid $(H_3BO_3)$ exhibits intermolecular hydrogen bonding,where hydrogen bonds form between different molecules to create a layered structure.
Therefore,only ortho-nitrophenol has intramolecular $H$-bonding.

(A) The strength of a hydrogen bond depends on the electronegativity difference and the charge density of the atoms involved.
$1$. The interaction $a$ $(F^{-}H...F^{-})$ is a symmetric,very strong hydrogen bond (often called a low-barrier hydrogen bond) due to the high electronegativity of $F$ and the negative charge,making it the strongest.
$2$. The interaction $b$ $(F^{-}H...F)$ involves a neutral $F$ atom,making it weaker than $a$ but stronger than interactions involving $O$ and $N$.
$3$. Comparing $c$ $(O^{-}H...N)$ and $d$ $(O^{-}H...O)$,the electronegativity of $O$ $(3.44)$ is greater than that of $N$ $(3.04)$. Thus,the $O-H...O$ interaction $(d)$ is stronger than the $O-H...N$ interaction $(c)$.
Therefore,the correct order of strength is $a > b > d > c$.

(C) Viscosity is primarily determined by the extent of intermolecular hydrogen bonding.
In $p$-nitrophenol,intermolecular hydrogen bonding is extensive,leading to association of molecules and higher viscosity.
In $o$-nitrophenol,intramolecular hydrogen bonding occurs,which restricts intermolecular association,resulting in lower viscosity.
Therefore,the correct order is $p$-nitrophenol > $o$-nitrophenol.
The given option $o$-nitrophenol > $p$-nitrophenol is incorrect.

(A) hydrogen bond is a weak electrostatic force of attraction (a type of dipole-dipole interaction) between a hydrogen atom and a small,highly electronegative atom. It is significantly weaker than a covalent bond.
The $O-H$ bond in water is a covalent bond,and its bond energy is much greater than that of a hydrogen bond.

(D) $1$. The extent of $H$-bonding depends on the number of $H$-bonds per molecule. $HF$ has $2$ $H$-bonds,$H_2O$ has $4$ $H$-bonds,and $H_2O_2$ has $6$ $H$-bonds. Thus,the order of extent is $HF < H_2O < H_2O_2$.
$2$. The strength of an $H$-bond depends on the electronegativity of the atom bonded to $H$. $F$ is more electronegative than $O$,so the $H$-bond strength order is $HF > H_2O > H_2O_2$.
$3$. Due to the open cage-like structure of ice $(H_2O(s))$ formed by $H$-bonding,its density is lower than that of liquid water $(H_2O(\ell))$. Thus,$H_2O(\ell) > H_2O(s)$ is correct.
$4$. Since all statements are correct,the answer is $D$.

(C) Intramolecular $H$-bonding occurs when a hydrogen atom is bonded to a highly electronegative atom (like $F, O, N$) and is also attracted to another electronegative atom within the same molecule,forming a stable ring structure.
In $o$-fluorophenol,the hydrogen atom of the $-OH$ group forms an intramolecular hydrogen bond with the fluorine atom at the ortho position,resulting in a stable five-membered ring.
Therefore,$o$-fluorophenol exhibits intramolecular $H$-bonding.

(D) Oxygen is highly electronegative and small in size,which allows $H_2O$ molecules to form intermolecular hydrogen bonds.
These hydrogen bonds hold the water molecules together in a liquid state at room temperature.
In contrast,sulphur has lower electronegativity and larger size,so $H_2S$ does not form intermolecular hydrogen bonds,existing as a gas.

(B) Hydrogen bonding occurs between a hydrogen atom covalently bonded to a highly electronegative atom (like $F, O, N$) and another highly electronegative atom with a lone pair of electrons.
In the pair $HCOOH$ (formic acid) and $CH_3COOH$ (acetic acid),both molecules contain a hydroxyl group $(-OH)$ and a carbonyl group $(C=O)$.
These groups allow for the formation of strong intermolecular hydrogen bonds between the hydrogen of the $-OH$ group of one molecule and the oxygen of the $C=O$ group of the other.
This leads to the formation of stable dimers,which represents a very strong form of hydrogen bonding compared to the other given options.

(B) Intermolecular $H$-bonding occurs between different molecules.
In the mixture of $H_2O$ and $ROH$ (alcohol),hydrogen bonding exists between the water molecule and the alcohol molecule.
$o$-Nitrophenol exhibits intramolecular $H$-bonding.
Chloral hydrate $(CCl_3CH(OH)_2)$ exhibits intramolecular $H$-bonding.
$[Ni(DMG)_2]$ also exhibits strong intramolecular $H$-bonding.

(C) $(C)$ Formic acid is more acidic than acetic acid due to the absence of the electron-donating $+I$ effect of the methyl group, which is present in acetic acid. Hydrogen bonding is responsible for the density of ice, the solvation/basicity of amines in water, and the dimerisation of carboxylic acids in non-polar solvents like benzene.