ChemistryQ351–400 of 843 questions
Page 8 of 11 · English
351
ChemistryEasyMCQMHT CET · 2025
Which of the following is an acidic oxide?
A
$CO$
B
$NO$
C
$N_2O$
D
$N_2O_5$
Solution
(D) An acidic oxide is an oxide that reacts with water to form an acid or with a base to form a salt.
$CO$,$NO$,and $N_2O$ are neutral oxides.
$N_2O_5$ is the anhydride of nitric acid $(HNO_3)$ and reacts with water as follows:
$N_2O_5 + H_2O \rightarrow 2HNO_3$.
Therefore,$N_2O_5$ is an acidic oxide.
$CO$,$NO$,and $N_2O$ are neutral oxides.
$N_2O_5$ is the anhydride of nitric acid $(HNO_3)$ and reacts with water as follows:
$N_2O_5 + H_2O \rightarrow 2HNO_3$.
Therefore,$N_2O_5$ is an acidic oxide.
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352
ChemistryEasyMCQMHT CET · 2025
Which of the following set of elements is present in apatite?
A
$Mg, P$ and $S$
B
$Ca, P$ and $O$
C
$Al, Mg$ and $P$
D
$Ca, S$ and $O$
Solution
(B) Apatite is a group of phosphate minerals,with the most common being fluorapatite,which has the chemical formula $Ca_5(PO_4)_3F$.
Other variations include hydroxyapatite,$Ca_5(PO_4)_3(OH)$.
In all these forms,the essential elements present are $Ca$ (calcium),$P$ (phosphorus),and $O$ (oxygen),along with $F$ or $OH$.
Therefore,the set of elements present in apatite is $Ca, P$ and $O$.
Other variations include hydroxyapatite,$Ca_5(PO_4)_3(OH)$.
In all these forms,the essential elements present are $Ca$ (calcium),$P$ (phosphorus),and $O$ (oxygen),along with $F$ or $OH$.
Therefore,the set of elements present in apatite is $Ca, P$ and $O$.
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353
ChemistryMediumMCQMHT CET · 2025
Which of the following statements is correct about $O_2$ and $O_3$ molecules?
A
$O_2$ and $O_3$ are paramagnetic
B
The enthalpy change during the formation of $O_3$ from $O_2$ is positive
C
The entropy change in the formation of $O_3$ from $O_2$ is positive
D
$O_3$ is more stable than $O_2$
Solution
(B) The formation of ozone from oxygen is represented by the reaction: $3O_2(g) \rightarrow 2O_3(g)$.
This reaction is endothermic,meaning the enthalpy change $(\Delta H)$ is positive $(+142 \ kJ \ mol^{-1})$.
Since the number of moles of gas decreases from $3$ to $2$,the entropy change $(\Delta S)$ is negative.
$O_2$ is more stable than $O_3$ because $O_3$ is an endothermic compound.
$O_2$ is paramagnetic due to the presence of two unpaired electrons in its antibonding molecular orbitals,whereas $O_3$ is diamagnetic.
This reaction is endothermic,meaning the enthalpy change $(\Delta H)$ is positive $(+142 \ kJ \ mol^{-1})$.
Since the number of moles of gas decreases from $3$ to $2$,the entropy change $(\Delta S)$ is negative.
$O_2$ is more stable than $O_3$ because $O_3$ is an endothermic compound.
$O_2$ is paramagnetic due to the presence of two unpaired electrons in its antibonding molecular orbitals,whereas $O_3$ is diamagnetic.
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354
ChemistryEasyMCQMHT CET · 2025
Which of the following statements is $NOT$ correct about ozone?
A
It is a strong reducing agent
B
It has an angular shape
C
It is a good bleaching agent
D
It absorbs harmful Ultra Violet radiation from the sun
Solution
(A) Ozone $(O_3)$ acts as a strong oxidizing agent because it easily releases nascent oxygen $(O_3 \rightarrow O_2 + [O])$.
It is not a reducing agent.
Ozone has an angular (bent) shape with a bond angle of approximately $117^{\circ}$.
It acts as a good bleaching agent due to its oxidizing property.
It protects the Earth's surface from harmful ultraviolet $(UV)$ radiation by absorbing it in the stratosphere.
It is not a reducing agent.
Ozone has an angular (bent) shape with a bond angle of approximately $117^{\circ}$.
It acts as a good bleaching agent due to its oxidizing property.
It protects the Earth's surface from harmful ultraviolet $(UV)$ radiation by absorbing it in the stratosphere.
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355
ChemistryMediumMCQMHT CET · 2025
Which of the following statements is false about oxygen and sulphur?
A
Atoms of oxygen and sulphur consist of $2$ unpaired electrons in their valence shell.
B
Oxygen and sulphur show $-2, +4$ and $+6$ oxidation states.
C
Oxygen is a gas while sulphur is a solid at room temperature.
D
The hydride of oxygen $(H_2O)$ is more stable than the hydride of sulphur $(H_2S)$.
Solution
(B) $1$. Oxygen has the electronic configuration $[He] 2s^2 2p^4$,which contains $2$ unpaired electrons in the $2p$ orbitals.
$2$. Sulphur has the electronic configuration $[Ne] 3s^2 3p^4$,which also contains $2$ unpaired electrons in the $3p$ orbitals.
$3$. Oxygen is highly electronegative and typically exhibits an oxidation state of $-2$. It does not show $+4$ or $+6$ oxidation states because it lacks $d$-orbitals.
$4$. Sulphur can show $-2, +2, +4,$ and $+6$ oxidation states due to the availability of vacant $d$-orbitals.
$5$. Therefore,the statement that both oxygen and sulphur show $-2, +4,$ and $+6$ oxidation states is false.
$2$. Sulphur has the electronic configuration $[Ne] 3s^2 3p^4$,which also contains $2$ unpaired electrons in the $3p$ orbitals.
$3$. Oxygen is highly electronegative and typically exhibits an oxidation state of $-2$. It does not show $+4$ or $+6$ oxidation states because it lacks $d$-orbitals.
$4$. Sulphur can show $-2, +2, +4,$ and $+6$ oxidation states due to the availability of vacant $d$-orbitals.
$5$. Therefore,the statement that both oxygen and sulphur show $-2, +4,$ and $+6$ oxidation states is false.
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356
ChemistryEasyMCQMHT CET · 2025
Identify the correct order of thermal stability of hydrides of $16$ group elements from the following.
A
$H_2S < H_2O < H_2Se < H_2Te$
B
$H_2O < H_2S < H_2Se < H_2Te$
C
$H_2Te < H_2Se < H_2S < H_2O$
D
$H_2Se < H_2Te < H_2O < H_2S$
Solution
(C) The thermal stability of hydrides of group $16$ elements decreases down the group as the size of the central atom increases.
As the size of the central atom increases,the bond length between the central atom and hydrogen increases,which leads to a decrease in bond dissociation enthalpy.
The order of thermal stability is $H_2O > H_2S > H_2Se > H_2Te$.
Therefore,the correct order is $H_2Te < H_2Se < H_2S < H_2O$.
As the size of the central atom increases,the bond length between the central atom and hydrogen increases,which leads to a decrease in bond dissociation enthalpy.
The order of thermal stability is $H_2O > H_2S > H_2Se > H_2Te$.
Therefore,the correct order is $H_2Te < H_2Se < H_2S < H_2O$.
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357
ChemistryEasyMCQMHT CET · 2025
Identify the correct molecular formula of 'Oleum' from the following.
A
$H_2S_2O_3$
B
$H_2S_2O_5$
C
$H_2S_2O_7$
D
$H_2S_2O_8$
Solution
(C) Oleum is also known as fuming sulfuric acid.
It is formed by dissolving sulfur trioxide $(SO_3)$ in concentrated sulfuric acid $(H_2SO_4)$.
The chemical formula for Oleum is $H_2S_2O_7$ (also written as $H_2SO_4 \cdot SO_3$).
It is formed by dissolving sulfur trioxide $(SO_3)$ in concentrated sulfuric acid $(H_2SO_4)$.
The chemical formula for Oleum is $H_2S_2O_7$ (also written as $H_2SO_4 \cdot SO_3$).
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358
ChemistryMediumMCQMHT CET · 2025
Identify the false statement from the following about fluorine.
A
It is a highly electronegative element.
B
It exhibits only $-1$ oxidation state.
C
It has high bond dissociation enthalpy among all halogens.
D
It forms only one oxoacid.
Solution
(C) Fluorine is the most electronegative element in the periodic table,which is a true statement.
It exhibits only $-1$ oxidation state because it lacks $d$-orbitals and is the most electronegative,making it unable to show positive oxidation states,which is a true statement.
Fluorine has the lowest bond dissociation enthalpy among all halogens due to the high inter-electronic repulsion between the lone pairs of the small $F$ atoms,making the statement 'It has high bond dissociation enthalpy among all halogens' false.
Fluorine forms only one oxoacid,which is $HOF$ (hypofluorous acid),which is a true statement.
It exhibits only $-1$ oxidation state because it lacks $d$-orbitals and is the most electronegative,making it unable to show positive oxidation states,which is a true statement.
Fluorine has the lowest bond dissociation enthalpy among all halogens due to the high inter-electronic repulsion between the lone pairs of the small $F$ atoms,making the statement 'It has high bond dissociation enthalpy among all halogens' false.
Fluorine forms only one oxoacid,which is $HOF$ (hypofluorous acid),which is a true statement.
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359
ChemistryMediumMCQMHT CET · 2025
Identify the correct decreasing order of boiling point for hydrides of halogens.
A
$HF > HCl > HBr > HI$
B
$HI > HBr > HCl > HF$
C
$HF > HI > HBr > HCl$
D
$HI > HF > HBr > HCl$
Solution
(C) The boiling point of hydrides of halogens depends on two main factors: intermolecular hydrogen bonding and van der Waals forces.
$HF$ has the highest boiling point due to strong intermolecular hydrogen bonding.
For the remaining hydrides ($HCl$,$HBr$,$HI$),the boiling point increases with increasing molecular mass due to the increase in the magnitude of van der Waals forces.
Thus,the order is $HI > HBr > HCl$.
Combining these,the overall decreasing order of boiling point is $HF > HI > HBr > HCl$.
$HF$ has the highest boiling point due to strong intermolecular hydrogen bonding.
For the remaining hydrides ($HCl$,$HBr$,$HI$),the boiling point increases with increasing molecular mass due to the increase in the magnitude of van der Waals forces.
Thus,the order is $HI > HBr > HCl$.
Combining these,the overall decreasing order of boiling point is $HF > HI > HBr > HCl$.
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360
ChemistryEasyMCQMHT CET · 2025
Which of the following statements is false about interhalogen compounds?
A
They are covalent in nature.
B
They are diamagnetic.
C
They are more reactive than halogens.
D
If these are represented as $XX^\prime$,the number of $X^\prime$ atoms is always even.
Solution
(D) Interhalogen compounds are represented by the general formula $XX^\prime_n$,where $X$ is the less electronegative halogen and $X^\prime$ is the more electronegative halogen,and $n = 1, 3, 5, 7$.
$1$. They are covalent in nature due to the small difference in electronegativity between the two halogens.
$2$. They are diamagnetic because all electrons are paired.
$3$. They are generally more reactive than individual halogens (except fluorine) because the $X-X^\prime$ bond is weaker than the $X-X$ or $X^\prime-X^\prime$ bonds.
$4$. The statement "If these are represented as $XX^\prime$,the number of $X^\prime$ atoms is always even" is false,as $n$ can be $1, 3, 5, 7$,which are odd numbers.
$1$. They are covalent in nature due to the small difference in electronegativity between the two halogens.
$2$. They are diamagnetic because all electrons are paired.
$3$. They are generally more reactive than individual halogens (except fluorine) because the $X-X^\prime$ bond is weaker than the $X-X$ or $X^\prime-X^\prime$ bonds.
$4$. The statement "If these are represented as $XX^\prime$,the number of $X^\prime$ atoms is always even" is false,as $n$ can be $1, 3, 5, 7$,which are odd numbers.
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361
ChemistryEasyMCQMHT CET · 2025
Which of the following halogens forms the maximum number of oxoacids?
A
$F$
B
$Cl$
C
$Br$
D
$I$
Solution
(B) Fluorine $(F)$ is the most electronegative element and has no $d$-orbitals,so it forms only one oxoacid,which is hypofluorous acid $(HOF)$.
Chlorine $(Cl)$,Bromine $(Br)$,and Iodine $(I)$ can exhibit higher oxidation states due to the presence of vacant $d$-orbitals.
Chlorine forms four types of oxoacids: hypochlorous acid $(HOCl)$,chlorous acid $(HOClO)$,chloric acid $(HOClO_2)$,and perchloric acid $(HOClO_3)$.
Among the halogens,Chlorine forms the maximum number of stable oxoacids.
Chlorine $(Cl)$,Bromine $(Br)$,and Iodine $(I)$ can exhibit higher oxidation states due to the presence of vacant $d$-orbitals.
Chlorine forms four types of oxoacids: hypochlorous acid $(HOCl)$,chlorous acid $(HOClO)$,chloric acid $(HOClO_2)$,and perchloric acid $(HOClO_3)$.
Among the halogens,Chlorine forms the maximum number of stable oxoacids.
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362
ChemistryEasyMCQMHT CET · 2025
Which of the following is the molecular formula of halous acid of chlorine?
A
$HClO$
B
$HClO_2$
C
$HClO_3$
D
$HClO_4$
Solution
(B) The oxoacids of chlorine are as follows:
$1$. Hypochlorous acid ($HOCl$ or $HClO$): Oxidation state of $Cl$ is $+1$.
$2$. Halous acid $(HClO_2)$: Oxidation state of $Cl$ is $+3$.
$3$. Halic acid $(HClO_3)$: Oxidation state of $Cl$ is $+5$.
$4$. Perhalic acid $(HClO_4)$: Oxidation state of $Cl$ is $+7$.
Therefore,the molecular formula of halous acid of chlorine is $HClO_2$.
$1$. Hypochlorous acid ($HOCl$ or $HClO$): Oxidation state of $Cl$ is $+1$.
$2$. Halous acid $(HClO_2)$: Oxidation state of $Cl$ is $+3$.
$3$. Halic acid $(HClO_3)$: Oxidation state of $Cl$ is $+5$.
$4$. Perhalic acid $(HClO_4)$: Oxidation state of $Cl$ is $+7$.
Therefore,the molecular formula of halous acid of chlorine is $HClO_2$.
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363
ChemistryEasyMCQMHT CET · 2025
What is the shape of an interhalogen compound where the central halogen exhibits a $+3$ oxidation state?
A
Tetrahedral
B
Bent '$T$' shape
C
Square pyramidal
D
Square planar
Solution
(B) Interhalogen compounds of the type $XX'_3$ (e.g.,$ClF_3$,$BrF_3$) have a central halogen atom in the $+3$ oxidation state.
According to $VSEPR$ theory,the central atom has $3$ bond pairs and $2$ lone pairs of electrons,resulting in a total of $5$ electron pairs.
This corresponds to $sp^3d$ hybridization with a trigonal bipyramidal electron geometry.
Due to the presence of $2$ lone pairs in the equatorial positions to minimize repulsion,the molecule adopts a bent '$T$' shape.
According to $VSEPR$ theory,the central atom has $3$ bond pairs and $2$ lone pairs of electrons,resulting in a total of $5$ electron pairs.
This corresponds to $sp^3d$ hybridization with a trigonal bipyramidal electron geometry.
Due to the presence of $2$ lone pairs in the equatorial positions to minimize repulsion,the molecule adopts a bent '$T$' shape.
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364
ChemistryEasyMCQMHT CET · 2025
Which of the following interhalogen compounds is in liquid state at room temperature?
A
$ICl$
B
$ClF_3$
C
$BrF_5$
D
$IF_7$
Solution
(A) Among the given interhalogen compounds,$ICl$ (iodine monochloride) exists as a ruby-red liquid at room temperature.
$ClF_3$ is a gas.
$BrF_5$ is a liquid,but $ICl$ is the most commonly cited example for this property in standard textbooks.
$IF_7$ is a gas.
$ClF_3$ is a gas.
$BrF_5$ is a liquid,but $ICl$ is the most commonly cited example for this property in standard textbooks.
$IF_7$ is a gas.
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365
ChemistryEasyMCQMHT CET · 2025
Which of the following is the only existing oxoacid of fluorine?
A
$HOF$
B
$HFO_2$
C
$HFO_3$
D
$HFO_4$
Solution
(A) Fluorine is the most electronegative element and has a small atomic size. Due to the absence of $d$-orbitals and its high electronegativity,it cannot exhibit positive oxidation states except in the case of $HOF$ (hypofluorous acid).
In $HOF$,the oxidation state of fluorine is $-1$,while oxygen is in the $-2$ state and hydrogen is in the $+1$ state. Other oxoacids like $HFO_2$,$HFO_3$,and $HFO_4$ do not exist because fluorine cannot expand its octet or accommodate positive oxidation states required for these structures.
In $HOF$,the oxidation state of fluorine is $-1$,while oxygen is in the $-2$ state and hydrogen is in the $+1$ state. Other oxoacids like $HFO_2$,$HFO_3$,and $HFO_4$ do not exist because fluorine cannot expand its octet or accommodate positive oxidation states required for these structures.
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366
ChemistryEasyMCQMHT CET · 2025
What is the number of oxygen atoms bonded to chlorine in its strongest oxoacid?
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(D) The strongest oxoacid of chlorine is perchloric acid,which has the chemical formula $HClO_4$.
In the structure of $HClO_4$,the chlorine atom is central and is bonded to four oxygen atoms.
Specifically,one oxygen atom is part of an $OH$ group,and the other three oxygen atoms are double-bonded to the chlorine atom $(Cl=O)$.
Therefore,the total number of oxygen atoms bonded to chlorine is $4$.
In the structure of $HClO_4$,the chlorine atom is central and is bonded to four oxygen atoms.
Specifically,one oxygen atom is part of an $OH$ group,and the other three oxygen atoms are double-bonded to the chlorine atom $(Cl=O)$.
Therefore,the total number of oxygen atoms bonded to chlorine is $4$.
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367
ChemistryEasyMCQMHT CET · 2025
Which of the following halogens does always show an oxidation state of $-1$?
A
$Cl$
B
$F$
C
$Br$
D
$I$
Solution
(B) Fluorine $(F)$ is the most electronegative element in the periodic table.
It has a small atomic size and lacks $d$-orbitals in its valence shell.
Due to its high electronegativity,it can only gain one electron to complete its octet and cannot exhibit positive oxidation states or expand its octet.
Therefore,it always exhibits an oxidation state of $-1$ in all its compounds.
It has a small atomic size and lacks $d$-orbitals in its valence shell.
Due to its high electronegativity,it can only gain one electron to complete its octet and cannot exhibit positive oxidation states or expand its octet.
Therefore,it always exhibits an oxidation state of $-1$ in all its compounds.
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368
ChemistryMediumMCQMHT CET · 2025
Identify from the following compounds the one in which the valence shell of $Xe$ consists of one lone pair of electrons.
A
$XeF_2$
B
$XeF_4$
C
$XeOF_2$
D
$XeOF_4$
Solution
(D) The valence shell of $Xe$ has $8$ electrons.
In $XeF_2$,$Xe$ forms $2$ bonds with $F$ atoms,leaving $6$ electrons ($3$ lone pairs).
In $XeF_4$,$Xe$ forms $4$ bonds with $F$ atoms,leaving $4$ electrons ($2$ lone pairs).
In $XeOF_2$,$Xe$ forms $2$ bonds with $F$ and $2$ bonds with $O$ (total $4$ bonds),leaving $4$ electrons ($2$ lone pairs).
In $XeOF_4$,$Xe$ forms $4$ bonds with $F$ and $2$ bonds with $O$ (total $6$ bonds),leaving $2$ electrons ($1$ lone pair).
Therefore,$XeOF_4$ is the compound where $Xe$ has one lone pair of electrons.
In $XeF_2$,$Xe$ forms $2$ bonds with $F$ atoms,leaving $6$ electrons ($3$ lone pairs).
In $XeF_4$,$Xe$ forms $4$ bonds with $F$ atoms,leaving $4$ electrons ($2$ lone pairs).
In $XeOF_2$,$Xe$ forms $2$ bonds with $F$ and $2$ bonds with $O$ (total $4$ bonds),leaving $4$ electrons ($2$ lone pairs).
In $XeOF_4$,$Xe$ forms $4$ bonds with $F$ and $2$ bonds with $O$ (total $6$ bonds),leaving $2$ electrons ($1$ lone pair).
Therefore,$XeOF_4$ is the compound where $Xe$ has one lone pair of electrons.
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369
ChemistryEasyMCQMHT CET · 2025
Which of the following inert gases is used for magnetic resonance imaging?
A
$He$
B
$Ne$
C
$Ar$
D
$Xe$
Solution
(A) Liquid $He$ (Helium) is used as a coolant in the superconducting magnets of $MRI$ (Magnetic Resonance Imaging) scanners to maintain the extremely low temperatures required for superconductivity.
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370
ChemistryEasyMCQMHT CET · 2025
What is the oxidation state of xenon in xenon monooxytetrafluoride?
A
$2$
B
$4$
C
$6$
D
$8$
Solution
(C) The chemical formula for xenon monooxytetrafluoride is $XeOF_4$.
Let the oxidation state of xenon be $x$.
The oxidation state of oxygen is $-2$ and the oxidation state of fluorine is $-1$.
Setting the sum of oxidation states equal to the charge of the molecule $(0)$:
$x + (-2) + 4(-1) = 0$
$x - 2 - 4 = 0$
$x - 6 = 0$
$x = +6$.
Therefore,the oxidation state of xenon in $XeOF_4$ is $+6$.
Let the oxidation state of xenon be $x$.
The oxidation state of oxygen is $-2$ and the oxidation state of fluorine is $-1$.
Setting the sum of oxidation states equal to the charge of the molecule $(0)$:
$x + (-2) + 4(-1) = 0$
$x - 2 - 4 = 0$
$x - 6 = 0$
$x = +6$.
Therefore,the oxidation state of xenon in $XeOF_4$ is $+6$.
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371
ChemistryMediumMCQMHT CET · 2025
Which of the following xenon compound has a structure similar to chlorine pentafluoride $(ClF_5)$?
A
$XeF_4$
B
$XeF_6$
C
$XeO_3$
D
$XeOF_4$
Solution
(D) The structure of $ClF_5$ is square pyramidal,which is based on $sp^3d^2$ hybridization with one lone pair of electrons on the central chlorine atom.
$XeOF_4$ has the central xenon atom bonded to four fluorine atoms and one oxygen atom (double bond).
Using the $VSEPR$ theory,the steric number for $Xe$ in $XeOF_4$ is $5$ (four single bonds + one double bond) plus one lone pair,resulting in $sp^3d^2$ hybridization.
Thus,$XeOF_4$ also exhibits a square pyramidal geometry,making it isostructural with $ClF_5$.
$XeOF_4$ has the central xenon atom bonded to four fluorine atoms and one oxygen atom (double bond).
Using the $VSEPR$ theory,the steric number for $Xe$ in $XeOF_4$ is $5$ (four single bonds + one double bond) plus one lone pair,resulting in $sp^3d^2$ hybridization.
Thus,$XeOF_4$ also exhibits a square pyramidal geometry,making it isostructural with $ClF_5$.
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372
ChemistryEasyMCQMHT CET · 2025
Which of the following polymers is obtained by the addition polymerisation method?
A
Nylon $6$
B
Terylene
C
Nylon $6, 6$
D
Teflon
Solution
(D) Addition polymerisation involves the repeated addition of monomer molecules possessing double or triple bonds to form a polymer without the loss of any small molecules.
$Teflon$ (polytetrafluoroethylene) is formed by the addition polymerisation of tetrafluoroethene $(CF_2=CF_2)$.
$Nylon$ $6$,$Terylene$,and $Nylon$ $6, 6$ are condensation polymers formed by the elimination of small molecules like $H_2O$ or $CH_3OH$.
$Teflon$ (polytetrafluoroethylene) is formed by the addition polymerisation of tetrafluoroethene $(CF_2=CF_2)$.
$Nylon$ $6$,$Terylene$,and $Nylon$ $6, 6$ are condensation polymers formed by the elimination of small molecules like $H_2O$ or $CH_3OH$.
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373
ChemistryEasyMCQMHT CET · 2025
Which of the following polymers requires a peroxide as an initiator for its preparation?
A
Nylon $6,6$
B
Polyacrylonitrile
C
Terylene
D
Bakelite
Solution
(B) Polyacrylonitrile is prepared by the free radical polymerization of acrylonitrile $(CH_2=CH-CN)$.
Free radical polymerization requires a radical initiator,such as a peroxide (e.g.,benzoyl peroxide),to initiate the chain reaction.
Nylon $6,6$ and Terylene are prepared by condensation polymerization,while Bakelite is prepared by step-growth polymerization involving phenol and formaldehyde.
Free radical polymerization requires a radical initiator,such as a peroxide (e.g.,benzoyl peroxide),to initiate the chain reaction.
Nylon $6,6$ and Terylene are prepared by condensation polymerization,while Bakelite is prepared by step-growth polymerization involving phenol and formaldehyde.
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374
ChemistryEasyMCQMHT CET · 2025
Which of the following polymers is classified as a fibre?
A
Nylon $6, 6$
B
Urea formaldehyde resin
C
Polystyrene
D
Neoprene
Solution
(A) Polymers are classified based on their intermolecular forces into elastomers,fibres,thermoplastics,and thermosetting polymers.
Fibres are thread-forming solids which possess high tensile strength and high modulus.
These characteristics are due to strong intermolecular forces like hydrogen bonding or dipole-dipole interactions.
Nylon $6, 6$ is a polyamide,which contains strong hydrogen bonding between the chains,making it a fibre.
Urea formaldehyde resin is a thermosetting polymer.
Polystyrene is a thermoplastic polymer.
Neoprene is an elastomer.
Therefore,the correct option is $A$.
Fibres are thread-forming solids which possess high tensile strength and high modulus.
These characteristics are due to strong intermolecular forces like hydrogen bonding or dipole-dipole interactions.
Nylon $6, 6$ is a polyamide,which contains strong hydrogen bonding between the chains,making it a fibre.
Urea formaldehyde resin is a thermosetting polymer.
Polystyrene is a thermoplastic polymer.
Neoprene is an elastomer.
Therefore,the correct option is $A$.
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375
ChemistryEasyMCQMHT CET · 2025
Which of the following polymers is obtained by the condensation polymerisation method?
A
Polythene
B
Nylon $6, 6$
C
Polyacrylonitrile
D
Teflon
Solution
(B) Condensation polymerization involves the repeated condensation reaction between two different bi-functional or tri-functional monomeric units,usually with the elimination of small molecules like $H_2O$,$HCl$,or $NH_3$.
$Nylon \ 6, 6$ is formed by the condensation of hexamethylenediamine $(H_2N(CH_2)_6NH_2)$ and adipic acid $(HOOC(CH_2)_4COOH)$.
Polythene,Polyacrylonitrile,and Teflon are all examples of addition polymers formed by the polymerization of unsaturated monomers.
$Nylon \ 6, 6$ is formed by the condensation of hexamethylenediamine $(H_2N(CH_2)_6NH_2)$ and adipic acid $(HOOC(CH_2)_4COOH)$.
Polythene,Polyacrylonitrile,and Teflon are all examples of addition polymers formed by the polymerization of unsaturated monomers.
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376
ChemistryEasyMCQMHT CET · 2025
Identify the thermoplastic polymer from the following.
A
Polyvinyl chloride
B
Bakelite
C
Terylene
D
Neoprene
Solution
(A) Thermoplastic polymers are linear or slightly branched long-chain molecules that can be repeatedly softened on heating and hardened on cooling.
$Polyvinyl \ chloride$ $(PVC)$ is a well-known thermoplastic polymer.
$Bakelite$ is a thermosetting polymer.
$Terylene$ is a polyester fiber (often considered a thermoplastic,but $PVC$ is the classic textbook example of a thermoplastic plastic).
$Neoprene$ is a synthetic rubber (elastomer).
Therefore,the correct answer is $Polyvinyl \ chloride$.
$Polyvinyl \ chloride$ $(PVC)$ is a well-known thermoplastic polymer.
$Bakelite$ is a thermosetting polymer.
$Terylene$ is a polyester fiber (often considered a thermoplastic,but $PVC$ is the classic textbook example of a thermoplastic plastic).
$Neoprene$ is a synthetic rubber (elastomer).
Therefore,the correct answer is $Polyvinyl \ chloride$.
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377
ChemistryEasyMCQMHT CET · 2025
Identify the monomers used to prepare glyptal.
A
$\beta-Hydroxy$ butyric acid and $\beta-Hydroxy$ valeric acid
B
Ethylene glycol and phenol
C
Ethylene glycol and Terephthalic acid
D
Ethylene glycol and phthalic acid
Solution
(D) Glyptal is a polyester resin formed by the condensation polymerization of ethylene glycol and phthalic acid.
The reaction involves the elimination of water molecules to form a polymer chain.
The monomers are $HO-CH_2-CH_2-OH$ (ethylene glycol) and $C_6H_4(COOH)_2$ (phthalic acid).
The reaction involves the elimination of water molecules to form a polymer chain.
The monomers are $HO-CH_2-CH_2-OH$ (ethylene glycol) and $C_6H_4(COOH)_2$ (phthalic acid).
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378
ChemistryEasyMCQMHT CET · 2025
Which of the following is a thermosetting polymer?
A
Polyvinyl chloride
B
Polystyrene
C
Urea-formaldehyde resin
D
Nylon $6,6$
Solution
(C) Thermosetting polymers are cross-linked or heavily branched molecules that undergo extensive cross-linking in molds during hardening,becoming infusible and insoluble.
Urea-formaldehyde resin is a classic example of a thermosetting polymer formed by the condensation polymerization of urea and formaldehyde.
Polyvinyl chloride,polystyrene,and Nylon $6,6$ are thermoplastic polymers,which soften on heating and harden on cooling.
Urea-formaldehyde resin is a classic example of a thermosetting polymer formed by the condensation polymerization of urea and formaldehyde.
Polyvinyl chloride,polystyrene,and Nylon $6,6$ are thermoplastic polymers,which soften on heating and harden on cooling.
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379
ChemistryEasyMCQMHT CET · 2025
Identify a copolymer from the following.
A
$Buna-S$
B
Polyacrylonitrile
C
Polypropene
D
Nylon $6,6$
Solution
(A) copolymer is a polymer formed from two or more different types of monomeric units.
$Buna-S$ is a copolymer formed from $1,3-butadiene$ and $styrene$.
$Nylon$ $6,6$ is also a copolymer formed from $hexamethylenediamine$ and $adipic$ $acid$.
However,in the context of standard chemistry questions,$Buna-S$ is the most common example of a synthetic rubber copolymer.
Note: Both $A$ and $D$ are technically copolymers,but $Buna-S$ is the classic example of an addition copolymer.
$Buna-S$ is a copolymer formed from $1,3-butadiene$ and $styrene$.
$Nylon$ $6,6$ is also a copolymer formed from $hexamethylenediamine$ and $adipic$ $acid$.
However,in the context of standard chemistry questions,$Buna-S$ is the most common example of a synthetic rubber copolymer.
Note: Both $A$ and $D$ are technically copolymers,but $Buna-S$ is the classic example of an addition copolymer.
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380
ChemistryEasyMCQMHT CET · 2025
Which of the following polymers is biodegradable?
A
Nylon $6$
B
Nylon $6,6$
C
Nylon $2-$nylon $6$
D
$HDP$
Solution
(C) Biodegradable polymers are those that can be broken down by microorganisms in the environment.
Nylon $2-$nylon $6$ is a polyamide copolymer of glycine and amino caproic acid,which is biodegradable.
Nylon $6$,Nylon $6,6$,and $HDP$ (High Density Polyethylene) are synthetic polymers that are not biodegradable.
Nylon $2-$nylon $6$ is a polyamide copolymer of glycine and amino caproic acid,which is biodegradable.
Nylon $6$,Nylon $6,6$,and $HDP$ (High Density Polyethylene) are synthetic polymers that are not biodegradable.
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381
ChemistryEasyMCQMHT CET · 2025
Which of the following is a thermosetting polymer?
A
Buna-$S$
B
Dacron
C
Polyvinyls
D
Bakelite
Solution
(D) Thermosetting polymers are cross-linked or heavily branched molecules that undergo extensive cross-linking in molds when heated,making them permanently hard and infusible.
$Bakelite$ (phenol-formaldehyde resin) is a classic example of a thermosetting polymer.
$Buna-S$ is an elastomer,while $Dacron$ and $Polyvinyls$ are thermoplastic polymers.
$Bakelite$ (phenol-formaldehyde resin) is a classic example of a thermosetting polymer.
$Buna-S$ is an elastomer,while $Dacron$ and $Polyvinyls$ are thermoplastic polymers.
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382
ChemistryEasyMCQMHT CET · 2025
Identify the cross-linked polymer from the following.
A
High density polyethylene
B
Low density polyethylene
C
Melamine
D
$PVC$
Solution
(C) Cross-linked polymers are formed from bi-functional and tri-functional monomers and contain strong covalent bonds between various linear polymer chains.
$Melamine$ is a classic example of a cross-linked polymer,which is a thermosetting plastic formed by the condensation polymerization of melamine and formaldehyde.
$High \ density \ polyethylene$,$Low \ density \ polyethylene$,and $PVC$ are examples of linear or branched thermoplastic polymers.
$Melamine$ is a classic example of a cross-linked polymer,which is a thermosetting plastic formed by the condensation polymerization of melamine and formaldehyde.
$High \ density \ polyethylene$,$Low \ density \ polyethylene$,and $PVC$ are examples of linear or branched thermoplastic polymers.
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383
ChemistryEasyMCQMHT CET · 2025
Which of the following polymers,when blended with cotton,forms terycot?
A
Dacron
B
Nylon $6$
C
Polyacrylonitrile
D
Nylon $6,6$
Solution
(A) Terycot is a blended fabric produced by mixing polyester with cotton. $Dacron$ (also known as $Terylene$) is a common polyester polymer. Therefore,when $Dacron$ is blended with cotton,it forms terycot.
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384
ChemistryEasyMCQMHT CET · 2025
Identify the monomer used in the preparation of polyacrylonitrile.
A
$H_2C=C(CH_3)-CH=CH_2$
B
$F_2C=CF_2$
C
$H_2C=CH-CN$
D
$H_2C=CH-CH=CH_2$
Solution
(C) Polyacrylonitrile $(PAN)$ is a polymer formed by the addition polymerization of acrylonitrile monomer.
The chemical formula for acrylonitrile is $CH_2=CH-CN$.
Therefore,the correct monomer is $H_2C=CH-CN$.
The chemical formula for acrylonitrile is $CH_2=CH-CN$.
Therefore,the correct monomer is $H_2C=CH-CN$.
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385
ChemistryEasyMCQMHT CET · 2025
From which of the following compounds is thermocol obtained?
A
Bisphenol
B
Vinyl chloride
C
Styrene
D
Butadiene
Solution
(C) Thermocol is a trade name for expanded polystyrene $(EPS)$.
It is produced by the polymerization of the monomer $Styrene$ $(C_6H_5CH=CH_2)$.
Therefore,the correct answer is $Styrene$.
It is produced by the polymerization of the monomer $Styrene$ $(C_6H_5CH=CH_2)$.
Therefore,the correct answer is $Styrene$.
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386
ChemistryEasyMCQMHT CET · 2025
Identify the use of glyptal.
A
To prepare gel used in electrophoresis.
B
To obtain paints.
C
To obtain decorative laminates.
D
To obtain food grade plastic container.
Solution
(B) Glyptal is a polyester polymer formed by the condensation polymerization of ethylene glycol and phthalic acid. It is primarily used in the manufacture of paints and lacquers.
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387
ChemistryEasyMCQMHT CET · 2025
Which of the following polymers is used to prepare shoe soles?
A
Polyacrylamide
B
Perspex
C
Buna-$N$
D
Glyptal
Solution
(C) Buna-$N$ is a synthetic rubber formed by the copolymerization of $1,3$-butadiene and acrylonitrile.
It is highly resistant to the action of petrol,lubricating oil,and organic solvents.
Due to these properties,it is widely used in the manufacturing of oil seals,tank linings,and shoe soles.
It is highly resistant to the action of petrol,lubricating oil,and organic solvents.
Due to these properties,it is widely used in the manufacturing of oil seals,tank linings,and shoe soles.
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388
ChemistryEasyMCQMHT CET · 2025
Identify the monomer of natural rubber.
A
Isoprene
B
Polyacrylonitrile
C
Chloroprene
D
Tetrafluoroethylene
Solution
(A) Natural rubber is a linear polymer of isoprene ($2$-methyl-$1,3$-butadiene).
Its chemical structure is $(CH_2=C(CH_3)-CH=CH_2)$.
It is a cis-$1,4$-polyisoprene.
Its chemical structure is $(CH_2=C(CH_3)-CH=CH_2)$.
It is a cis-$1,4$-polyisoprene.
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389
ChemistryEasyMCQMHT CET · 2025
Which of the following polymers contains the $-CO-NH-$ linkage?
A
Glyptal
B
Thermocol
C
Buna-$N$
D
Nylon-$6,6$
Solution
(D) The $-CO-NH-$ linkage is known as the amide linkage or peptide linkage.
Polymers containing this linkage are classified as polyamides.
Among the given options,$Nylon-6,6$ is a polyamide formed by the condensation polymerization of hexamethylenediamine and adipic acid,which results in the formation of $-CO-NH-$ bonds.
Glyptal is a polyester,Thermocol is polystyrene,and Buna-$N$ is a copolymer of butadiene and acrylonitrile.
Polymers containing this linkage are classified as polyamides.
Among the given options,$Nylon-6,6$ is a polyamide formed by the condensation polymerization of hexamethylenediamine and adipic acid,which results in the formation of $-CO-NH-$ bonds.
Glyptal is a polyester,Thermocol is polystyrene,and Buna-$N$ is a copolymer of butadiene and acrylonitrile.
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390
ChemistryEasyMCQMHT CET · 2025
Identify a monomer used in the preparation of neoprene.
A
$2-$methyl$-1,3-$butadiene
B
$2-$chloro$-1,3-$butadiene
C
Melamine
D
$1,3-$butadiene
Solution
(B) Neoprene is a synthetic rubber formed by the free radical polymerization of chloroprene.
Chloroprene is the common name for $2-$chloro$-1,3-$butadiene.
The chemical structure is $CH_2=C(Cl)-CH=CH_2$.
Therefore,the correct monomer is $2-$chloro$-1,3-$butadiene.
Chloroprene is the common name for $2-$chloro$-1,3-$butadiene.
The chemical structure is $CH_2=C(Cl)-CH=CH_2$.
Therefore,the correct monomer is $2-$chloro$-1,3-$butadiene.
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391
ChemistryMediumMCQMHT CET · 2025
What are the monomers used in the preparation of $PHBV$?
A
$3-Hydroxybutanoic$ acid and $3-Hydroxypentanoic$ acid
B
$HOOC-(CH_2)_4-COOH$ and $H_2N-(CH_2)_6-NH_2$
C
$C_6H_5OH$ and $CH_2O$
D
$NH_2-(CH_2)_5-COOH$
Solution
(A) $PHBV$ stands for $Poly-\beta-hydroxybutyrate-co-\beta-hydroxyvalerate$.
It is a biodegradable copolymer.
The monomers used in the preparation of $PHBV$ are $3-hydroxybutanoic$ acid $(CH_3-CH(OH)-CH_2-COOH)$ and $3-hydroxypentanoic$ acid $(CH_3-CH_2-CH(OH)-CH_2-COOH)$.
It is a biodegradable copolymer.
The monomers used in the preparation of $PHBV$ are $3-hydroxybutanoic$ acid $(CH_3-CH(OH)-CH_2-COOH)$ and $3-hydroxypentanoic$ acid $(CH_3-CH_2-CH(OH)-CH_2-COOH)$.
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392
ChemistryEasyMCQMHT CET · 2025
Which of the following polymers is used to obtain oil seals?
A
Teflon
B
Dacron
C
Nylon $6$
D
$LDP$
Solution
(A) The polymer used to obtain oil seals is $Buna-N$ (also known as $NBR$ or Nitrile rubber). However,among the given options,$Teflon$ $(PTFE)$ is widely used for gaskets and seals due to its chemical inertness and high thermal resistance. In many standard chemistry contexts,$Teflon$ is the intended answer for applications involving seals in harsh environments.
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393
ChemistryEasyMCQMHT CET · 2025
Which of the following is a polyester fibre?
A
$PHBV$
B
Urea formaldehyde
C
Thermocol
D
Polyacrylonitrile
Solution
(A) $PHBV$ (Poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate) is a biodegradable polyester.
It is a copolymer of $3$-hydroxybutanoic acid and $3$-hydroxypentanoic acid.
The ester linkages present in the polymer chain classify it as a polyester.
It is a copolymer of $3$-hydroxybutanoic acid and $3$-hydroxypentanoic acid.
The ester linkages present in the polymer chain classify it as a polyester.
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394
ChemistryEasyMCQMHT CET · 2025
Which of the following polymers is used to obtain rubber belts?
A
Buna-$N$
B
Perspex
C
$PVC$
D
Polycarbonate
Solution
(A) Buna-$N$ is a synthetic rubber copolymer of $1,3$-butadiene and acrylonitrile.
It is highly resistant to the action of petrol,lubricating oil,and organic solvents.
Due to these properties,it is extensively used in the manufacturing of oil seals,tank linings,and rubber belts.
It is highly resistant to the action of petrol,lubricating oil,and organic solvents.
Due to these properties,it is extensively used in the manufacturing of oil seals,tank linings,and rubber belts.
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395
ChemistryEasyMCQMHT CET · 2025
Identify the polyamide polymer from the following.
A
$SBR$
B
Melamine formaldehyde polymer
C
Bakelite
D
Nylon $6,6$
Solution
(D) Polyamides are polymers containing amide linkages $(-CONH-)$ in their backbone.
$Nylon \ 6,6$ is a condensation polymer formed by the reaction of hexamethylenediamine and adipic acid,which contains amide bonds.
$SBR$ (Styrene-butadiene rubber) is an addition copolymer.
Melamine formaldehyde polymer and Bakelite are thermosetting polymers formed by condensation,but they are not polyamides.
$Nylon \ 6,6$ is a condensation polymer formed by the reaction of hexamethylenediamine and adipic acid,which contains amide bonds.
$SBR$ (Styrene-butadiene rubber) is an addition copolymer.
Melamine formaldehyde polymer and Bakelite are thermosetting polymers formed by condensation,but they are not polyamides.
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396
ChemistryEasyMCQMHT CET · 2025
Identify the monomers used in the preparation of Nylon $2-$ nylon $6$.
A
Glycine and $\epsilon$-amino caproic acid
B
$\beta$-hydroxy butyric acid and $\beta$-hydroxy valeric acid
C
Ethylene glycol and phthalic acid
D
Acrylamide and vinyl chloride
Solution
(A) Nylon $2-$ nylon $6$ is a polyamide copolymer.
It is formed by the condensation polymerization of two monomers:
$1$. Glycine $(H_2N-CH_2-COOH)$
$2$. $\epsilon$-amino caproic acid $(H_2N-(CH_2)_5-COOH)$
Therefore,the correct option is $A$.
It is formed by the condensation polymerization of two monomers:
$1$. Glycine $(H_2N-CH_2-COOH)$
$2$. $\epsilon$-amino caproic acid $(H_2N-(CH_2)_5-COOH)$
Therefore,the correct option is $A$.
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397
ChemistryMediumMCQMHT CET · 2025
Which of the following household plastic materials is used to prepare drinking straws?
A
$PVC$
B
$LDPE$
C
$PS$
D
$PP$
Solution
(D) Drinking straws are commonly manufactured from $Polypropylene$ $(PP)$.
$PP$ is a thermoplastic polymer that is lightweight,flexible,and resistant to heat and chemicals,making it ideal for food-contact applications like straws.
$PP$ is a thermoplastic polymer that is lightweight,flexible,and resistant to heat and chemicals,making it ideal for food-contact applications like straws.
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398
ChemistryMediumMCQMHT CET · 2025
Which of the following polymers is used as a wool substitute?
A
Glyptal
B
Polyacrylonitrile
C
Terylene
D
Neoprene
Solution
(B) Polyacrylonitrile,also known as Orlon or Acrilan,is a synthetic polymer prepared by the addition polymerization of acrylonitrile $(CH_2=CH-CN)$.
It is widely used as a substitute for wool in the manufacture of commercial fibers,sweaters,and blankets because of its wool-like feel and warmth.
It is widely used as a substitute for wool in the manufacture of commercial fibers,sweaters,and blankets because of its wool-like feel and warmth.
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399
ChemistryEasyMCQMHT CET · 2025
What are the compounds used to obtain nylon salt?
A
Adipic acid and ammonia
B
Terephthalic acid and hexamethylenediamine
C
Adipic acid and hexamethylenediamine
D
$\beta$-hydroxybutyric acid and hexamethylenediamine
Solution
(C) Nylon $6,6$ is prepared by the condensation polymerization of hexamethylenediamine and adipic acid.
In the first step,these two monomers react to form a $1:1$ salt known as nylon salt.
The reaction is: $HOOC(CH_2)_4COOH + H_2N(CH_2)_6NH_2 \rightarrow [^-OOC(CH_2)_4COO^-][^+H_3N(CH_2)_6NH_3^+]$ (Nylon salt).
In the first step,these two monomers react to form a $1:1$ salt known as nylon salt.
The reaction is: $HOOC(CH_2)_4COOH + H_2N(CH_2)_6NH_2 \rightarrow [^-OOC(CH_2)_4COO^-][^+H_3N(CH_2)_6NH_3^+]$ (Nylon salt).
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400
ChemistryMediumMCQMHT CET · 2025
Which from the following polymers does $NOT$ contain either —$COO$— or —$CONH$— linkage in it?
A
Perspex
B
Polyacrylamide
C
Glyptal
D
Thermocol
Solution
(D) $1$. $Perspex$ (Polymethyl methacrylate) contains the ester linkage (—$COO$—).
$2$. $Glyptal$ (Polyethylene terephthalate/alkyd resin) contains the ester linkage (—$COO$—).
$3$. $Thermocol$ (Polystyrene) is a hydrocarbon polymer and does not contain any —$COO$— or —$CONH$— linkages.
$4$. $Polyacrylamide$ contains the amide linkage (—$CONH$—).
Therefore,the correct answer is $Thermocol$.
$2$. $Glyptal$ (Polyethylene terephthalate/alkyd resin) contains the ester linkage (—$COO$—).
$3$. $Thermocol$ (Polystyrene) is a hydrocarbon polymer and does not contain any —$COO$— or —$CONH$— linkages.
$4$. $Polyacrylamide$ contains the amide linkage (—$CONH$—).
Therefore,the correct answer is $Thermocol$.
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