Chemical Bonding and Molecular Structure MCQ Quiz - Objective Question with Answer for Chemical Bonding and Molecular Structure - Download Free PDF

Last updated on Jun 13, 2025

Latest Chemical Bonding and Molecular Structure MCQ Objective Questions

Chemical Bonding and Molecular Structure Question 1:

Match the LIST-I with LIST-II

LIST-I (Molecules/ion) LIST-II (Hybridisation of central atom)
A. PF₅ I. dsp²
B. SF₆ II. sp³d
C. Ni(CO)₄ III. sp³d²
D. [PtCl₄]²⁻ IV. sp³
Choose the correct answer from the options given below : 
 

  1. A-II, B-III, C-IV, D-I 
  2. A-IV, B-I, C-II, D-III 
  3. A-I, B-II, C-III, D-IV 
  4. A-III, B-I, C-IV, D-II 

Answer (Detailed Solution Below)

Option 1 : A-II, B-III, C-IV, D-I 

Chemical Bonding and Molecular Structure Question 1 Detailed Solution

CONCEPT:

Hybridization of Central Atoms

  • Hybridization: Hybridization is the concept where atomic orbitals mix to form new hybrid orbitals, which explain the molecular geometry and bonding in molecules.
  • The hybridization of the central atom depends on the number of bonds and lone pairs around it. The different hybridizations are:
    • sp³: Tetrahedral geometry, 4 bonds
    • sp³d: Trigonal bipyramidal geometry, 5 bonds
    • sp³d²: Octahedral geometry, 6 bonds
    • dsp²: Octahedral geometry with a metal center, 6 bonds

EXPLANATION:

  • Option A (PF₅): Phosphorus in PF₅ has five bonds and no lone pairs, so it undergoes sp³d hybridization, resulting in a trigonal bipyramidal geometry.
  • Option B (SF₆): Sulfur in SF₆ has six bonds and no lone pairs, so it undergoes sp³d² hybridization, resulting in an octahedral geometry.
  • Option C (Ni(CO)₄): Nickel in Ni(CO)₄ has four bonds and no lone pairs, so it undergoes sp³ hybridization, leading to a tetrahedral geometry.
    qImage684476512afc036f66a20bc1
  • Option D ([PtCl₆]²⁻): Platinum in [PtCl₆]²⁻ has six bonds, and in the presence of the ligand field, it undergoes dsp² hybridization, resulting in an octahedral geometry.
    qImage684476522afc036f66a20bc3

Therefore, the correct answer is A-II, B-III, C-IV, D-I.

Chemical Bonding and Molecular Structure Question 2:

Identify the correct orders against the property mentioned
A. H₂O > NH₃ > CHCl₃ – dipole moment
B. XeF₄ > XeO₃ > XeF₂ – number of lone pairs on central atom
C. O–H > C–H > N–O – bond length
D. N₂ > O₂ > H₂ – bond enthalpy
Choose the correct answer from the options given below:

  1. A, D only 
  2. B, D only 
  3. A, C only 
  4. B, C only

Answer (Detailed Solution Below)

Option 1 : A, D only 

Chemical Bonding and Molecular Structure Question 2 Detailed Solution

CONCEPT:

Dipole Moment, Lone Pairs, Bond Length, and Bond Enthalpy

  • Dipole Moment: The dipole moment depends on the electronegativity difference between bonded atoms and the molecular geometry.
  • Lone Pairs on the Central Atom: The number of lone pairs can be determined using the molecular structure and valence electrons of the central atom.
  • Bond Length: Bond length is inversely proportional to bond strength and directly related to the size of the atoms involved in the bond.
  • Bond Enthalpy: Bond enthalpy is the energy required to break a bond. Triple bonds are stronger than double bonds, which are stronger than single bonds.

EXPLANATION:

  • H₂O > NH₃ > CHCl₃ – Dipole Moment
    • H₂O has the highest dipole moment due to its bent shape and high electronegativity difference.
    • NH₃ has a lower dipole moment than H₂O, as it is pyramidal and less polar.
    • CHCl₃ has the lowest dipole moment because its geometry partially cancels out dipoles.
    • This order is correct.
  • XeF₄ > XeO₃ > XeF₂ – Number of Lone Pairs on Central Atom
    • XeF₄: Xenon has 4 bonds and 2 lone pairs.
    • XeO₃: Xenon has 3 bonds and 1 lone pair.
    • XeF₂: Xenon has 2 bonds and 3 lone pairs.
    • The correct order of lone pairs is XeF₂ > XeF₄ > XeO₃, so this option is incorrect.
  • O–H > C–H > N–O – Bond Length
    • O–H has the shortest bond length due to high bond strength and small atomic size.
    • C–H is longer than O–H but shorter than N–O.
    • N–O has the longest bond length due to weaker bond strength and larger atomic size.
    • This order is correct.
  • N₂ > O₂ > H₂ – Bond Enthalpy
    • N₂ has the highest bond enthalpy due to the strong triple bond.
    • O₂ has a lower bond enthalpy than N₂ due to its double bond.
    • H₂ has the lowest bond enthalpy due to its single bond.
    • This order is correct.

Therefore, the correct answer is A, D only.

Chemical Bonding and Molecular Structure Question 3:

Match the List-I with List-II.
 

List-I (Ion) List-II (Geometry)
A. XeO₃ I. sp³d; linear
B. XeF₂ II. sp³; pyramidal
C. XeOF₄ III. sp³d³; distorted octahedral
D. XeF₆ IV. sp³d²; square pyramidal

Choose the correct answer from the options given below:

  1. A-II, B-I, C-IV, D-III
  2. A-II, B-I, C-III, D-IV
  3. A-IV, B-II, C-III, D-I
  4. A-IV, B-II, C-I, D-III

Answer (Detailed Solution Below)

Option 1 : A-II, B-I, C-IV, D-III

Chemical Bonding and Molecular Structure Question 3 Detailed Solution

CONCEPT:

Geometry of Xenon Compounds Based on Hybridization

  • The geometry of xenon compounds can be predicted based on the hybridization of the central xenon atom and the presence of lone pairs.
  • The Valence Shell Electron Pair Repulsion (VSEPR) theory is used to determine the arrangement of bonds and lone pairs around the central atom.
  • The hybridization and molecular geometry of xenon compounds are as follows:
    • sp3 hybridization: Tetrahedral electronic geometry; molecular geometry depends on lone pairs.
    • sp3d hybridization: Trigonal bipyramidal electronic geometry; molecular geometry depends on lone pairs.
    • sp3d2 hybridization: Octahedral electronic geometry; molecular geometry depends on lone pairs.
    • sp3d3 hybridization: Pentagonal bipyramidal electronic geometry.

EXPLANATION:

  • A. XeO₃: Xenon is sp3 hybridized with one lone pair and three bonded oxygen atoms. This results in a pyramidal geometry.
    MBD EM CHE XII C07 E04 022 S01
  • B. XeF₂: Xenon is sp3d hybridized with three lone pairs and two bonded fluorine atoms. This results in a linear geometry.
    qImage68247a385c4f8d9e4ac5ddc7
  • C. XeOF₄: Xenon is sp3d2 hybridized with one lone pair and five bonded atoms (four fluorine and one oxygen). This results in a square pyramidal geometry.
    qImage68247a395c4f8d9e4ac5ddc9
  • D. XeF₆: Xenon is sp3d3 hybridized with one lone pair and six bonded fluorine atoms. This results in a distorted octahedral geometry.
    qImage68247a395c4f8d9e4ac5de44
  • Match the compounds with their geometries:
    • A. XeO₃: II (sp3; pyramidal)
    • B. XeF₂: I (sp3d; linear)
    • C. XeOF₄: IV (sp3d2; square pyramidal)
    • D. XeF₆: III (sp3d3; distorted octahedral)

Therefore, the correct answer is: A-II, B-I, C-IV, D-III.

Chemical Bonding and Molecular Structure Question 4:

Given below are two statements:
Statement I: A hypothetical diatomic molecule with bond order zero is quite stable.
Statement II: As bond order increases, the bond length increases.
In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are true
  2. Both Statement I and Statement II are false
  3. Statement I is true but Statement II is false
  4. Statement I is true but Statement II is true

Answer (Detailed Solution Below)

Option 2 : Both Statement I and Statement II are false

Chemical Bonding and Molecular Structure Question 4 Detailed Solution

CONCEPT:

Bond Order and Bond Stability

  • Bond Order is defined as the number of chemical bonds between a pair of atoms. It can be determined using Molecular Orbital Theory (MOT).
  • A bond order of zero means that there are equal numbers of bonding and antibonding electrons, resulting in no net bond formation.
  • A molecule with a bond order of zero is considered unstable and cannot exist under normal conditions.

Bond Order and Bond Length

  • Bond order is inversely related to bond length. As bond order increases, the bond strength increases, and the bond length decreases.
  • Higher bond order indicates stronger bonds, which are shorter in length.

EXPLANATION:

  • Statement I: "A hypothetical diatomic molecule with bond order zero is quite stable." This statement is false. A bond order of zero implies no bond formation, and the molecule would not exist as a stable entity.
  • Statement II: "As bond order increases, the bond length increases." This statement is also false. As bond order increases, bond strength increases, and bond length decreases.

Therefore, the correct answer is: Option 2: Both Statement I and Statement II are false.

Chemical Bonding and Molecular Structure Question 5:

Which of the following pairs of compounds is isoelectronic and isostructural?

  1. TeI2, XeF2 
  2. \(\mathrm{IBr}_{2}^{-}, \mathrm{XeF}_{2}\)

  3. IF., XeF,
  4. BeCl2, XeF2
  5. BeCl3, XeF2

Answer (Detailed Solution Below)

Option 2 :

\(\mathrm{IBr}_{2}^{-}, \mathrm{XeF}_{2}\)

Chemical Bonding and Molecular Structure Question 5 Detailed Solution

CONCEPT:

Isoelectronic and Isostructural Compounds

  • Isoelectronic compounds have the same number of electrons or the same electronic structure.
  • Isostructural compounds have the same spatial arrangement of atoms or the same molecular structure.

EXPLANATION:

  • For the compounds to be isoelectronic, they must have the same number of total electrons.
  • For the compounds to be isostructural, they must have the same geometry or shape.
  • Let's compare the given options:
    • Option 1: TeI2 and XeF2
      • TeI2 and XeF2 are not isoelectronic as they have different total electron counts.
    • Option 2: IBr2- and XeF2
      • IBr2- and XeF2 are isoelectronic, both having 22 electrons.
      • Both have a linear structure, making them isostructural.
        qImage679fa3dd51f33685400fc1a4qImage679fa3dd51f33685400fc1a9
    • Option 3: IF and XeF
      • IF and XeF are not isoelectronic as they have different total electron counts.
    • Option 4: BeCl2 and XeF2
      • BeCl2 and XeF2 are not isoelectronic as they have different total electron counts.

Therefore, the correct answer is option 2: IBr2- and XeF2.

Top Chemical Bonding and Molecular Structure MCQ Objective Questions

When is the Intramolecular hydrogen bond formed?

  1. When a hydrogen atom is in between the two highly electropositive atoms
  2. When Oxygen atom is in between the two highly electronegative atoms
  3. When a hydrogen atom is in between the two highly electronegative atoms
  4. When Oxygen atom is in between the two highly electropositive atoms

Answer (Detailed Solution Below)

Option 3 : When a hydrogen atom is in between the two highly electronegative atoms

Chemical Bonding and Molecular Structure Question 6 Detailed Solution

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Option 3 is correct, i.e. When a hydrogen atom is in between the two highly electronegative atoms.

  • There are two types of H bond and they are as follows:
    • Intramolecular hydrogen bond.
    • Intermolecular hydrogen bond.
  • Intramolecular hydrogen bond:
    • It is formed when a hydrogen atom is in between the two highly electronegative (F, O, N) atoms present within the same molecule. For example, in o-nitrophenol, the hydrogen is in between the two oxygen atoms.
  • Intermolecular hydrogen bond:
    • It is formed between two different molecules of the same or different compounds. For example, H-bond in case of HF molecule, alcohol or water molecules, etc.

The valency of sulphur in sulphur dioxide is-

  1. 3
  2. 4
  3. 2
  4. 1

Answer (Detailed Solution Below)

Option 2 : 4

Chemical Bonding and Molecular Structure Question 7 Detailed Solution

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The correct answer is 4.

Key Points

  • The number of electrons present in the valence shell of the atom is called valance electron.
  • The valency of an atom is defined as the combining capacity of an atom, In other word, the number of bonds formed by an atom is also called as valency of that atom.
  • The atomic wide variety of Sulphur is sixteen and the variety of electrons in its valence shell is 6.
  • Thus each oxygen forms two bonds with the Sulphur atom making its valency 4.

Additional Information

  • The valency of nitrogen is 3.
  • Magnesium which has an atomic number 2 has a valency of 2.
  • Elements with valency 1 are those elements that can either gain one electron or lose one electron in order to have a stable electronic configuration.
    Example- Hydrogen

Which of the following intermolecular force is also called as London force? 

  1. Dispersion Forces 
  2. Dipole–Induced Dipole Forces
  3. Dipole - Dipole Forces 
  4. Hydrogen bond forces

Answer (Detailed Solution Below)

Option 1 : Dispersion Forces 

Chemical Bonding and Molecular Structure Question 8 Detailed Solution

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The correct answer is Dispersion Forces.

Key Points

  • It is a type of Van der waal Force.
  • London force also known as dispersion forces is instantaneous dipole–induced dipole force.
  • London force is the weakest intermolecular force among Vanderwaal Forces.

Additional Information

  •  London Dispersion Forces Example
    • The unequal distribution of electrons about the nucleus in an atom can induce some dipole in the atom. When another atom or molecule comes in contact with this induced dipole, it can be distorted which leads to an electrostatic attraction between either atoms or molecules. 
  • Van Der Waals forces-
    • Van Der Waals forces are the interactions between atoms and molecules that result in a pull between them. These forces consist of weak intermolecular interacting with each the nearest possible distance. The molecules do not contain any charge.
  • Types of Vanderwaal Force-
    • London Dispersion Forces
    • dipole induced dipole
    • Hydrogen Bonding

The increasing order of ionic character is:

  1. BeCl2 < MgCl2 < CaCl2 < BaCl2
  2. BeCl2 < MgCl2 < BaCl2 < CaCl2
  3. BeCl2 < BaCl2 < MgCl2 < CaCl2
  4. BaCl2 < CaCl2 < MgCl2 < BeCl2

Answer (Detailed Solution Below)

Option 1 : BeCl2 < MgCl2 < CaCl2 < BaCl2

Chemical Bonding and Molecular Structure Question 9 Detailed Solution

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Concept:

Fajan's Rule:

  • Coulombic attraction between the cations and anions in certain cases leads to the deformation of the ions.
  • This deformation caused by one molecule to the other is called polarisation.
  • The extent to which the molecule is able to polarise the other is called its polarisation power.
  • The extent to which a molecule can get polarised is called its polarisability.
  • The rise in deformity of ions may give rise to increased electron density between the ions and this leads to a considerable amount of covalent bonding.

Explanation:

Factors affecting the Covalent / Ionic character of bonds:

  • The small size of cation:
    • The smaller the size of the cation, the greater is its covalent character.
  • The larger size of the anion: 
    • The larger the size of the anions, the less tightly their electrons will be held by the nucleus and it will be polarised more easily and will thus have more covalent character.
  • Larger charge on either of the two ions:
    • With the increase in charge on the ions, electrostatic attraction of the cation for the outer electrons of the anion also increases.
    • Consequently, the covalent character of the bond increases.
    • For example, the covalent character follows the order:  AlCl3 > MgCl2 > NaCl.
  • ​Electronic configuration of the ions:
    • Out of two ions having the same size and charge, the ion with a pseudo noble gas configuration will have higher polarising power and more covalent character than a cation with a noble gas configuration (i.e., ion having 8 electrons in the outermost shell).
  • Polarisability decreases along with a period and increases along with a group.
  • Low positive charge and large size of cation and a small charge on anion and small size of anion favor the formation of ionic compounds.
  • A high positive charge and small size of cation and a high charge on anion and large size of anion favor the formation of covalent compounds.
  • Among the given compounds BeCl2, BaCl2, MgCl2, CaCl2, the cations have the same charge +2, and anions have a net charge of -2.
  • All the cations Be+2, Ba+2, Mg+2, Ca+2 belong to the same group II of the periodic table.
  • As we move from Be, Mg, Ca, to Ba down the group, the polarisability decreases as well as the covalent character. Hence, the ionic character increases down the group.
  • The most ionic being BaCl2 and the least ionic being BeCl2.

Hence, the correct order of ionic character is: BeCl2 < MgCl2 < CaCl2 < BaCl2.

In Co-ordinate bond, the acceptor atoms must essentially contain in its valency shell an orbital:

  1. With single electron
  2. With no electron
  3. With three electron
  4. With paired electron

Answer (Detailed Solution Below)

Option 2 : With no electron

Chemical Bonding and Molecular Structure Question 10 Detailed Solution

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The correct answer is With no electron.

Key Points

  • In coordination chemistry, a coordinate covalent bond is also known as a dative bond, dipolar bond, or coordinate bond.
  • It is a kind of two-center, two-electron covalent bond in which the two electrons derive from the same atom.
  • The bonding of metal ions to ligands involves this kind of interaction.

Additional Information

  • There are three types of bonds:
    • Ionic bond
    • covalent bond
    • co-ordinate bond
  • Ionic bonds:
    • They are a particular kind of linkage created by the electrostatic attraction of dipolar ions in a chemical molecule.
    • When the valence (outermost) electrons of one atom are permanently transferred to another atom, a bond of this kind is created.
  • Covalent bond:
    • It is formed when two atoms exchange one or more pairs of electrons.
    • The two atomic nuclei are concurrently drawing these electrons to them.
    • When the difference between the electronegativities of two atoms is too tiny for an electron transfer to take place to create ions, a covalent bond is formed.

Which among the following is a covalent compound?

  1. Hydrogen chloride
  2. Magnesium hydroxide
  3. Calcium carbonate
  4. Sodium chloride

Answer (Detailed Solution Below)

Option 1 : Hydrogen chloride

Chemical Bonding and Molecular Structure Question 11 Detailed Solution

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The correct answer is Hydrogen chloride.Key Points

  • Covalent compounds are formed when two or more non-metal atoms share electrons to complete their outermost shell.
  • Hydrogen chloride is a covalent compound as it is formed by the sharing of electrons between hydrogen and chlorine atoms.
  • Magnesium hydroxide and calcium carbonate are both ionic compounds, as they are formed by the transfer of electrons from a metal atom (magnesium or calcium) to a non-metal atom (oxygen or hydroxide).
  • Sodium chloride is also an ionic compound, formed by the transfer of electrons between sodium and chlorine atoms.
  • Covalent compounds generally have lower melting and boiling points than ionic compounds, and are often gases or liquids at room temperature.
  • Hydrogen chloride is an important industrial chemical used in the production of PVC, refrigerants, and other chemicals.

Additional Information

  • Magnesium hydroxide is an important component of antacids and is also used in the production of flame retardants and wastewater treatment.
  • Calcium carbonate is found in many natural substances such as limestone, chalk, and marble, and is also used as a dietary supplement and in the production of paper, plastics, and paints.
  • Sodium chloride, also known as table salt, is commonly used as a seasoning and preservative in food, as well as in the production of chemicals and textiles.

The mode of hybridization of carbon in CO is

  1. sp
  2. sp2
  3. sp3
  4. None of these

Answer (Detailed Solution Below)

Option 1 : sp

Chemical Bonding and Molecular Structure Question 12 Detailed Solution

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Concept:

The geometry of a molecule:

  • The geometry of a molecule depends on the arrangement of bonds about its centre in space.
  • The arrangement further depends on the type of hybridization the centre atom is undergoing.
  • The orientation of the hybrid orbitals is different in different cases.
  • As bonds are formed via overlap of these orbitals, the bonds have directional nature.
  • Therefore, hybridization is directly linked to the geometry of the molecule.

Hybridization and bond angles:

  • According to VSEPR theory, the electron groups arrange themselves around each other so as to minimize repulsion.
  • The electron group includes the bond pairs as well as lone pairs of electrons.
  • If repulsion is more, the energy of the system is raised and the molecule becomes unstable.
  • So, the arrangement in which there is minimum repulsion and maximum attraction is the most stable structure.
  • The arrangement in space gives some angles between the central atom and the bonded atoms which are known as the bond angles.

Few types of hybridizationtheir modes of mixing, and geometry of molecules are-

H number Atomic orbitals Hybridization Geometry
2 S, p sp linear
3 S, p, p Sp2 trigonal planar
4 S, px, pz, py Sp3 tetrahedral
5 S, p, p, p, d Sp3d trigonal bipyramidal
6 S p, p, p,d, d Sp3d2 octahedral
7 S p, p, p, d, d, d Sp3d3 pentagonal bipyramidal


Explanation:

  • In CO, the structure is:
    : C  O :
  • There exist three bonds between carbon and oxygen out of which one is sigma and two are pi.
  • So, two pi bond corresponds to 'sp' hybridization.
  • The sigma bond between the two atoms is made by the head-on overlap of 2p - 2p orbitals of Carbon and Oxygen.
  • The pi bonds are formed by the lateral overlap between p orbitals of carbon and oxygen.


F1 Puja J Anil 24.03.21  D7

Hence, the mode of hybridization of carbon in CO is 'sp'.

Additional Informationsp Hybridisation of carbon:

  • The bond between two C atoms includes two pi and one sigma.
  • A pi bond is formed by pure p-p overlap.
  • One sigma bond exists between each carbon and hydrogen.
  • A total of two sigma bonds means sp hybridization and linear geometry.

sp2 Hybridisation of carbon:

  • The bond between two C atoms includes one pi and one sigma.
  • A pi bond is formed by pure p-p overlap.
  • This type of hybridization is seen in alkenes.
  • The geometry is trigonal planar.

sp3 Hybridisation of carbon:

  • Only sigma bonds are formed by the carbon atoms.
  • The geometry is tetrahedral.

Which one of the following statements is not correct?

  1. All carbons in diamond are linked by a carbon-carbon single bond.
  2. Graphite is a layered structure in which layers are held together by weak van der Waals forces.
  3. Graphite layers are formed by hexagonal rings of carbon atoms.
  4. Graphite layers are held together by carbon-carbon single bonds.

Answer (Detailed Solution Below)

Option 4 : Graphite layers are held together by carbon-carbon single bonds.

Chemical Bonding and Molecular Structure Question 13 Detailed Solution

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Graphite layers are held together by van der Waals forces and not by carbon-carbon single bonds. Hence, 2 is correct and 4 is incorrect.

  • Diamond is a covalent solid with each carbon atom covalently bonded to four other carbon atoms situated around it tetrahedrally, forming four single bonds. Hence, statement 1 is correct.
  • Graphite has a layered structure having rings of six carbon atoms that are arranged in a widely spaced horizontal plane. Hence, 2 and 3 are correct.

Among the following, the one which has maximum ionic character is

  1. NaCI
  2. KCI
  3. LiCI
  4. CsCI

Answer (Detailed Solution Below)

Option 4 : CsCI

Chemical Bonding and Molecular Structure Question 14 Detailed Solution

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Concept:

 According to Fajan's rule:

  • The smaller the cation, the bigger the anion, and the more the covalent character.
  • Bigger cations and smaller anions will favor ionic character.

Explanation:

In the given options all have the same anion (Cl-), the character of the compounds decided by cation only.

Na, K, Li, and Cs belong to I group and their order of atomic size is:

Li < Na < K < Cs.

According to Fajan's rule: Bigger cations and smaller anions will favor ionic character.

So, CsCl has maximum ionic character.

The total numbers of covalent bond in propane are ______.

  1. 7
  2. 12
  3. 8
  4. 10

Answer (Detailed Solution Below)

Option 4 : 10

Chemical Bonding and Molecular Structure Question 15 Detailed Solution

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The correct answer is 10.

Key Points

  • The atoms must be arranged in a precise way that will permit the orbitals to overlap in order for a covalent connection to form.
  • Due to the fact that sigma bonds are more powerful than pi- bonds, it is challenging to break one.
  • Sigma bonds are created by the alignment of atomic orbitals along the axis, and pi bonds are created by the alignment of two atomic orbital lobes.
  • Propane, C3H8, contains 10 sigma bonds altogether, consisting of 2C-C bonds and 8C-H bonds.
  • As seen in the picture, propane contains ten covalent bonds. 
  • chemical-formula-propane-800x800

Additional Information:

  • The three-carbon alkane propane has the chemical formula C3H8.
  • At room temperature and pressure, it is a gas, but it can be compressed into a liquid for transportation.
  • It is a by-product of the processing of natural gas and the refining of petroleum and is frequently utilised as a fuel in home, commercial, and low-emission transportation systems.
  • It was found in 1857 by French scientist Marcellin Berthelot, and by 1911 it was sold commercially in the US.
  • One of a category of liquefied petroleum gases is propane (LP gases). 
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