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

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.

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.
  
• B. XeF₂: Xenon is sp³d hybridized with three lone pairs and two bonded fluorine atoms. This results in a linear geometry.
  
• 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.
  
• D. XeF₆: Xenon is sp³d³ hybridized with one lone pair and six bonded fluorine atoms. This results in a distorted octahedral geometry.
  

• 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.

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.

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.
      
  • 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.

CONCEPT:

Bonding in Allyl Cyanide

• Allyl cyanide (CH₂=CH-CH₂-C≡N) consists of a combination of single, double, and triple bonds.
• Sigma (σ) bonds are single covalent bonds formed by the head-on overlap of atomic orbitals.
• Pi (π) bonds are formed by the side-by-side overlap of p orbitals and are present in double and triple bonds.
• Lone pairs are pairs of valence electrons not involved in bonding.

EXPLANATION:

• To determine the number of sigma and pi bonds and lone pairs in allyl cyanide:
  • Structure of Allyl Cyanide: CH₂=CH-CH₂-C≡N
  • Counting the bonds:
    • CH₂=CH- (1 C=C double bond: 1 sigma + 1 pi)
    • -CH₂- (2 C-H single bonds: 2 sigma, 1 C-C single bond: 1 sigma)
    • -C≡N (1 C≡N triple bond: 1 sigma + 2 pi)
    • -C-H (2 C-H single bonds: 2 sigma)
  • Total bonds:
    • Sigma bonds: 1 (C=C) + 2 (C-H) + 1 (C-C) + 1 (C≡N) + 2 (C-H) = 7 sigma bonds
    • Pi bonds: 1 (C=C) + 2 (C≡N) = 3 pi bonds
    • Lone pairs: 1 lone pair on nitrogen

Therefore, the allyl cyanide molecule contains 9 sigma bonds, 3 pi bonds, and one lone pair, making Option 2 the correct answer.

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 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

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.

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.

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.

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 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. 
• 

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). 

The correct answer is option 3.

Concept:

Ionic Compound:

• Ionic compounds are usually formed when metals react with non-metals.
• In other words, ionic compounds are held together by ionic bonds and are classed as ionic compounds.
• Elements can gain or lose electrons in order to attain their nearest noble gas configuration.
• The formation of ions (either by gaining or losing electrons) for the completion of octets helps them gain stability.
• In a reaction between metals and non-metals, metals generally lose electrons to complete their octet while non-metals gain electrons to complete their octet.
• Metals and non-metals generally react to form ionic compounds.
• The structure of an ionic compound depends on the relative sizes of the cations and anions.
• Ionic compounds include salts, oxides, hydroxides, sulfides, and the majority of inorganic compounds.
• Ionic solids are held together by the electrostatic attraction between the positive and negative ions.

Explanation:

• Carbon tetrachloride, sometimes known as CCl4, is not an ionic substance. Because non-metals (carbon and chlorine) share electrons rather than transfer them, it is a covalent compound, also referred to as a molecular compound.

So, out of the compounds you listed, CCl₄ (option 3) is the only one that is not an ionic compound.

Content:

The geometry of a molecule:

• The geometry of a molecule depends on the arrangement of bonds about its center in space.
• The arrangement further depends on the type of hybridization the center 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 are 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.
• There exist anomalies in predicted bond angles when there is the presence of lone pair electrons.

Explanation:

H2O:

• In water, the center atom oxygen is sp3 hybridized.
• Out of the four bond pairs, two are bond pairs and the other two are lone pairs.
• Due to the presence of lone electrons, there is repulsion between bond pairs and lone pairs.
• The repulsion between l.p is > b.p and thus pushes the two bonds to come closer decreasing the bond angle from ideal 109.8°.

H2S:

• In water, the center atom oxygen is sp3 hybridized.
• Out of the three bond pairs, two are bond pairs and the other two are lone pairs.
• Due to the presence of lone electrons, there is repulsion between bond pairs and lone pairs.
• The repulsion between l.p is > b.p and thus pushes the two bonds to come closer decreasing the bond angle from ideal 109.8°.
• As sulfur is a larger molecule than oxygen, its electron density is more dispersed. It is more easily distorted by the repulsion from the lone pair of electrons and the bond pairs come closer to each other.
• This decreases the bond angle more than in water.
• The bond angle in H₂S is 92.1⁰.

NH3:

• In ammonia, the center atom oxygen is sp3 hybridized.
• Out of the three bond pairs, three are bond pairs and the other is lone pair.
• Due to the presence of lone electrons, there is repulsion between bond pairs and lone pairs.
• The repulsion between l.p > b.p and thus pushes the two bonds to come closer decreasing the bond angle from ideal 109.8° to 107^o.

​

CH4

• CH4 molecule has sp3 hybridization.
• Each of these 4 sp3 hybrid orbitals overlaps with one 1s atomic orbital of hydrogen.
• The arrangement of the atoms gives tetrahedral geometry.
• There is no lone pair and the bond angle is ideal 109.5 °.

Hence, the minimum bond angle is present in H2S.

Important Points

• Distortion in shape occurs when there is presence of lone pair of electrons as the table shows:

Isoelectronic species :

• Those species having the same number of electrons but differ in nuclear charge are called isoelectronic species.
• Example: Mg2+, Na+, F–, and O2–  have 10 electrons each so they are isoelectronic.
• But, their radius is different due to differences in nuclear charges. 
• Ionic radius decreases along a period for Isoelectronic ions.

​Isostructural species:

• The species which have different atoms or elements in them but have the same hybridization and structure are called isostructural species.
• Examples are NF3 and NH3, they have both sp3 hybridization and pyramidal shape.

Explanation:

• The species and their structures and number of electrons are given below:

• From the table above we see that CH3-, NH3, H3O+ all have 10 electrons in them.
• Hence, they are isoelectronic species.

Thus the correct option is II, III, IV.