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

CONCEPT:

Xenon Fluoride Reactions and Formation of [Xe₂F₃]⁺

• Xenon (Xe) reacts with fluorine (F₂) in a 1:1 molar ratio under sunlight to form xenon difluoride (XeF₂), a linear molecule with 3 lone pairs on Xe.
• XeF₂ is a stable noble gas compound, which can further react with strong Lewis acids like AsF₅.
• When 2 equivalents of XeF₂ react with 1 equivalent of AsF₅, an ionic compound is formed:

  2 XeF₂ + AsF₅ → [Xe₂F₃]⁺ + [AsF₆]⁻

• The cation [Xe₂F₃]⁺ contains two Xe atoms connected via a bridging fluorine atom, forming a linear or angular structure.
• To calculate the total number of lone pairs on the cation, we need to consider all Xe and F atoms' lone pairs in [Xe₂F₃]⁺.

EXPLANATION:

  
 F–Xe–F–Xe–F. 

• Each Xe atom in XeF₂ typically has 3 lone pairs. In the cation [Xe₂F₃]⁺, both Xe atoms still retain 3 lone pairs due to their stable noble gas configuration:
  • Lone pairs on Xe (left) = 3
  • Lone pairs on Xe (right) = 3
• There are 2 terminal fluorine atoms and 1 bridging fluorine atom:
  • Each terminal F has 3 lone pairs → 2 × 3 = 6
  • Bridging F has 2 lone pairs (1 pair involved in bridging bond)
• Adding up all lone pairs in the cation [Xe₂F₃]⁺:
  • Xe atoms = 3 + 3 = 6
  • F (terminal) = 6
  • F (bridging) = 2
  • Total lone pairs = 6 + 6 + 2 = 14

Therefore, the total number of lone pairs present on the cation [Xe₂F₃]⁺ is 14.

CONCEPT:

Hybridization, Molecular Geometry, and Dipole Moment

• Hybridization describes the mixing of atomic orbitals on the central atom to form new hybrid orbitals in a molecule.
• The presence of lone pairs affects molecular geometry and the resultant dipole moment.
• Dipole moment depends on the geometry and the difference in electronegativities; molecules with symmetrical geometry or lone pairs arranged symmetrically often have zero dipole moment.
• Hybridizations considered here are sp², sp³, sp³d, and sp³d² corresponding to trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral geometries, respectively.
• Lone pairs on the central atom create asymmetry, often leading to non-zero dipole moments.

EXPLANATION:

Molecule	Hybridization	Dipole Moment	Lone Pairs on Central Atom
SO2	sp²	Non-zero	1
NF3	sp³	Non-zero	1
NH3	sp³	Non-zero	1
XeF2	sp³d	Zero	3
ClF3	sp³d	Non-zero	2
SF4	sp³d	Non-zero	1

• Among molecules with non-zero dipole moment, ClF₃ has the highest number of lone pairs (2) on the central atom.
• Its hybridization is sp³d.

Therefore, the correct answer is: sp³d (Option 4).

CONCEPT:

Polarity of B—C Bonds: Role of Carbon Hybridization

• The polarity of a B—C bond depends on the electronegativity difference between B and the C atom directly attached to it.
• Carbon's electronegativity increases with greater s-character:
  • sp (50% s) > sp² (33% s) > sp³ (25% s)
• Electronegativity difference (ΔEN) ∝ Bond polarity. So, the more electronegative the carbon, the more polar the B—C bond becomes.

EXPLANATION:

• P: B(C≡CH)₃ → Carbon attached to B is sp hybridized → highest electronegativity → highest polarity
• Q: B(CH=CH₂)₃ → Carbon attached to B is sp² → moderate electronegativity → moderate polarity
• S: B(CH₂C₆H₅)₃ → Carbon is sp³ but phenyl ring has mild electron-withdrawing resonance
• R: B(CH₂CH₃)₃ → Carbon is sp³ → lowest electronegativity → lowest polarity

Order of hybridization-based electronegativity (and hence B—C bond polarity):

sp (P) > sp² (Q) > sp³ with benzyl (S) > sp³ alkyl (R)

Therefore, the correct order of bond polarity is: R < S < Q < P.

The correct answer is Chlorine has the highest electron affinity, while fluorine has the highest electronegativity..

Key Points

• Chlorine has the highest electron affinity due to its optimal atomic size, which allows it to efficiently gain an electron and release energy in the process.
• Fluorine has the highest electronegativity (value of 3.98 on the Pauling scale), as it has a strong tendency to attract shared electrons in a chemical bond due to its small size and high nuclear charge.
• Electron affinity refers to the energy released when a neutral atom in the gas phase gains an electron to form a negative ion.
• Electronegativity, on the other hand, measures the ability of an atom to attract shared electrons in a chemical bond.
• The difference arises because fluorine's small size causes strong electron-electron repulsions, slightly reducing its electron affinity compared to chlorine.

Additional Information

• Electron Affinity:
  • The trend of electron affinity generally increases across a period and decreases down a group in the periodic table.
  • Chlorine's electron affinity is approximately 349 kJ/mol, the highest among all elements.
  • Fluorine's electron affinity is slightly lower at about 328 kJ/mol due to increased electron repulsion in its compact atomic structure.
• Electronegativity:
  • Electronegativity increases across a period and decreases down a group, following the periodic trend.
  • Fluorine, being the most electronegative element, plays a key role in various chemical reactions, including the formation of polar covalent bonds.
• Electron-Electron Repulsion:
  • Smaller atoms like fluorine experience higher repulsion among valence electrons, which can slightly reduce the energy released during electron addition.
  • This is why chlorine, despite being larger, has a higher electron affinity than fluorine.
• Applications in Chemistry:
  • Electronegativity is crucial in determining bond polarity and molecular geometry.
  • Electron affinity values are used to predict the likelihood of an atom forming negative ions in reactions.

Concept:

Molecular Orbitals in HCl Molecule

• In a diatomic molecule like HCl, molecular orbitals are formed by the linear combination of atomic orbitals (LCAO) from the individual atoms involved in bonding.
• Hydrogen has the 1s atomic orbital, while chlorine has several orbitals available for bonding. The most significant overlap occurs between the hydrogen 1s orbital and a chlorine orbital from the valence shell.
• The best combination for forming the molecular orbitals in HCl is the hydrogen 1s orbital and the chlorine 3px orbital. This is because the 3px orbital of chlorine is the one that has the proper symmetry to interact with the 1s orbital of hydrogen.

Explanation:

• The chlorine atom has a 3p orbital that interacts effectively with the 1s orbital of the hydrogen atom. This combination leads to the formation of molecular orbitals with bonding and antibonding character.
• The other options involving 1s, 2s, or 3s orbitals from chlorine would lead to less effective overlap compared to the 3px orbital, and thus are less likely to form strong molecular bonds in the HCl molecule.

Therefore, the correct answer is: 1s (H), 3px (Cl).

Explanation:- 

• Allotropes of carbon like Diamond, Graphite, and Fullerene show crystalline structures.
• Fullerene has infinite lattices of Diamond and Graphite.
• It is formed by nanotubes forming the discrete molecular structure.

Additional Information 

• Fullerene molecule consists of carbon atoms connected by single and double bonds.
• Common structures are C₆₀ and C_70.

Key Points

• The element that can form both a double bond and a single bond with a carbon atom is oxygen (O).
  • Oxygen can form a double bond with carbon by sharing two electrons and a single bond by sharing one electron.
• Fluorine (F) cannot form a single bond with carbon as it requires only one electron to complete its octet and carbon requires four electrons to complete its octet.
  • Therefore, fluorine can only form a single bond with carbon.
• Chlorine (Cl) and bromine (Br) can form a single bond with carbon but cannot form a double bond as they require two electrons to complete their octet and carbon requires four electrons to complete its octet.
• In addition to oxygen, nitrogen (N) can also form both a double bond and a single bond with carbon.
  • However, nitrogen requires three electrons to complete its octet and carbon requires four electrons to complete its octet.
  • Therefore, nitrogen can only form a triple bond or a single bond with carbon.

Additional Information

• It is important to note that the ability of an element to form a double or single bond with carbon depends on the number of valence electrons it has and its electronegativity.
• Fluorine (F) is the most electronegative element and can form only a single bond with carbon.
• Chlorine (Cl) and bromine (Br) have lower electronegativity than oxygen (O) and nitrogen (N) and can form a single bond with carbon.
• Oxygen (O) and nitrogen (N) are both highly electronegative elements and can form both single and double bonds with carbon.

Concept:

• The molecular structure of a molecule is the arrangement of surrounding atoms around the central atom.
• The molecular structure can be easily predicted by calculating the steric number of the molecule which is the total number of bond and lone pairs.
• Each steric number belongs to particular hybridization which further helps in  structure of molecule.

Explanation:

Structure of given molecular ions/species:

1. BrF₃

Steric number = 5

hybridization = sp³d (trigonal bipyramidal)

structure = T-shaped

2. FClO₂

steric number = 4

Hybridization = sp³

Structure = Trigonal pyramidal

3.  [XeF5]^-

steric number = 5

hybridization = sp³d³

Shape = Pentagonal planer

4. [ClF4]^-

steric number = 6

Hybridization = sp³d²

Structure = square planar

5. XeO₃

steric number = 4

hybridization = sp3

structure = trigonal pyramidal

Among given molecular species, [ClF4]-,  and [XeF5]^- have planar structure/shape. 

Conclusion:

Hence, the following set of ionic species will have planar geometry [ClF4]-,and [XeF5]^-

Explanation:

• All the three p orbitals of methylazide (\(C{H_3} - N = {N^ + } = {N^ - }\)) which form π-bonds of one carbon atom and two nitrogen atoms should lie in the same plane to have the effective overlap.
• In methylazide one negative charge and a conjugated double bond will contribute a total of 4 π-electrons.   
• From the MO theory, the three π atomic orbitals should result in 3 π molecular orbitals and they are arranged according to their energy content as shown below: \({\Psi _1}{\Psi _2}{\Psi _3}\) 

• The ground state (G.S) configuration of methylazide is \({\Psi _1}{\Psi _2}{\Psi _3}\), where \({\Psi _2}\) is the highest occupied molecular orbital (HOMO) of methylazide. It has one nodal plane and the Nodal plane passes through the N atom in the HOMO of methylazide.

Conclusion:

• The HOMO of π - molecular orbitals of methylazide is

Explanation:- 

• Iodine trichloride is a trigonal bipyramidal structure with three bonding pairs and two lone pairs of electrons.

 Additional Information

• ICl₃ is a polar molecule.
• Due to lone pair to lone pair repulsion, the electronegativity of Iodine Trichloride is high.
• ICl₃ acts as a dimer.
• The decomposition of ICl₃ gives Cl₂ gas.
• Being a polar molecule ICl₃ is a good conductor of electricity.

Concept:

VSEPR THEORY: 

• One can determine the shapes of a molecule based on the Valence shell electron repulsion theory(VSEPR Theory).
• According to the VSEPR Theory " electron pairs present in the molecule repeal with each other so that they will arrange the atoms in such way that, in which there is minimum repulsion between the electron pair."
• In order to find the shape of the molecule, the correct Lewis structure has to draw.

VSEPR Theory (Molecular Shape) 

A = Central atom, X = atom bonded to A, E = lone pair on A

Explanation:

• Xe belongs to group 18, hence it has 8 valence electrons.
• F belongs to group 17 has 7 valence electrons.
• So, here in the case of the molecule [XeF₅]^-, Xe is the central atom and the negative charge associated with the molecule is included in the valence shell of the central atom so that Xe will have 9 electrons for bonding.
• In [XeF5]-,there are five Xe-F single bonds so that five of the nine valence electrons of Xe are used for bonding. 
• So the central atom Xe have five bonding pair of electrons and two lone pair of electrons with pentagonal bipyramidal geometry. 
• But according to VSEPR Theory, due to the presence of two lone pairs, placed in an axial position in order to minimize the repulsion, the molecule  [XeF5]-  will adopt a pentagonal planar shape. 

The answer is Trigonal bipyramidal

Concept:

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 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 hybridization, their modes of mixing, and geometry of molecules are-

Explanation:-

• The geometry around Te in the symmetrical trimeric species of [TeO2F]- is Psedo Trigonal bipyramidal.

​

Conclusion:-

• Hence, the geometry around Te in the symmetrical trimeric species of [TeO2F]- is Trigonal bipyramidal.

Concept:

Bond dissociation energy -

• It is the amount of energy required to break a chemical bond between two atoms.
• Bond dissociation energy is the measure of the strength of the bond.
• The more the bond dissociation energy stronger will be the bond and vice-versa.

Factor affecting bond dissociation energy -

1. Atomic size - As the atomic size of the bonded atom increases bond length increases and less amount of energy is required to break this bond. Hence, bond dissociation energy decreases with atomic size.
2. Bond multiplicity - Bond dissociation energy increases as the multiplicity of the bond increases. For dissociation energy follow the order - Triple bond > Double bond > single bond.
3. Hybridization - More the number of hybrid orbitals lesser will the bond dissociation energy.
4. Electronegativity - The higher the electronegative difference between the bonded atom, the higher will be the bond strength, and hence, the higher will be the value of bond dissociation energy.

Explanation: -

We know that the more the s-character in the hybridization more will be the bond strength because the s orbital have more penetration than other orbitals hence making hybrid orbitals more acidic.

Let's check the percentage of s-character in all the given compounds: -

• HC ≡ C - H
  • ​We know that the triple-bonded carbon is sp hybridized.
  • 
  • Thus, the s-character is 50%
•   In benzene, we know all the carbons of the benzene ring are sp2 hybridized.

• Thus, the s-character is 33.33%
• In the case of 1,3-cyclopentadiene, the four carbons are sp2 hybrid and 1 is sp3 hybrid.

• Hence, the s-character of sp3 carbon is 25%.

In the case of cyclopropene, the asked carbon is also sp3 hybridized.

• But, there is high angular tension in cyclopropene. To reduce this angular strain the carbon in 3 member ring bends its bond by reducing the s-character from the sigma bonds of the ring and shifting it to the C-H bond, thus S-character is more than the usual sp3 hybridization in the C-H bond of cyclopropene.

• Further due to this angular stain, loses one of its hydrogens to become carbonation which is stabilized by resonance.

Thus, the order of s-character is Ethyne> Benzene > Cyclopropene > cyclopentadiene.

Conclusion:

We know bond strength is directly proportional to the s-character.

So, the correct match is a – iv; b – i; c – ii; d – iii

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.

• Few types of hybridization, their modes of mixing, and geometry of molecules are-

Concept:

In the given reactions, AsF₅ and ClF₃ react with HF to form cations and anions. The reaction involves the autoionization of HF, which plays a key role in these processes:

Autoionization of HF: HF can autoionize to form H₂F⁺ and F^-, enabling it to act as both a proton donor and acceptor in reactions:

\(2HF \leftrightarrow H_2F^+ + F^-\) 

• In the presence of strong Lewis acids like AsF₅ and ClF₃, HF autoionizes, and the fluoride anion coordinates with these acids to form complex cations and anions.

Explanation:

AsF5+2HF⇌[H2F]+(P)+[AsF6]−(Q)" id="MathJax-Element-11-Frame" role="presentation" style="position: relative;" tabindex="0">AsF5+2HF⇌[H2F]+(P)+[AsF6]−(Q)AsF5+2HF⇌[H2F]+(P)+[AsF6]−(Q)" id="MathJax-Element-73-Frame" role="presentation" style="position: relative;" tabindex="0">AsF5+2HF⇌[H2F]+(P)+[AsF6]−(Q)AsF5+2HF⇌[H2F]+(P)+[AsF6]−(Q) AsF5+2HF⇌[H2F]+(P)+[AsF6]−(Q) $\rm AsF_5 + 2HF \rightleftharpoons [H_2F]^+(P) + [AsF_6]^-(Q)$

• Statement A (P is bent): The cation [H₂F]⁺ has a bent geometry.

  • ​

• Statement B (Q is octahedral): The anion [AsF₆]^- has a structure resembling an octahedral arrangement.

  • ​

\(\rm CIF_3 + 2HF \rightleftharpoons [ClF_2]^+(R) + [HF_2]^- (S)\)

• Statement C (R is bent): The cation [ClF₂]⁺ has a bent geometry.

  • ​

• Statement D (S is linear): The anion [HF₂]^- has a linear geometry.

  • ​

Conclusion:

Since all statements A, B, C, and D are correct, the correct option is option 1.