Chemical Bonding MCQ Quiz in मल्याळम - Objective Question with Answer for Chemical Bonding - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept –

• Molecular orbital is the region in which probability of finding the electrons around all the nuclei constituting the molecule is maximum.
• Thus, they are polycentric and the identity of the atom is completely lost.

Explanation –

•  The wave function of symmetric bonding molecular orbital is denoted by Ψ_b for a diatomic molecule is as follows-

​Ψ_b = Ψ_A + Ψ_B

• The wavefunction of asymmetric antibonding molecular orbital is denoted by Ψ_b for a diatomic molecule is as follows-

Ψ_a = Ψ_A - Ψ_B

• Since the probability of the occurrence of an electron between the nuclei is proportional to Ψ². Therefore the charge density of the antibonding molecular orbital is given by-

Ψ_-² = (Ψ_A - Ψ_B)² = Ψ_A² +Ψ_B² – 2 Ψ_AΨ_B

Conclusion-

the charge density of the antibonding molecular orbital is given by-

Ψ_-² =Ψ_A² +Ψ_B² – 2 Ψ_A Ψ_B

Hence, the correct option is option D.

Concept:

→ More electronegative elements attract their valence shell electrons more strongly, so their AOs lie at lower energy than those of less electronegative elements

→ In a simple overlap of two AOs, the resulting bonding and antibonding MOs each resemble most the constituent AO nearest to it in energy

→ This means that the bonding MO most resembles the AO of the more electronegative partner

→ Bonding MOs are said to be polarised towards the more electronegative partner atom:   since bonding MOs are filled and antibonding MOs tend to be empty, this amounts to the classical definition of electronegativity

→ An antibonding MO in a simple overlap is composed of those amounts of AOs not used in making the bonding orbital, so the antibonding MO most resembles the AO of the less electronegative partner.   We say that antibonding MOs are polarised towards the less electronegative partner

Explanation:

→ Two atoms come together and interact to form a bond. 

→  The atomic orbitals mix together to form a new orbital.  The new orbitals formed are called molecular orbitals. 

→ Molecular orbitals are formed by atomic orbitals having the same energy levels and symmetry.  When molecular orbits are formed, atomic orbitals lose their identity.

→ The highest occupied molecular orbital of CO is a molecular orbital that puts significant electron density on the carbon atom.  The lowest unoccupied molecular orbital of CO is pi orbital.  Non-boding molecular orbitals of CO molecules have a slight bonding character.

• Molecular orbitals of isoelectronic molecules of N₂, CO. Electron density maps are drawn in a common scale. Atomic positions are from the left to the right, N₂, CO respectively.
• 1- 3 are occupied molecular orbitals, and 3 is the highest occupied molecular orbitals HOMO.
• 4-6 are unoccupied molecular orbitals, and 4 is the lowest unoccupied molecular orbitals LUMO.

• The electronic structure of the carbon monoxide (CO) molecule can be understood by analyzing its molecular orbitals, specifically its core and valence MOs that are presented in the above Figure.
• The calculated dipole moment contributions from each MO show that without considering the HOMO, CO would have a calculated dipole moment of μ = −8.03 D.
• The total dipole moment is calculated to be μ = 0.18 D, fairly aligning with the experimentally observed value of μ = 0.11 D, suggesting the simulations are accurately modeling reality.
• The calculation correctly identifies the direction of the dipole moment, with the carbon atom identified as the negative end, resulting from the 1σ and 2σ core orbitals having large dipole moments as electrons are localized at the oxygen and carbon atoms.
• All valence orbitals, apart from 7σ HOMO, display negative dipole moments, implying the electronic charge is distorted such that the negative end of the dipole moment is at the oxygen atom.
• The strong dipole moment component of the lone-pair HOMO (μ = 8.21 D) shifts the total dipole moment to have the negative end at the carbon atom, with the inspection of the 7σ HOMO reflecting a large charge concentration at the carbon atom positioned away from the oxygen.

Conclusion: Thus, due to polarization of charge equally on both the atom results into the formation of the dipole. Hence, the correct answer is option 2.

Concept:

The bond angle in a molecule is influenced by several factors, including the size of the substituents, the electronegativity of the atoms, and steric hindrance. Larger or more electronegative groups tend to increase bond angles as they repel adjacent atoms more strongly, while smaller or less electronegative groups allow for smaller bond angles.

Factors affecting bond angle:

• Electronegativity: More electronegative substituents tend to increase bond angles due to stronger repulsion of bonding electrons.

• Steric Effects: Larger groups create greater steric hindrance, increasing the bond angle to minimize crowding.

• Size of Substituents: Bulkier groups tend to increase bond angles as the atoms need to be farther apart to reduce repulsion.

Explanation:

In the molecules ClSO₂CH₃, ClSO₂CF₃, and ClSO₂CCl₃, the trend in bond angles is primarily influenced by the size and electronegativity of the substituents (CH₃, CF₃, and CCl₃).

• ClSO₂CH₃: The methyl group (CH₃) is smaller and less electronegative compared to the other groups, so it causes the smallest bond angle.

• ClSO₂CF₃: The trifluoromethyl group (CF₃) is highly electronegative, increasing the bond angle due to electron repulsion.

• ClSO₂CCl₃: The trichloromethyl group (CCl₃) is larger and causes significant steric repulsion, leading to the largest bond angle.

• 

• The electron-withdrawing groups will pull electron density away from the sulfur, reducing the repulsion between the lone pairs and the Cl–S and C–S bonds. This will allow for a smaller Cl–S–C bond angle.

• In order of increasing electron-withdrawing ability: CH₃ < CCl₃ < CF₃

  • Therefore, the Cl–S–C bond angle will be smallest in ClSO₂CF₃ and largest in ClSO₂CH₃.

• So, the correct trend is:

  • ClSO₂CH₃ > ClSO₂CCl₃ > ClSO₂CF₃

Conclusion:

The correct trend in bond angles is: ClSO₂CH₃ > ClSO₂CCl₃ > ClSO₂CF₃ (Option 4).

The correct answer is In CH3NCO, resonance involving nitrogen leads to a bond angle slightly above 120°, close to 125°. Conversely, SiH3NCO exhibits a bond angle at nitrogen of approximately 180° due to effective pπ-dπ back bonding facilitated by silicon's vacant d orbitals.

Concept:-

Explanation:-

This compound has Nitrogen as sp2 hybridized. Thus bond angle is expected to be 120°. But it is slightly greater than 120° because of the minor contributing resonance structure:

The minor contributing resonance structure has N as sp hybridized. Hence bond angle of the overall structure is found to be about 125°.

For SiH₃NCO, the structure is found to be planar because the lone pair of electrons on N are donated to the vacant d orbitals of Si through back bonding (pπ-dπ overlap).

Clearly, N is sp hybridized and so bond angle is 180°.

Conclusion:-

So, the correct statement is, CH3NCO, resonance involving nitrogen leads to a bond angle slightly above 120°, close to 125°. Conversely, SiH3NCO exhibits a bond angle at nitrogen of approximately 180° due to effective pπ-dπ back bonding facilitated by silicon's vacant d orbitals.

The answer is See-saw and square pyramidal

Concept:

The VSEPR theory identifies the molecular geometry by considering electron pair repulsions around the central atom. The number of bonding pairs versus lone pairs significantly influences the resultant shape.

Explanation:

SF₄:

For sulfur tetrafluoride (SF₄), sulfur is the central atom with four bonding pairs from fluorine atoms and one lone pair.

H = 1/2 [V + M - C + A] H = 1/2 [6 (S's valence electrons) + 4 (bonding F atoms) - 0 (charge) + 0 (additional)] = 5

An AX₄E molecule (4 bond pairs and 1 lone pair) adopts a see-saw geometry according to VSEPR theory.

BrF₅:

Bromine pentafluoride (BrF₅) involves a central bromine atom with five bonding pairs and one lone pair, making six regions of electron density.

H = 1/2 [V + M - C + A] H = 1/2 [7 (Br's valence electrons) + 5 (bonding F atoms) - 0 (charge) + 0 (additional)] = 6

An AX₅E molecule (5 bond pairs and 1 lone pair) adopts a square pyramidal geometry according to VSEPR theory.

Conclusion:

Hence, according to VSEPR theory, the shape of SF₄ and BrF₅, respectively, are See-saw and square pyramidal.

Concept:-

VSEPR theory:

Valence Shell Electron Pair theory is called as VSEPR theory.

• VSEPR theory is used to predict the shapes of the molecule based on the valence shell electron pairs of the central metal atom.
• According to VSEPR theory, there is always repulsion in the valence shell electrons of all the atoms and these atoms arrange themselves in a manner to minimize this repulsion.
• This arrangement of atoms in space is responsible for the geometry and shape of the molecule.

Prediction of the shape of the molecule -

• The least electronegative atom is set as the central atom.
• Count the valence shell electron of the central atom.
• Find out the number of hybridization based on the number of orbitals participating in bonding.
• Give the geometry of the molecule based on the hybridization.
• Count the number of bond pairs and lone pairs in the molecule.
• The idea of lone pairs gives the correct shape of the molecule.

Explanation:-

(A) H2O:

• The central atom is 'O'.
• The Outer cell electronic configuration of O is (Z =8)

= 2s2 2p4.

• It is surrounded by two 'O' atoms.
• In the case of H2O, the central atom O contributes 6 electrons to the hybridization and the two 'H' atoms contribute a total of 2 electrons.
• Total number of electrons available for bonding

​= 6 + 2

= 8

• The total number of electron pairs is 4, out of which there are 2 bonding pair and 2 lone pair.
• Hybridisation is sp3.
• Thus, the geometry of the molecule is Tetrahedral.
• As the shape of the molecule can be predicted by excluding the lone pairs on the central atom and the Number of bonding pairs of electrons is 2, thus the shape of H₂O is "Bent".
• The geometry of  H2O is shown below:

• Thus, statement A is incorrect.

(B) NO₂^- :

• The central atom is 'N'.
• The Outer cell electronic configuration of N is (Z =7)

= 2s2 2p3.

• It is surrounded by two 'O' atoms.
• In the case of NO2- ion, the central atom N contributes 5 electrons to the hybridization, the two 'O' atoms contribute a total of 4 electrons and one negative charge contributes one electron.
• Total number of electrons available for bonding

​= 5 + 4 + 1

= 10​

• Hybridization is sp2.
• Thus, the geometry of the molecule is trigonal planar, but the shape is bent-shape.

• Thus, statement B is incorrect.

(C) SF4 :

• The central atom is 'S'.
• The Outer cell electronic configuration of S is (Z =16)

= 3s2 3p4.

• It is surrounded by four 'F' atoms.
• In the case of SF4, the central atom S contributes 6 electrons to the hybridization and the four 'F' atoms contribute a total of 4 electrons.
• Total number of electrons available for bonding

​= 6 + 4

= 10

• The total number of electron pairs is 5, out of which there are 4 bonding pair and 1 lone pair.
• The number of bonding pairs of electrons = 4 and

the number of lone pairs of electrons = 1.

• Hybridisation is sp3d.
• Thus, the geometry of the molecule is Trigonal bipyramidal.
• As the shape of the molecule can be predicted by excluding the lone pairs on the central atom and the Number of bonding pairs of electrons is 4, thus the shape of SF4 is "see-saw".
• The shape of SF4 is shown below:

• Thus, statement C is correct.

(D) O3 :

• Central atom "O' is surrounded by other two O atoms.
• Electronic configuration of O(8) is  1s2 2s2 2p4.
• The total valence electron in O is 6.
• Hybridization in O3 is  sp2 and the shape of the molecule is bent shaped.

• Thus, statement D is incorrect.

Conclusion:-

• Hence, the incorrect statement/(s) from the following statements are A, B, and D only

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 the 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.

A few types of hybridization, their modes of mixing, and the geometry of molecules are-

• The schematic representation of dπ-pπ bonding is shown in the picture below:​

Explanation:

• dπ-pπ bonding is a special type of bonding where the p-orbital of one atom and the d-orbital of other atom take part in the π-bond formation.
• There must be vacant d-orbitals present in one of the elements.
• The π-bond is formed by sidewise overlapping of the orbitals.
• In the case of H3C-N=C=S molecule dπ-pπ bonding is not possible due to the unavailability of vacant d orbitals in the C atom.

• While, in the case of H3Si-N=C=S molecule dπ-pπ bonding is possible due to the presence of vacant d orbitals in the Si atom.
• The schematic representation of dπ-pπ bonding is shown in the picture below:

• Thus, the structures of H3C-N=C=S and H3Si-N=C=S respectively, are bent and linear.

​

Conclusion:-

• Hence, the structures of H3C-N=C=S and H3Si-N=C=S respectively, are bent and linear.

Concept:

• The shape of all the given molecules are open book shape.
• This shape is due to the hybridization in particular compounds.

Explanation:

Structure and shape of O₂F₂ molecule:

In O2F2 molecule, the O-atom undergoes sp³ hybridization which results in the tetrahedral geometry and therefore the shape is open book as below:

O₈ = 1s², 2s² 2p⁴

F₉ = 1s², 2s² 2p⁵

Structure and shape of H₂O₂ molecule:

Similar to the O2F2 molecule, in H2O2 there is sp³ hybridization that results in the open book structure as below:

O₈ = 1s², 2s² 2p⁴

H₁ = 1s¹

Structure and shape of S₂Cl₂ molecule: open book structure

S₁₆ = 1s², 2s² 2p⁶, 3s² 3p⁴

Cl₁₇ = 1s², 2s² 2p⁶, 3s² 3p⁵

Structure and shape of p-dimethoxy benzene: open book structure

Therefore, the shape of all the molecules given are open book and similar.

Conclusion:

Hence the correct answer is option 4.

Concept:

• Nitrosonium ion (cationic diatomic species) contains triple bond between N and O.
• It is considered more stable than nitric oxide.
• It can be obtained from nitriles on reaction with strong acids.
• weak nitrous acids also give nitrosonium ion on protonation by releasing water.
• Nitrosonium ion is used mainly for the formation of diazonium ion.

Explanation:

Molecular Orbital Diagram (MOT) of NO⁺:

(total number of valence electron = 5+6-1 =10)

LUMO (lowest unoccupied orbital) = \(\pi^*_{2px, 2py}\)

HOMO (Highest occupied orbital)  = \(\sigma_{2pz}\)

The incoming electron will occupy lowest unoccupied molecular orbital, which is \(\pi^*\) here.

Conclusion:

Hence, electron will be added first to the \(\pi^*\) orbital to change from NO⁺ to NO.

Concept:

• Atomic orbitals combines to form set of molecular orbitals. Linear combinations of two atomic orbitals gives set of two molecular orbitals. The combination can be addition as well as subtraction.
• Addition of atomic orbitals corresponds to constructive interference giving bonding molecular orbital, while, subtraction of atomic orbitals represents destructive interference giving antibonding molecular orbital.
• Bonding orbitals are more stable and have lower energy than antibonding orbital.

Explanation:

Combination of homonuclear diatomic molecules can be understood by considering few examples.

lets consider H₂ molecule.

electronic configuration of H = 1s¹

Molecular Orbital Diagram of H₂

gerade \(\sigma\) orbital is forming the bonding forming the bonding orbital

Conclusion:

Hence, the bonding orbital is σg of lowest energy in homonuclear diatomic molecule.