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

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.

The correct answer is Carbon.

Key Points

• Valency is the capacity of an atom to form covalent bonds with other atoms.
• In contrast, valence electrons are the number of electrons that are needed for a complete outer shell in a compound.
• Carbon has 4 electrons in its outer shell, thus needing four more for completion.
  • As a result, its valency and number of valence electrons are the same.

Additional Information

• The valency of Nitrogen is 3 and it has a total of 5 valence electrons.
• The valency of Sulphur is 2 and it has a total of 6 valence electrons.
• The valency of Oxygen is 2 and it has a total of 6 valence electrons.

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 polarization.
• 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:

• A) For a compound to have an ionic bond, a low positive charge on the cation and a large cation radius are preferred. This results in the cation having a low polarizing power, and the anion being less polarizable, preventing covalent character and resulting in an ion bond. Thus, A is correct.

  B) Conversely, for a compound to have a covalent bond, a small, highly positively charged cation is preferred as it has a large polarizing power and can polarize the electron cloud of the anion, resulting in some sharing of electrons that gives the bond a covalent character. Thus, B is also correct.

  C) This statement is incorrect. It is not the smaller cation with high charge that has less polarizing power, but rather the larger cation or the cation with a lower positive charge that would have less polarizing power.

  So Option 1) A and B is the correct answer according to Fajan’s rule.

  The concepts involved here include ionic bonding, covalent bonding, and the theory of polarization involving polarizing power and polarizability of ions. Fajan's rules help make the distinction between ionic and covalent character in compounds.

The correct answer is Lewis-base.

In News

• Lewis bases play a crucial role in many chemical reactions, including catalysis and coordination chemistry.

Key Points

• A Lewis base is a substance that can donate a pair of electrons to form a coordinate covalent bond.
• Lewis bases are often nucleophiles, meaning they are attracted to positive charges.
• Common examples include ammonia (NH₃) and water (H₂O).
• In a reaction with a Lewis acid, the Lewis base donates its electron pair to form a bond.

Additional Information

• Lewis Acid
  • A Lewis acid is a substance that can accept a pair of electrons to form a coordinate covalent bond.
  • Common examples include AlCl₃ and BF₃.
• Bronsted-Lowry Acid
  • A Bronsted-Lowry acid is a substance that can donate a proton (H⁺).
  • Examples include HCl and H₂SO₄.
• Bronsted-Lowry Base
  • A Bronsted-Lowry base is a substance that can accept a proton (H⁺).
  • Examples include OH^- and NH₃.

CONCEPT:

Molecular Orbitals in Conjugated Systems

• Molecular orbitals are formed by the linear combination of atomic orbitals in a molecule.
• π (pi) molecular orbitals are formed by the side-by-side overlap of p-orbitals, while σ (sigma) molecular orbitals are formed by the end-to-end overlap of atomic orbitals.
• An asterisk (*) denotes an antibonding molecular orbital, which has a higher energy level and an additional node compared to the corresponding bonding orbital.

EXPLANATION:

• The diagram shows a molecular orbital with nodes (areas where the electron probability is zero) and alternating signs (phases) of the lobes.

• This configuration represents a π* (pi-star) antibonding molecular orbital because it has more nodes than the corresponding bonding π orbital.
• Such molecular orbitals are characteristic of conjugated systems and are critical in determining the electronic properties of the molecule.

The correct answer is π*.

Concept:

The bond angle in a molecule is determined by the hybridization of the central atom and the repulsion between electron pairs as described by the Valence Shell Electron Pair Repulsion (VSEPR) theory. In a tetrahedral molecule like CH₂Cl₂, the ideal bond angle is 109.5º. However, variations can occur due to differences in bond repulsion between different atoms.

Explanation:

In CH₂Cl₂ (dichloromethane), the carbon atom is sp³ hybridized, leading to a tetrahedral geometry with an ideal bond angle of 109.5º. However, the presence of two chlorine atoms and two hydrogen atoms introduces differences in bond repulsion:

• Chlorine atoms are more electronegative and have larger atomic radii than hydrogen atoms, causing greater electron cloud repulsion.

• This increased repulsion between the two Cl atoms will slightly compress the H-C-H bond angles, making the Cl-C-Cl bond angle slightly larger than 109.5º.

• 

Conclusion:

Therefore, the Cl-C-Cl bond angle in CH₂Cl₂ is: > 109.5º

Concept:

Hybridization:

→ The process of intermixing the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shape.

For example, when one 2s and three 2p-orbitals of carbon hybridize, there is the formation of four new sp³ hybrid orbitals.

Types of hybridisation : sp , sp² , sp³ , sp³d , sp³d²  and sp³d³ .

Explanation:

→ PCl₅ : P configuration is 1s² 2s² 2p⁶ 3s² 3p³ . it is in the ground state. 

Excited configuration is 1s2 2s2 2p6 3s¹ 3p³ 3d¹ here 3s1 3p3 3d¹ are 5 half filled orbitals having almost the same energy will intermix and form new 5 hybrid orbitals of hybridisation sp³d. hence geometry is Trigonal bipyramidal.

→ SF₆  : S ground state configuration is 1s2 2s2 2p6 3s2 3p⁴ .

Excited configuration is 1s2 2s2 2p6 3s1 3p3 3d² here 3s1 3p3 3d² are 6  half filled orbitals having almost same energy will intermix and form new 6 hybrid orbitals of hybridisation sp3d². hence geometry is Octahedral.

→ BrF₅ : Br ground state configuration is [Ar] 3d¹⁰ 4s² 4p⁵ .

Excited state configuration is  [Ar] 3d10 4s2 4p³ 4d² .

Here 1 full filled and 5 half filled orbitals (total 6) having almost the same energy will hybridise with each other. This intermixing will form  sp3d² hybridization. 1 sp3d² orbital will occupy the lone pair and the other 5 will be occupied by F.

So here 5 bond pairs and one lone pair hence its shape is square pyramidal.

→ BF₃ : B  ground state configuration is 1s2 2s2 2p¹ .

Excited configuration is 1s2 2s¹ 2p² . Here 2s¹ 2p² are 3 half-filled orbitals having almost the same energy that will intermix and form new 3 hybrid orbitals of hybridization sp².

Hence geometry is Trigonal planar.

Hence answer is Option 2

Concept:

Molecular orbital:

• The electron has wave property as well as particle property. 
• The area where there is a maximum probability of finding the electron is called the molecular orbital.
• It is different for different orbitals.
• HOMO and LUMO are types of molecular orbitals.
• The full form of the HUMO is the highest occupied molecular orbital
• LUMO stands for lowest unoccupied molecular orbital.

According to molecular orbital theory, the energy order of molecules is:

• σ 1sσ1s*σ2sσ 2s*σ 2pzπ 2py = π 2px π 2py* = π 2px*σ 2pz*.
• For molecules till nitrogen, there is a slight change in the M.O energy order: 
  • σ 1sσ1s*σ2sσ 2sπ 2py = π 2pxσ 2pz π 2py ^*= *π 2px*σ 2pz*

​Explanation:

• The total number of electrons in HF = 1 + 9 = 10
• The M.O series of HF will be: σ 1s2σ1s2*σ2s2σ 2s2π 2py1 = π 2px1
• The last electron is going into a nonbonding orbital.
• Hence, the highest occupied molecular orbital or HOMO of HF is nonbonding.

Additional Information

• The LUMO of HF is antibonding.

Concept:

Bond Length:

• Bond Length is defined as the distance between the centers of two covalently bonded atoms.
• The length of the bond is determined by the number of bonded electrons. Generally, the length of the bond between two atoms is approximately the sum of the covalent radii of the two atoms.
• Bond order is \(\propto \frac{1}{{Bond\;length}}\)

Diamagnetic:

• Whenever two electrons are paired together in an orbital, or their total spin is 0, they are diamagnetic electrons. Atoms with all diamagnetic electrons are called diamagnetic atoms.
• Diamagnetic is a quantum mechanical effect that occurs in all materials, when it is the only contribution to the magnetism, the material is called diamagnetic.

Paramagnetic:

• Paramagnetic compounds are attracted to the magnetic field and also have unpaired electrons. So, all atoms with incompletely filled atomic orbitals are paramagnetic.
• Due to their spin, unpaired electrons have a magnetic dipole moment and act like a tiny magnet.

Calculation:

\(O_2^{2 - }\) is diamagnetic with bond order 1.

• \(\sigma 1{s^2},{\sigma ^{\rm{*}}}1{s^2},\sigma 2{s^2},{\rm{\;}}{\sigma ^{\rm{*}}}2s_1^{2\;},\sigma 2p_2^2\left[ {\pi 2p_x^2 = \pi 2p_y^2} \right]{\rm{\;}}\sigma 2p_x^1\left[ {\pi *2p_x^1 = \pi x2p_y^1} \right]n\)

\({\rm{Bond\;order}} = \frac{{10 - 8}}{2} = 1\)

• \(N_2^{2 - } \to \sigma 1{s^2},{\rm{\;}}{\sigma ^{\rm{*}}}1{s^2},{\rm{\;}}\sigma 2{s^2},{\sigma ^{\rm{*}}}2s_1^2\left[ {\pi 2p_x^2 = \pi 2p_y^2} \right]{\rm{\;}}\sigma 2p_2^2\left[ {\pi *2p_x^1 = \pi x2p_y^1} \right]\)

\({\rm{Bond\;order}} = \frac{{10 - 6}}{2} = 2\)

\(N_2^{2 - }\) is paramagnetic with bond order 2.

• \(C_2^{2 - } \to \sigma 1{s^2},{\sigma ^{\rm{*}}}1{s^2},\sigma 2{s^2},{\sigma ^{\rm{*}}}2{s^2}\left[ {\pi 2p_x^2 = \pi 2p_y^2} \right]\sigma 2{p_x}^2\)

\({\rm{Bond\;order}} = \frac{{10 - 4}}{6} = 3\)

\(C_2^{2 - }\) is diamagnetic with bond order 3.

Hence \(C_2^{2 - }\) has the least bond length and it is diamagnetic.

Hence the O₂ is paramagnetic with bond order 2.

• Then Bond order is \(\propto \frac{1}{{Bond\;length}}\)

CONCEPT:

Bond Dissociation Energy

• Bond dissociation energy (BDE) is the energy required to break a bond in a molecule to form neutral atoms.
• The bond dissociation energies of X₂, Y₂, and XY are given in a ratio of 1 : 0.5 : 1.

EXPLANATION:

• Given the enthalpy change (ΔH) for the formation of XY is -200 kJ mol^-1.
• Let the bond dissociation energy of X₂ be E. Then:
  • Bond dissociation energy of Y₂ = 0.5E
  • Bond dissociation energy of XY = E
• Using the given ΔH for the reaction:

  X₂ + Y₂ → 2XY

  • ΔH = (Bond dissociation energy of X₂ + Bond dissociation energy of Y₂) - 2 × Bond dissociation energy of XY
  • -200 kJ mol^-1 = (E + 0.5E) - 2E
  • -200 kJ mol^-1 = 1.5E - 2E
  • -200 kJ mol^-1 = -0.5E
  • E = 400 kJ mol^-1

Therefore, the bond dissociation energy of X₂ is 400 kJ mol^-1, which corresponds to option 3.