Molecular Orbital Theory MCQ Quiz in मल्याळम - Objective Question with Answer for Molecular Orbital Theory - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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 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:

Molecular orbital theory (MOT)-

• It states that atomic orbitals of comparable energies and proper symmetry combine to form molecular orbitals. So, molecular orbitals form when atomic orbitals combine in a manner of the linear combination of atomic orbitals.
• No. of molecular orbitals formed is always equal to the number of atomic orbitals combine.
• Molecular orbitals are further classified into bonding and antibonding molecular orbitals.
• MOT gives an idea about the following-
  • No. of electrons present in degenerated molecular orbitals.
  • Stability of the molecules.
  • Bond order
  • Magnetic behavior of the molecule.

Explanation:

→ Statement 1, "Be2 is not a stable molecule" is correct as -

According to molecular orbital theory, the molecular orbital configuration for Be2 is = σ1s2, σ*1s2, σ2s2, σ*2s2.

Formula to calculate bond order is:

Bond order=  \(\frac{Number\hspace{0.1cm} of \hspace{0.1cm} bonding \hspace{0.1cm} electrons - number \hspace{0.1cm} of \hspace{0.1cm} antibonding \hspace{0.1cm} electrons }{2}\)

So bond order in Be₂ is - \(\frac{4-4}{2}\) = 0

∴ Zero bond order indicates that the molecule Be₂ does not exist.

→ Statement 2, " He2 is not stable but \(\rm He^+_2\) is expected to exist" is correct as -

According to molecular orbital theory, the molecular orbital configuration for He2 is = σ1s2, σ*1s2.

Formula to calculate bond order is:

Bond order=  \(\frac{Number\hspace{0.1cm} of \hspace{0.1cm} bonding \hspace{0.1cm} electrons - number \hspace{0.1cm} of \hspace{0.1cm} antibonding \hspace{0.1cm} electrons }{2}\)

So bond order in He2 is - \(\frac{2-2}{2}\) = 0

∴ Zero bond order indicates that the molecule He2 does not exist.

But \(\rm He^+_2\) may exist as -  the molecular orbital configuration for \(\rm He^+_2\) is = σ1s2, σ*1s1.

Bond order=  \(\frac{Number\hspace{0.1cm} of \hspace{0.1cm} bonding \hspace{0.1cm} electrons - number \hspace{0.1cm} of \hspace{0.1cm} antibonding \hspace{0.1cm} electrons }{2}\)

So, bond order in \(\rm He^+_2\) is - \(\frac{2-1}{2}\) = 0.5

A non-zero bond order indicates that the molecule \(\rm He^+_2\) may exist.

→ Statement 3, "Bond strength of N2 is maximum amongst the homonuclear diatomic molecules belonging to the second period" is correct as -

According to molecular orbital theory, the molecular orbital configuration for N₂ is σ1s2, σ*1s2, σ2s2, σ*2s2, σ2pz2, 2px2π=2py2π.

Bond order=  \(\frac{Number\hspace{0.1cm} of \hspace{0.1cm} bonding \hspace{0.1cm} electrons - number \hspace{0.1cm} of \hspace{0.1cm} antibonding \hspace{0.1cm} electrons }{2}\)

So bond order in N2 is - \(\frac{10-4}{2}\) = 3

∴ The bond order in the N₂ molecule is 3 which is the maximum among the homonuclear diatomic molecules belonging to the 2nd period.

→ Statement 4, " The order of energies of molecular orbitals in N2 molecule is  σ2s < σ* 2s < σ2pz < (π2px = π2py) < (π*2px = π* 2py) < σ* 2pz" is incorrect as - The correct order of the energy of MO for N₂ is σ2s < σ* 2s < (π2px = π2py) < σ2pz < (π*2px = π* 2py) < σ* 2pz.

Conclusion:

The statement 4, "The order of energies of molecular orbitals in N2 molecule is  σ2s < σ* 2s < σ2pz < (π2px = π2py) < (π*2px = π* 2py) < σ* 2pz" is not correct from the view point of molecular orbital theory

Hence, the correct answer is option 4.

Concept: 

• A pair of electrons present in the valence orbital of the atom, which is not involved in the bonding, is lone pair of electrons.
• The two electrons involved in the bonding, form the bond pair of electrons.
• Nitrogen and oxygen both are p-block elements of 2nd period. According to periodic trends, electronegativity increases from left to right across a period.

Explanation:

• To calculate lone pairs and bond pairs, first identify the central atom.
• Generally, the central atom is always the one that is less electronegative. Here, the electronegativity of N is less than the electronegativity of O. Therefore, nitrogen is the central atom.
• In the question we are asked to find lone pairs and bond pairs of N only.

Structure of NO^-₃ is:     

Therefore, NO₃^- have Square planar geometry.

Conclusion:

• Here, 3 sigma bonds + 1 Pi bond = 4 bond pairs.
• There is no any lone pair on the Nitrogen therefore, 0 lone pair.

Concept:

Molecular Orbital theory - According to the theory, when atoms are combined to form molecules, formations of molecular orbitals takes place.

• No. of molecular orbitals is equal to the no. of atomic orbitals combined to form a molecule.
• Two types of molecular orbital are there in MOT, bonding and anti-bonding molecular orbital.
• Based on the configuration of the molecular orbital we can calculate the type of bonding and bond order present in the molecule.

Explanation:

Step 1: MOT configuration C₂ is:

\(\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\pi 2p_x^2 = \pi 2p_y^2\)

Step 2: Bond Order in C₂ = 8-4/2 = 2

Bond order in C₂ is two means a double bond is present in C₂.

Step 3: By observing configuration we can state that only degenerated Pi orbitals only participated in bonding having 4 electrons in total. Also, no sigma electron is participating in bonding. Hence no sigma bond is present in C₂.

So, C₂ molecule has 2 π bonds in its structure.

Hence, the correct answer is option 3.

Important Points To find the type of bond present in a diatomic molecule following steps have to be followed-

Step 1: Write down the configuration of the diatomic molecule according to MOT.

Step 2: Calculate the bond order using the formula - bonding electrons - antibonding electrons / 2

Step 3: Observe the MO configuration, and depending upon the type of electrons participating in bonding, give the type of bonding present in the molecule.

CONCEPT:

Paramagnetism

• Paramagnetic species contain unpaired electrons.
• Use Molecular Orbital Theory (MOT) and electronic configuration to determine the presence of unpaired electrons.

Explanation:-

• O₂:
  • According to MOT, O₂ has 2 unpaired electrons in π*2p.
  • Paramagnetic
• O₂⁺:
  • According to MOT, O₂⁺ has 1 unpaired electron in π*2p.
  • Paramagnetic
• O₂^–:
  • According to MOT, O₂^– has 1 unpaired electron in π*2p.
  • Paramagnetic
• NO:
  • Odd electron species.
  • Paramagnetic
• NO₂:
  • Odd electron species.
  • Paramagnetic
• CO:
  • No unpaired electrons.
  • Diamagnetic
• K₂[NiCl₄]:
  • Ni²⁺: 3d⁸ (weak field ligand, tetrahedral).
  • 2 unpaired electrons.
  • Paramagnetic
• [Co(NH₃)₆]Cl₃:
  • Co³⁺: 3d⁶ (strong field ligand, octahedral).
  • No unpaired electrons.
  • Diamagnetic
• K₂[Ni(CN)₄]:
  • Ni²⁺: 3d⁸ (strong field ligand, square planar).
  • No unpaired electrons.
  • Diamagnetic

Conclusion:-

Paramagnetic species: O₂, O₂⁺, O₂^–, NO, NO₂, K₂[NiCl₄] Total = 6

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 \(\propto \frac{1}{{Bond\;length}}\)

Electronic configuration of \(H_2^{2-} = (\sigma s^2)(\sigma ^* s^2)\)

Bond order of \(H_2^{2-} = \dfrac{N_b-N_a}{2} = \dfrac{2-2}{2} = 0\).

If a bond order of zero is obtained, that means that the molecule is too unstable and so it will not exist.

Hence, \(H_2^{2-}\) does not exist, due to zero bond order.

Option B is correct.

Bond order \(=\dfrac { Total\quad no.\quad of\quad bonding\quad electrons\quad -\quad Total\quad no.\quad of\quad anti\quad bonding }{ 2 }\)

Total no. of electrons in \(C_{2}^{2-}\) is \(14\)

\(\sigma { 1S }^{ 2 }{ \sigma }^{ * }{ 1S }^{ 2 }\sigma { 2S }^{ 2 }\sigma ^{ * }{ 2S }^{ 2 }\Pi { 2P }_{ y }^{ 2 }\Pi 2{ P }_{ 2 }^{ 2 }\sigma { 2P }_{ x }^{ 2 }\)

Bond order \(=\dfrac { 10-4 }{ 2 } =3\)

If bond order is in integers then molecule is diamagnetic.

(B) \(Bond\quad order\quad \alpha \quad \dfrac { 1 }{ Bond\quad length }\)

\(O_{2}^{2+}\) bond order is \(3\)

\(O_{2}\) bond order is \(2\)

(C) \(N_{2}^{+}\) and \(N_{2}^{-}\) have the same bond order.

CONCEPT:

Molecular Orbital Energy Levels in \(N_2\)

• The molecular orbital (MO) energy levels of diatomic molecules are determined by the linear combination of atomic orbitals (LCAO) from the two atoms forming the molecule.
• For molecules with atomic numbers ≤ 7 (like \(N_2\) ), the energy order of MOs differs due to less effective s-p mixing, which occurs because the energy difference between 2s and 2p orbitals is smaller.

Explanation:

The correct MO energy order for \(N_2\) is:

\(1\sigma_g < 1\sigma_u^* < 2\sigma_g < 2\sigma_u^* < 1\pi_u < 3\sigma_g < 1\pi_g^* < 3\sigma_u^*\)