Molecular Orbital Theory MCQ Quiz in తెలుగు - Objective Question with Answer for Molecular Orbital Theory - ముఫ్త్ [PDF] డౌన్లోడ్ కరెన్
Last updated on Apr 13, 2025
Latest Molecular Orbital Theory MCQ Objective Questions
Top Molecular Orbital Theory MCQ Objective Questions
Molecular Orbital Theory Question 1:
The highest occupied MO in N2 and O2+ respectively are (take x axis as internuclear axis):
Answer (Detailed Solution Below)
Molecular Orbital Theory Question 1 Detailed Solution
The correct answer is \(\sigma2p_x , \pi^*2p_y\)
Concept:-
\(\sigma2p_x , \pi^*2p_y\)
- Molecular Orbital Theory (MOT): This theory is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule. It involves the combination of atomic orbitals to form the same number of molecular orbitals, providing a quantum mechanical, more detailed depiction of molecules in comparison to the Lewis structure model or the valence bond theory.
- Molecular Orbitals: Molecular orbitals correspond to regions in a molecule where an electron is likely to be found. They can either be bonding or antibonding. Bonding orbitals result from the in-phase combination of atomic orbitals, have lower energy than the original atomic orbitals, and are indicated by Greek letter sigma (σ) or pi (π). Antibonding orbitals, denoted as σ* or π*, result from the out-of-phase combination of atomic orbitals, and have higher energy than the original atomic orbitals.
- Highest Occupied Molecular Orbital (HOMO): The HOMO refers to the molecular orbital that contains the highest energy electrons. Identifying the HOMO can be instrumental in predicting how a molecule will interact with others. Because the electrons in the HOMO have the highest energy in the molecule, they are typically the electrons involved in chemical reactions.
- Internuclear Axis: It is the straight line joining the nuclei of the atoms constituting a molecule. In the context of molecular orbitals, particularly in diatomic molecules, we often choose one axis (usually taken as the z-axis) as the internuclear axis. Then, px, py, and pz represent the orientations of the p-orbitals along the x, y, or z-axis, respectively.
Explanation:-
- Molecular orbital theory uses atomic orbitals from individual atoms to build up molecular orbitals for the entire system. These orbitals are not associated with individual atoms but instead belong to the whole molecule.
- The molecular orbital diagram of N2 and O2 can be constructed using molecular orbital theory. The ordering of molecular orbitals in these molecules will be as follows (taking axis as internuclear axis):
- N2 (Nitrogen): In the nitrogen molecule (N2), there are 14 electrons to distribute (7 from each nitrogen atom). The molecular orbital filling order is σ1s, σ1s, σ2s, σ2s, π2px = π2py, σ2pz, π2px = π2py, σ*2pz. After filling up these orbitals with nitrogen's 14 electrons, the highest occupied molecular orbital (HOMO) of nitrogen will be the σ2pz orbital.
- O2+ (Oxygen): Oxygen molecule ion (O2+) has 15 electrons to distribute (8 from each oxygen atom, but one less because of the positive charge). The molecular orbital filling order is similar to N2 (σ1s, σ1s, σ2s, σ2s, π2px = π2py, σ2pz, π2px = π2py, σ2pz) until you reach 14 electrons, which resides in π2px = π2py, but for the 15th electron, it's put into one of the π2px, π2py. So, the highest occupied molecular orbital (HOMO) of O2+ will be the π2px = π*2py orbital.
- Note, Here, π and σ represent the shapes and symmetry of the orbitals, while px, py, and pz represent the direction or orientation of the orbital (along the x, y, or z axis, respectively). Particularly, 'z' denotes the internuclear axis. '2p' is equivalent to the second electron shell and 'p' denotes the shape of the orbital. The '*' denotes an antibonding orbital
Conclusion:-
So, The highest occupied MO in N2 and O2+ respectively are (take x axis as internuclear axis) is \(\sigma2p_x , \pi^*2p_y\)
Molecular Orbital Theory Question 2:
Among the following molecules/ions,
\({\rm{C}}_2^{2 - },{\rm{\;N}}_2^{2 - },{\rm{O}}_2^{2 - },{{\rm{O}}_2}\)
which one is diamagnetic and has the shortest bond length?Answer (Detailed Solution Below)
Molecular Orbital Theory Question 2 Detailed Solution
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 \(\propto \frac{1}{{Bond\;length}}\)
Symbol |
No. of unpaired electron |
Bond Order |
Magnetic Character |
\(O_2^{2 - }\) |
0 |
1 |
Diamagnetic |
\(C_2^{2 - }\) |
0 |
3 |
Diamagnetic |
\(N_2^{2 - }\) |
2 |
2 |
Paramagnetic |
O2 |
1 |
2 |
Paramagnetic |
Molecular Orbital Theory Question 3:
Among the following molecules/ions,
\({\rm{C}}_2^{2 - },{\rm{\;N}}_2^{2 - },{\rm{O}}_2^{2 - },{{\rm{O}}_2}\)
which one is diamagnetic and has the shortest bond length?Answer (Detailed Solution Below)
Molecular Orbital Theory Question 3 Detailed Solution
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 \(\propto \frac{1}{{Bond\;length}}\)
Symbol |
No. of unpaired electron |
Bond Order |
Magnetic Character |
\(O_2^{2 - }\) |
0 |
1 |
Diamagnetic |
\(C_2^{2 - }\) |
0 |
3 |
Diamagnetic |
\(N_2^{2 - }\) |
2 |
2 |
Paramagnetic |
O2 |
1 |
2 |
Paramagnetic |
Molecular Orbital Theory Question 4:
Which of the following statement is not correct from the view point of molecular orbital theory?
Answer (Detailed Solution Below)
Molecular Orbital Theory Question 4 Detailed Solution
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.
Molecular Orbital Theory Question 5:
Which of the following diatomic molecular species has only π bonds according to Molecular Orbital Theory?
Answer (Detailed Solution Below)
Molecular Orbital Theory Question 5 Detailed Solution
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.
Molecular Orbital Theory Question 6:
Total number of molecules/species from following which will be paramagnetic is ______.
\(\mathrm{O}_{2}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2}^{-}\), NO, NO2, CO, K2[NiCl4], [Co(NH3)6]Cl3, K2[Ni(CN)4]
Answer (Detailed Solution Below) 6
Molecular Orbital Theory Question 6 Detailed Solution
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
Molecular Orbital Theory Question 7:
Among the following molecules/ions,
\({\rm{C}}_2^{2 - },{\rm{\;N}}_2^{2 - },{\rm{O}}_2^{2 - },{{\rm{O}}_2}\)
which one is diamagnetic and has the shortest bond length?Answer (Detailed Solution Below)
Molecular Orbital Theory Question 7 Detailed Solution
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 \(\propto \frac{1}{{Bond\;length}}\)
Symbol |
No. of unpaired electron |
Bond Order |
Magnetic Character |
\(O_2^{2 - }\) |
0 |
1 |
Diamagnetic |
\(C_2^{2 - }\) |
0 |
3 |
Diamagnetic |
\(N_2^{2 - }\) |
2 |
2 |
Paramagnetic |
O2 |
1 |
2 |
Paramagnetic |
Molecular Orbital Theory Question 8:
According to molecular orbital theory, which of the following will not be a viable molecule?
Answer (Detailed Solution Below)
Molecular Orbital Theory Question 8 Detailed Solution
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.
Molecular Orbital Theory Question 9:
According to Molecular Orbital Theory:
Answer (Detailed Solution Below)
Molecular Orbital Theory Question 9 Detailed Solution
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.
Molecular Orbital Theory Question 10:
The correct order of energy levels of the molecular orbitals of N2 is
Answer (Detailed Solution Below)
Molecular Orbital Theory Question 10 Detailed Solution
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:
\(1\sigma_g < 1\sigma_u^* < 2\sigma_g < 2\sigma_u^* < 1\pi_u < 3\sigma_g < 1\pi_g^* < 3\sigma_u^*\)