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Chemical Bonding and Molecular Structure
Molecular Orbital Theory
20 questions from 17 papers, each with its answer.
- questions
- 20
- papers
- 17
- years asked, 2017–2026
- 4 of 10
- of Chemical Bonding and Molecular Structure
- 18.52%
Appears in
Concepts tested
- Bond order 2
- Bond order of oxygen species 2
- Comparing bond orders 2
- Molecular Orbital Theory 2
- Bond order and bond length
- Bond order and stability
- Bond order from MO configuration
- Bond order of oxygen ions
- diatomic species with only pi bonds
- Electronic configuration of O2
- Magnetic behaviour of oxygen species
- magnetism and bond order of diatomic species
- MO energy order of N2
- Paramagnetism of S2
- Paramagnetism of superoxide
- Sideways overlap of p orbitals
All Questions
2026 Re-NEET 1 question
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The highest occupied molecular orbital for $Ne_2$ is$Ne_2$ has 20 electrons: $\sigma1s^2\,\sigma^*1s^2\,\sigma2s^2\,\sigma^*2s^2\,\sigma2p_z^2\,\pi2p_x^2=\pi2p_y^2\,\pi^*2p_x^2=\pi^*2p_y^2\,\sigma^*2p_z^2$. The highest occupied molecular orbital (HOMO) is $\sigma^*2p_z$.
2023 1 question
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The correct order of energies of molecular orbitals of $N_2$ molecule is:
For $N_2$ (fewer than 15 electrons) the $\pi2p$ orbitals lie below $\sigma2p_z$.
$\sigma1s<\sigma^*1s<\sigma2s<\sigma^*2s<(\pi2p_x=\pi2p_y)<\sigma2p_z<(\pi^*2p_x=\pi^*2p_y)<\sigma^*2p_z$ is the correct order of energy of MO for homo nuclear diatomic species $N_2$.
2023 Manipur 1 question
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Which one of the following statements is incorrect related to Molecular Orbital Theory?Molecular orbitals formed from $2P_x$ and $2P_y$ orbitals (perpendicular to the bond axis, typically taken as the z-axis) are actually unsymmetrical (not symmetrical) around the bond axis — they form $\pi$ molecular orbitals with electron density above and below the axis, not symmetric around it. So this statement is incorrect.
2022 1 question
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Which amongst the following is incorrect statement$O_2^+$ has 15 electrons, so it contains one unpaired electron. Hence it is paramagnetic, not diamagnetic – so this statement is incorrect. (Bond orders of $O_2^+$, $O_2$, $O_2^-$, $O_2^{2-}$ are 2.5, 2, 1.5, 1; $C_2$ has 4 electrons in two degenerate $\pi$ MOs; $H_2^+$ has one electron – all correct.)
2019 1 question
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Which of the following diatomic molecular species has only $\pi$ bonds according to Molecular Orbital Theory?MO configuration of $C_2$ (12 electrons): $\sigma1s^2\,\sigma^*1s^2\,\sigma2s^2\,\sigma^*2s^2\,\pi2p_x^2=\pi2p_y^2$ Bond order = 2, and both bonds are π bonds, so $C_2$ has only π bonds according to MOT.
2015 AIPMT-I 2 questions
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Which of the following options represents the correct bond order?Bond orders: $O_2^-$ = 1.5, $O_2$ = 2, $O_2^+$ = 2.5
So $O_2^- < O_2 < O_2^+$. -
The correct bond order in the following species is:Bond order $= \frac{N_b - N_a}{2}$, with 10 bonding electrons in each species.
$O_2^{2+}$ (14 e⁻): $\frac{10 - 4}{2} = 3$
$O_2^+$ (15 e⁻): $\frac{10 - 5}{2} = 2.5$
$O_2^-$ (17 e⁻): $\frac{10 - 7}{2} = 1.5$
So $O_2^- < O_2^+ < O_2^{2+}$.
2015 AIPMT-II 1 question
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Decreasing order of stability of $O_2$, $O_2^-$, $O_2^+$ and $O_2^{2-}$ is:Electrons: $O_2^+$ = 15, $O_2$ = 16, $O_2^-$ = 17, $O_2^{2-}$ = 18
Bond orders: $O_2^+$ = 2.5, $O_2$ = 2, $O_2^-$ = 1.5, $O_2^{2-}$ = 1
Stability ∝ bond order, so $O_2^+ > O_2 > O_2^- > O_2^{2-}$
2013 NEET 1 question
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Which of the following is paramagnetic?$O_2^-$ has 17 electrons, with one unpaired electron in an antibonding $\pi^*$ molecular orbital, so it is paramagnetic.
$NO^+$, CO and $CN^-$ each have 14 electrons, all paired.
2012 AIPMT-MAINS 2 questions
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During change of $O_2$ to $O_2^-$ ion, the electron adds on which one of the following orbitals?$O_2$ (16 e⁻): $\sigma1s^2\,\sigma^*1s^2\,\sigma2s^2\,\sigma^*2s^2\,\sigma2p_z^2\,\pi2p_x^2 = \pi2p_y^2\,\pi^*2p_x^1 = \pi^*2p_y^1$
The $\pi^*2p$ orbitals are only half filled, so the extra electron of $O_2^-$ enters a $\pi^*$ orbital. -
Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them:Bond order $= \frac{1}{2}(N_b - N_a)$
$He_2^+$ (3 e⁻): $\frac{1}{2}(2 - 1) = 0.5$
$O_2^-$ (17 e⁻): $\frac{1}{2}(10 - 7) = 1.5$
NO (15 e⁻): $\frac{1}{2}(10 - 5) = 2.5$
$C_2^{2-}$ (14 e⁻): $\frac{1}{2}(10 - 4) = 3$
Increasing order: $He_2^+ < O_2^- < NO < C_2^{2-}$
2012 AIPMT-PRE 2 questions
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Bond order of 1.5 is shown by:Bond order $= \frac{1}{2}(N_b - N_a)$
$O_2$ (16 e⁻): $\frac{1}{2}(10 - 6) = 2$
$O_2^+$ (15 e⁻): $\frac{1}{2}(10 - 5) = 2.5$
$O_2^-$ (17 e⁻): $\frac{1}{2}(10 - 7) = 1.5$
$O_2^{2-}$ (18 e⁻): $\frac{1}{2}(10 - 8) = 1$
So $O_2^-$ has bond order 1.5. -
The pair of species with the same bond order is$N_2$ = 3, $O_2$ = 2
$O_2^{2-}$ (18 e⁻) = $\frac{1}{2}(10 - 8) = 1$; $B_2$ (10 e⁻) = $\frac{1}{2}(6 - 4) = 1$
$O_2^+$ = 2.5, $NO^+$ = 3
NO = 2.5, CO = 3
So $O_2^{2-}$ and $B_2$ have the same bond order, 1.
2011 AIPMT-MAINS 1 question
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The pairs of species of oxygen and their magnetic behaviours are noted below. Which of the following presents the correct description?$O_2$ (16 electrons): $\sigma1s^2\ \sigma^*1s^2\ \sigma2s^2\ \sigma^*2s^2\ \sigma2p_z^2\ \pi2p_x^2 = \pi2p_y^2\ \pi^*2p_x^1 = \pi^*2p_y^1$; 2 unpaired electrons, paramagnetic.
$O_2^+$ (15 electrons): ... $\pi^*2p_x^1\ \pi^*2p_y^0$; 1 unpaired electron, paramagnetic.
$O_2^-$ (17 electrons): ... $\pi^*2p_x^2\ \pi^*2p_y^1$; 1 unpaired electron, paramagnetic.
$O_2^{2-}$ (18 electrons): ... $\pi^*2p_x^2\ \pi^*2p_y^2$; no unpaired electron, diamagnetic.
O atom ($1s^22s^22p_x^22p_y^12p_z^1$) has 2 unpaired electrons and $O^+$ ($1s^22s^22p_x^12p_y^12p_z^1$) has 3 unpaired electrons; both are paramagnetic.
Every pair containing $O_2^{2-}$ described as paramagnetic is wrong, and $O_2^-$ is not diamagnetic.
The correct description is $O_2^+$, $O_2$ – both paramagnetic.
2011 AIPMT-PRE 1 question
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Which of the following has the minimum bond length?MO configuration of $O_2$ (16 electrons): $\sigma1s^2\ \sigma^*1s^2\ \sigma2s^2\ \sigma^*2s^2\ \sigma2p_z^2\ \pi2p_x^2 = \pi2p_y^2\ \pi^*2p_x^1 = \pi^*2p_y^1$
Bond order = $\frac{N_b - N_a}{2}$. For $O_2$: $\frac{10 - 6}{2} = 2$
$O_2^+$ (one antibonding electron removed): $\frac{10 - 5}{2} = 2.5$
$O_2^-$ (one antibonding electron added): $\frac{10 - 7}{2} = 1.5$
$O_2^{2-}$ (two antibonding electrons added): $\frac{10 - 8}{2} = 1$
Bond length is inversely proportional to bond order.
$O_2^+$ has the highest bond order (2.5) and therefore the minimum bond length.
2010 AIPMT-PRE 1 question
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Which one of the following species does not exist under normal conditions?A species exists only if its bond order, $\frac{N_b - N_a}{2}$, is greater than zero.
$Li_2$ (6 electrons): $\sigma1s^2\ \sigma^*1s^2\ \sigma2s^2$; bond order = $\frac{4 - 2}{2} = 1$
$Be_2$ (8 electrons): $\sigma1s^2\ \sigma^*1s^2\ \sigma2s^2\ \sigma^*2s^2$; bond order = $\frac{4 - 4}{2} = 0$
$Be_2^+$ (7 electrons): $\sigma1s^2\ \sigma^*1s^2\ \sigma2s^2\ \sigma^*2s^1$; bond order = $\frac{4 - 3}{2} = 0.5$
$B_2$ (10 electrons): $\sigma1s^2\ \sigma^*1s^2\ \sigma2s^2\ \sigma^*2s^2\ \pi2p_x^1 = \pi2p_y^1$; bond order = $\frac{6 - 4}{2} = 1$
$Be_2$ has zero bond order, so it does not exist under normal conditions.
2009 AIPMT 1 question
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According to MO theory which of the following lists ranks the nitrogen species in terms of increasing bond order?$N_2$ (14 electrons): $\sigma1s^2\ \sigma^*1s^2\ \sigma2s^2\ \sigma^*2s^2\ \pi2p_x^2 = \pi2p_y^2\ \sigma2p_z^2$; bond order = $\frac{10 - 4}{2} = 3$
$N_2^-$ (15 electrons): the extra electron enters an antibonding $\pi^*2p$ orbital; bond order = $\frac{10 - 5}{2} = 2.5$
$N_2^{2-}$ (16 electrons): two extra electrons in antibonding $\pi^*2p$ orbitals; bond order = $\frac{10 - 6}{2} = 2$
Increasing bond order: $N_2^{2-}$ (2) < $N_2^-$ (2.5) < $N_2$ (3)
2008 AIPMT 1 question
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Four diatomic species are listed below in different sequences. Which of these presents the correct order of their increasing bond order ?Bond order $= \dfrac{1}{2}(N_b - N_a)$, where $N_b$ and $N_a$ are the numbers of electrons in bonding and antibonding molecular orbitals.
$He_2^+$ (3 electrons): $\sigma 1s^2\,\sigma^* 1s^1$; bond order $= \dfrac{1}{2}(2 - 1) = 0.5$
$O_2^-$ (17 electrons): one more antibonding electron than $O_2$ (bond order 2); bond order $= \dfrac{1}{2}(10 - 7) = 1.5$
$NO$ (15 electrons): bond order $= \dfrac{1}{2}(10 - 5) = 2.5$
$C_2^{2-}$ (14 electrons, isoelectronic with $N_2$): bond order $= \dfrac{1}{2}(10 - 4) = 3.0$
Bond orders: 0.5, 1.5, 2.5 and 3.0 respectively.
Increasing order of bond order: $He_2^+ \lt O_2^- \lt NO \lt C_2^{2-}$
2006 AIPMT-PRE 1 question
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The number of unpaired electrons in a paramagnetic diatomic molecule of an element with atomic number 16 is :-The element with atomic number 16 is sulphur. Its diatomic molecule is $S_2$.
Sulphur belongs to the same group as oxygen, so the molecular orbital configuration of $S_2$ in its valence shell is like that of $O_2$.
In $O_2$ the last two electrons enter the two degenerate antibonding orbitals $\pi^*2p_x$ and $\pi^*2p_y$ singly (Hund's rule).
Similarly, in $S_2$ the last two electrons occupy the two degenerate $\pi^*3p$ orbitals singly: $\pi^*3p_x^1 = \pi^*3p_y^1$.
So $S_2$ has two unpaired electrons according to molecular orbital theory, which makes it paramagnetic like $O_2$.
The number of unpaired electrons is 2.
2001 AIPMT 1 question
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Main axis of a diatomic molecule is z, atomic orbitals $p_x$ and $p_y$ overlap to form which of the following orbital : -When the molecular axis is z, the $p_z$ orbitals overlap head-on (along the axis) to form a $\sigma$ molecular orbital.
The $p_x$ and $p_y$ orbitals are perpendicular to the molecular axis. A $p_x$ orbital of one atom overlaps sideways with the $p_x$ orbital of the other atom (and $p_y$ with $p_y$), forming $\pi$ molecular orbitals ($\pi_{2p_x}$ and $\pi_{2p_y}$).
(Strictly, a $p_x$ orbital of one atom and a $p_y$ orbital of the other atom are at right angles, so their net overlap is zero and they cannot form a bond; the question refers to the $\pi$ molecular orbitals formed by the $p_x$ and $p_y$ orbitals.)
So, as per the answer key, $p_x$ and $p_y$ overlap to form $\pi$ molecular orbitals.