401. The molecular orbital configuration of \(\mathrm{He_2}\) may be written as \((\sigma_{1s})^2(\sigma_{1s}^*)^2\). Its bond order is
ⓐ. \(1\)
ⓑ. \(0\)
ⓒ. \(2\)
ⓓ. \(4\)
Correct Answer: \(0\)
Explanation: \( \textbf{Configuration:} \) \((\sigma_{1s})^2(\sigma_{1s}^*)^2\).
\( \textbf{Bonding electrons:} \) The \(\sigma_{1s}\) orbital contains \(2\) electrons, so \(N_b=2\).
\( \textbf{Antibonding electrons:} \) The \(\sigma_{1s}^*\) orbital contains \(2\) electrons, so \(N_a=2\).
\( \textbf{Bond-order formula:} \)
\[
\text{Bond order}=\frac{1}{2}(N_b-N_a)
\]
\( \textbf{Substitution:} \)
\[
\text{Bond order}=\frac{1}{2}(2-2)
\]
\( \textbf{Calculation:} \)
\[
\text{Bond order}=0
\]
\( \textbf{Final answer:} \) The bond order is \(0\). Bonding and antibonding effects cancel in this simple molecular orbital description.
402. A species with molecular orbital configuration \((\sigma_{1s})^2(\sigma_{1s}^*)^1\) has bond order
ⓐ. \(0\)
ⓑ. \(1\)
ⓒ. \(\frac{3}{2}\)
ⓓ. \(\frac{1}{2}\)
Correct Answer: \(\frac{1}{2}\)
Explanation: \( \textbf{Configuration:} \) \((\sigma_{1s})^2(\sigma_{1s}^*)^1\).
\( \textbf{Bonding electrons:} \) \(N_b=2\).
\( \textbf{Antibonding electrons:} \) \(N_a=1\).
\( \textbf{Bond-order relation:} \)
\[
\text{Bond order}=\frac{1}{2}(N_b-N_a)
\]
\( \textbf{Substitution:} \)
\[
\text{Bond order}=\frac{1}{2}(2-1)
\]
\( \textbf{Calculation:} \)
\[
\text{Bond order}=\frac{1}{2}
\]
\( \textbf{Final answer:} \) The bond order is \(\frac{1}{2}\). One antibonding electron weakens the bond but does not cancel the two bonding electrons completely.
403. In molecular orbital theory, a species is paramagnetic if it has
ⓐ. one or more unpaired electrons
ⓑ. only paired electrons in all orbitals
ⓒ. zero total electrons
ⓓ. no molecular orbitals at all
Correct Answer: one or more unpaired electrons
Explanation: Paramagnetism is associated with the presence of unpaired electrons. If a molecule or ion has one or more unpaired electrons in its molecular orbital configuration, it is attracted by a magnetic field. If all electrons are paired, the species is diamagnetic. Molecular orbital theory is especially useful because it can show whether electrons remain unpaired after filling molecular orbitals. This distinction explains why \(\mathrm{O_2}\) is paramagnetic even though a simple Lewis structure shows all electrons paired.
404. A molecular orbital diagram contains two singly occupied degenerate \(\pi^*\) orbitals. The species is expected to be
ⓐ. diamagnetic
ⓑ. non-magnetic because electrons are absent
ⓒ. paramagnetic
ⓓ. ionic only
Correct Answer: paramagnetic
Explanation: Singly occupied orbitals contain unpaired electrons. If two degenerate \(\pi^*\) orbitals each contain one electron, the species has two unpaired electrons. Unpaired electrons make a species paramagnetic. This is the molecular orbital explanation for the paramagnetic nature of \(\mathrm{O_2}\). The antibonding label \(\pi^*\) also affects bond order, but magnetism is decided by whether electrons are paired or unpaired.
405. Assertion: \(\mathrm{O_2}\) is paramagnetic according to molecular orbital theory.
Reason: Its molecular orbital configuration contains unpaired electrons in antibonding \(\pi^*\) orbitals.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Molecular orbital theory predicts that \(\mathrm{O_2}\) has two unpaired electrons. These unpaired electrons occupy degenerate antibonding \(\pi^*\) molecular orbitals. The presence of unpaired electrons makes oxygen paramagnetic. This is a major success of molecular orbital theory compared with a simple Lewis structure, which pairs all electrons in \(\mathrm{O=O}\). The Reason directly explains the magnetic behaviour stated in the Assertion.
406. For a diatomic molecule, increasing bond order usually corresponds to
ⓐ. decreasing bond strength and increasing bond length
ⓑ. no change in bond strength or bond length
ⓒ. increasing bond strength and decreasing bond length
ⓓ. removal of all antibonding electrons in every molecule
Correct Answer: increasing bond strength and decreasing bond length
Explanation: Bond order measures the net number of bonds predicted by molecular orbital theory. A higher bond order means a greater excess of bonding electrons over antibonding electrons. This usually gives a stronger bond because the net stabilizing effect is greater. A stronger bond generally holds nuclei closer together, so bond length decreases. The relation is a useful trend when comparing related diatomic species.
407. Interpret the graph description below.
A graph for related diatomic species shows bond order on the x-axis and bond length on the y-axis. The curve slopes downward as bond order increases.
The graph is best interpreted as showing that
ⓐ. higher bond order usually gives longer bond length
ⓑ. bond order and bond length are always identical quantities
ⓒ. antibonding electrons always shorten bonds
ⓓ. higher bond order usually gives shorter bond length
Correct Answer: higher bond order usually gives shorter bond length
Explanation: Bond order reflects the net bonding effect in a molecule or ion. As bond order increases, the attraction between bonded nuclei and bonding electrons becomes stronger. Stronger bonding usually pulls the nuclei closer. Therefore, bond length decreases as bond order increases in a related series. The graph expresses an inverse trend between bond order and bond length, not equality of the two quantities.
408. The molecular orbital filling of \(\mathrm{N_2}\) gives bond order \(3\). This supports the observation that \(\mathrm{N_2}\) has
ⓐ. no bond between nitrogen atoms
ⓑ. only one weak \(\pi\)-bond
ⓒ. two unpaired electrons in antibonding orbitals like \(\mathrm{O_2}\)
ⓓ. a very strong triple bond
Correct Answer: a very strong triple bond
Explanation: A bond order of \(3\) means there are three net bonding interactions between the two nitrogen atoms. This matches the common structural representation \(\mathrm{N\equiv N}\). A high bond order generally indicates a strong and short bond. The strength of the nitrogen-nitrogen triple bond helps explain the low reactivity of \(\mathrm{N_2}\) under ordinary conditions. Unlike \(\mathrm{O_2}\), \(\mathrm{N_2}\) has all electrons paired in its simple molecular orbital configuration.
409. A comparison of \(\mathrm{O_2}\), \(\mathrm{O_2^+}\), and \(\mathrm{O_2^-}\) is made using molecular orbital theory. Removing one electron from antibonding \(\pi^*\) orbitals to form \(\mathrm{O_2^+}\) will
ⓐ. increase bond order relative to \(\mathrm{O_2}\)
ⓑ. decrease bond order relative to \(\mathrm{O_2}\)
ⓒ. keep bond order exactly unchanged
ⓓ. remove one electron from a bonding orbital only
Correct Answer: increase bond order relative to \(\mathrm{O_2}\)
Explanation: In \(\mathrm{O_2}\), the highest occupied electrons include antibonding \(\pi^*\) electrons. Removing one electron to form \(\mathrm{O_2^+}\) removes an antibonding electron. Since antibonding electrons reduce bond order, removing one increases bond order. This makes \(\mathrm{O_2^+}\) stronger and shorter than \(\mathrm{O_2}\) in the molecular orbital comparison. The direction of change depends on whether the electron is removed from a bonding or antibonding orbital.
410. Adding one electron to \(\mathrm{O_2}\) to form \(\mathrm{O_2^-}\) places the electron in an antibonding molecular orbital. The expected effect is
ⓐ. higher bond order and shorter bond length than \(\mathrm{O_2}\)
ⓑ. lower bond order and longer bond length than \(\mathrm{O_2}\)
ⓒ. unchanged bond order but zero magnetic behaviour
ⓓ. conversion into a molecule with no oxygen atoms
Correct Answer: lower bond order and longer bond length than \(\mathrm{O_2}\)
Explanation: The added electron in \(\mathrm{O_2^-}\) enters an antibonding molecular orbital. Antibonding electrons oppose bonding and reduce bond order. A lower bond order usually corresponds to a weaker and longer bond. Therefore, \(\mathrm{O_2^-}\) has a lower bond order than \(\mathrm{O_2}\). This comparison shows how electron gain or loss can change bond strength in a predictable molecular orbital way.
411. Consider the following molecular orbital data for related species.
| Species | Bonding electrons \(N_b\) | Antibonding electrons \(N_a\) |
|---|
| P | \(8\) | \(4\) |
| Q | \(8\) | \(5\) |
| R | \(8\) | \(3\) |
The species with the highest bond order is
ⓐ. P
ⓑ. R
ⓒ. Q
ⓓ. P and Q equally
Correct Answer: R
Explanation: \( \textbf{Bond-order formula:} \)
\[
\text{Bond order}=\frac{1}{2}(N_b-N_a)
\]
\( \textbf{For P:} \)
\[
\frac{1}{2}(8-4)=2
\]
\( \textbf{For Q:} \)
\[
\frac{1}{2}(8-5)=\frac{3}{2}
\]
\( \textbf{For R:} \)
\[
\frac{1}{2}(8-3)=\frac{5}{2}
\]
\( \textbf{Comparison:} \) \(\frac{5}{2}\gt 2\gt \frac{3}{2}\).
\( \textbf{Final answer:} \) R has the highest bond order. With the same number of bonding electrons, fewer antibonding electrons give the stronger net bond.
412. A species has all its electrons paired in the molecular orbital diagram. The magnetic behaviour expected is
ⓐ. diamagnetic
ⓑ. paramagnetic
ⓒ. always ferromagnetic
ⓓ. decided only by molar mass
Correct Answer: diamagnetic
Explanation: Diamagnetic species have all electrons paired. Paired electrons have opposite spins, so their magnetic effects cancel. Such species are weakly repelled by a magnetic field. Paramagnetism requires at least one unpaired electron. Molecular orbital diagrams are useful because they show directly whether the highest occupied orbitals contain paired or unpaired electrons.
413. The bond order sequence for related oxygen species is best represented as
ⓐ. \(\mathrm{O_2^{2-}}\gt \mathrm{O_2^-}\gt \mathrm{O_2}\gt \mathrm{O_2^+}\)
ⓑ. \(\mathrm{O_2^+}\gt \mathrm{O_2}\gt \mathrm{O_2^-}\gt \mathrm{O_2^{2-}}\)
ⓒ. \(\mathrm{O_2}\gt \mathrm{O_2^+}\gt \mathrm{O_2^-}\gt \mathrm{O_2^{2-}}\)
ⓓ. \(\mathrm{O_2^-}\gt \mathrm{O_2^{2-}}\gt \mathrm{O_2}\gt \mathrm{O_2^+}\)
Correct Answer: \(\mathrm{O_2^+}\gt \mathrm{O_2}\gt \mathrm{O_2^-}\gt \mathrm{O_2^{2-}}\)
Explanation: In the oxygen series, electrons are removed from or added to antibonding \(\pi^*\) molecular orbitals. Removing one antibonding electron from \(\mathrm{O_2}\) to form \(\mathrm{O_2^+}\) increases bond order. Adding one antibonding electron to form \(\mathrm{O_2^-}\) decreases bond order. Adding two antibonding electrons to form \(\mathrm{O_2^{2-}}\) decreases it further. The order is controlled by antibonding electron count, not by total number of oxygen atoms.
414. If \(\mathrm{O_2}\) has bond order \(2\), the bond order of \(\mathrm{O_2^+}\) is
ⓐ. \(\frac{1}{2}\)
ⓑ. \(\frac{3}{2}\)
ⓒ. \(\frac{5}{2}\)
ⓓ. \(3\)
Correct Answer: \(\frac{5}{2}\)
Explanation: \( \textbf{Starting species:} \) \(\mathrm{O_2}\) has bond order \(2\).
\( \textbf{Electron removed:} \) Formation of \(\mathrm{O_2^+}\) removes one electron from an antibonding \(\pi^*\) molecular orbital.
\( \textbf{Effect of removing antibonding electron:} \) Antibonding electrons reduce bond order, so removing one increases bond order.
\( \textbf{Change in bond order:} \)
\[
\Delta \text{bond order}=+\frac{1}{2}
\]
\( \textbf{New bond order:} \)
\[
2+\frac{1}{2}=\frac{5}{2}
\]
\( \textbf{Final answer:} \) The bond order of \(\mathrm{O_2^+}\) is \(\frac{5}{2}\). Removing an electron from an antibonding orbital strengthens the bond instead of weakening it.
415. For \(\mathrm{O_2^-}\), one electron is added to an antibonding orbital of \(\mathrm{O_2}\). Its bond order is expected to be
ⓐ. \(\frac{1}{2}\)
ⓑ. \(\frac{3}{2}\)
ⓒ. \(1\)
ⓓ. \(\frac{5}{2}\)
Correct Answer: \(\frac{3}{2}\)
Explanation: \( \textbf{Reference species:} \) \(\mathrm{O_2}\) has bond order \(2\).
\( \textbf{Added electron:} \) The extra electron in \(\mathrm{O_2^-}\) enters an antibonding molecular orbital.
\( \textbf{Bond-order effect:} \) Each additional antibonding electron lowers bond order by \(\frac{1}{2}\).
\( \textbf{Calculation:} \)
\[
2-\frac{1}{2}=\frac{3}{2}
\]
\( \textbf{Bond-length implication:} \) A lower bond order means a weaker and longer \(\mathrm{O-O}\) bond compared with \(\mathrm{O_2}\).
\( \textbf{Final answer:} \) The bond order of \(\mathrm{O_2^-}\) is \(\frac{3}{2}\). The negative charge does not automatically mean a stronger bond; the orbital receiving the electron matters.
416. The oxygen species table below summarizes bond order and relative bond length.
| Species | Bond order | Expected relative bond length |
|---|
| P. \(\mathrm{O_2^+}\) | \(\frac{5}{2}\) | shortest |
| Q. \(\mathrm{O_2}\) | \(2\) | intermediate |
| R. \(\mathrm{O_2^-}\) | \(\frac{3}{2}\) | longer than \(\mathrm{O_2}\) |
| S. \(\mathrm{O_2^{2-}}\) | \(1\) | longest |
The table mainly shows that
ⓐ. bond length generally decreases as bond order increases
ⓑ. bond length generally increases as bond order increases
ⓒ. charge alone decides bond length without orbital occupancy
ⓓ. all oxygen species must have identical \(\mathrm{O-O}\) bond lengths
Correct Answer: bond length generally decreases as bond order increases
Explanation: In related species, bond order is a useful measure of net bonding strength. A higher bond order means more net bonding effect after antibonding electrons are considered. Stronger bonding pulls nuclei closer and usually shortens the bond. In the oxygen species series, \(\mathrm{O_2^+}\) has the highest bond order and shortest bond, while \(\mathrm{O_2^{2-}}\) has the lowest bond order and longest bond. The trend depends on molecular orbital electron occupancy, not merely on ionic charge.
417. A molecular orbital comparison gives the following data.
| Species | Unpaired electrons | Magnetic nature |
|---|
| P. \(\mathrm{O_2}\) | \(2\) | paramagnetic |
| Q. \(\mathrm{O_2^+}\) | \(1\) | paramagnetic |
| R. \(\mathrm{O_2^-}\) | \(1\) | paramagnetic |
| S. \(\mathrm{O_2^{2-}}\) | \(0\) | paramagnetic |
The row that needs correction is
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: Paramagnetism requires at least one unpaired electron. \(\mathrm{O_2}\) has two unpaired electrons in antibonding \(\pi^*\) orbitals, so it is paramagnetic. \(\mathrm{O_2^+}\) and \(\mathrm{O_2^-}\) each have one unpaired electron in the simple molecular orbital picture, so they are also paramagnetic. \(\mathrm{O_2^{2-}}\) has all electrons paired, so it is diamagnetic. Row S confuses a filled antibonding set with an unpaired-electron situation.
418. In the molecular orbital description of \(\mathrm{B_2}\), two electrons occupy degenerate \(\pi_{2p}\) orbitals singly. The molecule is therefore
ⓐ. diamagnetic with bond order \(0\)
ⓑ. diamagnetic with bond order \(2\)
ⓒ. paramagnetic with bond order \(1\)
ⓓ. paramagnetic with bond order \(3\)
Correct Answer: paramagnetic with bond order \(1\)
Explanation: \(\mathrm{B_2}\) has valence electrons that occupy the lower \(2s\)-based molecular orbitals and then the degenerate \(\pi_{2p}\) bonding orbitals in the usual ordering for lighter second-period diatomic molecules. The two electrons in the \(\pi_{2p}\) set occupy separate degenerate orbitals according to Hund's rule. This leaves two unpaired electrons, so \(\mathrm{B_2}\) is paramagnetic. The net excess of bonding electrons over antibonding electrons gives bond order \(1\). This result is an important success of molecular orbital theory because simple Lewis pairing alone does not predict the paramagnetism clearly.
419. The species \(\mathrm{C_2}\) is predicted by simple molecular orbital theory to have bond order \(2\) and no unpaired electrons. Its magnetic nature is
ⓐ. paramagnetic
ⓑ. ferromagnetic only
ⓒ. impossible to decide from electron pairing
ⓓ. diamagnetic
Correct Answer: diamagnetic
Explanation: Magnetic behaviour depends on whether electrons are paired or unpaired. In the simple molecular orbital configuration of \(\mathrm{C_2}\), all electrons are paired. A species with all electrons paired is diamagnetic. The bond order \(2\) indicates net bonding, but it does not by itself prove paramagnetism. For magnetism, the actual pairing pattern in the molecular orbitals must be checked.
420. For second-period homonuclear diatomic molecules up to \(\mathrm{N_2}\), the usual valence molecular orbital order places
ⓐ. \(\sigma(2p)\) below \(\pi(2p)\)
ⓑ. \(\sigma^*(2p)\) below all bonding orbitals
ⓒ. \(\pi(2p)\) below \(\sigma(2p)\)
ⓓ. \(\pi^*(2p)\) below \(\pi(2p)\)
Correct Answer: \(\pi(2p)\) below \(\sigma(2p)\)
Explanation: For lighter second-period homonuclear diatomic molecules such as \(\mathrm{B_2}\), \(\mathrm{C_2}\), and \(\mathrm{N_2}\), the usual ordering places the \(\pi(2p)\) molecular orbitals below the \(\sigma(2p)\) orbital. This ordering differs from that used for \(\mathrm{O_2}\) and \(\mathrm{F_2}\). The difference arises from interactions among molecular orbitals of suitable symmetry in the lighter molecules. Using the wrong order can give wrong predictions for electron pairing and magnetism. The order must be selected according to the molecule being considered.
