According to Molecular Orbital Theory,
(A) C2^2- is expected to be diamagnetic
(B) O2^2+ is expected

Q: Why does the energy sequence change for molecules with more than 14 electrons?

This is due to a phenomenon called $s-p$ mixing. In atoms like B, C, and N, the $2s$ and $2p$ orbitals are close enough in energy to interact (mix), which pushes the $\sigma_{2p_z}$ orbital higher in energy than the $\pi_{2p}$ orbitals. In O and F, the energy gap between $2s$ and $2p$ is too large for effective mixing, so $\sigma_{2p_z}$ remains lower in energy.

Q: Can a molecule with a fractional bond order actually exist?

Yes, absolutely! A fractional bond order, like $0.5$ in $He_2^+$ or $2.5$ in $N_2^+$, simply indicates that the net bonding force is equivalent to a fraction of a standard covalent bond. As long as the bond order is greater than zero, the molecule can exist, though it might be highly reactive.

Q: Why is $O_2$ paramagnetic even though it has an even number of electrons (16)?

This is one of the greatest triumphs of Molecular Orbital Theory! When you fill the 16 electrons of $O_2$ into the MO diagram, the last two electrons enter the degenerate (equal energy) $\pi_{2p_x}^*$ and $\pi_{2p_y}^*$ antibonding orbitals. According to Hund's Rule, they must occupy these orbitals singly with parallel spins. These two unpaired electrons make $O_2$ paramagnetic.

Q: If $N_2^+$ and $N_2^-$ have the same bond order, do they have the exact same bond length?

While their bond orders are identical ($2.5$), their bond lengths are slightly different. $N_2^-$ has more electrons, leading to greater interelectronic repulsion and a slightly less effective nuclear charge pulling the atoms together. Consequently, $N_2^-$ has a slightly longer bond length than $N_2^+$.

Q: What does a bond order of zero mean physically?

A bond order of zero means that the number of electrons in antibonding orbitals equals the number of electrons in bonding orbitals. The destabilizing effect perfectly cancels the stabilizing effect. Physically, this means no bond is formed, and the molecule (like $He_2$ or $Ne_2$) does not exist under normal conditions.

Question

According to Molecular Orbital Theory,

Options:

(A) $C_2^{2-}$ is expected to be diamagnetic

(B) $O_2^{2+}$ is expected to have a longer bond length than $O_2$

(C) $N_2^+$ and $N_2^-$ have the same bond order

(D) $He_2^+$ has the same energy as two isolated He atoms

Sigma Percentile · Accepted Answer

Molecular Orbital Theory (MOT) is one of the most elegant frameworks in chemistry, providing deep insights into the magnetic properties and bond strengths of molecules. In this problem, we are tasked with evaluating four different statements based on MOT. Let's embark on a journey to decode each option systematically.

### The Two Energy Sequences
Before diving into the options, it is crucial to recall that diatomic molecules of the second period follow two distinct energy filling sequences due to $s-p$ mixing:
1. **For molecules with $\le 14$ electrons** (like $B_2, C_2, N_2$): The $\sigma_{2p_z}$ orbital is pushed higher in energy than the $\pi_{2p_x}$ and $\pi_{2p_y}$ orbitals.
2. **For molecules with $> 14$ electrons** (like $O_2, F_2$): The $s-p$ mixing is negligible, and the $\sigma_{2p_z}$ orbital drops below the $\pi_{2p}$ orbitals.

### Analyzing Option (A): The Magnetic Nature of $C_2^{2-}$
Let's determine the total number of electrons in the $C_2^{2-}$ ion. A neutral $C_2$ molecule has $12$ electrons. The $2-$ charge indicates the addition of two extra electrons, bringing the total to **$14$ electrons**. This makes it isoelectronic with the $N_2$ molecule.

Filling these $14$ electrons into the appropriate energy sequence, we get:
$$\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^2 = \pi_{2p_y}^2 \sigma_{2p_z}^2$$

Observe the configuration carefully. Every single electron is paired up in its respective orbital. Since there are **zero unpaired electrons**, the species is strictly **diamagnetic**. Therefore, statement (A) is absolutely correct.

### Analyzing Option (B): Bond Length of $O_2^{2+}$ vs $O_2$
To compare bond lengths, we must first calculate the bond orders. The formula for bond order is:
$$	ext{Bond Order} = \frac{N_b - N_a}{2}$$
where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.

- **Neutral $O_2$** has $16$ electrons. Its bond order is $\frac{10 - 6}{2} = 2$.
- **$O_2^{2+}$** has lost two electrons from the highest occupied molecular orbitals, which are the antibonding $\pi_{2p}^*$ orbitals. With $14$ electrons, its bond order becomes $\frac{10 - 4}{2} = 3$.

A fundamental principle of chemical bonding states that **Bond Length is inversely proportional to Bond Order**. Since $O_2^{2+}$ has a higher bond order ($3$) compared to $O_2$ ($2$), it must have a **shorter** bond length, not longer. Thus, statement (B) is incorrect.

### Analyzing Option (C): Bond Orders of $N_2^+$ and $N_2^-$
Let's evaluate the bond orders for the ions of nitrogen.
- **Neutral $N_2$** has $14$ electrons and a bond order of $3$.
- **$N_2^+$** has $13$ electrons. It has lost one electron from the bonding $\sigma_{2p_z}$ orbital. Its bond order is $\frac{9 - 4}{2} = 2.5$.
- **$N_2^-$** has $15$ electrons. The extra electron enters the antibonding $\pi_{2p}^*$ orbital. Its bond order is $\frac{10 - 5}{2} = 2.5$.

Fascinatingly, whether you remove a bonding electron or add an antibonding electron to $N_2$, the net effect on the bond order is identical. Both $N_2^+$ and $N_2^-$ possess a bond order of **$2.5$**. Hence, statement (C) is correct.

### Analyzing Option (D): The Energy of $He_2^+$
The $He_2^+$ ion contains $3$ electrons. Its configuration is $\sigma_{1s}^2 \sigma_{1s}^{*1}$. 
Calculating the bond order gives:
$$	ext{Bond Order} = \frac{2 - 1}{2} = 0.5$$

Because the bond order is positive ($> 0$), a net attractive force exists, and a stable bond is formed. In thermodynamics, the formation of a stable bond releases energy. Therefore, the $He_2^+$ molecule sits in a lower energy well compared to the isolated $He$ and $He^+$ atoms. It does not have the same energy. Statement (D) is incorrect.

### Conclusion
Through rigorous application of Molecular Orbital Theory, we have deduced that statements **(A)** and **(C)** are the only correct options.

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: According to Molecular Orbital Theory,

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Visualized Solution

$M o l ec u l a r O r bi t a l T h eor y (M O T)$

$A na l y z in g O pt i o n (A) : C_{2}^{2 -}$

$M a g n e t i c N a t u r e o f C_{2}^{2 -}$

$A na l y z in g O pt i o n (B) : O_{2}^{2 +} v s O_{2}$

$B o n d L e n g t h C o m p a r i so n$

$A na l y z in g O pt i o n (C) : N_{2}^{+} an d N_{2}^{-}$

$B o n d O r d er C o n c l u s i o n$

$A na l y z in g O pt i o n (D) : H e_{2}^{+}$

$E n er g y o f H e_{2}^{+} v s I so l a t e d H e a t o m s$

$F ina l A n s w er$

The Sigma Insight: Molecular Orbital Theory

The Two Energy Sequences Before diving into the options, it is crucial to recall that diatomic molecules of the second period follow two distinct energy filling sequences due to $s - p$ mixing: 1. For molecules with $\leq 14$ electrons (like $B_{2}, C_{2}, N_{2}$ )

Analyzing Option (A)

Analyzing Option (B)

Analyzing Option (C)

Analyzing Option (D)

Conclusion

Similar Questions

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Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is(are)

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Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: According to Molecular Orbital Theory,

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Molecular Orbital Theory (MOT)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Option (A): C22−​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Nature of C22−​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Option (B): O22+​ vs O2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bond Length Comparison

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Option (C): N2+​ and N2−​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bond Order Conclusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Option (D): He2+​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy of He2+​ vs Isolated He atoms

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer