Among the following, the molecule expected to be stabilised by anion formation is C2, O2, NO, F2.
(a

Q: Why does the MO energy level order change for molecules with more than 14 electrons?

For molecules with 14 or fewer electrons (like $C_2$ and $N_2$), the $2s$ and $2p_z$ orbitals are close in energy and mix (s-p mixing), pushing the $\sigma_{2p_z}$ orbital higher in energy than the $\pi_{2p_x}$ and $\pi_{2p_y}$ orbitals. For $>14$ electrons (like $O_2$ and $F_2$), the energy gap is larger, so s-p mixing is negligible, and $\sigma_{2p_z}$ is lower in energy.

Q: Does an increase in bond order always mean increased stability?

Yes! A higher bond order means there are more bonding electrons relative to anti-bonding electrons, which pulls the atoms closer together, increases bond strength, and lowers the overall energy of the system, making it more stable.

Q: What happens if the incoming electron enters an anti-bonding orbital?

Anti-bonding orbitals have higher energy and nodes between the nuclei. Placing an electron there cancels out the stabilizing effect of a bonding electron, decreasing the bond order and destabilizing the molecule.

Q: Why is $C_2$ diamagnetic while $O_2$ is paramagnetic?

In $C_2$ (12 electrons), the highest occupied molecular orbitals are the degenerate $\pi_{2p_x}$ and $\pi_{2p_y}$, which are fully occupied with 4 electrons, leaving no unpaired electrons. In $O_2$ (16 electrons), the last 2 electrons enter the degenerate $\pi^*_{2p_x}$ and $\pi^*_{2p_y}$ orbitals singly (Hund's rule), making it paramagnetic.

Q: Can bond order be a fraction?

Absolutely! Since bond order is calculated as $(N_b - N_a)/2$, an odd number of total electrons will result in a fractional bond order, like 1.5 or 2.5. This simply indicates a bond strength intermediate between integer values.

Question

Among the following, the molecule expected to be stabilised by anion formation is $C_2, O_2, NO, F_2$.

Options:

(A) $C_2$

(B) $F_2$

(C) $NO$

(D) $O_2$

Sigma Percentile · Accepted Answer

The journey to understanding molecular stability through the lens of Molecular Orbital Theory (MOT) is one of the most fascinating adventures in chemistry. It allows us to peek into the quantum mechanical world and predict how molecules will behave when they gain or lose electrons.

### Analyzing the Setup

We are presented with four diatomic molecules: $C_2$, $O_2$, $NO$, and $F_2$. The question asks us to identify which of these molecules becomes more stable when it gains an electron to form an anion.

To answer this, we need a reliable metric for stability. In MOT, that metric is the **Bond Order**. The bond order gives us a direct measure of the net bonding interactions between two atoms. A higher bond order means a stronger, shorter, and more stable bond.

### The Master Equation

The formula for bond order is beautifully simple:

$$ 	ext{Bond Order} = \frac{N_b - N_a}{2} $$

Here, $N_b$ represents the number of electrons in bonding molecular orbitals, and $N_a$ represents the number of electrons in anti-bonding molecular orbitals.

When a molecule gains an electron, that electron must go into the lowest available unoccupied molecular orbital (LUMO). 
- If the LUMO is a **bonding orbital**, $N_b$ increases, the bond order increases, and the molecule is **stabilised**.
- If the LUMO is an **anti-bonding orbital**, $N_a$ increases, the bond order decreases, and the molecule is **destabilised**.

### Final Calculation

Let's evaluate our candidates one by one.

**1. The Carbon Molecule ($C_2$)**
A neutral $C_2$ molecule has 12 electrons. Its molecular orbital configuration is:
$$ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 = \pi_{2p_y}^2 $$
Calculating the bond order:
$$ 	ext{B.O.} = \frac{8 - 4}{2} = 2 $$

When $C_2$ gains an electron to form $C_2^-$, it has 13 electrons. The next available orbital is the bonding $\sigma_{2p_z}$ orbital.
$$ C_2^- : \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 = \pi_{2p_y}^2 \sigma_{2p_z}^1 $$
The new bond order becomes:
$$ 	ext{B.O.} = \frac{9 - 4}{2} = 2.5 $$
Since the bond order increased from 2 to 2.5, $C_2$ is **stabilised** by anion formation!

**2. The Other Candidates ($O_2$, $NO$, $F_2$)**
Let's quickly see why the others fail this test.
- **$O_2$ (16 electrons):** The highest occupied orbitals are the anti-bonding $\pi^*_{2p_x}$ and $\pi^*_{2p_y}$. An incoming electron will enter one of these anti-bonding orbitals, decreasing the bond order from 2 to 1.5.
- **$NO$ (15 electrons):** The last electron is already in an anti-bonding $\pi^*_{2p}$ orbital. A new electron will also enter this anti-bonding orbital, dropping the bond order from 2.5 to 2.
- **$F_2$ (18 electrons):** The highest occupied orbitals are the anti-bonding $\pi^*_{2p}$ orbitals. The next available orbital is the highly anti-bonding $\sigma^*_{2p_z}$. Adding an electron here plummets the bond order from 1 to 0.5.

In all these cases, the incoming electron enters an anti-bonding orbital, which destabilises the molecule. Therefore, $C_2$ stands alone as the correct answer.

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Among the following, the molecule expected to be stabilised by anion formation is $C_{2}, O_{2}, N O, F_{2}$ .

Select Answer:

Visualized Solution

$T h e S e t u p$

$B o n d O r d er an d S t abi l i t y$

$E l ec t r o ni c C o n f i g u r a t i o n o f C_{2}$

$B o n d O r d er o f C_{2} an d C_{2}^{-}$

$A na l y s i so f O_{2}, N O, F_{2}$

$C o n c l u s i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Molecular Orbital Theory

Analyzing the Setup

The Master Equation

Final Calculation

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