Match List-I with List-II
List-I (Molecule)
A. Ne2
B. N2
C. F2
D. O2
List-II (Bond order)
(i) 1
(ii)

Q: Why does N2 have a different MO filling order than O2?

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 that they interact (mix). This pushes the $\sigma 2p_z$ orbital higher in energy, placing it above the $\pi 2p_x$ and $\pi 2p_y$ orbitals. For O and F, the energy gap is too large, so no mixing occurs, and $\sigma 2p_z$ stays lower.

Q: Can a bond order be zero or negative?

Yes! A bond order of zero (like in $\text{Ne}_2$ or $\text{He}_2$) means the anti-bonding electrons completely cancel out the stabilizing effect of the bonding electrons. In reality, this means the molecule simply does not exist under normal conditions.

Q: How do I quickly find the bond order without writing the full configuration?

Use the 14-electron trick! Memorize that 14 electrons give a bond order of 3.0. For every single electron you add or subtract from 14, just subtract 0.5 from the bond order. For example, $\text{O}_2$ has 16 electrons (2 more than 14), so its bond order is $3.0 - 2(0.5) = 2.0$.

Q: Are anti-bonding electrons always bad for a molecule?

Yes, in terms of bond strength. Electrons in anti-bonding orbitals reside in regions of space that pull the nuclei apart, destabilizing the molecule and reducing the overall bond order.

Q: Why is O2 paramagnetic even though it has an even number of electrons (16)?

This is the greatest triumph of MO theory! When filling the last two electrons in $\text{O}_2$, they enter the degenerate (equal energy) $\pi^* 2p_x$ and $\pi^* 2p_y$ anti-bonding orbitals. According to Hund's Rule, they must occupy separate orbitals with parallel spins. These two unpaired electrons make liquid oxygen magnetic!

Question

Match List-I with List-II
Choose the correct answer from the options given below.

List-I:

P. $	ext{Ne}_2$

Q. $	ext{N}_2$

R. $	ext{F}_2$

S. $	ext{O}_2$

List-II:

1. 1

2. 2

3. 0

4. 3

Sigma Percentile · Accepted Answer

### The Magic of Molecular Orbital Theory

Have you ever wondered why liquid oxygen is attracted to a magnet, even though its Lewis dot structure shows all electrons perfectly paired up? This was one of the greatest mysteries in early chemistry, and it was beautifully solved by **Molecular Orbital (MO) Theory**.

Unlike Valence Bond Theory, which assumes electrons belong to specific bonds between atoms, MO theory treats electrons as belonging to the entire molecule. When atomic orbitals combine, they form **bonding molecular orbitals** (which stabilize the molecule) and **anti-bonding molecular orbitals** (which destabilize it).

The stability of a molecule is quantified by a simple yet powerful concept called **Bond Order**. The formula is elegantly straightforward:

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

Where $N_b$ is the number of electrons in bonding orbitals, and $N_a$ is the number of electrons in anti-bonding orbitals. Let's apply this to the molecules in our problem.

### Analyzing the Setup: Molecule by Molecule

#### The Case of Neon ($	ext{Ne}_2$)
Imagine trying to force two noble gas atoms together. A single Neon atom has $10$ electrons, so a hypothetical $	ext{Ne}_2$ molecule would have $20$ electrons.

If we fill the molecular orbitals, we get $10$ bonding electrons and $10$ anti-bonding electrons.

$$	ext{Bond Order} = \frac{10 - 10}{2} = 0$$

A bond order of zero means there is no net stabilizing force. The molecule simply does not exist! This perfectly explains why noble gases are monatomic. So, $	ext{Ne}_2$ matches with a bond order of $0$.

#### The Case of Nitrogen ($	ext{N}_2$)
Nitrogen is the workhorse of our atmosphere. With $7$ electrons per atom, $	ext{N}_2$ has $14$ electrons in total.

Filling the orbitals up to $14$ electrons gives us $10$ bonding electrons and only $4$ anti-bonding electrons (in the $\sigma^* 1s$ and $\sigma^* 2s$ orbitals).

$$	ext{Bond Order} = \frac{10 - 4}{2} = 3$$

A bond order of $3$ signifies a robust **triple bond**, making $	ext{N}_2$ incredibly stable and unreactive. Thus, $	ext{N}_2$ matches with a bond order of $3$.

#### The Case of Fluorine ($	ext{F}_2$)
Fluorine is highly reactive. An $	ext{F}_2$ molecule has $18$ electrons.

Compared to Nitrogen, the $4$ extra electrons are forced into the high-energy anti-bonding $\pi^* 2p$ orbitals. This increases $N_a$ to $8$, while $N_b$ remains at $10$.

$$	ext{Bond Order} = \frac{10 - 8}{2} = 1$$

A bond order of $1$ means a **single bond**. The high number of anti-bonding electrons makes this bond relatively weak and easy to break, explaining fluorine's fierce reactivity. $	ext{F}_2$ matches with a bond order of $1$.

#### The Case of Oxygen ($	ext{O}_2$)
Finally, the oxygen we breathe. $	ext{O}_2$ has $16$ electrons.

Here, $2$ electrons enter the anti-bonding $\pi^* 2p$ orbitals. According to Hund's rule, they occupy separate degenerate orbitals with parallel spins, which is exactly why oxygen is paramagnetic!

$$	ext{Bond Order} = \frac{10 - 6}{2} = 2$$

A bond order of $2$ indicates a **double bond**. $	ext{O}_2$ matches with a bond order of $2$.

### The 14-Electron Rule Shortcut

While writing out the full electronic configuration is rigorous, there is a brilliant shortcut for competitive exams like JEE.

**The 14-Electron Rule:** A diatomic molecule with exactly $14$ electrons (like $	ext{N}_2$ or $	ext{CO}$) always has a maximum bond order of **$3.0$**.

For every electron you add or remove from $14$, the bond order decreases by exactly **$0.5$**.

*   $14 e^- ightarrow 	ext{BO} = 3.0$
*   $15 e^- ightarrow 	ext{BO} = 2.5$
*   $16 e^- ightarrow 	ext{BO} = 2.0$ (Oxygen)
*   $18 e^- ightarrow 	ext{BO} = 1.0$ (Fluorine)
*   $20 e^- ightarrow 	ext{BO} = 0.0$ (Neon)

Using this trick, you can solve matrix match questions like this in under ten seconds!

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Match List-I with List-II Choose the correct answer from the options given below.

List-I

List-II

Select Matching Pairs:

Visualized Solution

$The Bond Order Formula$

$Bond Order of Ne_{2}$

$Bond Order of N_{2}$

$Bond Order of F_{2}$

$Bond Order of O_{2}$

$Final Matching$

$The 14-Electron Rule Shortcut$

The Sigma Insight: Molecular Orbital Theory

The Magic of Molecular Orbital Theory

Analyzing the Setup

# The Case of Neon ( $Ne_{2}$ ) Imagine trying to force two noble gas atoms together

# The Case of Nitrogen ( $N_{2}$ ) Nitrogen is the workhorse of our atmosphere

# The Case of Fluorine ( $F_{2}$ ) Fluorine is highly reactive

# The Case of Oxygen ( $O_{2}$ )

The 14-Electron Rule Shortcut

Similar Questions

In the following the correct bond order sequence is

Which of the following molecules/ions does not contain unpaired electrons?

Which one of the following constitutes a group of the isoelectronic species?

Among $H_{2}$ , $He_{2}^{+}$ , $Li_{2}$ , $Be_{2}$ , $B_{2}$ , $C_{2}$ , $N_{2}$ , $O_{2}^{-}$ , and $F_{2}$ , the number of diamagnetic species is - (Atomic number : H = 1, He = 2, Li = 3, Be = 4, B = 5, C = 6, N = 7, O = 8, f = 9)

Number of paramagnetic oxides among the following given oxides is $Li_{2} O, Na_{2} O_{2}, KO_{2}, HgO$ and $K_{2} O$

In which of the following pairs of molecules/ions both the species are not likely to exist?

Among the following, the molecule expected to be stabilised by anion formation is $C_{2}, O_{2}, N O, F_{2}$ .

During the change of $O_{2}$ to $O_{2}^{-}$ , the incoming electron goes to the orbital.

Two pi and half sigma bonds are present in

Assuming 2s-2p mixing is NOT operative, the paramagnetic species among the following is :

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Match List-I with List-II Choose the correct answer from the options given below.

List-I

List-II

Select Matching Pairs:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bond Order of Ne2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bond Order of N2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bond Order of F2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bond Order of O2​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The 14-Electron Rule Shortcut