Which one of the following properties is not shown by NO?
(a) It is diamagnetic in gaseous state
(b)

Q: Why is NO paramagnetic in the gaseous state but diamagnetic in the liquid/solid state?

In the gaseous state, NO exists as a monomer with one unpaired electron, making it paramagnetic. However, at low temperatures (liquid/solid state), it dimerizes to form $N_2O_2$. In this dimer, the unpaired electrons from two NO molecules pair up, making the resulting substance diamagnetic.

Q: How do we know the order of molecular orbitals for NO?

For molecules with 14 or fewer electrons (like $N_2$), the $\pi_{2p}$ orbitals are lower in energy than $\sigma_{2p_z}$ due to strong s-p mixing. However, for molecules with more than 14 electrons (like $O_2$, $F_2$, and NO), the energy gap between the 2s and 2p atomic orbitals is larger. This reduces s-p mixing, causing the $\sigma_{2p_z}$ orbital to drop below the $\pi_{2p}$ orbitals in energy.

Q: Is NO an acidic or basic oxide?

NO is a neutral oxide. It does not react with water to form an acid or a base. Other common examples of neutral oxides include nitrous oxide ($N_2O$) and carbon monoxide ($CO$).

Q: How exactly do we calculate the bond order here?

Bond order is calculated as half the difference between the number of bonding electrons ($N_b$) and antibonding electrons ($N_a$). For NO, if we count all electrons, $N_b = 10$ and $N_a = 5$. Plugging this into the formula gives $(10 - 5) / 2 = 2.5$.

Q: Why does NO react so easily with $O_2$ to form $NO_2$?

The unpaired electron in NO makes it a free radical, which is highly reactive. It readily combines with oxygen in the air to pair up that electron and form the more stable, brown-colored nitrogen dioxide ($NO_2$) gas.

Question

Which one of the following properties is not shown by NO?

Options:

(A) It is diamagnetic in gaseous state

(B) It is a neutral oxide

(C) It combines with oxygen to form nitrogen dioxide

(D) Its bond order is $2.5$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When we encounter a question asking about the properties of a specific molecule like Nitric Oxide ($	ext{NO}$), our first instinct should be to look at its electronic structure. The Molecular Orbital Theory (MOT) is the perfect tool for this.

Let's start by counting the total number of electrons in the $	ext{NO}$ molecule. Nitrogen contributes $7$ electrons, and Oxygen contributes $8$ electrons. This gives us a total of $15$ electrons. The fact that this is an odd number is a massive hint about the molecule's magnetic properties!

### The Master Equation: Molecular Orbital Configuration

Out of these $15$ electrons, the first $4$ will completely fill the inner shell orbitals, specifically the $\sigma 1s$ and $\sigma^* 1s$ orbitals. This leaves us with $11$ valence electrons to distribute among the higher energy molecular orbitals.

Following the Aufbau principle and Hund's rule, we start filling the valence molecular orbitals from lowest to highest energy:
1. The $\sigma_{2s}$ and $\sigma^*_{2s}$ orbitals take $2$ electrons each ($4$ electrons total).
2. Next, the $\sigma_{2p_z}$ orbital takes $2$ electrons.
3. The degenerate $\pi_{2p_x}$ and $\pi_{2p_y}$ orbitals take $2$ electrons each ($4$ electrons total).

At this point, we have successfully placed $10$ valence electrons. But we have one electron left! This final, lonely electron must go into the next available orbital, which is the anti-bonding $\pi^*_{2p_x}$ orbital.

### The Consequence of the Unpaired Electron

**This single unpaired electron in the $\pi^*_{2p_x}$ orbital is the key to the whole problem.**

Any molecule that contains one or more unpaired electrons is attracted by a magnetic field, a property known as **paramagnetism**. Therefore, in its gaseous state, $	ext{NO}$ is paramagnetic.

However, if we look at option (a), it claims that $	ext{NO}$ is diamagnetic in the gaseous state. This is a direct contradiction to our findings! Thus, option (a) is the incorrect statement we were looking for.

### Verifying the Other Properties

Just to be absolutely certain, let's quickly verify the other options.

To find the bond order, we use the formula:
$$	ext{Bond Order} = \frac{N_b - N_a}{2}$$

Where $N_b$ is the number of bonding electrons and $N_a$ is the number of anti-bonding electrons. Counting them up from our configuration, we have $10$ bonding electrons and $5$ anti-bonding electrons.
$$	ext{Bond Order} = \frac{10 - 5}{2} = 2.5$$
This confirms that option (d) is a correct statement.

Furthermore, chemically, $	ext{NO}$ is well-known as a **neutral oxide** (along with $	ext{N}_2	ext{O}$ and $	ext{CO}$), meaning it does not form an acid or base when reacted with water. It is also highly reactive due to its free-radical nature and readily combines with atmospheric oxygen to form the brown gas, nitrogen dioxide ($	ext{NO}_2$).

$$2	ext{NO} + 	ext{O}_2 ightarrow 2	ext{NO}_2$$

This confirms options (b) and (c) are also correct statements. Therefore, the only incorrect property listed is option (a).

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Which one of the following properties is not shown by NO?

Select Answer:

Visualized Solution

$T o t a l E l ec t r o n s in N O$

$I nn er S h e l l E l ec t r o n s$

$F i l l in g σ_{2 s} an d σ_{2 s}^{*}$

$F i l l in g σ_{2 p_{z}} an d π_{2 p}$

$T h e U n p ai r e d E l ec t r o n$

$M a g n e t i c N a t u r e$

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

$C h e mi c a l P r o p er t i es$

The Sigma Insight: Molecular Orbital Theory

Analyzing the Setup

The Master Equation

The Consequence of the Unpaired Electron

Verifying the Other Properties

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