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

We begin by recalling that nitric oxide, represented as $$NO$$, is a heteronuclear diatomic molecule containing a total of $$11$$ valence electrons: $$5$$ from nitrogen and $$6$$ from oxygen.

To understand its magnetic property, we write the molecular orbital (MO) configuration that results when the atomic orbitals of nitrogen and oxygen combine. The energy order for the MOs of a heteronuclear diatomic molecule in the second period is the same as for $$N_2$$ after the inversion point, namely

$$\sigma_{1s}^2\,\sigma_{1s}^{\*2}\,\sigma_{2s}^2\,\sigma_{2s}^{\*2}\, \sigma_{2p_z}^2\,\pi_{2p_x}^2 = \pi_{2p_y}^2\,\pi_{2p_x}^{\*1} = \pi_{2p_y}^{\*0}\, \sigma_{2p_z}^{\*0}$$

However, because $$NO$$ possesses one more electron than $$N_2$$, this eleventh electron occupies the $$\pi_{2p_x}^{\*}$$ or equivalently one of the degenerate $$\pi_{2p}^{\*}$$ orbitals. Thus there is an unpaired electron present.

The presence of a single unpaired electron makes $$NO$$ paramagnetic, that is, it is attracted by a magnetic field. Diamagnetism corresponds to the situation in which all electrons are paired. Therefore, the statement “It is diamagnetic in gaseous state” is false.

Next we verify the other three listed properties to see that they are indeed correct for $$NO$$.

We call an oxide neutral when it is neither distinctly acidic nor basic in water. Nitric oxide shows extremely weak interactions with water, does not produce $$H^+$$ or $$OH^-$$ ions in significant amount, and hence is classified as a neutral oxide. So the statement “It is a neutral oxide” is true.

We also know from elementary kinetics that gaseous $$NO$$ reacts rapidly with oxygen present in air to give brown nitrogen dioxide. The balanced equation is

$$2\,NO + O_2 \rightarrow 2\,NO_2$$

Because this combination indeed occurs, the statement “It combines with oxygen to form nitrogen dioxide” is true.

Finally, we determine the bond order. The general formula connecting bond order ($$B.O.$$) with the numbers of bonding electrons ($$N_b$$) and antibonding electrons ($$N_a$$) in the MO picture is first stated:

$$B.O. = \dfrac{N_b - N_a}{2}$$

From the configuration written above we count:

Bonding electrons $$N_b = 2\,( \sigma_{1s}) + 2\,( \sigma_{2s}) + 2\,( \sigma_{2p_z}) + 4\,( \pi_{2p_x}, \pi_{2p_y}) = 10$$

Antibonding electrons $$N_a = 2\,( \sigma_{1s}^{\*}) + 2\,( \sigma_{2s}^{\*}) + 1\,( \pi_{2p}^{\*}) = 5$$

Substituting these values into the formula we get

$$B.O. = \dfrac{10 - 5}{2} = \dfrac{5}{2} = 2.5$$

Thus the statement “Its bond order is 2.5” is true.

Among the four alternatives only the first one is incorrect, because in reality $$NO$$ is paramagnetic, not diamagnetic.

Hence, the correct answer is Option A.