Question

Order by bond length?$\mathrm{NO},{\mathrm{NO}}^{+},{\mathrm{NO}}^{-}$

Answer 1

Molecular orbital theory:

• Molecular orbitals embrace the nuclei of all bonded atoms in a molecule.
• Molecular orbitals are formed by the combination of atomic orbitals of comparable energy.
• The number of molecular orbitals formed is equal to the number of combined atomic orbitals.
• When two atomic orbitals combine we get two molecular orbitals, one is the bonding molecular orbital and the other is the antibonding molecular orbital.
• Each MO accommodates a maximum of two electrons with opposite spin.
• The shape of MO depends on the shape of the atomic orbitals.
• Electrons are filled in MOs in the increasing order of energy.
• The pairing of electrons in degenerate molecular orbitals cannot take place unless each degenerate molecular orbital is singly occupied(Hund's rule)

The molecular orbital electronic configuration:

• The order of increasing energy of molecular orbitals obtained by the combination of $1s,2s$and $2p$orbitals of two atoms is,

$\sigma 1s<\sigma *1s<\sigma 2s<\sigma *2s<\sigma 2p_z<\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}<\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{x}}=\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{y}}<\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{z}}$

• The molecular orbitals are filled in this order $\sigma 1s,\sigma *1s,\sigma 2s,\sigma *2s,\sigma 2p_z,\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}},\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{x}}=\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{y}},\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{z}}$.
• For diatomic molecules, this order is not correct that is $\sigma 2p_z$molecular orbital is higher in energy than $\sigma 2p_x$ and $\sigma 2p_y$ molecular orbitals.
• Therefore in diatoms, the molecular orbitals are filled in this order $\sigma 1s,\sigma *1s,\sigma 2s,\sigma *2s,\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}},\sigma 2p_z,\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{x}}=\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{y}},\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{z}}$.

Bond order:

• Bond order is defined as the number of covalent bonds in a molecule.
• It is equal to half of the difference between the number of electrons in bonding$\left(N_b\right)$ and antibonding orbitals$\left(N_a\right)$.

That is, the bond order $\mathrm{B}.\mathrm{O}=\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2}$

Part 1: Bond order of $\mathrm{NO}$

The total electron of $\mathrm{NO}$ molecules is $7+8=15$

The molecular orbital electronic configuration can be written as

$\sigma 1s^2,\sigma *1s^2,\sigma 2s^2,\sigma *2s^2,\sigma 2{p_z}^2,\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^2=\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2,\mathrm{\pi }*2{{\mathrm{p}}_{\mathrm{x}}}^1=\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{y}}$

That is, the number of electrons in bonding$\left(N_b\right)$ is 10

The number of electrons in antibonding$\left(N_b\right)$ is 5

So the bond order of $\mathrm{NO}$ is;

$$\begin{gathered} \mathrm{B}.\mathrm{O}=\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2} \\ \mathrm{B}.\mathrm{O}=\frac{10-5}{2} \\ \mathrm{B}.\mathrm{O}=\frac{5}{2} \\ \mathrm{B}.\mathrm{O}=2.5 \end{gathered}$$

Part 2: Bond order of ${\mathrm{NO}}^{+}$

The total electron of $\mathrm{NO}$ molecule is $7+8=15$

The positive charge indicates the loss of one electron

So the total electron of ${\mathrm{NO}}^{+}$ molecules is $15-1=14$

The molecular orbital electronic configuration can be written as

$\sigma 1s^2,\sigma *1s^2,\sigma 2s^2,\sigma *2s^2,\sigma 2{p_z}^2,\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^2=\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2$

That is, the number of electrons in bonding$\left(N_b\right)$ is 10

The number of electrons in antibonding$\left(N_b\right)$ is 4

So the bond order of ${\mathrm{NO}}^{+}$ is;

$$\begin{gathered} \mathrm{B}.\mathrm{O}=\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2} \\ \mathrm{B}.\mathrm{O}=\frac{10-4}{2} \\ \mathrm{B}.\mathrm{O}=\frac{6}{2} \\ \mathrm{B}.\mathrm{O}=3 \end{gathered}$$

Part 3: Bond order of ${\mathrm{NO}}^{-}$

The total electron of $\mathrm{NO}$ molecules is $7+8=15$

The negative charge indicates the gain of one electron

So the total electron of ${\mathrm{NO}}^{-}$ molecules is $15+1=16$

The molecular orbital electronic configuration can be written as

$\sigma 1s,\sigma *1s,\sigma 2s,\sigma *2s,\sigma 2p_z,\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}},\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{x}}=\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{y}},\mathrm{\pi }*2{\mathrm{p}}_{\mathrm{z}}$

$\sigma 1s^2,\sigma *1s^2,\sigma 2s^2,\sigma *2s^2,\sigma 2{p_z}^2,\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^2=\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2,\mathrm{\pi }*2{{\mathrm{p}}_{\mathrm{x}}}^1=\mathrm{\pi }*2{{\mathrm{p}}_{\mathrm{y}}}^1$

That is, the number of electrons in bonding$\left(N_b\right)$ is 10

The number of electrons in antibonding$\left(N_b\right)$ is 6

So the bond order of ${\mathrm{NO}}^{-}$ is;

$$\begin{gathered} \mathrm{B}.\mathrm{O}=\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2} \\ \mathrm{B}.\mathrm{O}=\frac{10-6}{2} \\ \mathrm{B}.\mathrm{O}=\frac{4}{2} \\ \mathrm{B}.\mathrm{O}=2 \end{gathered}$$

Bond length:

• The distance between the nuclei of two chemically bonded atoms in a molecule is known as the bond length,
• Bond length is inversely proportional to the bond order and stability.
• That is, as bond order increases, the bond length will decreases
• Therefore, the order by bond length is ${\mathrm{NO}}^{+}<\mathrm{NO}<{\mathrm{NO}}^{-}$