Find out sum of bond order of CO & NO⁺.

The correct answer is 6.

To find the sum of the bond orders of $CO$ and $N{O}^{+}$, we can analyze their electronic configurations using Molecular Orbital (MO) Theory.

Step 1: Determine the total number of electrons in each species
For carbon monoxide ($CO$):
Carbon (C) has an atomic number of 6 (6 electrons).
Oxygen (O) has an atomic number of 8 (8 electrons).
Total electrons in $CO$ = 6 + 8 = 14 electrons.

For the nitrosonium ion ($N{O}^{+}$):
Nitrogen (N) has an atomic number of 7 (7 electrons).
Oxygen (O) has an atomic number of 8 (8 electrons).
The positive charge (+) indicates the loss of 1 electron.
Total electrons in $N{O}^{+}$ = 7 + 8 - 1 = 14 electrons.

Since both $CO$ and $N{O}^{+}$ have 14 electrons, they are isoelectronic species. Isoelectronic molecules or ions have the same molecular orbital configuration and therefore have the same bond order.

Step 2: Write the Molecular Orbital configuration for 14 electrons
The filling sequence of molecular orbitals for species containing 14 electrons (like $N_2$, $CO$, etc.) is:
$\sigma 1s^2,{\sigma }^{*}1s^2,\sigma 2s^2,{\sigma }^{*}2s^2,\pi 2p_x^2=\pi 2p_y^2,\sigma 2p_z^2$

Step 3: Calculate the Bond Order
Bond order is given by the formula:
$\text{Bond Order}=\frac{N_b-N_a}{2}$
where:
$N_b$ is the number of bonding electrons.
$N_a$ is the number of antibonding electrons (denoted with an asterisk *).

From the configuration, we count the electrons:
Bonding electrons ($N_b$) = 2 (from $\sigma 1s$) + 2 (from $\sigma 2s$) + 4 (from $\pi 2p_{x,y}$) + 2 (from $\sigma 2p_z$) = 10 electrons.
Antibonding electrons ($N_a$) = 2 (from ${\sigma }^{*}1s$) + 2 (from ${\sigma }^{*}2s$) = 4 electrons.

Substituting these values into the formula:
$\text{Bond Order of CO}=\frac{10-4}{2}=3$

Since $N{O}^{+}$ is isoelectronic with $CO$, its bond order is also 3.

Step 4: Calculate the sum of the bond orders
$\text{Sum of Bond Orders}=\text{Bond Order of CO}+\text{Bond Order of}N{O}^{+}$
$\text{Sum of Bond Orders}=3+3=6$