Number of molecules having bond order 2 from the following molecules is \underline\hspace2cm . C_2, O_2, Be_2, Li_2, Ne_2, ~N_2, He_2

Chemical Bonding and Molecular Structure PYQ — JEE Main 2024 · 8 Apr · Shift 2 · Q30 with Solution

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Correct answer: 2

Use bond order formula

For diatomic molecules,

\text{Bond order} = \frac{N_b - N_a}{2}

where $N_b$ = number of bonding electrons and $N_a$ = number of antibonding electrons.

We check each given molecule.

$\mathrm{C}_2$

Each C has 6 electrons, so $\mathrm{C}_2$ has 12 electrons. For molecules up to $\mathrm{N}_2$ , MO order is:

\sigma(1s),\ \sigma^*(1s),\ \sigma(2s),\ \sigma^*(2s),\ \pi(2p_x)=\pi(2p_y),\ \sigma(2p_z)

Configuration:

(\sigma 1s)^2(\sigma^*1s)^2(\sigma 2s)^2(\sigma^*2s)^2(\pi 2p_x)^2(\pi 2p_y)^2

Bonding electrons = $8$ , antibonding electrons = $4$ . So,

\text{B.O.} = \frac{8-4}{2}=2

Hence, $\mathrm{C}_2$ has bond order 2.

$\mathrm{O}_2$

Each O has 8 electrons, so total = 16 electrons. For $\mathrm{O}_2$ , MO order is:

\sigma(1s),\ \sigma^*(1s),\ \sigma(2s),\ \sigma^*(2s),\ \sigma(2p_z),\ \pi(2p_x)=\pi(2p_y),\ \pi^*(2p_x)=\pi^*(2p_y)

Configuration gives bond order:

\text{B.O.} = 2

Hence, $\mathrm{O}_2$ has bond order 2.

$\mathrm{Be}_2$

Each Be has 4 electrons, so total = 8 electrons. Configuration:

(\sigma 1s)^2(\sigma^*1s)^2(\sigma 2s)^2(\sigma^*2s)^2

So,

\text{B.O.} = \frac{4-4}{2}=0

Not equal to 2.

$\mathrm{Li}_2$

Each Li has 3 electrons, so total = 6 electrons. Configuration:

(\sigma 1s)^2(\sigma^*1s)^2(\sigma 2s)^2

So,

\text{B.O.} = \frac{4-2}{2}=1

Not equal to 2.

$\mathrm{Ne}_2$

Each Ne has 10 electrons, so total = 20 electrons. All bonding and antibonding orbitals up to $2p$ are filled. Thus,

\text{B.O.} = 0

Not equal to 2.

$\mathrm{N}_2$

Each N has 7 electrons, so total = 14 electrons. Configuration gives:

\text{B.O.} = 3

Not equal to 2.

$\mathrm{He}_2$

Each He has 2 electrons, so total = 4 electrons. Configuration:

(\sigma 1s)^2(\sigma^*1s)^2

So,

\text{B.O.} = \frac{2-2}{2}=0

Not equal to 2.

Count molecules with bond order 2

These are:

$\mathrm{C}_2$
$\mathrm{O}_2$

So the number of molecules is

2

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View written solutionFree

Correct answer: 2

Use bond order formula

For diatomic molecules,

\text{Bond order} = \frac{N_b - N_a}{2}

where $N_b$ = number of bonding electrons and $N_a$ = number of antibonding electrons.

We check each given molecule.

$\mathrm{C}_2$

Each C has 6 electrons, so $\mathrm{C}_2$ has 12 electrons. For molecules up to $\mathrm{N}_2$ , MO order is:

\sigma(1s),\ \sigma^*(1s),\ \sigma(2s),\ \sigma^*(2s),\ \pi(2p_x)=\pi(2p_y),\ \sigma(2p_z)

Configuration:

(\sigma 1s)^2(\sigma^*1s)^2(\sigma 2s)^2(\sigma^*2s)^2(\pi 2p_x)^2(\pi 2p_y)^2

Bonding electrons = $8$ , antibonding electrons = $4$ . So,

\text{B.O.} = \frac{8-4}{2}=2

Hence, $\mathrm{C}_2$ has bond order 2.

$\mathrm{O}_2$

Each O has 8 electrons, so total = 16 electrons. For $\mathrm{O}_2$ , MO order is:

\sigma(1s),\ \sigma^*(1s),\ \sigma(2s),\ \sigma^*(2s),\ \sigma(2p_z),\ \pi(2p_x)=\pi(2p_y),\ \pi^*(2p_x)=\pi^*(2p_y)

Configuration gives bond order:

\text{B.O.} = 2

Hence, $\mathrm{O}_2$ has bond order 2.

$\mathrm{Be}_2$

Each Be has 4 electrons, so total = 8 electrons. Configuration:

(\sigma 1s)^2(\sigma^*1s)^2(\sigma 2s)^2(\sigma^*2s)^2

So,

\text{B.O.} = \frac{4-4}{2}=0

Not equal to 2.

$\mathrm{Li}_2$

Each Li has 3 electrons, so total = 6 electrons. Configuration:

(\sigma 1s)^2(\sigma^*1s)^2(\sigma 2s)^2

So,

\text{B.O.} = \frac{4-2}{2}=1

Not equal to 2.

$\mathrm{Ne}_2$

Each Ne has 10 electrons, so total = 20 electrons. All bonding and antibonding orbitals up to $2p$ are filled. Thus,

\text{B.O.} = 0

Not equal to 2.

$\mathrm{N}_2$

Each N has 7 electrons, so total = 14 electrons. Configuration gives:

\text{B.O.} = 3

Not equal to 2.

$\mathrm{He}_2$

Each He has 2 electrons, so total = 4 electrons. Configuration:

(\sigma 1s)^2(\sigma^*1s)^2

So,

\text{B.O.} = \frac{2-2}{2}=0

Not equal to 2.

Count molecules with bond order 2

These are:

$\mathrm{C}_2$
$\mathrm{O}_2$

So the number of molecules is

2

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