Question

The pair of molecules / ions with the same bond order value is

• A. B\(_2\), C\(_2\)
• B. O\(_2\), C\(_2\)
• C. O\(_2^+\), O\(_2^-\)
• D. H\(_2^+\), Li\(_2\)

Hint:

Bond Order = \( \frac{1}{2} (N_b - N_a) \), where \(N_b\) is number of electrons in bonding MOs, \(N_a\) in antibonding MOs. Memorize MO filling order for diatomic molecules of the second period: - For B\(_2\), C\(_2\), N\(_2\): \( \sigma_{2s}<\sigma^_{2s}<\pi_{2p_x}=\pi_{2p_y}<\sigma_{2p_z}<\pi^_{2p_x}=\pi^_{2p_y}<\sigma^_{2p_z} \) - For O\(_2\), F\(_2\), Ne\(_2\): \( \sigma_{2s}<\sigma^_{2s}<\sigma_{2p_z}<\pi_{2p_x}=\pi_{2p_y}<\pi^_{2p_x}=\pi^_{2p_y}<\sigma^_{2p_z} \) (The order of \( \sigma_{2p_z} \) and \( \pi_{2p} \) flips due to s-p mixing). Core \( \sigma_{1s}, \sigma^_{1s} \) are usually filled and cancel out for B.O. but contribute to N\(_b\), N\(_a\). Number of electrons: H=1, Li=3, B=5, C=6, O=8.

Answer 1

The Correct Option is B

We use Molecular Orbital Theory (MOT) to determine bond orders.
Bond Order (B.
O.
) = \( \frac{1}{2} (\text{No.
of electrons in Bonding MOs} - \text{No.
of electrons in Antibonding MOs}) \).
1.
B\(_2\): Boron (Z=5), 2 B atoms = 10 electrons.
MO configuration (up to \( \pi_{2p} \)): \( (\sigma_{1s})^2 (\sigma^_{1s})^2 (\sigma_{2s})^2 (\sigma^_{2s})^2 (\pi_{2p_x})^1 (\pi_{2p_y})^1 \).
(For B\(_2\), \( \pi_{2p} \) fills before \( \sigma_{2p_z} \)).
Bonding e\(^-\) (N\(_b\)) = 2 (from \( \sigma_{1s} \)) + 2 (from \( \sigma_{2s} \)) + 2 (from \( \pi_{2p} \)) = 6.
Antibonding e\(^-\) (N\(_a\)) = 2 (from \( \sigma^_{1s} \)) + 2 (from \( \sigma^_{2s} \)) = 4.
B.
O.
for B\(_2\) = \( \frac{1}{2}(6-4) = \frac{2}{2} = 1 \).
2.
C\(_2\): Carbon (Z=6), 2 C atoms = 12 electrons.
MO configuration: \( (\sigma_{1s})^2 (\sigma^_{1s})^2 (\sigma_{2s})^2 (\sigma^_{2s})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 \).
N\(_b\) = 2+2+4 = 8.
N\(_a\) = 2+2 = 4.
B.
O.
for C\(_2\) = \( \frac{1}{2}(8-4) = \frac{4}{2} = 2 \).
3.
O\(_2\): Oxygen (Z=8), 2 O atoms = 16 electrons.
MO configuration (for O\(_2\), \( \sigma_{2p_z} \) fills before \( \pi_{2p} \) in terms of energy, but filling order after \( \sigma^_{2s} \) is \( \sigma_{2p_z}, \pi_{2p_x}=\pi_{2p_y}, \pi^_{2p_x}=\pi^_{2p_y}, \sigma^_{2p_z} \)): \( (\sigma_{1s})^2 (\sigma^_{1s})^2 (\sigma_{2s})^2 (\sigma^_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^_{2p_x})^1 (\pi^_{2p_y})^1 \).
N\(_b\) = 2+2+2+4 = 10.
N\(_a\) = 2+2+2 = 6.
B.
O.
for O\(_2\) = \( \frac{1}{2}(10-6) = \frac{4}{2} = 2 \).
4.
O\(_2^+\): 15 electrons.
Remove one e\(^-\) from \( \pi^_{2p} \) of O\(_2\).
N\(_b\) = 10.
N\(_a\) = 5.
B.
O.
for O\(_2^+\) = \( \frac{1}{2}(10-5) = \frac{5}{2} = 2.
5 \).
5.
O\(_2^-\): 17 electrons.
Add one e\(^-\) to \( \pi^_{2p} \) of O\(_2\).
One \( \pi^_{2p} \) will be full, other half.
MO for O\(_2^-\) ends with \( (\pi^_{2p_x})^2 (\pi^_{2p_y})^1 \).
N\(_b\) = 10.
N\(_a\) = 7.
B.
O.
for O\(_2^-\) = \( \frac{1}{2}(10-7) = \frac{3}{2} = 1.
5 \).
6.
H\(_2^+\): Hydrogen (Z=1).
1 electron.
MO configuration: \( (\sigma_{1s})^1 \).
N\(_b\) = 1.
N\(_a\) = 0.
B.
O.
for H\(_2^+\) = \( \frac{1}{2}(1-0) = 0.
5 \).
7.
Li\(_2\): Lithium (Z=3).
2 Li atoms = 6 electrons.
MO configuration: \( (\sigma_{1s})^2 (\sigma^_{1s})^2 (\sigma_{2s})^2 \).
N\(_b\) = 2 (from \( \sigma_{1s} \)) + 2 (from \( \sigma_{2s} \)) = 4.
(Core electrons usually ignored for valence B.
O, but here total B.
O.
).
N\(_a\) = 2 (from \( \sigma^_{1s} \)).
B.
O.
for Li\(_2\) = \( \frac{1}{2}(4-2) = 1 \).
Comparing bond orders: % Option (1) B\(_2\) (B.
O.
=1), C\(_2\) (B.
O.
=2).
Different.
% Option (2) O\(_2\) (B.
O.
=2), C\(_2\) (B.
O.
=2).
Same.
% Option (3) O\(_2^+\) (B.
O.
=2.
5), O\(_2^-\) (B.
O.
=1.
5).
Different.
% Option (4) H\(_2^+\) (B.
O.
=0.
5), Li\(_2\) (B.
O.
=1).
Different.
The pair with the same bond order is O\(_2\) and C\(_2\).
This matches option (2).