First slide

Molecular Orbital Theory

Question

If the formation of diatomic boron molecule is predicted based on MO theory, the bond order would be

Easy

A

0.5
B

1.0
C

1.5
D

12.0

Solution

B2 = [5 + 5 = 10e^-] ⇒
M.O.E.order :
$\large \sigma 1{s^2}\,{\sigma ^}1{s^2}\;\sigma 2{s^2}\;{\sigma ^}2{s^2}\;\begin{array}{{20}{c}} {\pi 2{p_x}^1} \\ {\pi 2{p_y}^1} \end{array}\;\sigma 2pz\;\begin{array}{{20}{c}} {{\pi ^}2{p_x}} \\ {{\pi ^}2{p_y}} \end{array}{\sigma ^*}2{p_z}\$
$\large B.O = \frac{N_b - N_a}{2} = \frac{6 - 4}{2} = \frac{2}{2} = 1$
No of electrons 8 9 10 11 12 13 14 15 16 17 18 19 20
B.O 0 0.5 1.0 1.5 2.0 2.5 3.0 2.5 2.0 1.5 1.0 0.5 0

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