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QUESTION

# How to draw overlapping of pure or hybridized orbitals for Br2 and NO+? Explain the need for the orbital of an atom to hybridized based on the Lewis structures

There is no in either of these diatomic .

Note that in linear diatomic molecules, the p_z orbital always points along the internuclear axis, so it has to contribute to one of the sigma bonds.

I've drawn the overlaps below in the MO diagrams.

BROMINE (HOMONUCLEAR DIATOMIC)

For "Br"_2, it is the simpler of the two examples. It is a homonuclear diatomic, so all of its orbitals have a compatible partner: 4p_x with 4p_x, 4p_y with 4p_y, etc.

In this case, since there are only two atoms, they use their highest energy compatible orbitals to sigma bond on the internuclear axis. The energy of the 4s atomic orbital is -"24.37 eV", and the 4p atomic orbitals are -"12.49 eV" in energy (Inorganic Chemistry, Miessler et al., Table 5.2).

Each bromine would donate one \mathbf(4p_z) electron to form a sigma-bonding orbital.

As a result, there is no orbital hybridization here.

Here is the MO diagram below (I had to draw it myself since I couldn't find one online; the pi_(4px) and pi_(4py) orbitals---the 1b_(3u) and 1b_(2u)---are very close in energy to the sigma_(4pz), the 1b_(1u)):

Granted, that is not "Br"_2's highest-occupied molecular orbital (that would be the pi_(4px)^"*" and pi_(4py)^"*"---the 1b_(3u) and 1b_(2u)), but since both the bonding and antibonding pi molecular orbitals are occupied, it is the sigma_(4p_z) (1b_(1u)) that participates in the sigma bond.

You should notice that the 1b_(1u) orbital is the sigma_(4pz) bonding orbital, but the 2b_(1u)---the sigma_(4pz)^"*" antibonding orbital---has no electrons, so it doesn't contribute to the sigma bond. If it did, "Br"_2 would not exist.

Therefore, the sigma_(4pz) indeed is the molecular orbital that represents the single bond on "Br"_2.

NO""^(\mathbf(+)) BONDING (HETERONUCLEAR DIATOMIC)

"NO"^(+), on the other hand, is a heteronuclear diatomic. Since it is also diatomic, it also does not need to hybridize.

All of nitrogen's orbitals are compatible with oxygen's orbitals in energy (and in symmetry, but that is less crucial to our understanding for General Chemistry level education).

The MO diagram for neutral "NO" is as follows (Inorganic Chemistry, Miessler et al., Ch. 5, Answer Key):

(I superimposed some orbital depictions on the original diagram, and added symmetries and energies.)

If we consider "NO"^(+), we remove the electron from the highest-occupied molecular orbital, so we take out the one from the pi_(2px)^"*" antibonding orbital (2b_1) to form "NO"^(+).

At this point, its bonds have increased in strength. The bond order changed from:

(8 - 3)/2 = 2.5

to:

(8 - 2)/2 = 3

So we know it has a triple bond. That means it needs three orbitals contributed from each atom.

There are two electrons in the sigma_(2pz) molecular orbital (3a_1), and there are two electrons each in the pi_(2px) (1b_1) and pi_(2py) (1b_2) molecular orbitals.

"NO"^(+) therefore uses two 2p_x atomic orbitals, two 2p_y atomic orbitals, and two 2p_z atomic orbitals to bond.

As a result, there is no orbital hybridization here.

Each sigma-bonding pair contributes to a sigma bond, and each pi-bonding pair contributes to a pi bond. That accounts for the triple bond: one sigma and two pi bonds.