Hydrogen atom"s valence orbitals, before bonding, incorporate every orbital, and also all are the same power for a details #n#. At the point, there is only one electron.

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HAVING more THAN ONE ELECTRON SPLITS energy LEVELS

When presenting more electrons right into the system, i.e. With another hydrogen wanting to bond, the repulsion splits the power levels the the AOs of hydrogen.

They start out whereby each #n# offers subshells of equal energy, i.e. #2s# and #2p# space the same energy and also #3s#, #3p#, and #3d# space the exact same energy, but the #3l# orbitals are greater in power than the #2l# orbitals.

MOLECULAR orbit THEORY

The basic tenant of molecular Orbital concept (MO Theory) is the the number of MOs developed by a linear combination of atom orbitals (LCAO) is equal to the number of AOs used.

The energy dividing caused by electron/electron repulsion generates two MOs because of the one #1s# orbit per hydrogen the is bonding. They are called the bonding #sigma_(1s)# and also antibonding #sigma_(1s)^"*"# MOs.

The #2s#, #2p#, #3s#, and other AOs split to a very high relative energy that is the end of range of the MO chart we will be drawing, so we can emphasis on the #1s# AOs and also its MOs.

MO DIAGRAM for DIATOMIC HYDROGEN MOLECULE & ION

Now, allow us draw the MO diagram because that the #"H"_2# neutral molecule. Each hydrogen contributes one electron, which thus fills the lower-in-energy #sigma_(1s)# bonding orbital.

Add one electron, and you will obtain #"H"_2^(-)#, thus providing an electron in the antibonding #sigma_(1s)^"*"# MO. MO ELECTRON CONFIGURATIONS

Writing the MO electron configuration is relatively straightforward; read off of the diagram and also put superscripts because that how numerous electrons there are, analogous come the AO electron configuration.

The only distinction is the symbol supplied to show the MO rather of the AO and some parentheses.

For #"H"# atom, that would simply be #1s^1#.For #"H"_2# molecule, it would be #(sigma_(1s))^2#.For #"H"_2^(-)# ion, it would certainly be #(sigma_(1s))^2 (sigma_(1s)^"*")^1#.For #"H"_2^(2-)# ion, it would certainly be #(sigma_(1s))^2 (sigma_(1s)^"*")^2#.

BOND ORDER directly RELATES to BOND STRENGTH

The Bond Order, then, is the variety of bonding electrons minus the variety of antibonding electrons, the complete divided through #2#. If you recognize which orbitals space which, that is reasonably simple for #"H"_2#.

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For #"H"_2#, we get:

#color(blue)("Bond Order") = ("No. Of Bonding e"^(-) - "No. That Antibonding e"^(-))/2#

#= (2 - 0)/2 = color(blue)("1")#

which makes sense since #"H"_2# has actually one solitary bond.

For #"H"_2^(-)#, us have:

#color(blue)("Bond Order") = ("No. That Bonding e"^(-) - "No. The Antibonding e"^(-))/2#

#= (2 - 1)/2 = color(blue)("1/2")#

The donation of one electron in the antibonding orbital decreases the bond order, definition that the shortcut in #"H"_2^(-)# is weaker 보다 the bond in #"H"_2#.

If you"re adhering to the pattern, girlfriend should attract the MO diagram because that #"H"_2^(2-)# and also see that we"d get:

#color(blue)("Bond Order") = ("No. That Bonding e"^(-) - "No. The Antibonding e"^(-))/2#

#= (2 - 2)/2 = color(blue)("0")#

The donation of one electron in the antibonding orbit decreases the shortcut order, an interpretation that the bond in #"H"_2^(2-)# is even weaker 보다 the bond in #"H"_2^(-)#.

So weak, in fact, that due to the fact that the bond order is #0#, this molecule doesn"t exist! (You have the right to do the same for #"Be"_2# and show the it doesn"t exist either.)