[remeday86]

I learned how to do electron configuration from general chemistry and I thought it was easy.

Now that I'm taking pchem, I'm a bit confused.

For example:
O = 1s²2s²2p⁴

But in pchem, it would be:

Let sigma = * and pi = #  (Sorry I don't know how to code in sigma and pi symbol)

O₂ = K K (*_g2s)²(*_u2s)²(*_g2p_z)²(#_u2_x)²(#_u2p_y)²(#_g2p_x)¹(#_g2_y)¹

with bond order of 3.

I understand what "g" and "u" means which is the bonding and antibonding, respectively.
I understand what "K" means which is filled n= 1 shell which is (*_g1s)²(*_u1s) (I think?)
I understand bond order = 1/2 (# of electrons in bonding ortibals - # of electrons in antibonding orbitals)

I DON'T understand why they are including the bonding/antibonding , g and u.
I DON'T understand what the x, y, and z subscript are for.
I DON'T understand how they MADE that electron configuration.
I DON'T understand how they got a bond order of 3.


Second example:
N₂ = K K (*_g2s)²(*_u2s)²(#_u2p_x)²(#_u2p_y)²(*_g2p_z)²


with bond order of 3.


I DON'T understand how they got a sigma (*) orbital at the end and why it's a ground state :/

Third example:
P₂= K K L L (*_g3s)²(*_u3s)²(#_u3p)⁴(*_g3p_z)²



I DON'T understand what "L" stand for except it represents the filled n = 2 shell.
I DON'T understand why the "p" doesn't have a subscript and it has 4 electrons.
I DON'T understand why the other "p" does have subscript "z".
I DON'T understand how to find this bond order.



I know it is a lot of questions. Any help or hints would be much appreciated.
Thank you in advance.

[iolzizlyi]

Oxygen does not have a bond order of 3, it should have a bond order of two unless it's O₂²⁺.  Since the 1s and 2s MOs are filled, only the 2p MOs will be used in determining the bond order.  Note that all 3 of the 2p bonding orbitals -- sigma gerade, pi ungerade (x), and pi ungerade (y) -- are filled completely.  Note also that both of the pi gerade antibonding orbitals contain one electron each.  6 bonding electrons - 2 antibonding electrons = 4 and so 4/2 = 2 = bond order.

g and u do not mean bonding and antibonding.  They have to do with the symmetry of the orbitals.  If the orbital changes sign upon inversion through the center of the molecule, the symmetry is ungerade.  If the orbital does not change sign, the symmetry is gerade.  Pictures would really help here.  g and u can however help you deduce whether the orbital is bonding or antibonding.

K does mean what you think it means.

The x, y, z subscripts indicate spatial orientations of the orbitals.  Usually the bond axis is referred to as the z-axis.  This is why you see a sigma bonding orbital with a z subscript.

L does mean what you think it means.  It's just shorthand.

In the P₂ example, the 4 electrons is another kind of shorthand.  It is understood that the p orbitals in the x and y directions are degenerate so they combined them into one term representing 2 orbitals.  There are no antibonding electrons in this example (hint).

[remeday86]

Thank you iolzizlyi so much!!!

One more question about bond order:

N₂⁺ = K K (*g2s)²(*u2s)²(#u2px)²(#u2py)²(*g2pz)¹

The bond order is 2*1/2 = 2.5

How did they get that?
From my understanding to your post, then there are
(4 bonding electrons - 1 antibonding)/2 = 3/2  

[remeday86]

nvm! sorry I got it!

Another question that involves metals:

Let $ = delta

Cr2 = K K L L M M (*g4s)2 (*g3d)2(#u3d)4($g3d)4

Bond order is 6.

My guess:
[(2+2+4+4) bonding - 0 antibonding] /2 = 12/2 = 6?