# What shapes do d orbital represent?

## Main Question or Discussion Point

In d orbital there are 5 degenerates.
How $d_z^2$ represents something like a donut shape.
Shouldn't it be a parabola with respect to z axis, like when plotting graph of $x^2$.
Also have confusion about shapes of other 4 degenerates.
For s it is spherical and simple.
For p there are 3 degenerates directed along x,y,z axes.

Have confusion about shapes of these orbitals. Can somebody help?

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mfb
Mentor
How $d_z^2$ represents something like a donut shape.
Shouldn't it be a parabola with respect to z axis, like when plotting graph of $x^2$.
The "2" is an index, not an exponent.

The orbitals are solutions to the Schrödinger equation. They just happen to look like that. What images usually show are the regions with a high probability to find the electron.

So what is 2 representing here as an index?
Also in square planar geometry having hybridisation dsp2 , there are certain orbital degenerates involved, totally 4. What are these and what is their reason?

DrDu
The "2" is an index, not an exponent.
No, it is an exponent, not an index. However, it is an abreviation for $z^2-(x^2+y^2+z^2)/3$. Specifically, you can form 6 quadratic expressions: xy, xz, yz, $x^2$, $y^2$, $z^2$, but the combination $x^2+y^2+z^2$ transforms like an s-orbital, hence you subtract it.

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mfb
Specifically, you can form 6 quadratic expressions: xy, xz, yz, $x^2$, $y^2$, $z^2$, but the combination $x^2+y^2+z^2$ transforms like an s-orbital, hence you subtract it.
Isn't then x2 +1 a quadratic expression and so on? How total six then?
How then $x^2+y^2+z^2$ transforms like an s-orbital?

Also in square planar geometry having hybridisation dsp2 , there are certain orbital degenerates involved, totally 4. What are these and what is their reason?
What is the answer for this?

DrDu
Isn't then x2 +1 a quadratic expression and so on? How total six then?
How then $x^2+y^2+z^2$ transforms like an s-orbital?

What is the answer for this?
To 1: "1" isn't a quadratic expression in x, y or z.
$x^2+y^2+z^2=r^2$ i.e. the square of the radial distance. It is rotational isotropic, i.e. transforms as an s orbital and doesn't carry angular momentum.

2. Formally, you can always form hybrids of d, s and p orbitals, irrespective of whether they are degenerate or not, but they will be relevant for bonding only if the s, p and d orbitals are nearly energetically degenerate.

2. Formally, you can always form hybrids of d, s and p orbitals, irrespective of whether they are degenerate or not, but they will be relevant for bonding only if the s, p and d orbitals are nearly energetically degenerate.
But why in dsp2 hybridisation orbitals like if look at their orientation,
dx2 -y2, px,py and s are only involved?
Why not other d orbital orientations and what about pz?

DrDu
Because these are the only functions which give your orbitals lying in the xy plane. If you would like to construct dsp2 orbitals lying in some other plane, you would have to use another set of orbitals.

Because these are the only functions which give your orbitals lying in the xy.
But why one needs orbitals lying in xy ?

DrDu
You were talking about square planar geometry, weren't you?

Yes. I see for tetrahedral some other 4 orbitals. But is there any specific reason that how to know which orientations for which geometry by any trick or something?

DrDu
As the molecular coordinate system is arbitrary, to chose the xy plane for planar compounds is convention. Which hybrids correspond to which geometries is not so easy to see.

As the molecular coordinate system is arbitrary, to chose the xy plane for planar compounds is convention. Which hybrids correspond to which geometries is not so easy to see.
So we analyze that by experiments?
Is there not any mathematical proof?
Do we have to rote memorize it for different geometries?

DrDu
No, orbitals are theoretical constructs, and not directly amenable to experiments.
The rationale behind the choice of the hybrid orbitals is due to Linus Pauling's boo: The nature of the chemical bond.
Have a look at it, it is a very good book. However, by now, quite some concepts are outdated or disproven.

I would try to see the book later. Thanks for providing so much help.