# What shapes do d orbital represent?

1. Jan 28, 2015

### Raghav Gupta

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?

2. Jan 28, 2015

### Staff: Mentor

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.

3. Jan 29, 2015

### Raghav Gupta

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?

4. Jan 29, 2015

### DrDu

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.

Last edited: Jan 29, 2015
5. Jan 29, 2015

### Raghav Gupta

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?

6. Jan 30, 2015

### DrDu

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.

7. Jan 30, 2015

### Raghav Gupta

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?

8. Jan 30, 2015

### 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.

9. Jan 30, 2015

### Raghav Gupta

But why one needs orbitals lying in xy ?

10. Jan 30, 2015

### DrDu

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

11. Jan 30, 2015

### Raghav Gupta

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?

12. Jan 30, 2015

### 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.

13. Jan 30, 2015

### Raghav Gupta

So we analyze that by experiments?
Is there not any mathematical proof?
Do we have to rote memorize it for different geometries?

14. Jan 30, 2015

### 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.

15. Jan 30, 2015

### Raghav Gupta

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