Question

how many d orbitals have n = 6?

can someone tell me how to get this?

Answer 1

20 d orbitals

Answer 2

*how*

Answer 3

and btw 20 is wrong

Answer 4

just see in this image for n =6 u get 4 d orbits
3d,4d,5d and 6d, each d orbit contains 5d orbitals hence 20 d orbitals u get

Answer 5

if u want interms of orbits it will be four

Answer 6

it's actually 5..

Answer 7

how tell me

Answer 8

Remember that 4-part quantum descriptor I posted yesterday for you? ;-)
\[\large( n , \ell , m_\ell , m_s )\]

Well...
\(\huge\cdot\) The value for \(\large \ell\) determines an orbital's shape
\(\huge\cdot\) The value for \(\large m_\ell\) its orientation; but because some orbitals are described by complex numbers (that a+b\(i\) stuff) it sometimes has an effect on shape too

How to get this? Um, I'm going to go ahead an assume you're talking in generality because the "how" is a large topic of a whole science called quantum mechanics. I am not going to give you a entire written description of how quantum mechanics works because of some very good reasons:

1. I don't have the time, and the rewards for all that effort doesn't justify it. I'm willing to help, but you need to narrow the focus of your question, like a lot. Ask something specific

2. I don't personally feel qualified to even consider myself an expert on quantum mechanics. And quantum mechanics itself is incredibly complex, being research by people farm more experienced and intelligent than myself.

3. Quantum mechanics deals with stuff that can't be seen directly and is mostly theoretical (i.e.: untested or unobserved in real-world experiment) or that experimental data is based on the best estimations we can do at this time, (i.e.: compare \(\pi\cdot \sqrt{-2}\) to a decimal? haha ok best of luck to you with that one). We're starting to find that the universe is built on special types of complex numbers and that our "convenient" decimal number system and Cartesian coordinates are an oversimplified version of the TRUE complexity of the universe.

4. I would probably encounter the character limit. In my response.

^ read me please.


With that said, D-orbitals are a combination of the following based on electron pairs (which have opposite spins). What does this look like visually? Glad you asked, and thank you Wikimedia for having this:
http://upload.wikimedia.org/wikipedia/commons/thumb/e/e1/D_orbitals.svg/2000px-D_orbitals.svg.png

The also have a computer approximation of what n={1 to 6} for hydrogen (1s\(^1\)) where you'd be most likely to find the electron:
http://en.wikipedia.org/wiki/File:HydrogenOrbitalsN6L0M0.png

Keep in mind electrons sometimes are like Cartman from Southpark going, "F U guys I'm going home!" (this is formally known has the Heisenberg Uncertainty Principle) It's a probability the electron will be there, not a guarantee.

So...

If you look at the image, it's 4+4+4+4+4 = 4 * 5 = 20 orbital "lobes"

Answer 9

it's still a while ago for me haha

Answer 10

Again @lgbasallote , you'd do well to narrow the focus of your question ;-)

Answer 11

Makes me wonder if you actually read what I answered for you yesterday too :-(

Answer 12

well i just wrote what's in the paper...that's the exact question and yes i read it...i got some of it...some didnt register..sadly

Answer 13

haaaa nice!!!

Answer 14

exact question is how many d orbitals have n = 6?
a) 2 b) 5 c) 10 d) 7 e) 18

Answer 15

and i also dont get what quantum mechanics has to do with this..

Answer 16

Ah you're probably looking at 6d\(^{something}\)

Answer 17

could be..

Answer 18

it hink its only 6d then i can say any Nd orbit will have 5 orbitals!!!!

Answer 19

wait no...the answer is 5

Answer 20

how did you get it @chmvijay ?

Answer 21

See? The way the question is worded, and posting the choice option if it's multiple choice is VERY necessary on a site like this. We don't have the problem. We have to guess the subject matter when thinking of an answer. For this I would refer you to the periodic table... (hold on I'll get a good link)

Answer 22

hey it is not for 6 d if you consider any d orbit it contains definately 5 orbitals

Answer 23

oh btw.. @agentx5 no offense..you have a habit of using complex words :S im not experienced in chemistry so i dont know your terms sometimes

Answer 24

question setting has little bit not understandable , since 2o is not there we should go for 5 i think
their intension of asking might be different

Answer 25

i didnt know choices were important here lol

Answer 26

http://www.meta-synthesis.com/webbook/34_qn/janet_number.png

D = 5 orbital combinations, 2 pairs of electrons per orbital, in the 5 formations I showed you above in my first mega-post

For example can you name how many orbital combinations you'd find for the F-orbital block?

Answer 27

seven

Answer 28

oh wait does it come from -2, -1, 0, 1, 2? that thingy?

Answer 29

no no i dont think so

Answer 30

@chmvijay is correct! You have 7 different formations for orbitals in the F-block ;-)

14 columns / 2 pairs of electrons = 7 formations

Answer 31

Isn't the periodic table handy? ;-)

Answer 32

but it's d block l

Answer 33

lol*

Answer 34

Yes and if you're filling electrons in the D-block, how many combinations are there by the time you've got to 3d\(^{10}\) (for understanding, can you figure out which element that is, if you're assuming it's neutral/uncharged?)

Answer 35

i dont know what you're talking about sorry

Answer 36

am i right though? the answer is 5 because there are 5 orientations for d?

Answer 37

i have to go sleep now =_= ill come back to this in the morning....

Answer 38

For D orbitals:

5 orbital orientations (again, the wikipedia picture diagram)
10 electrons to fill a D-orbital up completely (5 * 2 e- per pair)
20 orbital lobes (quantum probability curves)

Answer 39

I agree 5 is most likely the best answer for how the question was worded. Try it :-)

Answer 40

A "d" orbital has angular momentum quantum number l = 2. (s = 0, p = 1, d = 2, f = 3, g = 4 and so forth). The number of d orbitals is given by how many values of the azimuthal quantum number m_l there are. m_l always goes from -l to +l, e.g. in the case of l = 2 from -2 to +2 (-2,-1,0,+1,+2), for a total of 2*l + 1, e.g. in the case of l=2 a total of 5 orbitals.

The value of the principal quantum number n doesn't matter except that it must be greater than 2, because n = 1 and n = 2 don't have d orbitals at all. (The allowed values of l are 0 to n-1.)

Answer 41

@Carl_Pham I got a syntax helpful tip for ya! :-D

\ell = \(\ell\)

subscript\large_{\ell} = subscript\(\large_{\ell}\)

He_2^4 = He\(_2^4\)

To make the syntax active code you need to put a \( before it

Answer 42

and a \) after it (I can't put this in the same post or it activates)

Answer 43

Thanks!

Answer 44

Give it a try @Carl_Pham :-D

Answer 45

It's incredible helpful for writing stuff in chemistry and physics, two sections you seem to frequent.

Answer 46

okay thanks..it's beginning to make some sense now :) just a few more practice!! \m/