Chemistry homework pls help

[TNT333]

Q3 pls thank you.

[n0el]

a) 1s
b) 3p
c)4d

[deleted_user]

Yeah just google more: https://www.topperlearning.com/doubts-s ... 3-w6n4m2tt

[Horsemen]

this reminds me of breaking bad

[Ashvin]

why would someone ask a chem question here?! kids these days.

[Sargsyan]

why would someone ask a chem question here?! kids these days.

why not kiddo

[Kaiserklein]

This literally takes 1 minute to find on google right

[RefluxSemantic]

The three quantum numbers (n, l, and m) that describe an orbital are integers: 0, 1, 2, 3, and so on.
The principal quantum number (n) cannot be zero. The allowed values of n are therefore 1, 2, 3, 4, and so on.
The angular quantum number (l) can be any integer between 0 and n - 1. If n = 3, for example, l can be either 0, 1, or 2.
The magnetic quantum number (m) can be any integer between -l and +l. If l = 2, m can be either -2, -1, 0, +1, or +2.

This is the basis behind these shells. For the first shells, n = 1, so l can only be zero and then m can also only be zero. For n=1 we thus have only the s-shell, which can contain two electrons because the electrons can have two spins.

The basis of this is the pauli exclusion princinple: Electrons are what we call fermions, which cannot occupy the same quantum state. These three quantum numbers form the quantum state, together with the spin (as the electron can have both spin up and spin down, which are then different quantum states).

The naming convention is sometimes a bit confusing, but we use the number to indicate the value of n and then the letters to indicate the value of l. The naming convention is as follows:

s: l = 0
p: l = 1
d: l = 2
f: l = 3

Armed with this knowledge, we can answer your question. The orbital corresponding to n=1 and l=0 would then be 1s. The orbital corresponding to n=3 and l=1 would be 3p and the orbital corresponding to n=4 and l=2 would be 4d. It's actually rather simple compared to the other questions.