Question

Which of the following is a possible set of quantum numbers for an electron?
Select one:
a. (1, 1, 0, +½)
b. (2, 1, 2, +½)
c. (3, 2, 0, -½)
d. (3, -2, 1, -½)

Answer 1

@Photon336 @alphadxg

Answer 2

What do you think @simplymarie_x ?

Answer 3

I think it's C?

Answer 4

\[m_{s} |fourth~quantum~number~spin| +\frac{ 1 }{ 2 } or ~ -\frac{ 1 }{ 2 }\] so this one doesn't really matter. so it's not the deciding factor.

Answer 5

the first quantum number is n that's the energy level. then there is l which is the shape/ then mL which I think of as orientation. mL can't be greater than l so right away B is eliminated. D is eliminated because L can't be negative.

Answer 6

let's look at a and c

our first principle quantum number n = 1. right

and l = 1 and mL = 0.

Well, the last three numbers are okay but there is something about this that seems weird.
it's the fact that n = 1 and l = 1

if l = 1 that means that there are three possible orbitals {-1,0,1}

this would be the configuration of a p orbital but that's not the case, p orbitals start at n= 2. so that's wrong.


for C we've got n = 3 and l = 2 mL = 0

if l = 2, mL can be (-2,-1,0,1,2) which means this is a d orbital.

d orbitals start at n = 3.

therefore, C makes the most sense.