## Homework Problem Help

AChoudhry_1L
Posts: 55
Joined: Sat Sep 07, 2019 12:17 am

### Homework Problem Help

Could someone explain this problem me?

1D23) How many orbitals can have the following quantum numbers in an a tom:(a)n=2, l=1;(b)n=4, l=2, ml=-2; (c)n=2;(d)n=3, l=2, ml=+1?

Jacob Puchalski 1G
Posts: 81
Joined: Fri Aug 09, 2019 12:16 am

### Re: Homework Problem Help

For a, you know it's the second period and a p orbital. There's six elements in the second period with a p orbital, so there's six possible orbitals. That's how I'd go about each question.

EvanWang
Posts: 81
Joined: Sat Sep 07, 2019 12:16 am

### Re: Homework Problem Help

An orbital must have 3 quantum numbers, and no two orbitals can have the same quantum numbers. Knowing this, b and d both indicate only 1 orbital because they list 3 quantum numbers.

For a and c, since all 3 quantum numbers are not given we must list the other possible numbers to find the number of orbitals. For a, since it lists n=2 and l=1, that means we could have ml=-1,0,1. This lists 3 possible orbitals.

For b, since it only lists n=2, that means l=0,1. For l=0, ml can only be 0. for l=1, ml can be -1,0,1. Together, those are 4 other possible orbitals.

One trick is that if only you know the principle quantum number, n, then the total number of orbitals in that shell is n2. So for shell n=2, there are 4 orbitals.

RichBollini4G
Posts: 75
Joined: Wed Sep 18, 2019 12:18 am

### Re: Homework Problem Help

Does anyone know what section problems we should be turning in for homework #4? Is it still quantum?

AChoudhry_1L
Posts: 55
Joined: Sat Sep 07, 2019 12:17 am

### Re: Homework Problem Help

RichBollini3C wrote:Does anyone know what section problems we should be turning in for homework #4? Is it still quantum?

Yes, homework #4 can be on quantum since we were still doing the quantum unit this week.

AChoudhry_1L
Posts: 55
Joined: Sat Sep 07, 2019 12:17 am

### Re: Homework Problem Help

EvanWang wrote:An orbital must have 3 quantum numbers, and no two orbitals can have the same quantum numbers. Knowing this, b and d both indicate only 1 orbital because they list 3 quantum numbers.

For a and c, since all 3 quantum numbers are not given we must list the other possible numbers to find the number of orbitals. For a, since it lists n=2 and l=1, that means we could have ml=-1,0,1. This lists 3 possible orbitals.

For b, since it only lists n=2, that means l=0,1. For l=0, ml can only be 0. for l=1, ml can be -1,0,1. Together, those are 4 other possible orbitals.

One trick is that if only you know the principle quantum number, n, then the total number of orbitals in that shell is n2. So for shell n=2, there are 4 orbitals.

Does this trick always work or can it only be used in specific problems?