Gen Chem Question

[Master Yoda]

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An atom of an element has an electron configuration ending in 3p4. Which of the following statements about the element's electron configuration is incorrect?
A) There are eight electrons in the 2nd shell.
B) The 3rd shell needs two more electrons to be completely filled.
C) There are six electrons in the 3rd shell.
D) Five different subshells contain electrons.

Can someone explain this to me?

 

[schmoob]

I think it's B but I could be wrong. With 2 more elecrons, the S orbital will be full. But the D orbital will still empty, and that has room for 10 more.
Again, I would double check but it's my best guess.

 

[Incis0r]

I agree with @schmoob . Answer should be B. It needs 12 more electrons for the 3rd shell to be completely filled (n = 3 means l = 2 which means we have s, p, and d orbitals).

D may be confusing you- there are five subshells with electrons:

1s
2s
2p
3s
3p

 

[Master Yoda]

awesome. thank you

 

[CLAZO121]

Why wouldn't the answer be A?

 

[CLAZO121]

OHHH Never mind i just caught my mistake! If N=2, that means l=1, which means we have p and s orbitals. There are 2 electrons in the S and 6 in the P

2+6=8 😉

 

[Mhines24]

I agree with @schmoob . Answer should be B. It needs 12 more electrons for the 3rd shell to be completely filled (n = 3 means l = 2 which means we have s, p, and d orbitals).

D may be confusing you- there are five subshells with electrons:

1s
2s
2p
3s
3p

There would actually be 6 subshells because 4s fills before 3p

 

[Incis0r]

There would actually be 6 subshells because 4s fills before 3p

4s fills before 3d, but after 3p.

 

[acesofseven]

4s fills before 3d, but after 3p.

Agreed. But remember that the 4s electrons will be removed first if it means it means the 3d orbitals gets completely or half filled.

 

[Mhines24]

4s fills before 3d, but after 3p.

Your right using the n+l rule 3p and 4s both equal 4 but 3p is a lower energy level so it will fill before 4s, my apologies!