0
$\begingroup$

Which statement is correct about electron orbitals and energy levels?

A. Yttrium, $\ce{Y}$ ($Z = 39$) is the first element in the periodic table with an electron in a f sub-level.

B. The maximum number of electrons in one d orbital is 10.

C. The maximum number of electrons in the 4th main energy level is 18.

D. In a main energy level, the sub-level with the highest energy is labelled f.

I thought B and D are all possible solutions to this answer.

B--the maximum number of electrons in 1 d orbital is 10.

D--in main energy level, the sub-levels in order of increasing energy is s $<$ p $<$d $<$f

So why is the answer only D?

Please advise and sorry in advance for the wrong tags or title name, as I am new to the group and am trying to improve on it.

$\endgroup$
  • 5
    $\begingroup$ You mixed up a d orbital with the d subshell. One d subshell contains five d orbitals. $\endgroup$ – orthocresol Dec 31 '15 at 4:23
  • $\begingroup$ @orthocresol, so for B, the maximum number of electrons in one d orbital should be 2? $\endgroup$ – CCC Dec 31 '15 at 4:26
  • 1
    $\begingroup$ Meowth, that's right! $\endgroup$ – orthocresol Dec 31 '15 at 4:43
  • 4
    $\begingroup$ If you really want to blow your mind (and, incidentally, prove D as technically sort-of incorrect), look into g-orbitals, and h-orbitals, and i-orbitals, and .... $\endgroup$ – hBy2Py Dec 31 '15 at 4:47
1
$\begingroup$

Although @orthocresol got this first, and it should be credited to him, I'm posting it as an answer.

In option (B), the statement is incorrect as it refers to $d$ orbitals, which can (according to Pauli's Exclusion Principle) hold 2 electrons each, rather than the $d$ subshell, which can hold 10 electrons, as it has a total of 5 $d$ orbitals which can hold 2 electrons each, and $5*2=10$.

Thus, only option (D) is correct.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.