anonymous
  • anonymous
What are the possible values of the four quantum numbers for a 2p electron in boron?
OCW Scholar - Introduction to Solid State Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
n is for the shell, l is subshell, ml is the subshell shape, and ms is the spin
anonymous
  • anonymous
so i'll start it off
anonymous
  • anonymous
we're only looking at 2p

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
so first: n=2, l=1, m=-1, and s=1/2
anonymous
  • anonymous
It can have a fourth quantum number s that be either +1/2 or -1/2...every shell (n number) has sub-shell (l number) oriented in one of the possible orientations (m number) that only allows two electrons with opposite magnetic spin (s number).
anonymous
  • anonymous
The neutral boron atom has the following shell configuration:\[[Ne]2s^{2}p^{1}\]2p corresponds to principle quantum number n=2 (2 in 2p), azimuthal quantum number l=1 (p in 2p). I'm almost positive that the choice of magnetic quantum number m and spin quantum number s are arbitrary in this case. To the extent the choice of m and s are arbitrary, any of the following work:\[n=2, l=1, m_{l}=-1, s=+1/2\]\[n=2, l=1, m_{l}=-1, s=-1/2\]\[n=2, l=1, m_{l}=0, s=+1/2\]\[n=2, l=1, m_{l}=0, s=-1/2\]\[n=2, l=1, m_{l}=1, s=+1/2\]\[n=2, l=1, m_{l}=1, s=-1/2\] I recall reading that by convention the lowest available m is assigned first, but +1/2 s electrons are assigned before -1/2 s electrons. I'm having trouble figuring out where I read that. In any case, all of this works out to exactly what BohrMachine posted, above: \[n=2, l=1, m_{l}=-1, s=+1/2\]
anonymous
  • anonymous
I found my source. Lecture Notes #1 (LN-1), page 9, second paragraph after the bullet points. The Aufbau Principle.

Looking for something else?

Not the answer you are looking for? Search for more explanations.