abb0t
  • abb0t
Weekend Challenge Question: (General Chemistry/ Inorganic)
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!
abb0t
  • abb0t
Sketch an MO energy diagram for the the M.O (molecular orbitals) of the \(valence\) electrons in cyanide, \(CN\). Keeping in mind to show the atomic and molecular orbitals of CN (that means x, y, & z orbitals). Make sure to also keep in mind the ordering of the energy states of each orbital. Additionally, which of the orbitals have a nodal plane along the \(internuclear\) (bond) \(axis\)?
anonymous
  • anonymous
|dw:1378031539365:dw| Well this is the sigma bonding part of the diagram, what you do is take the electrons from the carbon and from the nitrogen and distribute them amongst the levels of sigma and pi the s electrons go into the sigma bonding and antibonding levels, So carbon has 6 electrons and nitrogen has 7. So we place that on the board. We are going to be pairing them off in the different energy levels the astericks stands for antibonding, so basically what I did was to convert the atomic orbitals into the bonding and nonbonding molecular orbitals.
aaronq
  • aaronq
you misplaced the last electron into the \(\pi^*\) orbital when it should've gone into the \(\sigma\) orbital. Good job, though!

Looking for something else?

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

More answers

abb0t
  • abb0t
Yes, it does look correct. Just the energy differences between C and N look about the same. But yes, v ery nicely done. I didn't think many people knew how to draw M.O. diagrams :-)
anonymous
  • anonymous
ok my question might look stupid but I don't get it, shouldn't the total number of electrons be even? why is there a single electron up in the last orbital?
aaronq
  • aaronq
Only valence electrons are depicted, 4 e from C, 5 e from N, total of 9.
aaronq
  • aaronq
Technically the question should've asked for the MO of \(CN^-\), so the molecule wouldn't be a radical.
anonymous
  • anonymous
alright! thank you @aaronq :)

Looking for something else?

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