If three electrons are available to fill three empty 2p atomic orbitals, how will the electrons be distributed in the three orbitals?A) one electron in each orbitalB) two electrons in one orbital, one in another, none in the third orbitalC) three in one orbital, none in the other twoD) none of the aboveWhich of the following will have the largest third ionization energy: Aluminum, Strontium, or Lead? Explain.Which is larger, a lithium atom or a lithium ion? Explain.Arrange the following in order of decreasing ionic radii:a) Cs^+1, Fr^+1, Pb^+2, Ba^+2b)O^-2, Te^-2, Se^-2, F^-Thank you so much for all your help!
i would use butter paper in the tin
should be fine but spray the inside liberally
Hi Brittani, You could line the outside of the tinfoil with brown paper, just to be on the safe sideI just use a paper bagI hope this helps,all the bestRab
Choice A, Hund's rule states that an orbital set loads with one electron in each orbital first, then starts pairing upYou are removing three electrons from each of these atoms Aluminum has three electrons in its outer energy level, so remove all three and leave the next lower energy level complete with a stable octet of electrons Strontium has 2 electrons in its outer energy level (it is in group 2) but after you remove these two electrons you would be cracking into a stable octet of electrons in order to remove the third electron, requiring lots of energy, so this one has the highest third ionization energyLead has 4 electrons in the outer energy level, the 6s2 and 6p2 so would have a lower ionization energy to remove the third electrons as you have not reached the more stable 5 d l0 orbital set Any metal ion is smaller than its respective atom because it has lost its outer energy levela Fr +, Cs + Ba+2 Pb+2 bTe-2 Se-2 0-2 F-1
