An electron will have highest energy in which set ? A. 3,2,1,1/2 B. 4,2,-1,1/2 C. 4,1,0,-1/2 D. 5,0,0,1/2

#(n + l) ∝ "Energy"#

Greater the #(n + l)# value greater is the energy of electron.

If two electrons have same #(n + l)# value then the one with greater #n# value is of more energy.

A. #(n + l) = (3 + 2) = 5#

B. #(n + l) = (4 + 2) = 6#

C. #(n + l) = (4 + 1) = 5#

D. #(n + l) = (5 + 0) = 5#

Therefore, B is the correct option.

We must identify the orbital first...

Then we would find that the #5s# are higher in energy than the #4d# MOST of the time. The only exception I know is yttrium, which has the #4d# orbitals SLIGHTLY higher than the #5s# (by #"0.21 eV"#, or #"4.84 kcal/mol"#).

More info on quantum numbers is already given here.

#A)# #(n,l,m_l,m_s) = (3,2,1,1/2)#

This indicates a #3d# orbital of some sort, with a spin-up electron.

#B)# #(n,l,m_l,m_s) = (4,2,-1,1/2)#

This indicates a #4d# orbital of some sort, with a spin-up electron.

#C)# #(n,l,m_l,m_s) = (4,1,0,-1/2)#

This indicates a #4p_z# orbital, with a spin-down electron.

#D)# #(n,l,m_l,m_s) = (5,0,0,1/2)#

This indicates a #5s# orbital, with a spin-up electron.

Data from Inorganic Chemistry (Miessler, Fischer, and Tarr) shows that the #5s# are clearly higher in energy than the #4d# for MOST transition metals that have valence electrons in those orbitals.

Only yttrium apparently has #4d# orbitals SLIGHTLY higher in energy than the #5s#. The rest of the transition metals, zirconium through cadmium, have the #5s# orbital higher in energy than the #4d#.