Which of the following is a #2p# orbital?

The #2p# orbital is #D#.

• #A# is an #ns# orbital, probably #3s# or #4s#, though it depends on their size.
• #B# is a #3d_(z^2)# orbital.
• #C# is a #3d_(xz)# orbital.

However, do not trust the image. #B#, #C#, and #D# are wrong because all the lobe signs are presented to be the same whereas they aren't.

The #3d_(z^2)# orbital ring is the opposite sign, the #3d_(xz)# has diagonal lobes of the same sign (#yz# and #xy# nodal planes), and the #2p_z# orbital's second lobe is the opposite sign.

The #2p# orbital has:

• #n = 2#, for the principal quantum number. #n# can be one number in the set #{1,2,3, . . . }#.
• #l = 1#, for the angular momentum quantum number since #l = {0,1,2,3, . . . } harr {s,p,d, f, . . . }#.

If you recall:

• The total number of nodes (radial or angular regions of zero electron density) is equal to #n - 1#.
• The total number of angular nodes (nodal planes or conic nodes) is #l#.
• The total number of radial nodes is #n - l - 1#.

Since #l = 1#, there is one angular node, and since #n - 1 = 1#, it is the only node.

On either side of a nodal plane is a lobe of the opposite sign, and thus, there are two lobes on a #2p# orbital. That means it has to be either #B# or #D#.

The #B# orbital has #2# conic nodes, which are angular nodes, and corresponds to how #l = 2# corresponds to a #d# orbital.

#B# is actually a #bb(3d_(z^2))# orbital, like so:

Both lobes on #B# are actually the same sign, whereas it's the ring that is the opposite sign.

The #2p# orbital is #color(blue)(D)#.