Which atomic orbitals of which subshells have a dumbbell shape?

1 Answer
Jan 8, 2017

Well, of the simpler orbitals (i.e. not #(n-2)f# or #(n-3)g# orbitals), the #bb(np_(x,y,z))# and #bb((n-1)d_(z^2))# orbitals are shaped like dumbbells, because their wave functions have...

  • for the #np_x,np_y,np_z# case, one nodal plane perpendicular to the principal axis of that orbital.
  • for the #(n-1)d_(z^2)# case, two conical nodes (each a special kind of nodal "plane").

If that doesn't make sense, that's OK. Examples that fit the bill for the above orbitals are shown below so you can see.

http://2012books.lardbucket.org/

http://wiki.chemprime.chemeddl.org/

Source: I drew it

I go into specific mathematical proofs below (simpler than one might expect) to show why or how the above-described nodal plane or conical nodes are present.


The wave function #psi_(nlm_l)(r,theta,phi)# for atomic orbitals, a function of the radial coordinate #r# and two possible angular coordinates #theta,phi#, can be separated into the #R_(nl)(r)Y_(l)^(m_l)(theta,phi)# component functions, where #n,l,m_l# are the familiar quantum numbers.

https://upload.wikimedia.org/

We'll focus on #Y#.

When #Y_(l)^(m_l)(theta,phi)# contains #trig(theta)trig(phi)#, #trig(theta)#, or #trig(phi)# such that:

  • #Y(theta,phi) = 0# for a specific plane perpendicular to the principal axis (for a nodal plane), OR
  • #Y(theta) = 0# for two different values of #theta# (or #phi# for #Y(phi)#) while the other angular coordinate is constant (for two conical nodes),

...we have a dumbbell shape.

Here are examples of such functions for the #2p_x#, #3p_y#, and #3d_(z^2)# orbitals:

#Y_(2p_x)(theta,phi) prop sinthetacosphi#
#Y_(3p_y)(theta,phi) prop sinthetasinphi#
#Y_(3d_(z^2))(theta) prop (3cos^2theta - 1)#

  • #Y_(2p_x)(theta,phi) = 0# when #theta = 0,pi, . . . # and/or #phi = pi/2,(3pi)/2, . . . #. By the definition of #theta# and #phi# above, that forms a node along the #z# axis and the #y# axis, giving a nodal #bb(yz)# plane for the #bb(2p_x)# orbital. Indeed, the #yz# plane is perpendicular to the principal, #bb(x)# axis.

So, the #2p_x# orbital is dumbbell-shaped, just like the #2p_y# and #2p_z#.

http://2012books.lardbucket.org/

  • #Y_(3p_y)(theta,phi) = 0# when #theta = 0,pi, . . . # and/or #phi = 0,pi, . . . #, which marks nodes along the #z# axis for #theta = 0,pi# and the #x# axis for #phi = 0,pi#, giving us a nodal #bb(xz)# plane, which is indeed perpendicular to the principal, #bb(y)# axis.

So, the #3p_y# orbital is dumbbell-shaped, just like the #3p_x# and #3p_z#.

http://wiki.chemprime.chemeddl.org/

  • #Y_(3d_(z^2))(theta) = 0# when #3cos^2theta - 1 = 0#. It turns out that solving that gives #theta = arccos(pm1/sqrt3)#, which corresponds to an axis about #bb(54.7^@)# from the #z# axis.

Since #Y_(3d_(z^2))# is not a function of #phi# though, #phi# is constant, and the angle from the #z# axis is revolved around the #z# axis, generating the conical nodes (with a decline of #~~54.7^@# and #~~234.7^@#, respectively, from the #bb(z)# axis).

One conical node is from #theta = arccos(1/sqrt3)#, and the other is from #theta = arccos(-1/sqrt3)#.

So, the #3d_z^2# orbital is dumbbell-shaped, as shown below with the conical nodes:

Source: I drew it