For the angular circulation function #Y_(l)^(m_l)(theta,phi)#, #m_l# represents the projection of the angular momentum #l#. As it transforms out, it is usually the form of the orbitals because it synchronizes to the #z# angular momentum of the orbital.

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THE MAGNETIC QUANTUM NUMBER, because that THE 4F ORBITAL

#m_l#, the magnetic quantum number, bring away on the values #0, pm1, pm2, . . . , pm l#, where #l# is an integer. Because that each #l#, there coincides an orbital name, so #l = 0 -> s#, #l = 1 -> p#, #l = 2 -> d#, #l = 3 -> f#, #l = 4 -> g#, etc.

For the #4f# orbital, #n = 4# and #l = 3#, definition that #m_l = 0, pm1, pm2, pm3#.

Therefore, there are #7# #m_l# values for the #4f# orbital, and also there wake up to be 7 different #4f# orbitals:

#4f_(y(3x^2-y^2)): m_l = -3##4f_(z(x^2-y^2)): m_l = -2##4f_(yz^2): m_l = -1##4f_(z^3): m_l = 0##4f_(xz^2): m_l = +1##4f_(xyz): m_l = +2##4f_(x(x^2-3y^2)): m_l = +3#

But why are there #7#? Yes, #m_l# correlates with just how there space #7#, yet there are other reasons why there room #7# distinctive #f# orbitals.

MORE-OR-LESS WHY THERE are #mathbf(2l+1)# ORBITALS POSSIBLE

Consider the #2p# orbitals as a easier example.

When you have actually #m_l = -1, 0, +1#, you understand you get three orbitals: #2p_x#, #2p_y#, and #2p_z#, but #m_l# just correlates with 3 orbitals without further explanation.

You could also take the #2p_z# orbital and also think the it one more way. If you want the angular inert in the #z# direction, you have the right to examine the following angular part of the Schrodinger equation in spherical coordinates:

#hatL_zY_l^(m_l)(theta,phi) = m_lℏY_l^(m_l)(theta,phi)#

resembling #hatHpsi = Epsi#.

Without obtaining into as well many complex details (you don"t have to know the three expressions because that the #Y_(l)^(m_l)(theta,phi)#), you have the right to assert the the #z# angular momentum #L_z# have the right to be expressed because that three different #m_l# values, which trace out the #2p_z# orbit in units of #ℏ#, wherein #ℏ = h/(2pi)# and also #h# is Planck"s constant. You have three pertinent #m_l# projections shown (ignore the #pm2ℏ#; that"s due to the fact that the diagram is of a #d# orbital, but we can apply it come the #2p_z# orbital):

#m_l = -1 -> L_z => -ℏ# (forming the bottom lobe, labeled #-ℏ#)#m_l = 0 -> L_z => 0# (forming the central dot, or node, labeling #0#)#m_l = +1 -> L_z => ℏ# (forming the height lobe, labeling #ℏ#)

Naturally, the height lobe is symmetrically-shaped in relationship to the bottom lobe.

THE ORTHOGONALITY CONDITION

Now, a quantum mechanically postulate/requirement because that an orbital is the it be orthogonal (perpendicular) to every the other feasible orbitals.

Two mathematical means of conveying that are the dot product and the cross product:

#hatxcdothaty = > cdot > = 0##hatxcdothatz = > cdot > = 0##hatycdothatz = > cdot > = 0##hatx xxhaty = > xx > = > = hatz##haty xxhatz = > xx > = > = hatx##hatz xxhatx = > xx > = > = haty#

Since the #2p_z# lies follow me the #z# axis and since the #z# axis is the #2p_z# orbital"s axis that symmetry, the only method the other orbitals can be orthogonal is the they lie follow me the #mathbf(x)# and #mathbf(y)# axes (as you may have recognized in the sample from the above dot and also cross products).

Clearly, the #x#, #y#, and also #z# axes, as you"ve seen many times in #xyz# name: coordinates systems, space perpendicular.

As a an outcome of #m_l = -1, 0, +1# as fine as the orthogonality condition, there space three orbitals possible: #2p_x#, #2p_y#, and #2p_z#.

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Similarly, due to the fact that there space #7# values of #m_l# for the #4f# orbitals, castle are built to be an especially shaped follow to the over angular inert Schrodinger equation, in together a means that they are all orthogonal to each other.

There just happens to it is in #7# different orbital shapes that space all orthogonal to each other, just in more complicated ways than the #2p# example.