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):

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.