WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, welcome back to www.educator.com.
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Welcome back to Physical Chemistry.
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Today, we are going to continue with our example problems for the hydrogen atom.
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Let us get started.
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Our first problem is the following, the energy of a given orbital is a function of the primary quantum number N.
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The relation is as follows, E sub N = - E²/ 8 π A₀ A₀ N².
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Again, this is the permittivity of free space, the A₀ A₀ is the Bohr radius and N is the primary quantum number.
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With the exception of N = 1, each primary energy level is degenerate and for N = 2, N = 3, N = 4, so on.
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In other words, they are difference states with the same energy.
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Derive a formula for the degeneracy of a given level N.
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That is, for each value of N, how many states have energy E sub M.
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We want you to derive the formula for that.
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Let us see what we have got.
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We know that 4 each N, the L value 0,1, 2, and so on, all the way to N -1.
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For each L, we know that M is equal to 0, + or -1, + or -2, all the way to + or – L.
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I will go to the next slide here.
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For each L value, there are 2 L + 1 states.
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Therefore, we form the following sum.
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The sum as L goes from 0 to N -1 of 2 L + 1.
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For every L value, we have 0, + or -1, + or -2, + or -, whatever all the way to + or – L.
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For every L, there is 2 L + 1 state.
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L is going to run from 0 all the way to N -1.
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Whatever N we choose, we are going to add 2 L + 1 state for L0, 2 L + 1 states for 1,
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2 L + 1 state for 2, 2 L + 1 all the way to N -1.
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This is the sum that we need.
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This is the number of states that have the particular energy E sub M.
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The sum forms an arithmetic sequence.
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If you remember what an arithmetic sequence is, it is a sequence of numbers
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where each number differs from the one before it by a fixed number.
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+ 5, 13579 is an arithmetic sequence and the difference is 2.
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1 to 3 is 2, 3 to 5 is 2, 5 to 7 is 2, and you are adding which is why we call it arithmetic.
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When you are multiplying like for example 2, 4, 8, 16, in the case you are multiplying by 2,
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each term was called a geometric sequence.
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This is an arithmetic sequence.
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The sum is actually equal to, when L = 0, we get 1.
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When L = 1, we get 3.
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When L = 2, we get 5.
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We are going to add them all the way to, N -1 we put N -1 in here.
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It is going to be 2 × N -1 + 1.
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L is equal to 1 + 3 + 5 +… + 2N - 2 + 1.
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The final term of the sequence is going to be 2 N -1.
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We have a closed form formula for the sum of an arithmetic sequence.
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The sum of an arithmetic sequence with N terms is equal to the first term + the last term ×
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the total number of terms divided by 2.
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In this particular sequence, the first term is 1.
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The last term is 2 N -1.
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The number of terms we have is N terms, from 0 to N -1 is N terms.
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Our sum is equal to 1 which is the first term, + 2 N -1 × N/ 2.
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That is equal to N²/ 2 is equal to N².
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There you go.
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For each N, there are N² states having energy E sub M, when this particular molecule is not in a magnetic field.
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Let us go ahead and say it now.
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When you place something in a magnetic field, the magnetic field actually ends up splitting the energy.
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Now, the energy levels are going to degenerate.
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You end up actually separating the energy levels which is why the spectrum often shows,
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when you put in a magnetic field several lines instead of just one line.
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For each N, there are N² states having energy E sub N when not in a magnetic field.
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I will go ahead and write what I just said.
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Recall that when an atom is placed in a magnetic field, the field splits E sub N into 2 L + 1 levels.
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Let us see.
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For the angular portion of the ψ 321 32 -1 orbital, use the following linear combinations
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to represent the spherical harmonics S21 and S2 -1 as functions of the real variables θ and φ.
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Let us talk about what is going on.
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For the angular portion of the ψ orbital 321, use the following linear combinations
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to represent the spherical harmonics as functions of the real variables θ and φ.
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Let us just recall it a little bit of our theory here.
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I will use ψ 321 as the example.
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321, that is made up of a radial function 32, which is a function of R.
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Our spherical harmonic 21, which is a function of θ and φ.
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The problem is, when M is not equal to 0, the spherical harmonics they are complex.
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They are real in θ but they are complex in φ, E ⁺I φ E ⁻I φ.
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Mathematically, this is does represent a problem but what chemist decided to do is,
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they decided to see if there are ways to actually change these complex functions and just represent them as real functions.
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What they decided to do, I think we did not talk about this in the previous lesson, they took the spherical harmonic.
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When we say angular portion, this is the radial portion of the wave function.
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This is the angular portion of the wave function.
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Radial, angular, or spherical harmonic, basically when we say angular portion.
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They decided to see if they can represent these as real functions, real functions of θ and φ as opposed to complex functions.
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They did that by taking linear combinations of them so that the complex part would cancel out.
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They tried a whole bunch of linear combinations until they got one that is actually ended up working.
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The reason they are able to do this was because we just said that these things are degenerate.
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For functions that have the same energy, if I take a linear combination of them, they have the same energy.
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In other words, they still satisfied the Schrӧdinger equation.
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Basically, what we did is we ended up creating, taking S21 and then taking S2 -1,
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adding them together or subtracting them depending and coming up with a new orbital that actually is a real function.
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Instead of just dealing with S21 directly, we decided to fiddle around with it and
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come up with a new orbital that is just a function of real variables θ and φ.
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Not a function of complex variables.
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That is all that is happening here, they still satisfy the Schrӧdinger equation.
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They still have the same energy.
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They are still Eigen functions of the Schrӧdinger operator, of the angular momentum² the angular momentum operator.
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For all practical purposes, they are kind of the same function but they behave better.
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That is all we have done, we have taken a function, we have created a new function
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from a pair that happen to have the same energy to create a function that behaves better.
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Once again, use the following linear combinations to represent a spherical harmonics as functions of the real variables θ and φ.
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This is one thing we are going to do, this is the other thing we are going to do.
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We are going to take S21, we are going to add it to S2 -1.
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We are going to multiply it by 1/ √ 2.
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We are going to see what function we come up with.
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That is going to be the DXZ orbital.
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Let us go ahead and see if we can do this.
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Let us go ahead and talk about what S2 and S1 are so we can look them up in our book.
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S21 is equal to 15/ 8 π¹/2 sin θ cos θ E ⁺I φ.
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We see this is a complex function and S2 -1 is equal to 15/ 8 π ^½.
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This part is the same, sin θ cos θ except this E ⁻I φ.
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Again, a complex function.
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Chemists, for one reason or another decided they want one or more complex functions so they decide to add these,
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subtract them, subject them to some linear combination of these 2,
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to see if they can come up with functions that behave better.
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These are the linear combinations that they came up with.
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Let us go ahead and see what we can do.
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DXZ , we are going to do this one first.
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I think I will go back to black.
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Let us go ahead and do blue.
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DXZ = this thing, 1/ √ 2 × S1 S2.
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That is equal to 1/ √ 2 ×, I’m going to pullout this and this are the same.
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It was 15/ 8 π ^½ , the sin θ and the cos θ are the same.
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I’m going to write E ⁺I φ + E ⁺I φ.
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I'm going to write this as 1/ √ 2 15/ 8 π ½ sin θ cos θ ×, I’m going to express these in there.
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Polar form, I’m going to write this as cos φ + I sin φ, Euler’s formula as an expression E ⁺I something.
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E ⁺I θ = cos θ + I sin θ and then + E ^-θ is going to end up equaling cos θ - I sin θ + cos φ - I sin φ.
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This and this cancel out.
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What I'm left with is, 1/4 ^½ 15/ 4 π ^½.
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I have taken a π and I have written it to make a little bit easier to deal with.
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Sin θ cos θ × 2 cos of φ.
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1/4 is ½, the 1/2 and 2 cancel and what I end up with is 15/ 4 π ^½ sin θ cos θ cos of φ.
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This is DXZ, this is the orbital that they came up with.
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Notice, S21 S2 -1 had E ⁺I φ and E ⁻I φ, the complex went away.
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I have this orbital which is sin θ cos θ of φ.
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Real variable and real variable.
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I took orbitals that are complex, I combine them to create an orbital which is real.
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That is all I have done.
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I personally, do not particularly like that they have done this.
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What this does is when you actually plot this out using a polar plot,
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this is what gives you the famous dumbbell sheet orbitals and all the orbital shapes that you see in your general chemistry book.
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They are also in your physical chemistry book.
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The lobe is this way, that is what these are.
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When they plot these, it gives a certain directional character in space along the X axis, along the ZY axis,
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and between the X and Y axis, and between the Y and Z axis.
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As we get these lobes, they actually give you some information.
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I did not know, some× they do, some× they do not.
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But that is what is happening here.
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They want to turn complex orbital into real orbitals.
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OK so let us go ahead and do DYZ.
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DYZ we said is equal to 1/ π √ 2 × S21, this time - S2-1.
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This is going to equal 1/ I √ 2, do the exact same thing as I did before.
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15/8 π ^½, this is sin θ cos θ.
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It is actually nice to go through these which is why I included them in example problem.
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I think it is nice to actually go through the mathematics and see where these functions come from.
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The complex version and also the real version of the end up plotting.
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Some books it just feels as though there is a sort of throwing these equations out like which one do I use.
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You can use the complex but chemists tend to use the real version, the real variable versions.
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This is θ and this is going to be E ⁻I φ.
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This is going to end up equaling 1/ I √ 2 15/8 π ^½ sin θ cos θ.
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Now, we have cos φ + I sin φ, then we have - cos φ - I sin φ.
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- +, we end up with 1/ I √ 2 × 15/8 π ^½ sin of θ cos of θ.
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Cos of φ - cos φ they go away, I sin φ.
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You end up with 2 I sin of φ.
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You end up with 1/2 I × 15/4 π ^½ sin θ cos θ × 2 I sin φ.
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The 2 I cancel and you are left with 15/ 4 π ^½ sin θ cos θ sin φ.
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This is equal to the DYZ orbital.
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There you go.
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Let me see if I have another page here that I can use.
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I do, let me go ahead and do that.
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Let us talk about what is actually happened here.
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Sorry, I want to make sure you understand what is happening globally.
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Ψ 321, I think that is what we had right and ψ 32-1.
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We took the 21 and 2 -1 portions.
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We took the spherical harmonic S21 and S2 -1, both of these were complex.
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What we did is we found a linear combination, 1/ √ 2 adding these.
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Basically, what I want to do is I want to find a way to turn this into a real function, a function of real variables.
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I want to turn this into a function of real variables, instead of complex and complex.
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By taking 1/ √ 2, this + this, I end up converting this one into the version that has the sin θ cos θ cos φ.
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I ended up by taking 1/ √ 2 I × this – this.
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I end up converting it into the version that has the sin θ cos θ sin φ.
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I have 2 spherical harmonics.
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I ended up converting them to 2.
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Now, I call this one DXZ and I call this one DYZ.
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Start off with 2, I ended up with 2, but now they are functions of the real.
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Θ θ and φ, there is no complex number to be found.
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That is all that is that I have done.
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I sort of change the way they look but it is still the same function.
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I still end up getting the same numbers out, whatever calculations with them.
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That is all I have done, I have dressed them up.
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Repeat the procedure for example 2, except this time with S22 and S2 – 2.
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Usind the following linear combinations, which you should notice are the same as for example number 2.
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Now, using the spherical harmonics S22 and S2 -2, can I find the linear combination of them which I have right here.
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We are actually giving it to them explicitly.
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1/ √ 2 × the sum of them and 1/ √ 2 × the difference of them.
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We got a little bit of an issue here.
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This is 2, this is – 2, this is 2, and this is -2, sorry about that.
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It is the same linear combination, I'm just taking 1/ √ 2 × the sum of I/ √ 2 × the difference.
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Again, converting this through this into a function of real variables.
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That is all that is happening here.
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Let us go ahead and take care of these.
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We have got a DXZ and it is not DXZ actually, this one and this one, what happened here?
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This one was actually going to be the DX² - Y² orbital and this one is actually going to end up being the DXY orbital.
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There we go, now we are good.
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We can actually do the problems.
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Let us start off with the DX² - Y².
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DX² - Y² = 1/ √ 2.
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I will rewrite this, S22 + S2 -2 that is going to equal 1/ √ 2.
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Let us actually write out what S22 S2 -2 are.
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S22 is equal to 15/ 32 π ^½, this is going to be sin² θ E ⁺I φ.
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S2 -2 = 15/ 32 π ^½ sin² θ E ⁺I – 2 φ.
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This is S22, this is S2 -2.
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It is going to equal , the 15/ 32 π we have to include that.
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The sin² θ is the same so I pulled that one out as a factor.
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And I'm left with E ⁺I 2 φ + E ⁻I 2 φ.
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That is going to equal, DX² - Y² is going to equal,
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I’m going to write that as 1/ 2.
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I pulled out, this is now going to be 16 π.
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I pulled out a 2 out of there.
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½ sin² θ, exactly the same as before.
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Now, it is cos 2 φ + I sin 2 φ + cos 2 φ - I sin 2 φ.
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That cancels with that and I'm going to end up getting 15/16 π¹/2 sin² θ × the cos of 2 φ.
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Actually, 2 × the cos of 2 φ.
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This and this cancel and I'm left with my final DX² - Y² is equal to 15/ 16 π ^½ sin² θ cos of 2 φ.
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There we go, that takes care of that one. I'm going to do D of XY.
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This was going to be 1/ I √ 2 × S22 - S2 -2.
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And this is going to equal 1/ I √ 2 15/ 32 π ^½.
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The sin² θ comes out, we have E ⁺I2 φ – E ⁻I 2 φ.
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Let us go ahead and pull out a 2 here.
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√ 2 cancels √ 2, so I get 1/2 I × 15/ 16 π ^½ sin² θ.
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The same thing, this time we have cos 2 φ + I sin of 2 φ - cos of 2 φ - I sin of 2 φ.
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Cos of 2 φ - cos of 2 φ, this and this they add together.
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I end up with 1/ 2 I × 15/ 16 π ^½ sin² θ × 2 I sin 2 φ.
00:29:22.900 --> 00:29:49.200
2 I cancels the 2 I, and I'm left with the D of XY orbital is equal to 15/ 16 π ^½ × sin² θ × sin of 2 φ.
00:29:49.200 --> 00:29:50.300
There we go.
00:29:50.300 --> 00:30:09.700
Basically, I have 2 complex spherical harmonics S22 and S2-2,
00:30:09.700 --> 00:30:15.200
I need functions of real variables not functions of real and complex variables.
00:30:15.200 --> 00:30:19.400
I have taken this one and the combination with this one,
00:30:19.400 --> 00:30:26.100
I have transformed it into the one that involves the sin² θ cos 2 φ.
00:30:26.100 --> 00:30:32.700
I have taken this one and by combining it with this one, in a linear combination that ended up working out really nicely,
00:30:32.700 --> 00:30:37.900
I ended up with the one that gave us the sin² θ sin 2 φ.
00:30:37.900 --> 00:30:44.300
Of course, there are the constants, the 15/ 16 π in front, that is all I have done.
00:30:44.300 --> 00:30:53.300
I have taken complex spherical harmonics and I turned them into real functions and end up giving me the same values.
00:30:53.300 --> 00:30:55.700
I have dressed up the function, I have made it look better.
00:30:55.700 --> 00:30:59.400
I make it easier to handle, as far as real variable is concerned.
00:30:59.400 --> 00:31:04.400
Mathematically, the fact of the matter is, it is easier to handle than this one.
00:31:04.400 --> 00:31:11.100
It is not any more difficult to handle a complex variable function as opposed to a real variable function.
00:31:11.100 --> 00:31:16.900
In fact, in my personal opinion, complex variable functions are easier to handle in a lot of ways.
00:31:16.900 --> 00:31:22.400
Again, for one reason or another, historically, chemists prefer to deal with just functions of real variables.
00:31:22.400 --> 00:31:27.900
These have the advantage of you can plot them.
00:31:27.900 --> 00:31:35.100
The plots that you see, those bell shaped plots, all of the polar plots for the densities of the electrons,
00:31:35.100 --> 00:31:49.000
the DC², the XY, the YZ, the lobes here and there, you are actually seeing plots of these polar plots of those.
00:31:49.000 --> 00:31:53.100
Let us move on to the next one.
00:31:53.100 --> 00:31:56.000
A lot of information here.
00:31:56.000 --> 00:32:00.100
It is not difficult, it is just extra information.
00:32:00.100 --> 00:32:05.900
Here is when you actually have to look back on your book to find the functions that you need.
00:32:05.900 --> 00:32:11.100
The following are the change of variable relations between the Cartesian and spherical coordinates.
00:32:11.100 --> 00:32:13.000
X is equal to R sin θ cos φ.
00:32:13.000 --> 00:32:15.900
Y is equal to R sin θ sin φ.
00:32:15.900 --> 00:32:19.100
Z is equal to R cos θ.
00:32:19.100 --> 00:32:24.000
The following are the real variable representations of the angular portions of
00:32:24.000 --> 00:32:28.900
the spherical harmonics of the hydrogen atom wave functions for L = 2.
00:32:28.900 --> 00:32:30.900
In other words, the D orbitals.
00:32:30.900 --> 00:32:33.600
Pretty much what we just did.
00:32:33.600 --> 00:32:39.700
We did this one, this one, this one, and this one.
00:32:39.700 --> 00:32:47.500
The one for DZ³², the one for ψ 320.
00:32:47.500 --> 00:32:54.500
In other words, the S20 that one did not need to be converted into a functional real variable
00:32:54.500 --> 00:32:57.400
because it is already a function of a real variable as it stands.
00:32:57.400 --> 00:33:01.100
I do not have to subject it to any linear combination.
00:33:01.100 --> 00:33:07.700
It is the others, the 21 2 -1, the 22 2-2, spherical harmonics that I had to subject
00:33:07.700 --> 00:33:11.700
to linear combinations to convert them to these equations.
00:33:11.700 --> 00:33:13.800
The ones in real variables.
00:33:13.800 --> 00:33:15.800
This on, I can just take as is.
00:33:15.800 --> 00:33:22.200
These are the real variable representations of the D orbitals.
00:33:22.200 --> 00:33:25.200
Ψ 320, we call it DZ².
00:33:25.200 --> 00:33:27.800
Ψ 321, we call it DXZ.
00:33:27.800 --> 00:33:30.400
Ψ 32 -1, we call it DYZ.
00:33:30.400 --> 00:33:34.700
322, re call it DX ⁻Y².
00:33:34.700 --> 00:33:40.800
32 -2, we call it DXY.
00:33:40.800 --> 00:33:44.900
What you are charged with doing is the following.
00:33:44.900 --> 00:33:57.100
Use the information given to explain why these orbital functions are actually given the labels DZ², DXZ, DYZ, DXY², and DXY.
00:33:57.100 --> 00:34:04.100
Why we call it ψ 320, 321, 32-1, 322, 32 -2.
00:34:04.100 --> 00:34:07.000
We do but this is how we refer to them.
00:34:07.000 --> 00:34:13.600
Why do we refer to them like this?
00:34:13.600 --> 00:34:15.700
Let us see what we have got.
00:34:15.700 --> 00:34:20.300
Let us go ahead and take the first one.
00:34:20.300 --> 00:34:37.400
Let us go ahead and take the first one which is the ψ 320, that function is 3 cos² -1.
00:34:37.400 --> 00:34:49.100
I actually need to write my things down here.
00:34:49.100 --> 00:34:56.300
X = R sin θ cos φ.
00:34:56.300 --> 00:35:01.000
Y = R sin Θ sin φ.
00:35:01.000 --> 00:35:05.600
Z = R cos θ.
00:35:05.600 --> 00:35:12.300
I'm going to change this and I'm going to express it in terms of X, Y, and Z.
00:35:12.300 --> 00:35:17.800
3 cos I² θ -1.
00:35:17.800 --> 00:35:22.800
Cos² Θ is nothing more than Z²/ R².
00:35:22.800 --> 00:35:29.700
I’m going to write this as 3 Z²/ R² -1.
00:35:29.700 --> 00:35:41.900
There you go, there is my Z² because the spherical coordinate version,
00:35:41.900 --> 00:35:51.500
once I convert it using the conversions between spherical and Cartesian version which is 3 Z²/ R²,
00:35:51.500 --> 00:35:58.100
because it is something proportional to Z², I end up calling this DZ².
00:35:58.100 --> 00:36:02.200
That is where the DZ² comes from.
00:36:02.200 --> 00:36:04.200
Let us do the next one.
00:36:04.200 --> 00:36:15.800
Let us go ahead and do the sin θ cos θ cos φ.
00:36:15.800 --> 00:36:20.800
We can consider that one to be the ψ 321.
00:36:20.800 --> 00:36:28.000
Sin θ, I’m going to convert this to XYZ form.
00:36:28.000 --> 00:36:30.900
Let us see what it turns out to be.
00:36:30.900 --> 00:36:35.300
Sin θ, I'm going to use this one.
00:36:35.300 --> 00:36:45.400
It is going to be X/ R cos φ.
00:36:45.400 --> 00:36:51.800
The cos θ is going to be Z/ R.
00:36:51.800 --> 00:37:04.000
The cos φ, I’m going to take this one cos φ is equal to X/ R sin θ.
00:37:04.000 --> 00:37:26.600
This is equal to XZ X/ R² × R sin θ cos φ.
00:37:26.600 --> 00:37:30.100
R sin θ cos φ is nothing more than X.
00:37:30.100 --> 00:37:42.300
This is X φ X/ R² X, those cancel and I'm left with X Z/ R².
00:37:42.300 --> 00:37:46.300
There is my XZ, something proportional to 1/ R².
00:37:46.300 --> 00:37:55.000
This orbital, that is why this is called D sub XZ.
00:37:55.000 --> 00:37:57.300
Let us do our next one.
00:37:57.300 --> 00:38:01.400
This one we can consider the 32 -1 orbital.
00:38:01.400 --> 00:38:20.900
That one has sin θ, I’m not using the constants in front of 15/ 16 π with those orbital, R sin θ cos θ sin of φ.
00:38:20.900 --> 00:38:24.400
Let me write it here again for reference.
00:38:24.400 --> 00:38:29.900
X = R sin θ cos φ.
00:38:29.900 --> 00:38:34.200
Y = R sin Θ sin φ.
00:38:34.200 --> 00:38:39.200
Z = R cos θ.
00:38:39.200 --> 00:38:41.800
Θ, φ, and R all over the place.
00:38:41.800 --> 00:38:44.400
Sin θ cos θ sin φ.
00:38:44.400 --> 00:38:59.500
Sin θ is equal to Y/ R × the sin of φ.
00:38:59.500 --> 00:39:03.600
Cos θ is equal to Z/ R.
00:39:03.600 --> 00:39:10.200
I’m converting it back to the Cartesian form.
00:39:10.200 --> 00:39:17.200
Sin of φ is equal to Y/ R sin Θ.
00:39:17.200 --> 00:39:36.400
This is equal to YZY/ R² × R sin θ sin φ which is equal to YZY/ R².
00:39:36.400 --> 00:39:46.900
R sin θ sin φ is just Y so that goes away, leaving me YZ/ R².
00:39:46.900 --> 00:39:52.700
There is my YZ, this is my DYZ orbital.
00:39:52.700 --> 00:40:01.300
That is why I call it DYZ because ψ 321 orbital, when I express those, the sin and cos,
00:40:01.300 --> 00:40:09.400
and things like that, in terms of XYZ, I end up getting something which is 1/ R² × YZ.
00:40:09.400 --> 00:40:14.700
We call it DYZ.
00:40:14.700 --> 00:40:32.300
What else have we got?
00:40:32.300 --> 00:40:48.800
Let us say the ψ 322, and that one the trigonometric portion is sin² θ × the cos of 2 φ.
00:40:48.800 --> 00:41:12.900
For the cos of 2 φ, let us use the identity cos² φ - sin² φ.
00:41:12.900 --> 00:41:31.200
We can write this as sin² θ × cos² φ - sin² φ.
00:41:31.200 --> 00:41:34.600
Again, using these conversions I end up with the following.
00:41:34.600 --> 00:41:44.200
This ends up being equal to X²/ R² cos² φ.
00:41:44.200 --> 00:41:56.700
The cos² φ is equal to X²/ R² sin² θ.
00:41:56.700 --> 00:42:11.500
The sin² of φ is going to be Y²/ R² sin² θ.
00:42:11.500 --> 00:42:36.100
This is going to equal X² × X² - Y²/ R² × R² sin² θ cos² φ.
00:42:36.100 --> 00:42:51.800
This is equal to X² × X² - Y²/ R² × X².
00:42:51.800 --> 00:43:01.300
The X² cancel and I'm left with X² - Y²/ R².
00:43:01.300 --> 00:43:06.200
It is 1/ R² × X² - Y².
00:43:06.200 --> 00:43:11.700
This is where we get our DX² - Y² label.
00:43:11.700 --> 00:43:19.200
This is what we call it the X²- Y² orbital.
00:43:19.200 --> 00:43:21.600
Let us do our final one.
00:43:21.600 --> 00:43:36.600
I think it was ψ 32 -2 and that one was sin² θ sin of 2 φ.
00:43:36.600 --> 00:43:39.000
Let us go ahead and use an identity here.
00:43:39.000 --> 00:43:41.300
We will use the identity.
00:43:41.300 --> 00:43:55.800
Sin² θ × 2 sin φ cos φ, pretty standard double angle identity first.
00:43:55.800 --> 00:43:58.500
Let us go ahead and write again.
00:43:58.500 --> 00:44:06.000
We had on this page, X = R sin θ cos φ.
00:44:06.000 --> 00:44:11.000
Y = R sin θ sin φ.
00:44:11.000 --> 00:44:14.200
Z = R cos of θ.
00:44:14.200 --> 00:44:33.600
When we put these into these, we end up getting X² / R² cos² φ × 2 × the sin of φ
00:44:33.600 --> 00:45:07.200
which is Y/ R sin θ × cos of φ, which is X × R sin θ which is equal to X² × 2 XY/ R² × R² sin² θ cos² φ,
00:45:07.200 --> 00:45:17.400
which is equal to X² × 2 XY/ R² × X².
00:45:17.400 --> 00:45:25.800
The X² cancel and I'm left with 2 XY/ R².
00:45:25.800 --> 00:45:31.300
XY proportional to the 2/ R².
00:45:31.300 --> 00:45:39.100
Therefore, this orbital is why we actually call it the DXY orbital.
00:45:39.100 --> 00:45:40.500
I hope that makes sense.
00:45:40.500 --> 00:45:42.700
Thank you so much for joining us here at www.educator.com.
00:45:42.700 --> 00:48:33.000
We will see you next time, bye.