Consider the following two wavefunctions of the sp2 hybrid...

Consider the following two wavefunctions of the sp2 hybrid orbitals. ?1 = N1(?2s + ?2?2pz) ?2 = N2(?2?2s - ?2pz + ?3?2px) (a) (20pt) Determine the normalization constant N1 and N2 for the above two wavefunctions of sp² hybrid orbital. (b) (20pt) Find the normalized wavefunction for the 3rd sp² hybrid orbital: ?3 = (a?2s + b?2pz + c?2px). Determine the coefficients a, b, and c. (c) (20pt) Following the appendix, rewrite the hybrid orbital wavefunction ?1 by using the hydrogenic atom wavefunctions. By assuming Z = 3, r = 2.7 a0 and ? = ?/2, at what polar angle ?, ?1 will have the global maximum (from graphics and first derivative)? (d) (20pt) Repeat the above calculation for ?2.

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00:01 Hello, so in this question values for two wave function that is si 1 s and si 2 s are given and we need to prove that this two are orthogonal to each other okay so we need to prove that this two are orthogonal to each other using the conditions for orthogonality okay so using the equations that is integration from zero to infinity for si 1 s s and si 2 s so that will be equal to integration from 0 to infinity a1 s e raise to power minus of zr divided by a0 multiplied by a2s times 1 minus zr divided by 2a0 times e raise to power minus of zr divided by 2a0 times 4 pi r r or so square we are okay so this 4 pi a 2 s and a 1 s gets common because they are constant okay so this will be equal to 4 pi a 1 s a 2 s multiplied by integration from 0 to infinity for the following that is e -r -to -power but here it is r okay so this r square goes into the bracket to r square e -rae -rae -rae -res -to -power minus of 3 zr divided by 2a -0 zr divided by 2a -0 minus integration from 0 to infinity for zr -r -cube divided by 2a -0 times e -ra -to -power minus of 3 zrrrr divided by 2a -0 multiplied by dr okay so this will be the expression for the following now if we use the mathematical form we are going to use the mathematical form that is integration from zero to infinity for x raised to power n minus 1 times o to power minus of a x dot dx will be equal to under root or we can say square root of n divided by a raise to power n so this will be the expression or we can say mathematical form that we are going to use okay so if we apply this mathematical form in the above equation then this will be equal to 4 pi a 1 s a 2s times under root 3 divided by 3 z divided by 2 z divided by 2 a 0 whole cube minus under root 4 divided by 3 z divided by 2 a not raise to power 4 times z divided by 2 a not okay so now if we solve this expression further then it will be equal to 4 pi a 1 s a 2 s times 2 multiplied by 2 a not divided if we take common two okay if we take common two in this equation then two times two a not divided by 3 z this is 3 z whole to power 3 minus of 2 2 a not divided by 3 z whole cube so as we can see this minus this will result into 0 and 0 multiplied by this will also be 0 therefore the conditions for orthogonality that will be si star 1 s and si2s will be equal to 0.

04:35 So this is the conditions for orthogonality...

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