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Chemistry LibreTexts

Solutions 1

  • Page ID
    52456
  • Q1

    Write out the complete time-independent Hamiltonians for (each term is explicitly given):

    1. the helium atom, electrons 1,2: \[ -\dfrac{\hbar^2}{2m_e}(\nabla^2_1 + \nabla^2_2) - \dfrac{Ze^2}{4\pi\epsilon}(\dfrac{1}{r_1}+\dfrac{1}{r_2}) + \dfrac{e^2}{4\pi\epsilon r_{12}}\]
    2. the \(H_2^+\) ion, nuclei A,B:   \[ -\dfrac{\hbar^2}{2m_p}(\nabla^2_A + \nabla^2_B) - \dfrac{Ze^2}{4\pi\epsilon}(\dfrac{1}{r_A}+\dfrac{1}{r_B}) + \dfrac{e^2}{4\pi\epsilon r_{AB}}\]
    3. the \(H_2\) molecule, nuclei A,B electrons 1,2: \[ -\dfrac{\hbar^2}{2m_p}(\nabla^2_A + \nabla^2_B) -\dfrac{\hbar^2}{2m_e}(\nabla^2_1 + \nabla^2_2) - \dfrac{e^2}{4\pi\epsilon} (\dfrac{1}{r_{1A}}+\dfrac{1}{r_{1B}} + \dfrac{1}{r_{2A}}+\dfrac{1}{r_{2B}}  -\dfrac{1}{r_{AB}}  - \dfrac{1}{r_{12} })\] 

    Q2

    What is the Born-Oppenheimer approximation and what do we use it for? When would it fail?

    The Born-Oppenheimer approximation assumes that the nuclei in a system are stationary relative to the faster moving electrons. This allows separation of the wavefunction into a nuclear contribution and an independent electronic contribution. The approximation fails when nuclear motion is non-neglible (i.e., if the nuclei are moving faster than expected such as if very vibrational excited).

    Q3

    Confirm two s-p orbitals are orthonormal. The two \(|sp \rangle\) orbitals are: 

    \[ | sp_1 \rangle = \dfrac{1}{\sqrt {2}}( |2s\rangle + |2p \rangle)\]

    \[ | sp_2 \rangle = \dfrac{1}{\sqrt {2}}( |2s\rangle - |2p \rangle)\]

    Orthonormal means fulfilling both orthogonality and normality.

    Normality means showing:

    \[ \langle sp_1 |  sp_1 \rangle =  \langle sp_2 |  sp_2 \rangle  = 1\]

    Orthogonality means showing: 

    \[ \langle sp_1 |  sp_2 \rangle =  \langle sp_2 |  sp_1 \rangle = 0\]

    We must keep in mind that s and p orbitals are orthonormal:

    \[ \langle s|s \rangle = \langle p|p \rangle = 1 \]

    and 

    \[ \langle s|p \rangle = \langle p|s \rangle = 0 \]

    Therefore using the decomposition of the hybrid orbitals....

    \[ \langle sp_1 |  sp_1 \rangle = \dfrac{1}{\sqrt {2}}( \langle2s| + \langle2p |)  \dfrac{1}{\sqrt {2}}( |2s\rangle + |2p \rangle) = \dfrac{1}{2}( \langle s|s \rangle + \langle s|p \rangle + \langle p|s \rangle +  \langle p|p \rangle  )  = \dfrac{1}{2}( 1 + 0 + 0 + 1) =  1 \] 

    \[ \langle sp_2 |  sp_2 \rangle = \dfrac{1}{\sqrt {2}}( \langle2s| - \langle2p |)  \dfrac{1}{\sqrt {2}}( |2s\rangle - |2p \rangle) = \dfrac{1}{2}( \langle s|s \rangle - \langle s|p \rangle - \langle p|s \rangle +  \langle p|p \rangle  )  = \dfrac{1}{2}( 1 - 0 - 0 + 1) =  1 \] 

    \[ \langle sp_1 |  sp_2 \rangle = \dfrac{1}{\sqrt {2}}( \langle2s| + \langle2p |)  \dfrac{1}{\sqrt {2}}( |2s\rangle - |2p \rangle) = \dfrac{1}{2}( \langle s|s \rangle - \langle s|p \rangle - \langle p|s \rangle -  \langle p|p \rangle  )  = \dfrac{1}{2}( 1 - 0 + 0 - 1) =  0 \]

    \[ \langle sp_2 |  sp_1 \rangle = \dfrac{1}{\sqrt {2}}( \langle2s| - \langle2p |)  \dfrac{1}{\sqrt {2}}( |2s\rangle + |2p \rangle) = \dfrac{1}{2}( \langle s|s \rangle - \langle s|p \rangle - \langle p|s \rangle -  \langle p|p \rangle  )  = \dfrac{1}{2}( 1 + 0 - 0 - 1) =  0 \]

    Q4

    Show \(|sp^2\rangle = \dfrac{|s\rangle + \sqrt{2} |p \rangle}{\sqrt{3}}\)  is normalized if \(|s \rangle\) and \(|p \rangle\) are normalized. This means showing that \( \langle sp^2|sp^2\rangle = 1 \):

    \[ \langle sp^2|sp^2\rangle = (\dfrac{\langle s | + \sqrt{2} \langle p |}{\sqrt{3}} )( \dfrac{|s\rangle + \sqrt{2} |p \rangle}{\sqrt{3}})  = \dfrac{1}{3}(\langle s | s \rangle + \sqrt{2} \langle s|p \rangle = \sqrt{2} \langle p | s \rangle + 2 \langle p|p \rangle) = \dfrac{1}{3}(1 + 0 + 0 + 2) = 1\]

    Q5

    What is the average energy of a Hydrogen atom \(|sp^2 \rangle\) hybrid orbital if energy of \( |s \rangle \) is \(E_s \) and energy of \(|p \rangle \) is \( E_p \)?

    We are given that \( \langle s| \hat{H} | s \rangle  =  E_s \) and \( \langle p| \hat{H} | p \rangle = E_p \).

    It is also useful to realize that 

    \[ \langle s| \hat{H} | p \rangle  = \langle p| \hat{H} | s \rangle = 0 \]

    \[ E =  \dfrac{ \langle sp^2| \hat{H} | sp^2 \rangle}{ \langle sp^2|sp^2 \rangle} =   \left(\dfrac{\langle s | + \sqrt{2} \langle p |}{\sqrt{3}} \right) \hat{H} \left( \dfrac{|s\rangle + \sqrt{2} |p \rangle}{\sqrt{3}} \right) \]

    \[= \dfrac{1}{3}( \langle s| \hat{H} | s \rangle  + \sqrt{2}\langle s| \hat{H} | p \rangle + \sqrt{2}\langle p| \hat{H} | s \rangle + 2 \langle p| \hat{H} | p \rangle   = \dfrac{1}{3}(E_s + 0 + 0 + 2 E_p)   \]

    = \[ \dfrac{1}{3}E_s + \dfrac{2}{3}E_p \]

    Q6

    When one s orbital and two p orbitals are used to generate hybrid orbitals, a total of three orbitals are generated.