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Chemistry LibreTexts

Solutions 3

  • Page ID
    52460
  • Q1:

     

    Molecule # Valence Electrons Geometry Electronic Configuration
    \(LiH_2^+\) 2 linear \( (2\sigma_g)^2\)
    \(BeH_2^+\) 3 linear \( (2\sigma_g)^2(1\sigma_u)^1\)
    \(BH_2\) 5 bent \((2a_1)^2(1b^2)(3a_1)^1\)
    \(NH_2\) 7 linear \( (2\sigma_g)^2(1\sigma_u)^2(1\pi_u)^2(3\sigma_g)^1\)
    \(OH_2\) 8 bent \((2a_1)^2(1b^2)(3a_1)^1(1b_2)^2\)
    \(NeH_2\) 10 linear \( (2\sigma_g)^2(1\sigma_u)^2(1\pi_u)^4(3\sigma_g)^2\)
    Valence electrons Molecule angle HAH linear, D∞h angular, C2v

    Q2:

     

    a)

    Molecule # Valence Electrons Geometry Electronic Configuration
    \(BH_3\) 6 pyramidal \( (2a_1)^2(1e_1)^2(3a_1)^2\)
    \(CH_3^+\) 6 pyramidal \( (2a_1)^2(1e_1)^2(3a_1)^2\)
    \(CH_3\) 7 planar \( (2a_1)^2(1e_1)^2(3a_1)^2\)
    \(CH_3^-\) 8 planar \( (2a'_1)^2(1e'_1)^2(1a''_1)^2(3a'_1)^2\)
    \(NH_3\) 8 planar \( (2a'_1)^2(1e'_1)^2(1a''_1)^2(3a'_1)^2\)
    \(OH_3^+\) 8 plandar \( (2a'_1)^2(1e'_1)^2(1a''_1)^2(3a'_1)^2\)

    b) \(CH_3\) has seven valence electrons and planar geometry, but with fewer electrons in high energy orbitals, it has a lower equilibrium energy. Therefore, assuming the peak planar energy is the same, \(CH_3\) has greater inversion energy.

     

    Q3:

     

    The peaks labeled from left to right correspond to the 1s, 2s, and 2p orbitals respectively.

     

    Q4:

     

    The fundamental features of the spectrum reflecting the energy levels and populations of the MOs are

    \(N_2\):

    3 small peaks, with a large peak in between the first and second small peaks for ascending energy:    I I I I

    \(Cl_2\):

    2 large peaks followed by 3 small peaks:   II I I I

     

    Q5:

     

    The prominent features of the spectrum are the presence of two completely different peaks, suggesting incomplete hybridization. For complete hybridization, we would expect one peak corresponding to one energy of equal orbitals. Within the low energy peak, there are further differences, possibly suggesting higher energy p orbitals alongside hybridized orbitals.

    Q6:

     

    Drawing the MO diagram, we can calculate bond order and fine it to be 3 for \(N_2\) and 2.5 for \(N^-_{2}\). Higher bond order implies shorter bond distance, so \(N_2\) has a shorter bond.

    Q7:

     

    Distribution of even electrons among orbitals results in lone pairs only when the last two electrons are placed in degenerate orbitals. This only occurs for the \(\pi_{2p}\) for \(B_2\) and \(O_2\). All other diatomics have the last two electrons distributed among non-degenerate orbitals or distributed among partially filled degenerate orbitals.