Consider a \(\mathrm{N}_{2}\) molecule in its first excited electronic state; that is, when an electron in the highest occupied molecular orbital is promoted to the lowest empty molecular obital. (a) Identify the molecular orbitals involved and sketch a diagram to show the transition. (b) Compare the bond order and bond length of \(\mathrm{N}_{2}{ }^{*}\) with \(\mathrm{N}_{2}\), where the asterisk denotes the excited molecule. (c) Is \(\mathrm{N}_{2}{ }^{*}\) diamagnetic or paramagnetic? (d) When \(\mathrm{N}_{2} *\) loses its excess energy and converts to the ground state \(\mathrm{N}_{2}\), it emits a photon of wavelength \(470 \mathrm{nm}\), which makes up part of the auroras lights. Calculate the energy difference between these levels.

Step by step solution

01

Identify Molecular Orbitals involved and sketch the transition

The highest occupied molecular orbital for nitrogen molecule (\(\mathrm{N}_{2}\)) is \(\pi_{2p}\), whereas the lowest unoccupied molecular orbital is \(\sigma_{2p}^{*}\). Thus the transition involves an electron being promoted from \(\pi_{2p}\) to \(\sigma_{2p}^{*}\). A diagram can be drawn showing this electron transition.

02

Compare Bond order and Bond length

In the ground state \(\mathrm{N}_{2}\), the bond order can be calculated as (number of electrons in bonding MOs - number of electrons in antibonding MOs) / 2, which is (10 - 2) / 2 = 4. In the excited state, it is (10 - 3) / 2 = 3.5. Therefore, the bond order decreases upon excitation. As bond length is inversely proportional to bond order, the bond length in the excited state is greater than the ground state.

03

Determine Magnetic Character

In the ground state, \(\mathrm{N}_{2}\) is diamagnetic as all electrons are paired. However, when an electron is promoted to a higher energy level, there becomes an unpaired electron in the molecule. Thus, the the excited state \(\mathrm{N}_{2}{ }^{*}\) is paramagnetic.

04

Calculate Energy Difference

The energy of a photon is given by the equation \(E = \dfrac{hc}{\lambda}\), where \(h\) is the Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength. Substituting given values, \(E = \dfrac{(6.626 \times 10^{-34} Js)(3.00 \times 10^{8} m/s)}{470 \times 10^{-9}m}\) gives the energy difference of around \(4.23 \times 10^{-19} Joules\).

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