?Consider the molecular orbital diagram shown below, a diagram that is often app
ID: 1010109 • Letter: #
Question
?Consider the molecular orbital diagram shown below, a diagram that is often applied to the first row diatomic molecules. Note that this diagram comprises orbitals derived from the n = 2 valence shell of orbitals (2s and 2p). Use this diagram to determine (1) if the following diatomic molecules or ions will be diamagnetic (all electrons paired) or paramagnetic (one or more unpaired electrons), and (2) for paramagnetic species, the number of unpaired electrons.
O2, N2, CO, NO, NO-, NO+, CN-, FO-
ot 2P 2p tk: 2s 2sExplanation / Answer
The 2s orbital has 2 electrons and the 2s* has 2. There are a total of (4 + 4) = 8 p electrons which go into the 2p, and * orbitals. The 2p has 2 electrons, x has 2 and y has two. The remaining two electrons go into the x* and y* orbitals, each orbital containing 1 electron. Thus, O2 has 2 unpaired electrons and is paramagnetic.
N2: Formed by combination of two N atomic orbitals, each having configuration [He]2s22p3.
We follow the same approach as before and note that the 2s and 2s* orbitals are completely filled. The remaining 6 electrons go into 2p, x and y. Each of these orbitals houses 2 electrons each; there are no unpaired electrons and hence N2 is diamagnetic.
CO: Numbers of electrons in the outermost orbitals of C and O atoms are 4 (C:[He]2s22p2) and 6 (O:[He]2s22p4) – total number of electrons is 10. 4 electrons go into the 2s and 2s* orbitals (just like O2 in 1 above). There are 6 electrons remaining and they fill up the 2p, x and y orbitals (like N2 in 2 above). Hence there are no unpaired electrons and thus CO is diamagnetic.
NO: Total number of electrons in the outermost molecular orbitals = 5 (from N atom) + 6 (from O atom) = 11. Just like the three examples above, 4 electrons go into the 2s and 2s* orbitals. There are 7 electrons remaining, out of which 6 go into 2p, x and y orbitals (like N2 above). The remaining 1 electron go into either x* or y* orbital; thus NO is paramagnetic due to 1 unpaired electron.
NO-: Same as NO, expect that there are 12 electrons now (1 extra due to negative charge). The extra electron occupies x* or y* orbitals (both the non-bonding x* and y* orbitals must be singly occupied as per Hund’s rule). There are 2 unpaired electrons and hence NO- is paramagnetic.
NO+: Same as NO, except that there are 10 electrons (1 electron short due to positive charge). Total is 10 electrons that fill up 2s, 2s*, 2p, x and y orbitals completely. There is no unpaired electron and hence NO+ is diamagnetic.
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