Problem 9.69

Using Figures 9.37 and 9.45 as guides, give the molecular orbital electron configuration for each of the following cations. In each case, indicate whether the addition of an electron would increase or decrease the stability of the species.

Here's the Big Idea: we draw the MO diagram for the diatomics, dump the electrons in, and calculate the bond orders for the cations. Then we add an electron into the first vacant MO and recalculate the bond order. If we add an electron to a bonding MO the bond order will increase and the species will be more stable than the starting cation, and if we add an electron to an antibonding MO, the bond order will decrease and the species will be less stable than the starting cation. Got it?

(a)B₂⁺

Here, we have to consider the formation of molecular orbitals from atomic p-type orbitals in addition to MO formation from s-type atomic orbitals. We will not concern ourselves with the 1s electrons of B in this problem (they are considered to be core electrons and as such do not participate in bonding.) From the B 2s orbitals, we get two sigma-type MOs (s_2s and s_2s*). Now, boron has an occupied 2p subshell; how do we combine the three 2p atomic orbitals on each boron to get MOs? We have six total 2p orbitals to consider; we shall therefore get six MOs. Recall that a sigma MO is formed from end-to-end overlap of two 2p orbitals; thus, we will form 2 sigma-type MOs, s_2p and s_2p*. (Let us say that the 2pz orbitals on each B atom overlapped and formed these sigma MOs.) We are left with two 2p atomic orbitals (the 2py and 2px orbitals) on each atom; the two 2px orbitals can combine to form two pi type MOs p_2p and p_2p*, and the two 2py atomic orbitals can combine to give another set of pi-type MOs of the same energy as those formed from combining the 2px AOs. We display this graphically as follows:

The ordering of these MOs is something you should memorize. Remember: for B₂ to N₂, the p_2p MO is lower in energy than the s_2p MO.

Now, we place the 2s and 2p electrons of B into the atomic orbitals (we are drawing one B as an ion since we are working with B₂⁺. We take the total of 5 valence electrons and place them into the MOs using Hund's rules:

The bond order of B₂⁺ is 1/2(3-2) = 1/2; if we add an electron, we place it into the p_2p MO, and the bond order becomes 1/2(4-2) = 1; thus, the addition of an electron stabilizes B₂ relative to B₂⁺ because the bond order increases.

(b)Li₂⁺

To begin, we form sigma-type molecular orbitals from the 1s and 2s atomic orbitals of Li:

Note that the 1s atomic orbitals mix with each other, as do the 2s atomic orbitals (in actuality, all atomic orbitals with the same l-value will mix, but we won't concern ourselves with that here.) Also note that, since we combined four atomic orbitals, we obtained four molecular orbitals.

Now, we count up the number of electrons in Li₂⁺. We have 3 + 3 -1 = 5 electrons (this is a cation); one Li atom has its entire complement of three electrons, and the other Li atom has only two electrons. We place the five electrons of the cation into the molecular orbitals according to Hund's rule:

So we have the MO configuration of Li₂⁺. Now, to determine if addition of an electron would increase or decrease the stability of this species, we calculate the bond order. This is just 1/2(the number of bonding electrons - the number of antibonding electrons). The antibonding electrons are those in the molecular orbitals marked with a star (*). Here we have 1/2(3-2) = 1/2; this means that the bond in Li₂⁺ ion is not quite as strong as a single bond (bond order=1). If we add an electron to Li₂⁺, we place it into the first available MO; we have a spot available in the s_2s orbital. By adding an electron, we change the bond order; it is now 1/2(4-2) = 1. Addition of the electron thus stabilizes this species.

(c)C₂⁺ (oops, I worked this ion instead of N₂⁺. This one should still be somewhat educational, but work the nitrogen one also. The answer is in the back of your text. The p_2p MO is lower in energy than the s_2p MO in C₂ as well as N₂ so the MO diagram doesn't change.)

Here, we have 4+4-1=7 electrons (showing only the 2s and 2p electrons of each C atom.) The MO diagram is:

(d)Ne₂²⁺

We have 8+8-2=14 electrons (2s and 2p only); to make the +2 cation we remove two electrons, so we have 12 electrons to place in the MOs. Remember that the p_2p MO is higher in energy than the s_2p MO for O₂ - Ne₂ (cf. Fig. 9.46, p 376). Draw the diagram, and verify that the bond order of Ne₂²⁺ is 1/2(8-6)=1; if we add an electron, we put it into the s_2p* MO, and the bond order becomes 1/2(8-7)=1/2; thus, Ne₂⁺ is unstable relative to Ne₂²⁺.

whew. long one.