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A student studies a diatomic species formed by two oxygen atoms…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A student studies a diatomic species formed by two oxygen atoms. She has the valence molecular orbitals of the species with every valence electron in place, as shown in the figure. The levels follow the energy order that NCERT gives for $\text{O}_2$ , in which $\sigma 2p_z$ lies below the pair $\pi 2p_x$ and $\pi 2p_y$ . The $\sigma 1s$ and $\sigma^* 1s$ orbitals are full but are not drawn. Each arrow is one electron.

A molecular orbital energy ladder for the valence electrons of a diatomic species. An arrow on the left shows energy rising upwards. From the bottom the levels are: sigma 2s, a single box with two electrons, one arrow up and one down; sigma star 2s, a single box with two electrons; sigma 2p z, a single box with two electrons; pi 2p x and pi 2p y, two boxes side by side at the same height, each with two electrons; pi star 2p x and pi star 2p y, two boxes side by side, the left one with two electrons and the right one with a single upward arrow; and at the top sigma star 2p z, a single empty box.

The student must find out which species this is. Then she must predict its behaviour from molecular orbital theory. Count the arrows with care. A half-filled orbital matters for the magnetic nature, and every electron counts in the bond order.

How many electrons does the species have in all, the 1s electrons included?
What is the charge on the species? Give a negative charge with a minus sign.
What is the bond order of the species?
How many unpaired electrons does it have?
A species formed by two fluorine atoms has exactly the same filled orbitals as the figure, the 1s orbitals too. What is its charge?
The species of the figure loses two electrons. What is the bond order of the species formed?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. The valence orbitals with their electrons. From the bottom: $\sigma 2s$ has 2, $\sigma^* 2s$ has 2, $\sigma 2p_z$ has 2, the $\pi 2p$ pair has 4, the $\pi^* 2p$ pair has 3, and $\sigma^* 2p_z$ is empty.

Plan. Count the electrons and compare with two neutral O atoms. Then use bond order $= \tfrac{1}{2}(N_b - N_a)$ and look for half-filled orbitals.

Step 1. (a) All electrons. The figure shows $2 + 2 + 2 + 4 + 3 = 13$ valence electrons. The full $\sigma 1s$ and $\sigma^* 1s$ add 4 more: $17$ electrons in all.

Step 2. (b) The charge. Two neutral O atoms have $2 \times 8 = 16$ electrons. The species has one extra electron, so its charge is $-1$ . It is the superoxide ion, $\text{O}_2^-$ .

Step 3. (c) Bond order. Bonding: $\sigma 1s$ , $\sigma 2s$ , $\sigma 2p_z$ and the $\pi 2p$ pair hold $N_b = 10$ electrons. Antibonding: $\sigma^* 1s$ , $\sigma^* 2s$ and the $\pi^*$ pair hold $N_a = 7$ . Bond order $= \tfrac{1}{2}(10 - 7) = 1.5$ .

Step 4. (d) Unpaired electrons. The $\pi^*$ pair holds 3 electrons. One orbital is full and one is half-filled. So there is $1$ unpaired electron, and the ion is paramagnetic.

Step 5. (e) The fluorine species. Two F atoms have $2 \times 9 = 18$ electrons. To have $17$ , the species must lose one electron. It is $\text{F}_2^+$ , charge $+1$ .

Step 6. (f) Two electrons removed. The highest filled orbitals are the $\pi^*$ pair. Removing two electrons leaves one there, with $15$ electrons in all. Then $N_b = 10$ and $N_a = 5$ , so the bond order is $2.5$ . That species is $\text{O}_2^+$ .

Answer: (a) $17$ electrons. (b) $-1$ . (c) $1.5$ . (d) $1$ . (e) $+1$ . (f) $2.5$ .

Check. Use only the drawn valence electrons: 8 bonding and 5 antibonding give $\tfrac{1}{2}(8 - 5) = 1.5$ , the same bond order. Species with the same number of electrons and the same order of orbitals have the same bond order, so $\text{F}_2^+$ also has $1.5$ .

Watch out. Removing electrons from $\text{O}_2^-$ makes the bond stronger, not weaker. The electrons leave antibonding orbitals.

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A student studies a diatomic species formed by two oxygen atoms…

The worked solution

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