Tricky structures and formal charges

Formal charge, a quick recap

The formal charge of an atom compares the electrons it owns in a Lewis structure with the electrons of the free atom:

Here $V$ is the number of valence electrons of the free atom, $L$ the number of electrons in its lone pairs, and $B$ the number of bonds it forms. The formal charges of all the atoms add up to the charge of the molecule or ion.

When several structures are possible, the best one follows three rules:

1. The smallest charges. Keep the sum of the sizes of the formal charges as small as possible.
2. Minus on the most electronegative atom. A negative charge belongs on the atom that pulls electrons hardest.
3. No like charges side by side. Two neighbouring atoms should not carry charges of the same sign.

Choosing between resonance structures

Worked example. Laughing gas, $\text{N}_2\text{O}$, has the atoms in the order N, N, O and $16$ valence electrons. Three structures are possible:

1. $\text{N=N=O}$. End N: $5 - 4 - 2 = -1$. Middle N: $5 - 0 - 4 = +1$. O: $6 - 4 - 2 = 0$. Sum of sizes: $2$.
2. $\text{N}{\equiv}\text{N-O}$. End N: $5 - 2 - 3 = 0$. Middle N: $+1$. O: $6 - 6 - 1 = -1$. Sum of sizes: $2$.
3. $\text{N-N}{\equiv}\text{O}$. End N: $5 - 6 - 1 = -2$. Middle N: $+1$. O: $6 - 2 - 3 = +1$. Sum of sizes: $4$.

Structures 1 and 2 tie on rule 1. Structure 2 puts the minus charge on oxygen, so it is the larger part of the blend. Structure 3 hardly counts.

Expanded octets

Atoms from period 3 down, like phosphorus, sulfur, chlorine, xenon and iodine, can hold more than eight electrons around them. They are larger, and they have empty $d$ orbitals close in energy. Carbon, nitrogen, oxygen and fluorine never go past eight: they have only $2s$ and $2p$ orbitals.

Worked example: the sulfate ion. $\text{SO}_4^{2-}$ has $6 + 4 \times 6 + 2 = 32$ valence electrons.

1. Four single bonds. S: $6 - 0 - 4 = +2$. Each O: $6 - 6 - 1 = -1$. Sum of sizes: $6$.
2. Two double bonds. S: $6 - 0 - 6 = 0$. Two O with double bonds: $0$. Two O with single bonds: $-1$ each. Sum of sizes: $2$.

The formal charge rules prefer two double bonds, with twelve electrons around sulfur. Resonance spreads the six bond pairs over four oxygens, so every $\text{S-O}$ bond is the same, with bond order $\dfrac{6}{4} = 1.5$.

Careful calculations suggest that sulfur keeps a real positive charge, so the bonds are partly ionic. Exams still take the structure with the smallest formal charges as the best Lewis structure.

Xenon tetrafluoride. $\text{XeF}_4$ has $8 + 4 \times 7 = 36$ valence electrons. Four bonds use $8$, and each fluorine takes three lone pairs, another $24$. The last $4$ electrons are two lone pairs on xenon. Xenon holds $12$ electrons, and every formal charge is zero.

The triiodide ion. $\text{I}_3^-$ has $3 \times 7 + 1 = 22$ valence electrons. Two bonds use $4$, and each end iodine takes three lone pairs, another $12$. The last $6$ go to the middle iodine as three lone pairs, so it holds $10$ electrons. Its formal charge is $7 - 6 - 2 = -1$. The ends have $7 - 6 - 1 = 0$. So the extra electron sits, formally, on the middle atom.

Odd-electron molecules

When the valence electrons add up to an odd number, one electron cannot pair. Nitrogen monoxide, NO, has $11$; nitrogen dioxide, $\text{NO}_2$, has $17$; chlorine dioxide, $\text{ClO}_2$, has $19$.

Nitrogen dioxide. Draw $\text{O=N-O}$ with the odd electron on nitrogen. N: $5 - 1 - 3 = +1$. The double-bonded O: $6 - 4 - 2 = 0$. The single-bonded O: $6 - 6 - 1 = -1$. A second structure swaps the bonds, so both $\text{N-O}$ bonds are equal, with bond order $1.5$.

The unpaired electron makes $\text{NO}_2$ paramagnetic and brown. On cooling, two molecules pair their odd electrons into colourless $\text{N}_2\text{O}_4$.

Chlorine dioxide. The best structure is $\text{O=Cl=O}$, with one lone pair and the odd electron on chlorine. Every formal charge is zero, and chlorine holds $11$ electrons. This yellow gas bleaches paper pulp and cleans drinking water.

Back bonding in BF3

Boron in $\text{BF}_3$ has only six electrons. It has an empty $2p$ orbital, standing at right angles to the plane of the molecule. A filled $2p$ orbital of each fluorine lies parallel to it. So a lone pair of fluorine can spread into the empty orbital of boron. This sideways $\pi$ bond, from fluorine back to boron, is called back bonding.

• The evidence. The $\text{B-F}$ bond is $130\ \text{pm}$ long, shorter than a $\text{B-F}$ single bond should be. It has some double bond character.
• A surprise it explains. Fluorine is the most electronegative halogen, so $\text{BF}_3$ should be the strongest Lewis acid of the boron halides. It is the weakest: $\text{BF}_3 < \text{BCl}_3 < \text{BBr}_3$.
• Why. Two $2p$ orbitals of the same size overlap well. The $3p$ and $4p$ orbitals of chlorine and bromine are larger, so they overlap poorly with boron's $2p$. Back bonding fills boron's empty orbital most in $\text{BF}_3$, so it accepts a new pair least eagerly.

Three-centre bonds in diborane

$\text{BH}_3$ would leave boron with six electrons, like $\text{BF}_3$, but hydrogen has no lone pair to give back. So two $\text{BH}_3$ units join into diborane, $\text{B}_2\text{H}_6$.

Count the electrons: $2 \times 3 + 6 \times 1 = 12$. A structure like that of ethane would need seven two-electron bonds, which is $14$ electrons. There are not enough.

1. Four terminal bonds. Four hydrogens join one boron each by ordinary $\text{B-H}$ bonds, $119\ \text{pm}$ long. They lie in one plane with the two borons, and use $8$ electrons.
2. Two bridges. The other two hydrogens sit above and below that plane, each joined to both borons. Each $\text{B-H-B}$ bridge holds one pair of electrons, spread over three atoms. That uses the last $4$ electrons.

A bridge is a three-centre two-electron bond, also called a banana bond for its curved shape. Each bridging $\text{B-H}$ link has only half a pair, so it is longer, $133\ \text{pm}$. There is no bond between the two borons.

Try it in Python

The program finds the formal charges in the two drawings of the sulfate ion, and the sum of their sizes.

Common mistakes

• Counting bonding electrons instead of bonds. In $V - L - B$, each bond counts once.
• Giving carbon, nitrogen, oxygen or fluorine more than eight electrons. Only atoms from period 3 down can expand their octet.
• Stopping at rule 1. When two structures tie, the minus charge goes on the more electronegative atom.
• Forcing every electron into a pair. A molecule with an odd number of valence electrons keeps one unpaired electron.
• Drawing diborane like ethane. It has no $\text{B-B}$ bond. Two hydrogen bridges hold the borons together.

Remember these

• Formal charge $= V - L - B$. The charges add up to the charge of the species.
• Best structure: the smallest charges, minus on the most electronegative atom, no like charges side by side.
• Only atoms from period 3 down can hold more than eight electrons.
• $\text{NO}$, $\text{NO}_2$ and $\text{ClO}_2$ have an odd number of electrons, so they are paramagnetic.
• Back bonding makes $\text{BF}_3$ the weakest Lewis acid of the boron halides.
• Diborane has four terminal $\text{B-H}$ bonds and two three-centre two-electron $\text{B-H-B}$ bridges.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. In the phosphate ion, $\text{PO}_4^{3-}$, drawn with one $\text{P=O}$ bond and three $\text{P-O}$ bonds, what is the formal charge on phosphorus?

2. Phosphorus pentachloride, $\text{PCl}_5$, exists. Can $\text{NCl}_5$ exist?

3. Why is nitrogen dioxide attracted into a magnetic field?

4. How many electrons does each $\text{B-H-B}$ bridge of diborane hold, and how many atoms share them?