Question

Draw the Lewis structure with lowest formal charges, and determine the charge of each atom in (a) \(\mathrm{CN}^{-} ;\) (b) \(\mathrm{ClO}^{-}\).

Short answer

(a) The best structure for \(\mathrm{CN}^{-}\) is \( \mathrm{:C} \equiv \mathrm{N}:^{-} \) with formal charges 0 and -1. (b) For \(\mathrm{ClO}^{-}\), it is \( \mathrm{:Cl}-\mathrm{O}:^{-} \) with formal charges 0 and -1.

Step-by-step solution

Determine the total number of valence electrons

Calculate the total number of valence electrons for each molecule, including any additional electrons due to negative charges.

Sketch the skeletal structure

Draw a rough structure showing bonds between atoms.

Distribute electrons to atoms

Distribute the valence electrons around the atoms, starting from the outer atoms and moving towards the central atom.

Add multiple bonds if necessary

If any atom does not have an octet, form double or triple bonds as necessary.

Calculate formal charges

Calculate the formal charge of each atom using the formula: \[ \text{Formal Charge} = (\text{valence electrons}) - (\text{non-bonding electrons}) - \frac{1}{2}(\text{bonding electrons}) \].

Verify and choose the best structure

Ensure the final structure has the lowest formal charges on each atom and the correct net charge.

Draw the final Lewis structure for \(\mathrm{CN}^{-}\)

1. For \(\mathrm{CN}^{-}\), count the valence electrons: \text{C (4 valence electrons)} + \text{N (5 valence electrons)} + 1 \text{ (for the negative charge)} = 10.2. Draw C-N with a triple bond and place lone pairs to satisfy the octet rule: \[ :\mathrm{C} \equiv \mathrm{N}:^{-} \].3. Calculate formal charges: - For C: \[ 4 - 0 - \frac{1}{2}(8) = 0 \]. - For N: \[ 5 - 2 - \frac{1}{2}(6) = -1 \]. Final structure: \( \mathrm{:C} \equiv \mathrm{N}:^{-} \).

Draw the final Lewis structure for ClO⁻

1. For \(\mathrm{ClO}^{-}\), count the valence electrons: \text{Cl (7 valence electrons)} + \text{O (6 valence electrons)} + 1 \text{ (for the negative charge)} = 14.2. Draw Cl-O with a single bond and distribute lone pairs to satisfy the octet rule: \[ :\mathrm{Cl}-\mathrm{O}:- \].3. Calculate formal charges: - For Cl: \[ 7 - 6 - \frac{1}{2}(2) = 0 \]. - For O: \[ 6 - 6 - \frac{1}{2}(2) = -1 \]. Final structure: \( \mathrm{:Cl}-\mathrm{O}:^{-} \).

Key concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom. These electrons play a major role in chemical bonding, as they are the ones involved in forming bonds with other atoms. For instance, in the exercise provided, we need to count the valence electrons for each molecule before we can draw the Lewis structure.

In \(\text{CN^{-}}\), carbon has 4 valence electrons, nitrogen has 5, and we add one more for the negative charge. That totals 10 valence electrons. Similarly, in \(\text{ClO^{-}}\), chlorine has 7 valence electrons, oxygen has 6, and again we add one more for the negative charge, summing to 14 valence electrons. Knowing how to count valence electrons is essential for predicting how atoms bond and how molecules are structured.

Formal Charge

The formal charge of an atom in a molecule helps us to understand the distribution of electrons and to determine the most stable Lewis structure. Formal charge is calculated using the formula:

\[ \text{Formal Charge} = (\text{valence electrons}) - (\text{non-bonding electrons}) - \frac{1}{2}(\text{bonding electrons}) \]

This equation helps identify the charge that an atom would have if all bonding electrons were shared equally between the atoms. For example:

• For carbon in \(\text{CN^{-}}\):
  \ 4 - 0 - \frac{1}{2}(8) = 0 \


• For nitrogen in \(\text{CN^{-}}\):
  \ 5 - 2 - \frac{1}{2}(6) = -1 \

This shows that the nitrogen atom carries a negative charge, which aligns with the overall charge of the molecule. The goal is to make the formal charges as close to zero as possible for stability.

Octet Rule

The octet rule is a crucial principle in chemistry that states: 'Atoms tend to form bonds until they are surrounded by eight valence electrons'. This concept helps explain the shapes and stability of molecules.

• In the case of \(\text{CN^{-}}\), carbon and nitrogen form a triple bond allowing both atoms to have eight electrons in their valence shell. Carbon gains its octet by sharing three pairs of electrons with nitrogen, which also satisfies its octet.


• For \(\text{ClO^{-}}\), a single bond between Cl and O is enough. Both atoms achieve their complete octet by sharing electrons and accommodating lone pairs.

However, there are exceptions to the octet rule, especially for molecules with an odd number of electrons or those involving heavier elements.

Multiple Bonds

Sometimes, a single bond is not enough to satisfy the octet rule for both atoms involved. In such cases, atoms can form double or triple bonds, sharing two or three pairs of electrons respectively.

• For example, in \(\text{CN^{-}}\), a triple bond is necessary between carbon and nitrogen. The structure \(\text{:C} \equiv \text{N:}^{-}\) allows both atoms to complete their octet and minimizes formal charges.


• For the \(\text{ClO^{-}}\) molecule, a single bond is adequate to satisfy the octets of both chlorine and oxygen, so a multiple bond is not required.

Forming the correct number of bonds ensures that the molecule is both stable and correctly represented according to its chemical properties.