Question

\\(\text{clo}_4^-\\)

draw the molecule by placing atoms on the grid and connecting them with bonds. include all lone pairs of electrons. show the formal charges of all nonhydrogen atoms.

Explanation:

Calculate total valence electrons

The chemical formula shown partially on the left is \(\text{ClO}_4^-\) (perchlorate ion).

• Chlorine (\(\text{Cl}\)) has 7 valence electrons.
• Oxygen (\(\text{O}\)) has 6 valence electrons each: \(4 \times 6 = 24\).
• The negative charge (\(-\)) adds 1 electron.
• Total valence electrons: \(7 + 24 + 1 = 32\).

Determine central atom and skeleton

Chlorine is less electronegative than oxygen, so \(\text{Cl}\) is the central atom.

• Place \(\text{Cl}\) in the center and connect it to four \(\text{O}\) atoms with single bonds.
• This uses 8 electrons (4 single bonds), leaving \(32 - 8 = 24\) electrons.

Distribute remaining electrons

Distribute the remaining 24 electrons as lone pairs to complete the octets of the four outer oxygen atoms.

• Each of the four \(\text{O}\) atoms receives 3 lone pairs (6 electrons).
• All 32 valence electrons are now placed.
• In this single-bonded structure, each \(\text{O}\) has a formal charge of \(-1\), and \(\text{Cl}\) has a formal charge of \(+3\).

Minimize formal charges

Chlorine is in Period 3 and can expand its octet to minimize formal charges.

• To reduce formal charges, move lone pairs from three of the oxygen atoms to form double bonds with chlorine.
• This results in three \(\text{Cl}=\text{O}\) double bonds and one \(\text{Cl}-\text{O}\) single bond.

Assign final formal charges

Calculate the formal charges for the minimized structure:

• Central \(\text{Cl}\): \(7 - 7 \text{ bonds} - 0 \text{ lone pair electrons} = 0\).
• Double-bonded \(\text{O}\) atoms (three of them): \(6 - 2 \text{ bonds} - 4 \text{ lone pair electrons} = 0\).
• Single-bonded \(\text{O}\) atom (one of them): \(6 - 1 \text{ bond} - 6 \text{ lone pair electrons} = -1\).
• This structure minimizes formal charges while maintaining a net charge of \(-1\).

Answer:

To draw the best Lewis structure for the perchlorate ion (\(\text{ClO}_4^-\)):

1. Central Atom: Place Chlorine (\(\text{Cl}\)) in the center.
2. Surrounding Atoms: Place four Oxygen (\(\text{O}\)) atoms around the central \(\text{Cl}\).
3. Bonds:

• Draw three double bonds (\(\text{Cl}=\text{O}\)) between the central \(\text{Cl}\) and three of the \(\text{O}\) atoms.
• Draw one single bond (\(\text{Cl}-\text{O}\)) between the central \(\text{Cl}\) and the remaining \(\text{O}\) atom.

1. Lone Pairs:

• Place two lone pairs (4 electrons) on each of the three double-bonded \(\text{O}\) atoms.
• Place three lone pairs (6 electrons) on the single-bonded \(\text{O}\) atom.
• The central \(\text{Cl}\) has zero lone pairs.

1. Formal Charges:

• The single-bonded \(\text{O}\) atom has a formal charge of \(-1\).
• The central \(\text{Cl}\) and the three double-bonded \(\text{O}\) atoms have a formal charge of \(0\).