🎯⭐ INTERACTIVE LESSON

Lewis Structures and Formal Charge

Learn step-by-step with interactive practice!

Lewis Structures and Formal Charge - Complete Interactive Lesson

Part 1: Drawing Lewis Structures

Valence electrons are the electrons in the outermost energy level (shell) of an atom. These are the electrons that participate in chemical bonding and determine an element's chemical properties.

For main group elements (Groups 1, 2, and 13–18), the number of valence electrons equals the group number:

Group	Valence $e^{-}$	Examples
1	1	H, Li, Na
2	2	Be, Mg, Ca
13	3	B, Al
14	4	C, Si
15	5	N, P
16	6	O, S
17	7	F, Cl, Br
18	8	Ne, Ar (except He = 2)

Knowing the number of valence electrons is the first step in drawing any Lewis structure.

Part 1 of 7 — Drawing Lewis Structures

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 1

Understanding the core concepts covered in Part 1
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

How many valence electrons does a nitrogen (N) atom have?

The octet rule states that atoms tend to gain, lose, or share electrons until they have eight electrons in their valence shell. This gives them the same electron configuration as the nearest noble gas.

Key points:

Most atoms want 8 valence electrons (an octet)
Hydrogen is an exception — it only needs 2 electrons (a duet) to match helium
Atoms achieve octets by forming covalent bonds (sharing electrons) or ionic bonds (transferring electrons)

Why does the octet rule work?

Noble gases (Group 18) are extremely stable because their valence shells are completely filled. Other atoms "want" to achieve this same stable configuration.

For example, oxygen has 6 valence electrons and needs 2 more to complete its octet. It can share 2 electrons with another atom by forming bonds.

🔑 Key Concept: Atoms gain, lose, or share electrons to achieve 8 valence electrons (2 for hydrogen). This drives all chemical bonding.

Apply the octet rule to determine bonding behavior.

Before drawing a Lewis structure, you must count the total number of valence electrons in the molecule or ion.

Steps:

Count the valence electrons for each atom
Add them all together
For anions (negative charge): add electrons equal to the charge
For cations (positive charge): subtract electrons equal to the charge

Example 1: $H_{2}O$

H: 1 valence $e^{-}$ × 2 atoms = 2

Count the total valence electrons for each species.

Select the correct number of valence electrons for each element.

Counting valence electrons in polyatomic ions requires adjusting for the charge.

Part 2: Octet Rule & Exceptions

Drawing Lewis structures follows a systematic algorithm. Master these steps and you can draw the structure for any molecule or ion.

Step 1: Count the total valence electrons

Sum valence $e^{-}$ for all atoms
Add $e^{-}$ for negative charges, subtract for positive charges

Step 2: Identify the central atom

Usually the least electronegative atom (not H or F)
H and F are always terminal (outer) atoms

Step 3: Draw single bonds from the central atom to each surrounding atom

Part 3: Formal Charge

Sometimes, after placing all single bonds and distributing lone pairs to the outer atoms, the central atom still doesn't have an octet. When this happens, you must form multiple bonds.

The fix: Convert one or more lone pairs from an adjacent atom into bonding pairs (additional bonds).

Converting 1 lone pair → double bond (4 shared electrons)
Converting 2 lone pairs → triple bond (6 shared electrons)

How to know when you need multiple bonds: After Step 4 of the algorithm, check the central atom's electron count. If it's less than 8, you need to form multiple bonds.

Important: Multiple bonds are most commonly formed between:

C, N, O, and S (second-period and some third-period elements)
These atoms are small enough for effective side-by-side (pi) orbital overlap

Let's draw the Lewis structure for $O_{2}$ .

Step 1: Total valence electrons = 6 + 6 = 12

Step 2: Neither atom is "central" — it's a diatomic molecule: O—O

Part 4: Resonance Structures

Formal charge (FC) is a bookkeeping tool that helps us determine which Lewis structure is the most reasonable when multiple structures are possible.

Formal charge tells us the hypothetical charge on each atom if all bonding electrons were shared perfectly equally between bonded atoms.

$\boxed{FC = V - N - \frac{B}{2}}$

Part 5: Expanded & Incomplete Octets

Sometimes, a single Lewis structure is not sufficient to describe the actual electron distribution in a molecule. When electrons can be delocalized (spread out) across multiple positions, we draw resonance structures.

Resonance structures are two or more valid Lewis structures for the same molecule that differ only in the placement of electrons (not atoms).

Key points:

The atoms stay in the same positions
Only electrons (bonds and lone pairs) move
We connect resonance structures with a double-headed arrow (↔)
The actual molecule is a resonance hybrid — an average of all resonance structures
No single resonance structure is "correct" on its own

💡 Tip: Resonance structures differ only in electron placement — the atoms never move. The real molecule is a blend (hybrid) of all structures.

The resonance hybrid has characteristics intermediate between all contributing structures. For example, bonds that are single in one structure and double in another are actually intermediate (bond order between 1 and 2).

Part 5 of 7 — Expanded & Incomplete Octets

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 5

Understanding the core concepts covered in Part 5
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

Ozone $($ is a classic example of resonance.

Part 6: Problem-Solving Workshop

While the octet rule works for most molecules, there are three important categories of exceptions:

1. Incomplete Octets — fewer than 8 electrons around the central atom

Elements: Be $(4 e^{-})$ , B $(6 e^{-})$ , Al $($

Part 7: Synthesis & AP Review

Drawing Lewis structures on the AP exam requires combining all the skills from this unit. Here's your complete checklist:

Step 1: Count total valence electrons (adjust for charges) Step 2: Identify the central atom (least electronegative, not H or F) Step 3: Draw single bonds to all terminal atoms Step 4: Distribute remaining electrons as lone pairs (outer atoms first) Step 5: Check octets — form multiple bonds if needed Step 6: Calculate formal charges and choose the best structure Step 7: Check for resonance — draw all equivalent structures Step 8: Consider exceptions (incomplete octets, expanded octets, radicals)

AP Exam Tips:

Always show lone pairs on your drawings
Put brackets and the charge around ions: [structure] ${}^{2-}$
When asked to "justify" your structure, discuss formal charges
Know that bond order from resonance affects bond length and strength

Part 7 of 7 — Synthesis & AP Review

Bringing It All Together

This comprehensive review connects every concept from Parts 1–6 with AP-style problems. The questions are designed to mirror what you'll see on the actual exam — multi-step, multi-concept, and requiring clear written explanations.

O: 6 valence

e^{-}

× 1 atom = 6

Total = 2 + 6 = 8 valence electrons

Example 2: $CO_{3}^{2-}$ (carbonate ion)

C: 4 valence $e^{-}$ × 1 = 4
O: 6 valence $e^{-}$ × 3 = 18
Add 2 for the 2− charge = +2
Total = 4 + 18 + 2 = 24 valence electrons

Each single bond uses 2 electrons

Step 4: Distribute remaining electrons as lone pairs

Fill octets on outer atoms first
Place any leftover electrons on the central atom

Step 5: Check — does every atom have an octet?

If the central atom lacks an octet, convert lone pairs on outer atoms into multiple bonds

💡 Tip: The central atom is usually the least electronegative atom. H and F are always terminal atoms.

Part 2 of 7 — Octet Rule & Exceptions

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 2

Understanding the core concepts covered in Part 2
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

The central atom is a critical first decision in drawing Lewis structures.

Problem: Draw the Lewis structure of water $(H_{2}O)$ .

Solution:

Let's apply the algorithm to water $(H_{2}O)$ .

Step 1: Count valence electrons

O: 6 $e^{-}$ , H: 1 $e^{-}$ each → Total = 6 + 2(1) = 8 $e^{-}$

Step 2: Central atom

Oxygen is the central atom (H is always terminal)

Step 3: Draw single bonds

H—O—H uses 2 bonds = 4 electrons
Remaining: 8 − 4 = 4 electrons

Step 4: Distribute remaining electrons

H atoms already have their duet (2 $e^{-}$ from the bond)
Place remaining 4 $e^{-}$ as 2 lone pairs on oxygen

Step 5: Check octets

O: 2 bonds $(4 e^{-})$ + 2 lone pairs $(4 e^{-})$ = 8 $e^{-}$ ✓

The Lewis structure of water shows oxygen with two bonding pairs and two lone pairs, giving it a bent shape.

Work through the algorithm for carbon tetrachloride $(CCl_{4})$ .

Problem: Draw the Lewis structure for hydrogen cyanide (HCN).

Solution:

Let's draw the Lewis structure for hydrogen cyanide (HCN).

Step 1: Count valence electrons

H: 1, C: 4, N: 5 → Total = 10 $e^{-}$

Step 2: Central atom

Carbon is the central atom (least electronegative, H is always terminal)
Arrangement: H—C—N

Step 3: Draw single bonds

H—C and C—N single bonds use 4 electrons
Remaining: 10 − 4 = 6 electrons

Step 4: Distribute remaining electrons

Place lone pairs on the outer atom (N) first
N gets 3 lone pairs $(6 e^{-})$ → N has 2 (bond) + 6 = 8 ✓
But C only has 4 $e^{-}$ (2 bonds) — not enough!

Step 5: Form multiple bonds

Convert 2 lone pairs from N into 2 additional bonds
Result: H—C≡N (a triple bond between C and N)
C: 2 (H bond) + 6 (triple bond) = 8 ✓
N: 6 (triple bond) + 2 (1 lone pair) = 8 ✓

HCN has a triple bond between carbon and nitrogen, with one lone pair on nitrogen.

Test your understanding of the Lewis structure drawing process.

For each molecule, select the correct central atom.

Apply the algorithm to nitrogen trifluoride $(NF_{3})$ .

Step 3: Single bond: O—O uses 2 $e^{-}$ . Remaining: 10 $e^{-}$

Step 4: Distribute remaining electrons:

Each O gets 3 lone pairs first attempt
But wait — 3 lone pairs × 2 atoms = 12 $e^{-}$ , and we only have 10
Give each O as many lone pairs as possible: that's 5 $e^{-}$ each... but $e^{-}$ come in pairs
Give 3 lone pairs to one O $(6 e^{-})$ and 2 lone pairs to the other $(4 e^{-})$ → uses 10 ✓
But the second O only has 2 + 4 = 6 $e^{-}$ — no octet!

Step 5: Convert 1 lone pair from the first O into a second bond:

O=O (double bond)
Each O: 4 $e^{-}$ (double bond) + 4 $e^{-}$ (2 lone pairs) = 8 ✓

Total electrons used: 4 (double bond) + 2 × 4 (lone pairs) = 12 ✓

Test your understanding of the $O_{2}$ Lewis structure.

Nitrogen gas $(N_{2})$ contains one of the strongest bonds in chemistry — a triple bond.

Step 1: Total valence electrons = 5 + 5 = 10

Step 3: Single bond N—N uses 2 $e^{-}$ . Remaining: 8 $e^{-}$

Step 4: Distribute to fill octets:

With a single bond, each N has 2 $e^{-}$ from the bond
Each N needs 6 more → total needed = 12 $e^{-}$ , but only 8 remain
Not enough for single bond!

Step 5: Form multiple bonds:

Try double bond: each N has 4 (bond) + needs 4 more = 8 total. Remaining after double bond: 10 − 4 = 6 $e^{-}$ . Give 3 $e^{-}$ to each → only 1.5 lone pairs each. Not ideal.
Try triple bond: N≡N uses 6 $e^{-}$ . Remaining: 10 − 6 = 4

Final structure: :N≡N: with 1 lone pair on each nitrogen.

Carbon dioxide $(CO_{2})$ has the arrangement O—C—O. Work through the algorithm.

Multiple bonds differ from single bonds in both strength and length:

Bond Type	Bond Order	Relative Strength	Relative Length
Single (—)	1	Weakest	Longest
Double (=)	2	Moderate	Moderate
Triple (≡)	3	Strongest	Shortest

Key trends:

More shared electrons → stronger bond → shorter bond
N≡N bond energy ≈ 946 kJ/mol (very strong!)
This is why $N_{2}$ is so unreactive — it takes enormous energy to break the triple bond

Bond order = number of bonding pairs between two atoms

Single bond: bond order = 1
Double bond: bond order = 2
Triple bond: bond order = 3

Select the correct bond type for each molecule.

Apply your knowledge of bond order, strength, and length.

V = number of valence electrons (from the periodic table)
N = number of nonbonding (lone pair) electrons on the atom
B = number of bonding electrons around the atom

Alternatively, since $\frac{B}{2}$ equals the number of bonds:

$\boxed{FC = V - N - \text{(number of bonds)}}$

Important: The sum of all formal charges in a molecule must equal the overall charge of the molecule or ion.

Part 4 of 7 — Resonance Structures

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 4

Understanding the core concepts covered in Part 4
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

Problem: Calculate the formal charge on each atom in $H_{2}O$ .

Solution:

Let's calculate the formal charge on each atom in $H_{2}O$ .

The Lewis structure: H—Ö—H (oxygen has 2 lone pairs)

Oxygen:

V = 6 (Group 16)
N = 4 (2 lone pairs = 4 electrons)
B = 4 (2 bonds × 2 electrons each)
FC = 6 − 4 − 4/2 = 6 − 4 − 2 = 0

Each Hydrogen:

V = 1 (Group 1)
N = 0 (no lone pairs)
B = 2 (1 bond = 2 electrons)
FC = 1 − 0 − 2/2 = 1 − 0 − 1 = 0

Check: Sum of formal charges = 0 + 0 + 0 = 0 ✓ (neutral molecule)

All formal charges are zero — this is the ideal situation and confirms this is a good Lewis structure.

Practice using the formal charge formula.

Calculate the formal charge for each atom using FC = V − N − B/2.

When multiple valid Lewis structures exist, use formal charges to pick the best one:

Rule 1: The structure with formal charges closest to zero on all atoms is preferred.

Rule 2: If formal charges cannot all be zero, negative formal charges should be on the more electronegative atoms.

Rule 3: Structures where adjacent atoms have formal charges of the same sign are unfavorable (like charges repel).

Rule 4: The sum of all formal charges must equal the molecule's overall charge.

Example: $CO_{2}$ — comparing two structures

Structure A: O=C=O → FC: all zeros ✓ Best! Structure B: O≡C—Ö → FC: O(≡) = +1, C = 0, O(—) = −1

Structure A is preferred because all formal charges are zero.

🔑 Key Concept: Minimize formal charges first (Rule 1), then place any remaining negative charges on the most electronegative atoms (Rule 2).

Use formal charge rules to select the best structure.

Select the correct formal charge for each described atom.

Apply Rule 2 of formal charge analysis.

Step 1: Total valence electrons = 3 × 6 = 18

Step 2: Central oxygen, with two terminal oxygens: O—O—O

Drawing Structure 1:

After applying the algorithm, one valid structure is: Ö=O—Ö:
Central O: double bond left, single bond right
Left O: 2 lone pairs, Right O: 3 lone pairs

Drawing Structure 2:

Equally valid: :Ö—O=Ö
The double bond is on the other side

Resonance structures: Ö=O—Ö: ↔ :Ö—O=Ö

The resonance hybrid:

Both O—O bonds are identical in reality
Bond order = 1.5 (average of single and double)
Bond length is between a single and double bond
Each terminal oxygen carries −½ formal charge on average

Experiments confirm that both O—O bonds in ozone are the same length — proving the resonance hybrid model.

Clarify common misconceptions about resonance.

The nitrate ion is a perfect example of resonance with three equivalent structures.

Total valence electrons: 5 (N) + 3 × 6 (O) + 1 (charge) = 24

Three resonance structures:

In each structure, nitrogen forms:

One double bond to one oxygen
Two single bonds to the other two oxygens

The double bond "rotates" among the three positions:

Structure 1: O=N(— $O^{-}$ )(— $O^{-}$ ) Structure 2: ( ${}^{-}O$ —)N=O(— $O^{-}$ ) Structure 3: ( ${}^{-}O$ —)N(— $O^{-}$ )=O

Formal charges in each structure:

N: FC = 5 − 0 − 8/2 = +1
O (double bonded): FC = 6 − 4 − 4/2 = 0
O (single bonded): FC = 6 − 6 − 2/2 = −1
Sum: +1 + 0 + 2(−1) = −1 ✓ (matches the ion charge)

The resonance hybrid:

All three N—O bonds are identical
Bond order = 4 bonds ÷ 3 positions = 1.33
Each oxygen carries a formal charge of −2/3
The negative charge is evenly distributed

Answer questions about the nitrate ion resonance structures.

Resonance occurs when:

A lone pair is adjacent to a multiple bond — the lone pair can be delocalized
Multiple equivalent positions exist for a double or triple bond
There is a p orbital that can overlap with adjacent p orbitals

Common molecules/ions with resonance:

Ozone $(O_{3})$ : 2 resonance structures
Nitrate $(NO_{3}^{-})$ : 3 resonance structures
Carbonate $(CO_{3}^{2-})$ : 3 resonance structures
Benzene $(C_{6}H_{6})$ : 2 major resonance structures
Acetate $(CH_{3}COO^{-})$ : 2 resonance structures

Resonance does NOT occur when:

All bonds are single bonds with no adjacent lone pairs
The structure has no possible way to rearrange electrons
Moving electrons would violate the octet rule

Determine whether each molecule or ion exhibits resonance.

Apply resonance concepts to predict molecular properties.

Example:

BF_{3}

— boron has only 6 electrons

2. Expanded Octets — more than 8 electrons around the central atom

Only elements in Period 3 and beyond (they have empty d orbitals)
Examples: $PCl_{5}$ $(10 e^{-})$ , $SF_{6}$ $(12 e^{-})$ , $XeF_{2}$ $(10 e^{-})$
Elements from Period 2 (C, N, O, F) can NEVER exceed 8

3. Odd-Electron Species — molecules with an odd number of total electrons

At least one atom cannot have an octet
Examples: NO $(11 e^{-})$ , $NO_{2}$ $(17 e^{-})$
These are called free radicals

Recognizing which exception applies is a critical AP Chemistry skill.

🔑 Key Concept: The three octet rule exceptions are: incomplete octets (Be, B, Al), expanded octets (Period 3+ elements), and odd-electron species (free radicals).

Part 6 of 7 — Problem-Solving Workshop

Practice Makes Perfect

This workshop features multi-step problems that mirror the AP Chemistry exam format. Each problem requires you to combine concepts from previous parts and show your work clearly.

🔑 Why this matters: The AP Chemistry exam rewards students who can apply concepts to unfamiliar problems — structured practice is the best preparation.

What You'll Master in Part 6

Working through complete multi-step problems from start to finish
Building problem-solving strategies you can apply on the AP exam
Identifying which concepts to apply and in what order

Boron trifluoride $(BF_{3})$ is the classic example of an incomplete octet.

Step 1: Total valence electrons = 3 (B) + 3 × 7 (F) = 24

Step 2: Boron is the central atom

Step 3: Three B—F single bonds use 6 $e^{-}$ . Remaining: 18 $e^{-}$

Step 4: Give each F 3 lone pairs (9 lone pairs = 18 $e^{-}$ ). All remaining electrons used.

Step 5: Check octets:

Each F: 2 (bond) + 6 (lone pairs) = 8 ✓
B: only 6 $e^{-}$ (3 bonds) — incomplete octet!

Could we form a double bond? Yes, we could draw F=B with a double bond to give B an octet. But this would put a +1 formal charge on F (very electronegative!) and −1 on B. This is unfavorable.

Best structure: Three single bonds with boron having only 6 electrons. Boron is electron-deficient and acts as a Lewis acid (electron pair acceptor).

This is why $BF_{3}$ readily reacts with molecules that can donate an electron pair (Lewis bases like $NH_{3}$ ).

Identify molecules with incomplete octets.

Elements in Period 3 and beyond can accommodate more than 8 electrons because they have access to empty d orbitals. $n\cdot \cdot PCl_{5}$ (Phosphorus Pentachloride):**

Total valence $e^{-}$ : 5 + 5 × 7 = 40
P forms 5 single bonds to Cl $(uses 10 e^{-})$
Each Cl gets 3 lone pairs $(30 e^{-})$
P has 10 electrons around it — expanded octet!
Geometry: trigonal bipyramidal $n\cdot \cdot SF_{6}$ (Sulfur Hexafluoride):**
Total valence $e^{-}$ : 6 + 6 × 7 = 48
S forms 6 single bonds to F $(uses 12 e^{-})$
Each F gets 3 lone pairs $(36 e^{-})$
S has 12 electrons around it — expanded octet!
Geometry: octahedral

Critical rule for the AP exam: Only atoms from Period 3+ (P, S, Cl, Br, I, Xe, etc.) can have expanded octets. Period 2 atoms (C, N, O, F) NEVER exceed 8 electrons.

⚠️ Warning: Only atoms from Period 3 and beyond can have expanded octets. Period 2 elements (C, N, O, F) can NEVER exceed 8 electrons — this is a common AP exam trap!

Work through expanded octet examples.

When a molecule has an odd number of total valence electrons, it's impossible for every atom to have an octet. At least one atom will have an unpaired electron.

Example: NO (Nitric Oxide)

Total valence $e^{-}$ : 5 + 6 = 11 (odd!)
Best Lewis structure: :N̈=Ö· (with an unpaired electron on N)
Nitrogen has 7 electrons around it — one short of an octet

Example: $NO_{2}$ (Nitrogen Dioxide)

Total valence $e^{-}$ : 5 + 2 × 6 = 17 (odd!)
The unpaired electron sits on nitrogen
This is why $NO_{2}$ is a reactive, brown gas

Properties of free radicals:

Extremely reactive (they want to pair that lone electron)
Often colored (absorb visible light)
Paramagnetic (attracted to magnetic fields due to unpaired electrons)
Many are important in atmospheric chemistry and biology

On the AP exam, if you count an odd number of total electrons, immediately recognize it as a radical species.

Identify which type of octet exception each species represents.

Determine which atoms can have expanded octets.

🔑 Why this matters: AP Chemistry exam questions rarely test one concept in isolation — success requires connecting ideas across topics.

What You'll Master in Part 7

Solving AP-style questions that integrate multiple concepts from this unit
Writing clear, concise explanations using proper chemistry terminology
Identifying and avoiding common AP exam traps and mistakes

Problem: Draw the best Lewis structure for $SOCl_{2}$ (S is central).

Solution:

Let's work through a complex molecule: $SOCl_{2}$ (S is central).

Step 1: Total valence electrons

S: 6, O: 6, Cl: 7 × 2 = 14
Total = 6 + 6 + 14 = 26 $e^{-}$

Step 2: Sulfur is the central atom

Step 3: Draw single bonds: S—O, S—Cl, S—Cl $(uses 6 e^{-})$

Remaining: 26 − 6 = 20 $e^{-}$

Step 4: Fill octets on outer atoms:

O gets 3 lone pairs $(6 e^{-})$ , each Cl gets 3 lone pairs (6 $e^{-}$ each)
Total distributed: 6 + 6 + 6 = 18 $e^{-}$

Step 5: Check S: 3 bonds $(6 e^{-})$ + 1 lone pair $(2 e^{-})$ = 8 ✓

Step 6: Formal charges:

S: 6 − 2 − 6/2 = +1
O: 6 − 6 − 2/2 = −1
Each Cl: 7 − 6 − 2/2 = 0

Alternative structure with S=O double bond:

S: 6 − 2 − 8/2 = 0, O: 6 − 4 − 4/2 = 0
All formal charges zero — this is preferred!
S can have expanded octet (Period 3), so 10 $e^{-}$ around S is allowed.

Apply the full algorithm to a challenging molecule.

Work through the Lewis structure for $POCl_{3}$ (P is central, bonded to one O and three Cl atoms).

On the AP exam, you may be given multiple Lewis structures and asked which is "best" or "most reasonable." Here's a decision framework:

Priority order for evaluating structures:

All atoms have octets (or appropriate exceptions)
Formal charges are minimized (as close to 0 as possible)
Negative FC on more electronegative atoms
No adjacent atoms with same-sign formal charges

Common AP question types:

"Draw the Lewis structure for X" — use the full algorithm
"Which structure is most stable?" — compare formal charges
"How many resonance structures?" — count equivalent placements
"Predict bond order" — total bonds ÷ number of positions
"Why is this bond length between single and double?" — resonance
"Explain why $BF_{3}$ acts as a Lewis acid" — incomplete octet

The cyanate ion $(OCN^{-})$ can have several Lewis structures. Determine the best one.

Select the correct answer for each AP-style concept question.

Tackle a multi-concept AP-style question.

Each H: 1 bond

(2 e^{-})

= 2

e^{-}

✓ (duet rule)

Give each N 1 lone pair (2

e^{-}

each): 4

e^{-}

used ✓

Each N: 6 (triple bond) + 2 (lone pair) = 8 ✓

Remaining: 20 − 18 = 2

e^{-}

→ 1 lone pair on S