Placing Bonds and Lone Pairs

Introduction

After choosing a skeleton, the next Lewis task is to spend the electron budget exactly. Start with one pair for each provisional bond, give terminal atoms the electrons they need, and then put any remaining electrons on the central atom. The process is mechanical enough to check, yet it must remain tied to chemistry: hydrogen takes a duet, and a short central octet may signal a multiple bond rather than extra electrons.

Core explanation

A single bond consumes two valence electrons from the total budget. If a skeleton has three single bonds, six electrons have been assigned before any lone pairs. Terminal hydrogen needs no further dots once it has one bond, because the shared pair is its duet. A terminal fluorine with one bond commonly receives three lone pairs, adding six nonbonding electrons and completing its octet. The same pattern applies to many terminal halogens in simple diagrams.

Consider NH₃. N contributes five valence electrons and three H atoms contribute one each, for eight total. The skeleton has three N–H single bonds, using six. The two electrons left become one lone pair on N. Every H has a duet and N has an octet: six electrons in bonds plus two in its lone pair. If the lone pair were put on a hydrogen instead, that H would exceed its duet and N would still have only six around it.

Consider H₂O. Its eight-electron budget is allocated as four electrons in two O–H bonds and four electrons in two lone pairs on O. Water and ammonia demonstrate why lone pairs matter even when a formula states only bonded atoms. The difference between two and one lone pair around the centers helps explain their later geometry and polarity. A Lewis diagram with correct bonds but missing lone pairs is incomplete.

For a molecule with terminal non-hydrogen atoms, first complete those terminals after drawing the skeleton. In CO₂, O–C–O uses four of sixteen electrons. Each terminal O can be given three lone pairs, using twelve more, so the budget is exhausted. Each O now has an octet, but carbon has only four electrons around it. The correct response is not to add nonexistent electrons. A later step converts a lone pair from each oxygen into an additional C–O bonding pair, yielding O=C=O with the same total sixteen electrons and a carbon octet.

The order of placement is a procedure, not a law that forbids exceptions. Electron-deficient molecules, odd-electron species and expanded-valence representations need separate judgment. The basic method is most reliable for straightforward main-group molecules and common ions. The final check always reconciles total electrons, local counts and net charge.

Step-by-step reasoning

1. Write the total valence-electron budget above the skeleton. 2. Draw single bonds between connected atoms and subtract two electrons for each. 3. Add lone pairs to terminal non-hydrogen atoms until their ordinary octets are met. 4. Place any remaining electrons on the central atom. 5. Check H duets, common octets and exact total; if the center is short, investigate multiple bonds without adding electrons.

Visual explanation

Make a ledger beside an H₂O drawing: total 8; two O–H bonds use 4; remaining 4 become two O lone pairs; remaining 0. Below it, show CO₂'s preliminary ledger: total 16; two single bonds use 4; oxygen terminal lone pairs use 12; remaining 0, yet carbon has only 4 around it. Add arrows that move oxygen lone pairs into C–O bonding regions, keeping the ledger total unchanged.

Real-world analogy

Allocating a fixed number of seats among rooms resembles distributing a fixed electron budget: assigning a seat to one use means it cannot be added again elsewhere. This analogy helps prevent inventing dots. In real molecules, electrons are not stationary objects assigned to rooms, so the picture is only bookkeeping.

Real-world example

Ammonia's nitrogen lone pair is chemically important. It can be donated to H⁺ to form NH₄⁺, changing N from three N–H bonds plus a lone pair to four N–H bonds in a common Lewis drawing. The electron pair did not appear from nowhere; it was already included in NH₃'s eight-electron budget.

Why?

Why do the remaining two electrons in NH₃ go on nitrogen rather than on a hydrogen? Each H already counts the pair in one N–H bond and has its full duet. Nitrogen has only six electrons around it until the remaining pair is placed there.

Common misconception

“If the center lacks an octet, add another lone pair even after the budget reaches zero.” That would change the species' electron count. Instead, first see whether a terminal lone pair can be represented as an additional bond or whether the species is an octet exception.

Worked example

Construct the first-pass Lewis diagram for OF₂. Oxygen contributes 6 and two fluorines contribute 2 × 7 = 14, for twenty electrons. Put O in the center with F–O–F single bonds, using four electrons. Give each terminal F three lone pairs, using twelve more; four electrons remain. Put those four as two lone pairs on O. Now each F has six lone-pair electrons plus two bonding electrons, and O has four lone-pair plus four bonding electrons. All three atoms have octets, and 4 + 12 + 4 = 20 electrons are shown. No double bonds are needed.

Quick check

1. Where do the last two electrons go after drawing three N–H bonds in NH₃? Answer: They form one lone pair on nitrogen, completing its ordinary eight-electron count.

Exam focus

Write a running electron ledger rather than placing dots by appearance. Finish H at two electrons, common terminal nonmetals at eight and then inspect the center. Never exceed the total budget to repair a drawing.

Advanced insight

In a real molecule, “placing” an electron pair on one atom is a formal representation of electron distribution. Lone-pair density can influence reactivity and shape, while molecular orbital descriptions may spread electron density more broadly. The Lewis allocation is still a powerful local bookkeeping model.

Summary

Each bond line consumes two electrons and each lone pair consumes two. Completing terminal atoms and allocating the remainder to the center produces valid simple Lewis diagrams when the total budget and local counts agree. An incomplete central octet may require bond-order revision, not extra electrons.

Practice questions

1. How many electrons do three single bonds use? Answer: Six, because each line represents a pair of electrons. 2. How many lone pairs are on oxygen in the usual H₂O Lewis diagram? Answer: Two lone pairs, using four of water's eight valence electrons. 3. Why is the first-pass O–C–O diagram incomplete for CO₂? Answer: After terminal octets use all electrons, carbon still has only four around it. 4. Does F–O–F require a multiple bond to satisfy ordinary octets? Answer: No. Single bonds and the correct lone-pair allocation satisfy its electron budget.