Back to Topic 2.2 — The covalent model
2.2.2Chemistry SL10 flashcards

Molecular shapes (VSEPR)

Practice Flashcards

Flip to reveal answers
Card 1 of 102.2.2
2.2.2
Question

What does VSEPR stand for?

Click to reveal answer

Track your progress — Sign up free to save your progress and get smart review reminders based on spaced repetition.

All 10 Flashcards — Molecular shapes (VSEPR)

Sign up free to track progress and get spaced-repetition review schedules.

Card 1definition

Question

What does VSEPR stand for?

Answer

**V**alence **S**hell **E**lectron **P**air **R**epulsion.

Card 2definition

Question

What is an electron domain?

Answer

Any group of electrons around the central atom — a single/double/triple **bond (each = 1 domain)** or a **lone pair**.

Card 3concept

Question

Shape for 2 domains, 0 lone pairs?

Answer

**Linear**, 180° (e.g. CO_{2}, HCN).

Card 4concept

Question

Shape for 3 domains, 0 lone pairs?

Answer

**Trigonal planar**, 120° (e.g. BF_{3}).

Card 5concept

Question

Shape for 4 domains, 0 lone pairs?

Answer

**Tetrahedral**, 109.5° (e.g. CH_{4}).

Card 6concept

Question

Shape for 3 bonds + 1 lone pair?

Answer

**Trigonal pyramidal**, ~107° (e.g. NH_{3}).

Card 7concept

Question

Shape for 2 bonds + 2 lone pairs?

Answer

**Bent**, ~104.5° (e.g. H_{2}O).

Card 8concept

Question

How do lone pairs affect bond angle?

Answer

Lone pairs repel **more** than bonding pairs, so they **reduce** the bond angle.

Card 9concept

Question

Why is CO_{2} linear despite double bonds?

Answer

Each double bond is **one** electron domain; 2 domains, 0 lone pairs → linear, 180°.

Card 10comparison

Question

Order of bond angle: CH_{4}, NH_{3}, H_{2}O?

Answer

CH_{4} (109.5°) > NH_{3} (107°) > H_{2}O (104.5°) — angle falls as lone pairs increase.

Track your progress with spaced repetition

Sign up free — Aimnova tells you exactly which cards to review and when, so you remember everything before your IB exam.

Start Free