Anand Classes provides complete Class 11 Chemistry notes on the Shape of BeF2 (Linear), BF3 (Trigonal Planar), and CH4 (Tetrahedral) Molecules based on VSEPR Theory. Learn how electron pair repulsion explains the geometry and bond angles of these molecules, along with diagrams and step-by-step reasoning. These notes also include solved Q&A, conceptual MCQs, and practice problems for NEET, JEE, and CBSE board exams. Perfect for quick revision and exam preparation. Click the print button to download study material and notes.
What is the Shape of BeF2 Molecule? : Linear
In BeF2, the central Be atom (Z = 4; 1s2 2s2) has two electrons in the valence shell. In the formation of BeF2, each of these two valence electrons is shared by two fluorine atoms.
As a result, the Be atom is surrounded by two bond pairs of electrons. Therefore, the geometry of BeF2 molecule is linear and the bond angle is 180o.
pairs around Be-atom.
Other molecules such as BeCl2, ZnCl2, HgCl2 have linear shape.
What is the Shape of BF3 Molecule? : Trigonal Planar
The central atom boron (Z = 5; 1s2 2s2 2p1) has three valence electrons. In the formation of BF3 molecule, each electron in the valence shell of the B atom forms a bond pair with a fluorine atom.
As a result, the central boron atom is surrounded by three bond pairs and the molecule adopts trigonal planar geometry.
In this geometry, all the F–B–F bond angles are of 1200. This geometry is planar because the three F atoms and the B atom lie in the same plane.
pairs around B-atom
Molecules such as BCl3, AlCl3, etc. have the same shape.
What is the Shape of CH4 Molecule? : Tetrahedral
The central atom carbon (Z = 6; 1s2 2s2 2p2) has four valence electrons. All the four valence electrons are bonded to four hydrogen atoms, forming four bond pairs around the central carbon atom.
These four electron pairs, trying to remain as far apart as possible, adopt a tetrahedral structure.
In this geometry, all the H–C–H bond angles are 109o28′ (≈ 109.5o).
Other examples of tetrahedral molecules are SiF4, CCl4, SiH4, NH4+, etc.
Short Answer Conceptual Type Questions (SAT)
Q1. Why does BeF2 have a linear shape?
Answer: BeF2 has two bond pairs around the central Be atom. To minimize repulsion, these are arranged at 180o, giving a linear geometry.
Q2. Explain why BF3 is trigonal planar.
Answer: BF3 has three bond pairs around boron and no lone pair. They orient themselves at 120o in one plane, resulting in a trigonal planar geometry.
Q3. Why is CH4 tetrahedral and not square planar?
Answer: Carbon has four bond pairs with hydrogen. The repulsion among four pairs is minimized in 3D space when they are oriented at 109.5o in a tetrahedral arrangement, not in a 2D square planar form.
Q4. State the bond angles in BeF2, BF3, and CH4.
Answer:
- BeF2 → 180o (linear)
- BF3 → 120o (trigonal planar)
- CH4 → 109.5o (tetrahedral)
Q5. Give two examples of molecules with tetrahedral geometry.
Answer: CH4, SiF4, CCl4, NH4+.
Multiple Choice Questions (MCQs)
Q1. The shape of BeF2 is:
(a) Trigonal planar
(b) Tetrahedral
(c) Linear
(d) Bent
Answer: (c) Linear
Explanation: Two bond pairs are oriented at 180o.
Q2. Which molecule has bond angles of 120o?
(a) CH4
(b) BeF2
(c) BF3
(d) NH3
Answer: (c) BF3
Explanation: BF3 has three bond pairs arranged in a plane at 120o.
Q3. In CH4, the H–C–H bond angle is approximately:
(a) 90o
(b) 109.5o
(c) 120o
(d) 180o
Answer: (b) 109.5o
Explanation: Four bond pairs form a tetrahedral arrangement.
Q4. Which of the following is an example of a linear molecule?
(a) BF3
(b) SiF4
(c) BeCl2
(d) CH4
Answer: (c) BeCl2
Explanation: BeCl2 has two bond pairs giving a linear geometry.
Assertion–Reason Type Questions
Q1.
Assertion (A): BeF2 is linear.
Reason (R): Two bond pairs arrange themselves at 1800 to minimize repulsion.
Answer: Both A and R are true, and R is the correct explanation of A.
Q2.
Assertion (A): BF3 has a bond angle of 109.5o.
Reason (R): Boron is surrounded by three bond pairs.
Answer: A is false, R is true.
Explanation: BF3 has bond angles of 120o, not 109.5o.
Q3.
Assertion (A): CH4 adopts a tetrahedral structure.
Reason (R): Four bond pairs repel equally and orient in 3D space at 109.5o.
Answer: Both A and R are true, and R is the correct explanation of A.
Case Study Based Questions
Passage:
VSEPR theory predicts molecular shapes based on electron pair repulsion. Molecules with only bond pairs adopt simple geometries. $BeF_{2}$ has two bond pairs, $BF_{3}$ has three bond pairs, and $CH_{4}$ has four bond pairs. Their shapes are linear, trigonal planar, and tetrahedral, respectively, with bond angles of $180^{circ}$, $120^{circ}$, and $109.5^{circ}$.
Questions:
Q1. Which molecule among $BeF_{2}$, $BF_{3}$, and $CH_{4}$ has the smallest bond angle?
Answer: $CH_{4}$ ($109.5^{\circ}$).
Q2. Why is $BF_{3}$ planar?
Answer: Because it has three bond pairs that orient themselves in a single plane at $120^{\circ}$.
Q3. What is the difference in geometry between $BeF_{2}$ and $CH_{4}$?
Answer: $BeF_{2}$ is linear with bond angle $180^{\circ}$, while $CH_{4}$ is tetrahedral with bond angle $109.5^{\circ}$.
Q4. Give one more example of a molecule with trigonal planar geometry.
Answer: $BCl_{3}$ or $AlCl_{3}$.
