Q.80 Chemistry ⭐ 360° Option Elimination Active

NEET 2021 Chemistry Question 80 Solution

Topic: Chemical Bonding and Molecular Structure · Official NTA Booklet Code M1

Question

BF3 is planar and electron deficient compound. Hybridization and number of electrons around the central atom, respectively are:

(1)sp3 and 4
(2)sp3 and 6
(3)sp2 and 6
(4)sp2 and 8

✓

Verified Against NTA Final Answer Key

Option (3) : sp^2 and 6

📚 Chemical Bonding and Molecular Structure Easy

📋 Given Parameters & Problem Setup

Molecule Boron trifluoride (BF3)

Geometry Planar (trigonal planar)

Electronic Nature Electron deficient

⚡

The 30-Second Exam Hall Shortcut: ⚡ 3 Bonds + 0 Lone Pairs = sp² Hybridized, 3 × 2 = 6 Electrons

⏱️ Save ~15 seconds

Boron in $BF_3$ forms 3 single $\sigma$ -bonds with fluorine and has zero lone pairs: Steric number $= 3 \implies sp^2$ hybridization (trigonal planar). Three shared covalent bond pairs contain $3 \times 2 = 6$ valence electrons around the central boron atom (incomplete octet/sextet)! Match Option (3)!

1 Boron valence electrons

= 3

2 Forms 3 single covalent bonds with 3 fluorine atoms: 3 bond pairs, 0 lone pairs.

3 Steric number

= 3 \implies sp^2

hybridization.

4 Total electrons around central boron atom

= 3\text{ pairs} \times 2 = 6

electrons.

5 Respectively:

sp^2

and

6

6 Match Option (3).

🛡️

Why You Can 100% Trust This Shortcut: Canonical hybridization and valence octet deficit in Group 13 trihalides.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Electronic Configuration of Central Boron Atom

Boron ( $Z = 5$ ) has the ground-state electronic configuration $1s^2 2s^2 2p^1$ . In the excited state, one $2s$ electron is promoted to an empty $2p$ orbital: $1s^2 2s^1 2p_x^1 2p_y^1 2p_z^0$ .

Step 2

Hybridization of Boron in BF3

The one $2s$ orbital and two $2p$ orbitals mix to form three equivalent $sp^2$ hybrid orbitals directed towards the corners of an equilateral triangle at bond angles of $120^\circ$ (trigonal planar geometry). Each hybrid orbital overlaps with a $2p$ orbital of fluorine to form a $\sigma$ -bond.

\text{Hybridization} = sp^2 \quad (\text{Trigonal planar})

Step 3

Counting Electrons Around the Central Atom

The central boron atom forms three single $\sigma$ -bonds with fluorine and possesses no lone pairs. Each single covalent bond contains 2 shared electrons:

\text{Total electrons around Boron} = 3 \times 2 = 6\text{ electrons}

Step 4

Conclusion

The hybridization is $sp^2$ and the number of electrons around the central atom is 6, corresponding to Option (3).

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 1

✗ Incorrect

Verdict & Scientific Reason:

Incorrect tetrahedral hybridization and incorrect electron count.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming 4 electron pairs.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Assigns $sp^3$ hybridization to a strictly planar molecule.

⚠️

Student Fallacy (Why candidates pick this trap): Assigning

sp^3

hybridization to a 3-coordinate planar molecule.

3 3

✓ 100% Correct

Verdict & Scientific Reason:

Exact parameters: $sp^2$ hybridization with 6 valence electrons (incomplete octet).

💡

Why this is airtight: None; textbook chemical bonding analysis.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Claims boron has a complete octet (8 electrons), directly contradicting that $BF_3$ is electron-deficient.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming the octet rule is satisfied in

BF_3

⚠️

NTA Trap Alert: Octet Assumption Trap (8 Electrons)

Students who assume every stable molecule must satisfy the octet rule choose Option (4) ( $sp^2$ and 8). The prompt itself explicitly mentions "BF3 is planar and ELECTRON DEFICIENT compound", confirming an incomplete octet with only 6 electrons.

🛡️ How to avoid the -1 mark: Read the prompt: "electron deficient"

\implies 6

electrons, NOT 8!

📖

Core NEET Syllabus Standard

NCERT Chemistry · Chemical Bonding and Molecular Structure Standard Benchmark

Core Scientific Principle Tested:

"In BF3, boron is sp2 hybridized with 3 bond pairs and an unhybridized empty 2p orbital, surrounded by 6 valence electrons."

Syllabus Focus: Hybridization and Incomplete Octet of Central Atom

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Chemistry Question 80 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ 3 Bonds + 0 Lone Pairs = sp² Hybridized, 3 × 2 = 6 Electrons

Step-by-Step Rigorous Workflow & Mathematical Proof

Electronic Configuration of Central Boron Atom

Hybridization of Boron in BF3

Counting Electrons Around the Central Atom

Conclusion

360° Option Elimination Engine Key Feature

NTA Trap Alert: Octet Assumption Trap (8 Electrons)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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