(A) BF₃

(B) [BF₄]⁻

(C) H₃N·BF₃

(D) (CH₃)₂O·BF₃

The species with the shortest B–F bond distance among the given options is BF₃ (Option A).

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Correct Answer: BF₃ Has the Shortest B–F Bond

Among BF₃, [BF₄]⁻, H₃N·BF₃ and (CH₃)₂O·BF₃, the shortest B–F bond distance is found in BF₃. This is due to strong pπ–pπ back bonding from fluorine to the vacant p-orbital of boron, which gives the B–F bond partial double-bond character and therefore shortens the bond length.

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Concept: Back Bonding and B–F Bond Length

• In BF₃, boron is electron-deficient and has an empty p-orbital, while each fluorine has lone pairs available for pπ–pπ back donation.

• This back bonding increases B–F bond order above 1, imparting partial double-bond character and decreasing the B–F bond length to about 130 pm (≈1.30 Å).

• When boron’s electron deficiency is reduced by coordination to Lewis bases (like NH₃ or ethers) or by forming [BF₄]⁻, back bonding decreases, and the B–F bonds become more like pure single bonds, so the bond length increases.

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Option-wise Explanation

Option A: BF₃

• BF₃ is a planar, electron-deficient molecule with boron in sp² hybridization and an empty p-orbital.

• Lone pairs on fluorine atoms donate electron density back into this empty p-orbital (pπ–pπ back bonding), giving significant double-bond character to B–F and thus the shortest B–F bond distance among the given species.

✔ Correct option

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Option B: [BF₄]⁻

• In [BF₄]⁻, boron is surrounded by four fluorine atoms in a tetrahedral geometry with sp³ hybridization, and all B–F bonds are essentially pure single bonds.

• The negative charge is delocalized over the fluorine atoms, but boron is no longer strongly electron deficient, so effective pπ–pπ back bonding is greatly reduced, leading to longer B–F bonds than in BF₃.

✖ B–F bond length is longer than in BF₃

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Option C: H₃N·BF₃ (Ammonia–BF₃ Adduct)

• In H₃N·BF₃, NH₃ donates a lone pair to boron, forming a coordinate (dative) N→B bond and satisfying boron’s electron deficiency.

• Because boron becomes less electron deficient, F→B back bonding decreases, so the B–F bonds lose double-bond character and become closer to normal single bonds, increasing their length relative to free BF₃.

✖ Adduct formation lengthens B–F relative to BF₃

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Option D: (CH₃)₂O·BF₃ (Dimethyl Ether–BF₃ Adduct)

• Dimethyl ether acts as a Lewis base and donates a lone pair from oxygen to boron, again forming an O→B coordinate bond and reducing boron’s electron deficiency.

• As in the NH₃ adduct, reduced electron deficiency at boron weakens F→B back bonding, so the B–F bonds become more single-bond-like and longer than in BF₃.

✖ B–F bonds are longer than in BF₃

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Key Takeaway for Exams

• Maximum pπ–pπ back bonding ⇒ higher bond order ⇒ shorter B–F bond.

• Free BF₃ shows the strongest F→B back bonding and therefore has the minimum B–F bond distance among BF₃, [BF₄]⁻, H₃N·BF₃ and (CH₃)₂O·BF₃.