Question

Select the most appropriate statement. In BF$_3$

• A. All the bonds are completely ionic
• B. The B–F bond is partially ionic
• C. B–F bond has partial double bond character
• D. Bond energy and bond length data indicates single bond character of the B–F bond
• E. All the bonds are covalent

Hint:

Electron-deficient compounds like BF$_3$ stabilize via back bonding.

Answer 1

The Correct Option is C

Concept:
BF$_3$ is an electron-deficient molecule where boron has only 6 electrons in its valence shell. To stabilize itself, it undergoes $p\pi$-$p\pi$ back bonding with fluorine atoms, giving the B–F bond partial double bond character.

Step 1: Electronic configuration and bonding.
Boron (Z = 5) forms three $\sigma$ bonds using sp$^2$ hybridisation and has an empty p-orbital.

Step 2: Back bonding explanation.
Fluorine atoms have lone pairs in p-orbitals. These electrons are donated into the empty p-orbital of boron: \[ F(p) \rightarrow B(p) \] This is called $p\pi$-$p\pi$ back bonding.

Step 3: Effect of back bonding.

• Shorter bond length than expected for single bond
• Higher bond energy
• Partial double bond character

Step 4: Why other options are incorrect.

• (A) Incorrect: Bond is not ionic; electronegativity difference not large enough
• (B) Incorrect: Though some polarity exists, this is not the most accurate description
• (D) Incorrect: Experimental data shows bond length shorter than single bond
• (E) Incorrect: Though covalent, it ignores partial double bond nature Final Conclusion:
B–F bonds exhibit partial double bond character due to back bonding.