In molecular geometry and chemical bonding, which choice best describes the bond polarity and overall molecular polarity of bromine pentafluoride, BrF5?

Introduction / Context:
Understanding molecular polarity requires looking at both the polarity of individual bonds and the shape of the molecule as a whole. Bromine pentafluoride, BrF5, is a classic example where VSEPR theory is used to determine geometry and then relate it to polarity. This question checks whether you can combine knowledge of electronegativity differences with molecular geometry to decide if BrF5 is polar or nonpolar and whether its individual bonds are polar.

Given Data / Assumptions:

• Molecule: bromine pentafluoride, formula BrF5.
• Central atom: bromine (Br), surrounding atoms: five fluorine (F) atoms.
• Electronegativity: fluorine is more electronegative than bromine.
• We apply VSEPR theory to determine the shape.

Concept / Approach:
A bond is polar if there is a significant difference in electronegativity between the two bonded atoms, leading to partial charges. In BrF5, fluorine is more electronegative than bromine, so each Br F bond is polar. The molecular geometry is determined by six electron pairs around bromine (five bonding pairs and one lone pair), which leads to an octahedral electron pair arrangement but a square pyramidal molecular shape. Because of this asymmetry, the dipole moments of the Br F bonds do not cancel completely, so the overall molecule is polar.

Step-by-Step Solution:
Step 1: Determine bond polarity. Fluorine has higher electronegativity than bromine, so each Br F bond is polar with electron density shifted toward fluorine. Step 2: Determine the electron domain geometry around bromine. There are six electron domains (five bonding, one lone pair), suggesting an octahedral arrangement of domains. Step 3: Use VSEPR theory to find molecular shape. With one lone pair out of six domains, BrF5 adopts a square pyramidal geometry. Step 4: Analyze dipole cancellation. In a square pyramidal shape, the lone pair and arrangement of Br F bonds prevent complete cancellation of dipole moments. Step 5: Conclude that bonds are polar and the overall molecule is also polar, matching the description polar bonds / polar molecule.

Verification / Alternative check:
You can compare BrF5 with related molecules such as SF6. Sulfur hexafluoride, SF6, has six identical S F bonds in a perfect octahedral shape, so the bond dipoles cancel, and the molecule is nonpolar despite having polar bonds. In contrast, BrF5 has only five F atoms around bromine plus one lone pair, which distorts symmetry. Because of this unsymmetrical structure, vector addition of dipoles does not give zero. This reasoning confirms that BrF5 is polar, not nonpolar.

Why Other Options Are Wrong:
Option A, nonpolar bonds / nonpolar molecule, is wrong because the Br F bonds are definitely polar due to the high electronegativity of fluorine. Option B, nonpolar bonds / polar molecule, is inconsistent because nonpolar bonds cannot combine to form a polar molecule in this context. Option D, polar bonds / nonpolar molecule, would be correct for a highly symmetrical molecule like SF6 but not for BrF5, where the lone pair breaks symmetry. Only option C correctly states that the bonds are polar and the molecule is also polar.

Common Pitfalls:
A common mistake is to assume that any molecule with several identical bonds is automatically nonpolar, without checking the presence of lone pairs or distortions in shape. Another pitfall is to confuse electron domain geometry with molecular geometry and assume that an octahedral domain arrangement always means a nonpolar molecule. Remember that lone pairs often create asymmetry and lead to net dipole moments. Carefully applying VSEPR theory for both bond polarity and molecular shape prevents these errors.

Final Answer:
Bromine pentafluoride, BrF5, has polar bonds and is an overall polar molecule.