The least polar bond is found in a molecule of HI, HF, HCl, HBr.
The correct answer and explanation is:
Correct Answer: HI (hydrogen iodide)
Explanation (300 words):
To determine the least polar bond among the molecules HI, HF, HCl, and HBr, we need to understand bond polarity, which depends on the difference in electronegativity between the two bonded atoms.
Electronegativity is a measure of an atom’s ability to attract electrons in a chemical bond. The greater the difference in electronegativity between two atoms, the more polar the bond.
Hydrogen (H) has an electronegativity of 2.1. The halogens involved have the following electronegativities:
- F (fluorine): 4.0
- Cl (chlorine): 3.0
- Br (bromine): 2.8
- I (iodine): 2.5
Now, let’s calculate the electronegativity differences:
- HF: |4.0 – 2.1| = 1.9 → Most polar
- HCl: |3.0 – 2.1| = 0.9
- HBr: |2.8 – 2.1| = 0.7
- HI: |2.5 – 2.1| = 0.4 → Least polar
Therefore, HI has the smallest electronegativity difference, making it the least polar bond among the given molecules.
Why Polarity Matters:
Polarity affects many physical and chemical properties such as boiling point, solubility, dipole moment, and reactivity. More polar molecules (like HF) often have higher boiling points and stronger intermolecular forces due to stronger dipole-dipole interactions.
In contrast, a molecule like HI, with a less polar bond, has weaker dipole interactions and behaves more like a nonpolar molecule in certain conditions.
In summary, HI has the least polar bond because the electronegativity difference between hydrogen and iodine is the smallest among the four options.