Question

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Answer 1

Answer:

HF ≥ H₂O because the HF bond is more polar than the OH bond.

Explanation:

Electrons in the bond will not be shared equally. Hydrogen electronegativity value is 2,1 but Fluorine is 4, so electrons in the covalent bond will be more attracted to Fluorine. The electron cloud will be more distorted towards Fluorine.

I hope this is correct, and as always, I am joyous to assist anyone at any time.

Answer 2

Answer: Option (d) is the correct answer.

Explanation:

Florine is much more electronegative than oxygen atom. And, more is the electronegativity difference between the combining atoms more will be the strength of bond.

This is because then greater will be the polarity of compound due to which its atoms are strongly held together.

So, in HF the difference in electronegativity will be as follows.

          Electronegativity difference = electronegativity of fluorine - electronegativity of hydrogen

                                  = 4 - 2.1

                                  = 1.9

Whereas difference in electronegativity of hydrogen and oxygen atoms of $H_{2}O$ are as follows.

             Electronegativity difference = electronegativity of oxygen - electronegativity of hydrogen

                                         = 3.5 - 2.1

                                         = 1.4

Since, electronegativity difference in HF is greater.

Therefore, we can conclude that the statement H F ≥ H2O because the H F bond is more polar than the H O bond, is true about the strength of the hydrogen bonding in H F and $H_{2}O$.