4. Which of the following has the bonds arranged in order of decreasing polarity? — Bonding Chemistry Question

STEPS:

1. Understand the concept of bond polarity: Bond polarity is directly determined by the difference in electronegativity ($\Delta\text{EN}$) between the two bonded atoms. The greater the electronegativity difference, the more polar the bond. A difference of zero (between two identical atoms) results in a completely nonpolar bond.
2. Recall the periodic trends for electronegativity: Electronegativity generally increases going from left to right across a period and decreases going down a group. Fluorine ($\text{F}$) is the most electronegative element on the periodic table ($\text{EN} = 4.0$), followed closely by oxygen ($\text{O}$) and nitrogen ($\text{N}$ is $\approx 3.0$). Hydrogen ($\text{H}$) has a moderate electronegativity of $2.1$.
3. Analyze Option A ($\text{H–F} > \text{N–F} > \text{F–F}$):
* $\text{H–F}$ bond: Fluorine ($4.0$) and hydrogen ($2.1$) have a very large difference:
$$\Delta\text{EN} = 4.0 - 2.1 = \mathbf{1.9}$$
This is a highly polar covalent bond.
* $\text{N–F}$ bond: Fluorine ($4.0$) and nitrogen ($3.0$) are both highly electronegative, leading to a moderate difference:
$$\Delta\text{EN} = 4.0 - 3.0 = \mathbf{1.0}$$
This bond is polar, but significantly less so than $\text{H–F}$.
* $\text{F–F}$ bond: Two identical fluorine atoms share electrons completely equally:
$$\Delta\text{EN} = 4.0 - 4.0 = \mathbf{0}$$
This is a completely nonpolar bond.
* Comparing the differences ($1.9 > 1.0 > 0$), we see that Option A is correctly arranged in order of decreasing bond polarity, which confirms it is the correct answer.

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WHY_OTHERS_WRONG:

• Option B is incorrect ($\text{H–I} > \text{H–Br} > \text{H–F}$): Electronegativity of the halogens decreases down the group ($\text{F} > \text{Cl} > \text{Br} > \text{I}$). Since hydrogen is constant, the bond polarity actually increases from $\text{H–I}$ to $\text{H–F}$ ($\text{H–I} < \text{H–Br} < \text{H–F}$), which is the reverse of decreasing polarity.
• Option C is incorrect ($\text{O–N} > \text{O–S} > \text{O–Te}$): Oxygen is highly electronegative ($3.5$). As you go down a group (from $\text{N}$ to $\text{S}$ and $\text{Te}$), electronegativity decreases. Consequently, the electronegativity difference with oxygen increases down the series ($\text{O–N} \approx 0.5 < \text{O–S} \approx 1.0 < \text{O–Te} \approx 1.4$), representing an order of increasing polarity.
• Option D is incorrect ($\text{Sb–I} > \text{Sb–Te} > \text{Sb–Cl}$): Antimony ($\text{Sb}$) has an electronegativity of $\approx 2.1$. Comparing the other atoms: $\text{Cl}$ ($3.0$) yields $\Delta\text{EN} \approx 0.9$, $\text{I}$ ($2.5$) yields $\Delta\text{EN} \approx 0.4$, and $\text{Te}$ ($2.1$) yields $\Delta\text{EN} \approx 0.0$. This sequence goes from moderate to zero to high polarity, which is not in a strictly decreasing order.