Question

Using just a periodic table (not a table of electronegativities), decide which of these is likely to be the most polar bond. Explain your answer. (a) \(\mathrm{C}-\mathrm{F}\) (b) \(S-F\) (c) \(\mathrm{Si}=\mathrm{F}\) (d) \(O-F\)

Short answer

The \(\mathrm{C}-\mathrm{F}\) bond is likely the most polar.

Step-by-step solution

Understand Bond Polarity

Bond polarity arises due to differences in electronegativity between two atoms. When one atom attracts shared electrons more strongly, the bond becomes polar. More significant differences in electronegativity result in more polar bonds.

Use Position in the Periodic Table to Estimate Electronegativity

Identify the positions of the elements in the periodic table to assess general trends in electronegativity. Electronegativity typically increases across a period from left to right and decreases down a group.

Compare Electronegativity Trends for Relevant Elements

Examine the position of C, Si, O, S, and F in the periodic table: - Carbon (C) and Silicon (Si) are in Group 14. - Oxygen (O) and Sulfur (S) are in Group 16. - Fluorine (F) is in Group 17. Fluorine is the most electronegative, whereas C and Si are less electronegative than O and S.

Analyze Each Bond for Potential Polarity

Estimate the expected polarity of each bond based on element positions:- (a) \(\mathrm{C}-\mathrm{F}\): Large difference as F is very electronegative.- (b) \(\mathrm{S}-\mathrm{F}\): F is more electronegative, but less than C-F.- (c) \(\mathrm{Si} = \mathrm{F}\): Similar to C-F but with Si being less electronegative than C.- (d) \(\mathrm{O}-\mathrm{F}\): Difference is smaller than C-F as O is also quite electronegative.

Determine the Most Polar Bond

Based on the largest difference in electronegativity with respect to F, \(\mathrm{C}-\mathrm{F}\) is more polar compared to \(\mathrm{S}-\mathrm{F}\), \(\mathrm{Si} = \mathrm{F}\), and \(\mathrm{O}-\mathrm{F}\), because carbon is significantly less electronegative than fluorine.

Key concepts

Electronegativity

Electronegativity is a measure of how strongly an atom attracts the shared electrons in a chemical bond. When we look at the periodic table, we can observe specific trends in electronegativity:
- Electronegativity increases as we move across a period from left to right. - It decreases as we move down a group.
For example, in the same period, Oxygen is more electronegative than Nitrogen. Meanwhile, in the same group, Oxygen is more electronegative than Sulfur.

These trends can be explained by the increasing positive charge of the nucleus, which attracts the electrons more strongly as we move across a period. Down a group, the effect of added electron shells distances the valence electrons from the nucleus, reducing the atom's ability to attract electrons. This makes electronegativity a crucial concept in determining the type and polarity of bonds formed between atoms.

Bond Polarity

Bond polarity occurs when there is an uneven distribution of electrons between two bonded atoms, leading to one end of the bond being slightly negative and the other positive. Typically, this happens when two atoms have differing electronegativities. The atom with the higher electronegativity will attract the electrons more, making the bond polar.

• If the difference in electronegativity is large, the bond is more polar.
• Conversely, if the difference is small, the bond is less polar.

An easy way to visualize bond polarity is by using dipole arrows, pointing towards the more electronegative atom. Therefore, understanding the concept of bond polarity is crucial in predicting the behavior and properties of molecules, including their interactions and solubility.

Fluorine

Fluorine is a unique element, renowned for its extreme electronegativity, which is the highest of all elements on the periodic table. This high electronegativity means that Fluorine has a strong ability to attract electrons from other elements when forming bonds.

• In chemical bonds, because Fluorine is so electronegative, it typically pulls electron density towards itself, creating highly polar bonds.
• The polar bonds involving Fluorine often have significant implications in chemical reactivity and stability.

In the exercise's context, the bond between Carbon (C) and Fluorine (F) (\( \mathrm{C}-\mathrm{F} \) bond), is the most polar because of the stark difference in electronegativity between Carbon and Fluorine. As a result, understanding Fluorine’s electronegativity is essential in explaining why it forms such polar bonds compared to other elements, like Oxygen or Sulfur.