The factors affecting intramolecular covalent bonds strength are: bond order, size of the atom, & polarity. My question is whether there is one factor more significant than the other, and if so, do we need to know it for the imat?
for example: O2 has a double bond(sigma &pi), but no polarity. HF has only one bond(sigma) but is polar & both atoms in it are smaller than Oxygen. So, which bond is harder to break? is there a regularity?
Your question is an excellent one and delves deep into the world of chemical bonding for the IMAT exam, it shows that you understood the concept of intramolecular forces.
The strength of a covalent bond, and thus the energy required to break it, depends on several factors, including bond order, atom size, and polarity.
- Bond Order: Higher bond order generally means a stronger bond. For example, a triple bond is stronger than a double bond, which in turn is stronger than a single bond. This is due to the increased electron density between the bonded atoms, which leads to a stronger electrostatic attraction.
- Atom Size: As a general rule, smaller atoms form stronger covalent bonds due to their smaller size, which allows the electron clouds of the atoms to be closer together, strengthening the bond.
- Polarity: Polarity itself does not directly affect the strength of a single covalent bond. However, in a molecule, polar bonds can lead to intermolecular forces like hydrogen bonding, which can affect the physical properties of the substance, including boiling and melting points.
Now, to address your specific question about the O2 vs. HF bond. The bond dissociation energy (the energy required to break a bond) of an O2 molecule (double bond, nonpolar) is approximately 495 kJ/mol​1​. On the other hand, the bond dissociation energy of an HF molecule (single bond, polar) is about 562 kJ/mol​2​. Even though O2 has a double bond and HF has a single bond, the bond in HF is harder to break. This discrepancy arises mainly because of the high electronegativity of fluorine, resulting in a highly polar bond, and the small size of both H and F atoms, which allows them to form a very strong bond.
Keep in mind, though, that the bond enthalpies are often average values, especially when a molecule has multiple identical bonds. For example, in a methane (CH4) molecule, even though there are four identical C-H bonds, the energy to break each of them is not the same because the environment changes when each hydrogen atom is removed. In such cases, the bond enthalpy quoted is an average value​3​.
All these factors contribute to the complexity of predicting bond strength. It’s not as straightforward as saying one factor is always more significant than the other; it’s more accurate to say that these factors interact in complex ways to determine the overall bond strength.
During our next 2-3 classes, we’ll delve into this topic in more detail, exploring how these factors interplay to affect the strength of covalent bonds.
Thank you! I thought that the polarity has an affect on a specific individual molecule, because the electronegativity differences pulls the more positive atom towards the more negative one, bringing them closer together. is it not true?
Electronegativity places a rule as the charge of the atoms (now slightly negative or positive) attract each other, as one of them will have “less” electrons and the other more electrons, so they are slightly polar and attract each other.
Usually, on the exam, they will ask you questions about the bond length, which directly results from things such as the bond’s size and strength.
OK, but isn’t it true to state that polarity in covalent bonds shortens the bond, and thus making is stronger?(in a specific individual molecule). Before you wrote that polarity itself does not directly affect the strength of a single covalent bond, and I’m not sure why.
It’s not directly, the polarity is the result of the electronegativity difference, so it’s more correct to say that it’s the electronegativity difference causing the bond to be shorter and stronger. Hope it makes sense!
1 Like