Electronegativity refers to the ability of an atom to attract electrons toward itself. A more electronegative atom has higher ability to attract electrons towards itself.

Electronegativity depends on two factors

1. Effective Nuclear Charge (ENC)
2. Distance from nucleus

To summarize:

1. ↑ Effective Nuclear Charge (ENC) = ↑ Electronegativity
2. ↑ Distance from nucleus = ↓ Electronegativity

Effective Nuclear Charge ≈ No. of protons – No. of inner shell shielding electrons

This is an intuitive formula. The protons in the atom are trying to attract electrons (unlike charges attract) while the inner shell electrons are trying to repel away electrons that are being added to the valence shell (like charges repel). Thus, ENC measures the net attraction that an atom has on the valence electrons.

For example, Fluorine (electronic arrangement: 2,7) is more electronegative than Oxygen (2,6).

ENC of F = 9 – 2 = 7

ENC of O = 8 – 2 =6

Therefore, it can be concluded from the above example that electronegativity increases across a period, from Group I to Group VII. Group 0 is excluded as it has stable electronic arrangement, hence will not form bonds.

How about elements from the same group? Don’t they have the same ENC?

Yes, you are right! Elements of the same group have the same ENC. For instance Fluorine (2,7) and Chlorine (2,8,7)

ENC of F = 9 – 2 = 7

ENC of Cl = 17 – 10 = 7

To compare elements in the same group, we have to use distance from the nucleus. Chlorine has 3 electron shells as compared to 2 for Fluorine. Thus, the valence shell of Chlorine is further away from the positive nucleus as compared to Fluorine.

The greater distance between the positive nucleus and the valence shell makes it more difficult for Chlorine to attract electrons as compared to Fluorine. Hence, Chlorine is less electronegative as compared to Fluorine.

Therefore, it can be concluded from the above example that electronegativity decreases down the group!

Putting these two factors together, you should be able to conclude for yourself that Fluorine is the most electronegative atom while Francium is the least electronegative! (For A level students, I am basing this on the Pauling scale.)

Readers of previous posts will (hopefully) remember that the covalent bond is defined as the electrostatic force of attraction between the localized shared pair of electrons and the 2 positively charged nucleus.

Since electrostatic forces of attraction depend on the difference in charges, a double bond is going to be stronger than a single bond and a triple bond is going to be stronger than a double bond. For instance, MgO has a higher m.p. as compared to NaF because there is a greater difference in the charges between the Mg²⁺ and O^2– ions as compared the Na⁺ and F^– ions, hence the electrostatic forces of attraction between the Mg²⁺ and O^2– ions are stronger and require more energy to break, resulting in higher m.p.

The same holds true for covalent bonds. Thus, the C ≡ C bond is stronger than the C = C bond. In turn, the C = C bond is stronger than the C – C bond. See diagram below.

However, closer inspection of the bond energies indicates that the increase in bond strength is non-linear in nature. In fact, it is increasing at a decreasing rate!

Tell me why!

As the number of electrons in the same space increases, there would be more repulsion between the electrons (since like charges repel), resulting in a lower than proportionate increase in the bond strength as the number of bonds increases.

Note: JC students should know that the “proper explanation” should include a discussion about the sigma, pi bonds and their different degrees of overlap of the electron cloud. However, the basic idea remains the same. There is less overlap for pi bonds because electrons are arranged in such a way so as to minimize repulsion. Even then, the 2 pi bonds are not of equal strength, suggesting that there remains some degree of electron repulsion.

As you guys know, the product of bromination should be drawn like this:

Instead of something like:

The reason for this lies in the way the addition reaction takes place

To be precise, this mechanism is known as “electrophilic addition” and applies to addition of electrophiles such as halogens (e.g. bromine) and water to alkenes. This mechanism does not apply to catalytic hydrogenation which will be covered in another post.

What are “electrophiles”?

Electrophiles simply refer to molecules/ions that “like electrons”. On the other hand, an electrophobe is a molecule/ion that “dislikes electrons”.

Mechanism

The C = C portion of the alkene is said to have “high electron density” / “electron-rich” as there are 4 electrons that are clustered around this region. Therefore when a bromine molecule is brought near to the C = C portion of the alkene, the electrons on the bromine atom that is closer to C =C bond will be repelled to the other bromine atom because like charges repel.

The bromine atom that is nearer to the C = C bond acquires a partial positive charge (δ +) while the other bromine atom becomes partially negative (δ – ). This is illustrated in the diagram below.

The partially positive bromine atom is attracted to the “electron-rich” C = C double bond. A pair of electrons is transferred from the C = C bond to the partially positive bromine atom forming a C – Br bond with one of the carbon atoms. The C = C bond becomes a C – C bond. This results in the formation of an intermediate known as a carbocation (think carbon – cation) and a negative bromine ion. See diagram below.

Finally, the negative bromine ion is attracted to the carbocation forming the final product, 1,2-dibromoethane!

Tell me why! (A pair of electrons are transferred from the C = C bond to the partially positive bromine atom..)

As you guys know, most chemical reactions are driven by the quest for greater stability.  The C = C bond [BE=  610kJ/mol] while stronger than the C – C bond [BE = 350kJ/mol] is less than twice as strong. You can artificially decompose the C = C bond into two C – C bonds as shown below.

Why is one bond weaker than the other? Since like charges repel each other,  you can think of the reason for this as due to higher levels of repulsion as there are more electrons that are “clustered” in the same space. [JC students should think in terms of sigma and pi bonds. Just saying.]

Hence when the partially positive bromine atom that is deficient in electron approaches the C = C bond, the pair of electrons from the C = C bond which is less stable [BE = 260kJ/mol] will “defect” to it, forming the C – Br bond [BE = 280kJ/mol], which is stronger, hence achieving a higher degree of stability!

Okay.. Minor things first.. As you guys know, for simplicity’s sake, the oxidation of alcohols to carboxylic acids using an oxidizing agent such as acidified potassium dichromate (K₂Cr₂O₇) is represented by an equation that looks like this:

However, I detected dissent from people in class.. Like.. Why don’t we write the full equation? What so difficult about it? Tell me why?! That is because the full equation looks something like:

I’m sure u guys agree that it is better that we stick to the simplified form.. no?

Anyways.. Moving on.. Why must the oxidizing agent be in excess?

That is because the oxidation of alcohols to carboxylic acids takes place via a 2 stage process.

In stage 1:

Alcohols are oxidized to form aldehydes. In the example below, ethanol is oxidized to an aldehyde called ethanal.

In stage 2

The aldehyde is then oxidized to form the carboxylic acid. In the example below, ethanal is oxidized to form ethanoic acid.

If there is insufficient oxidizing agent, the reaction stops at the aldehyde instead of forming the desired carboxylic acid! Similarly, reflux condition is needed to prevent the ethanal from escaping before it is completely oxidized to form the carboxylic acid!

Note: For JC students, this mechanism applies to primary alcohols. Often, secondary alcohols do not fully oxidize to form carboxylic acids but stop at ketones.

Covalent bond between atoms in a molecule: electrostatic force of attraction between the localized shared pair of electrons and the 2 positively charged nucleus. See diagram below

Intermolecular bonds: as the name suggests are bonds between the molecules. See diagram below

How do intermolecular bonds come about?

Even though we draw the dot-cross diagrams as shown above, please remember that electrons are not stationary but are always in constant random motion!

(Diagram above shows an electron cloud orbiting a nucleus.. Note that the continuously changing pattern shows random motion of the electrons! – For some reason, the gif is not moving..  will find a better pic soon..)

Therefore, at any point in time, there will always be an uneven distribution of electrons in a molecule. This results in one part of the molecule having a positive charge while the other part must have a negative charge. (The fancy A level term for this is instantaneous dipole)

The instantaneous dipole in one molecule can cause the formation of dipoles in nearby unpolarized molecules.

As a result, a weak electrostatic attraction forms between the positive portion of one molecule and the negative portion of another molecule. This type of attraction is called the intermolecular bond/ Van Der Waals forces of attraction (fancy A level term = instantaneous dipole-induced dipole)

*Please note that these intermolecular bonds are fleeting in nature because the electrons are in constant motion. However, the process of bond forming and bond breaking keeps repeating such that on average, there will be attractive forces between the molecules.

How does Mr affect the strength of the intermolecular bonds?

1. The higher the Mr, the stronger the intermolecular bond (more important reason at O levels)

When Mr increases, number of protons and hence number of electrons increases. Hence, the dipoles formed by the random motion of electrons will, on average have larger +ve and larger  –ve charges.

This increases the electrostatic forces of attraction between the positive portion of one molecule and the negative portion of the other molecule resulting in a stronger intermolecular bond.

2.  Molecules with larger Mr have higher surface area and hence more dipole interactions (less important at O levels)

When Mr increases, the molecular size tends to increase due to increasing number of electron shells. Therefore, molecules have larger area in contact with other molecules. This increases the amount of the dipole interactions, hence forming more intermolecular bonds.

In turn, how does Mr affect m.p. and b.p. of a simple covalent substance?

Therefore, from the explanation above, higher Mr results in stronger intermolecular bonds which will require more energy to break. Therefore, m.p. and b.p. increases when Mr increases!

However, this is not the entire story. In the next post, I will discuss why some molecules with lower relative molecular masses have higher m.p./b.p. than molecules with higher relative molecular masses.

*Other Notes:

1. The above explanation only applies to simple covalent substances

2. Covalent bonds between the atoms are never broken during melting or boiling!!! (i.e. H₂O after boiling still remains as H₂O molecules just that the molecules are more widely spaced. It does not become hydrogen and oxygen atoms!) Therefore, simple covalent molecules have low m.p./b.p. because only the weak intermolecular bonds are broken during melting and boiling. The covalent bonds between the atoms which are very strong are not broken.