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Why does hydrogen fluoride (Mr = 20) have a higher m.p./b.p. as compared to fluorine gas (Mr=38) even though both are simple covalent molecules?

As we learnt in the previous post, simple covalent molecules with higher relative molecular masses (Mr) should have higher m.p./b.p. as compared to those with lower Mr.  However, this is not the case for all simple covalent molecules.

When two atoms of different elements are joined together in covalent bonding, the sharing of electrons is not always equal, creating permanent partial charges on each atom. (Fancy A level term = permanent dipole). These types of molecules are known as polar molecules.

Why are the electrons not shared equally?

The atom that exhibits a higher affinity for electrons would “pull” the shared pair of electrons closer to itself, creating a partial negative charge on itself, leaving the other atom with a positive charge.

The relative affinity for electrons in turn depends on the electronegativity of the atom / effective nuclear charge (ENC) of the atom.

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.

Using this concept, we calculate the ENC of Hydrogen and Fluorine. As a reminder, Hydrogen has 1 proton (electronic arrangement: 1) while Fluorine has 9 protons (electronic arrangement: 2,7).

ENC of H = 1 – 0 = 1

ENC of F = 9 – 2 = 7

Since Fluorine has higher ENC as compared to Hydrogen, the shared pair of electrons would be pulled closer to the Fluorine atom, leaving the Fluorine atom with a permanent partial negative charge while Hydrogen has a permanent partial positive charge. See diagram below.

                       HF

It can also be represented like this:

dipole moment

Dipole moments are useful as they help us determine if polyatomic molecules (E.g. CO2, H2O, NH3) are polar or not.

On the other hand, the electrons in a fluorine molecule (F2) are shared equally between the two atoms as the ENC of both atoms are equal. Thus, fluorine molecule will not have any permanent partial positive or partial negative charges. Thus, the intermolecular bonds in Fluorine are the result of the instantaneous-dipole, induced dipole interactions (id-id).

 

How does all of this explain the difference in b.p./m.p.?

The intermolecular bonds between HF molecules are the result of the electrostatic forces of attraction between the permanent partial positive portion of a molecule with the permanent negative portion of another molecule (fancy A level term permanent dipole-permanent dipole interaction (pd-pd)).

pd-pd

This is stronger than the id-id interactions that exist between the F2 molecules, since in HF, the dipoles are permanent as opposed to fleeting. Since the intermolecular bonds are stronger in HF, they require more energy to break and HF will have a higher m.p./b.p. than F2.

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Must polar bonds give rise to polar molecules? And, why is water a polar molecule?

As you have read in the previous post, when two atoms of different elements are joined together in covalent bonding, the sharing of electrons is not always equal, creating permanent partial charges on each atom. These types of molecules are known as polar molecules.

However, this is insufficient for determining if polyatomic molecules (E.g. CO2, H2O etc) are polar in nature. To determine if these molecules are polar, what matters most is the net dipole moment.

Consider CO2, which has a linear molecular shape. Because oxygen is more electronegative than carbon (check by calculating the ENC of oxygen and carbon), this gives rise to 2 dipole moments pointing in opposite directions. Hence, the dipole moments cancel each other and CO2 is said to be non-polar.

CO2

However, water has a bent molecular shape. Oxygen is more electronegative than hydrogen. Thus this gives rise to 2 dipole moments as seen below.

H2O

But, in this case, water is a polar molecule because of its bent structure. The horizontal components of both dipole moments are in the opposite direction and hence will cancel out each other. But, the vertical components of the dipole moments are pointing in the same direction, giving rise to a net dipole moment as seen in the diagram below. Thus water is said to be a polar molecule.

Recall that the dipole moment points from the partial positive end to the partial negative end. Thus water is a polar molecule with a permanent partial negative charge (δ – ) on the oxygen atom and permanent partial positive charges (δ + )  on the hydrogen atoms.

h2o net dipole moment

*Note: I have not addressed the issue of why CO2 is linear but H2O is bent. This has to do with the Valence Shell Electron Pair Repulsion (VSEPR) theory which is too long to write about in a short blog post. Furthermore, it will be covered in great detail by your JC teachers. For those of you who are really interested or have too much free time, you can read about it here.

Is esterification a redox reaction?

A few weeks ago, T2 (you know who you are) asked me the above question. At the time, I gave a rather brusque reply that esterification is not a redox reaction as the oxidation state did not change. (That was because I was feeling devastated after a (non-chemistry) lesson.. Don’t worry, if you’re reading this post, I am most probably not referring to you.. But I digress.) Anyways, the question deserves to be answered in greater detail! So here it is!

The diagram below shows an esterification reaction between methanol and ethanoic acid to form methyl ethanoate. We will use this example to illustrate why esterification is not a redox reaction.

Image

Eagle-eyed observers would note that methanol has lost a hydrogen (i.e. oxidized), this should imply that ethanoic acid is reduced and hence the reaction is a redox reaction. However, the best method of determining if a reaction is redox is to check the oxidation states/determine if electrons are gained or lost. The other two definitions are approximations used at O levels to make your life simplier. They fail at explaining complex scenarios such as this. In another (yet to be written) post, I will explain why it is generally ok to use the other two definitions in most situations.

Rules for assignment of oxidation states

As you know, in an ionic compound, the charge of the ion is the oxidation state. For example, the o.s. of Na in NaCl is +1.

However, for covalent compounds, the more electronegative atom is assigned a negative oxidation state (because it has a higher affinity for electrons) while the less electronegative atom is assigned a positive oxidation state. If they form a single covalent bond, the more electronegative atom is assigned an o.s. of -1 while the other atom is assigned an o.s. of +1.

Take for example, a compound formed between fluorine and chlorine (FCl). Fluorine being more electronegative will be assigned an oxidation state of -1 while chlorine being less electronegative is assigned an oxidation state of +1.

 

So how does this apply to esterification?

To begin with, please note that the oxidation state of the oxygen atom on methanol is “-1” as it is more electronegative (ENC of oxygen = 6, ENC of hydrogen = 1) than the hydrogen atom that it is bonded to. See diagram below.

Image

On the other hand, for ethanoic acid, the oxidation state of the carbon atom on the carbonyl group is “+1” as it is bonded to a more electronegative oxygen atom (ENC of oxygen =6, ENC of carbon =4). See diagram below.

Image

After esterification, the oxygen from the alcohol forms a single covalent bond with the carbonyl functional group of the ethanoic acid as shown below.

Image

Note that oxygen is more electronegative than carbon. Hence, the o.s. of oxygen remains as “-1” while the o.s. of carbon remains as “+1”. Since there is no change in o.s., esterification is not considered a redox reaction!

Electronegativity!

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.)

The curious case of the carbon – carbon bond strengths

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 O2– ions as compared the Na+ and F ions, hence the electrostatic forces of attraction between the Mg2+ and O2– 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.

post5d1

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!

post5d2

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.

Why can’t both bromine atoms be added to the same carbon atom during addition reaction with alkenes?

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

Post4 d1

Instead of something like:

post4d2

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.

post4d3

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.

post4d4

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

post4d5

 

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.

post4d6

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!

Why must the oxidizing agent be in excess during the oxidation of alcohol to carboxylic acid?

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:

post6d1

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:

post6d2

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.

post6d3

In stage 2

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

post6d4

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.