An element is defined as a substance which cannot, by known chemical means, be split up into two or more simpler substances. A compound is a pure substance, containing of two or more elements chemically combined together. A mixture is a substance that contains two or more substances which are physically together but are not chemically reacted with each other.
Chalk is a mixture with the main component consisting of calcium carbonate or calcium sulfate(VI).
Air is a mixture consisting of nitrogen, oxygen, carbon dioxide, water vapor, and others.
Calcium carbonate is a compound consisting of the elements: calcium, carbon, and oxygen.
Iodine is an element.
Starch is a mixture with the main components consisting of amylose and amylopectin in varying amounts.
Zinc is an element.
Do you know?
| Similarities between an element and a compound | Differences between an element and a compound |
(a) Both are made up of particles (atoms) which are the building blocks. |
(a) An element is made up of only one type of atom. |
(b) Both are pure substances. |
(b) An element can be metallic (calcium) or non-metallic (oxygen). |
(c) Both have fixed melting and boiling points. |
(c) An element cannot be chemically split into simpler substances. |
(b) Which compound in part (a) has metallic, ionic, and covalent bonding? Explain each of this type of bonding.
Zinc has metallic bonding, calcium carbonate has ionic bonding, while iodine has covalent bonding.
Metallic bonding:
Within a metal, atoms partially lose their loosely bound valence electrons. These electrons are mobile and delocalized, not belonging to any one single atom and yet not completely lost from the lattice. A metal can thus be viewed as a rigid lattice of positive ions surrounded by a sea of delocalized electrons. What holds the lattice together is the strong metallic bonding — the electrostatic attraction between the positive ions and the delocalized valence electrons.
Ionic bonding:
Ionic compounds are generally formed between metals (with one, two, or three valence electrons for Groups 1, 2 and 13, respectively) and nonmetals (with five, six, or seven valence electrons). The metal atoms lose electrons from its outermost valence shell, forming positively charged ions (cations). In contrast, the non-metal atoms receive these electrons, place them in the outermost valence shell, and form negatively charged ions (anions). The resultant electrostatic attraction between the cations and anions is known as the ionic bond.
Covalent bonding:
A covalent bond is formed by the sharing of electrons from the valence shell between two atoms. These shared electrons are known as “bonding electrons.” The shared electrons are localized between the two nuclei, within the inter-nuclei region, in contrast to the non-bonding electrons, which move three-dimensionally around the nucleus of its own atom. The following shows the formation of a covalent bond between two similar hydrogen atoms to form a hydrogen molecule using the electron cloud model:
The covalent bond is the resultant electrostatic attraction between the localized shared electrons and the two positively charged nuclei.
| Q | So does this mean that as long as the reaction is between a metal and a non-metal, ionic bonds must be formed? But if the reaction is between two non-metals, then covalent bonds must be formed? |
A: Not necessarily! A metal reacting with a non-metal to give ionic compounds is a very useful GUIDELINE, but it is not always true. For example, Al is a metal and Cl2 is a non-metal, but AlCl3 is a covalent compound. Ultimately, what makes a reaction occur is determined by whether the energy change is favorable and whether there is sufficient energy for the particles to react. Thus, we can safely say that when aluminum metal and chlorine gas react, the energy change for the reaction would “preferably” result in the formation of a covalent compound rather than an ionic compound.
| Q | What makes a metal atom more likely to lose electrons while a non-metal atom more likely to gain electrons? |
A: A metal atom is more likely to lose electrons than a non-metal atom is because the net electrostatic attractive force acting on the valence shell of a metal atom is much weaker than that for a non-metal atom. Now, since the net electrostatic attractive force acting on the valence shell of a non-metal atom is much stronger, if the non-metal atom gains electrons, these extra electrons would also be strongly attracted like the rest of the valence electrons that originally belongs to the non-metal atom.
| Q | Since the net attractive force acting on the valence shell of a non-metal atom is quite strong, does this also explain why non-metals “prefer” to form covalent bonds? |
A: Yes, you are right! If two non-metal atoms “prefer” not to lose electrons, then the “best type” of bonding would involve the sharing of electrons. It is a “compromise”!
Do you know?
— Strong and non-directional. Therefore, when a force is applied across a piece of metal, the metal atoms can slide over one another without the breaking of the metallic bonds. This accounts for the malleability (able to be deformed into different shapes) and ductility (able to be drawn into wires) of metals.
(i) The number of valence electrons available for bonding. (This factor is useful for explaining the increase in metallic bond strength across a period of metals. For example, from Na to Mg to Al, the metallic bond strength increases.)
(ii) The size of the metal cation. (This factor is useful for explaining the decrease in metallic bond strength down a group of metals. For example, from Na to K to Rb to CS, the metallic bond strength decreases.)
— Ionic compound consists of charged cations and anions bonded together by ionic bonds, which are difficult to break under normal conditions. Hence, when a melting or boiling process is carried out, the heat energy is used to overcome the strong ionic bond. This accounts for the high melting or boiling point.
— A giant ionic lattice structure consists of ions rigidly bonded, hence these charged particles cannot move when in the solid state. But when melted into the molten state, these ions can act as charge carriers. This accounts for the electrical conductivity of molten ionic compound but not for the solid compound.
— In the ionic lattice, each cation is surrounded by a number of anions and NOT just by only one anion, and vice versa. The number of cations around an anion or the number of anions around a cation is known as coordination number.
(i) The charges of the cation and anion. For example, the ionic bond in MgO is stronger than that in Na2O because Mg2+ is doubly positively charged. Or, the ionic bond in MgO is stronger than that in MgF2 because O2− is doubly negatively charged.
(ii) The sizes of the cation and anion. For example, the ionic bond in MgO is stronger than that CaO because Mg2+ has a smaller cationic radius than Ca2+, therefore it has a greater charge density 
Or, the ionic bond in MgO is stronger than that in MgS because O2− has a smaller anionic radius than S2−.
| Q | So, the stronger ionic bond for MgO as compared to Na2O is not related to the number of cations and anions in the formula unit (referring to the MgO or Na2O)? |
A: Of course not. The strength of ionic bond, which is measured by the melting and boiling points, is dependent on the electrostatic attractive force between a cation and an anion, instead of the number of ions that are present in the formula unit. As a cation is usually surrounded by a number of anions, and vice versa, there is more than one ionic bond formed for a cation or anion!
| Q | Is the stronger ionic bond in MgO than in Na2O contributed by the smaller cationic size of Mg2+ than Na+? |
A: Certainly! Both the Mg2+ and Na+ are isoelectronic (i.e., they have the same number of electrons), but since Mg2+ ion has a higher nuclear charge (i.e., more protons) than the Na+ ion, the electron cloud of the Mg2+ ion is more strongly attracted. Hence, the distance of separation (i.e., inter-ionic separation) between the Mg2+ and O2− ions is smaller than that between the Na+ and O2− ions. Thus, the ionic bond in MgO is stronger than that in Na2O.
| Q | Since the ionic bond is strong, why is an ionic compound brittle in nature? |
A: Although an ionic compound is hard, because of the strong ionic bond, it is also brittle, meaning it can be broken easily. Why? This is because when a force is applied across a plane of ions, it would displace it in such a way that the cations would now face each other while the anions would also face each other. This would repel the two planes, causing the lattice to crack.
Do you know?
— Covalent bond can be formed between two similar atoms of the same element or between two dissimilar atoms of different elements. Depending on the number of extra electrons each atom needs to fulfill the octet rule, there can be a single bond (e.g., H−H, Cl−Cl, H−Cl), double bond (e.g., O=O, O=C=O, C=C), or triple bond (e.g., N≡N, H−C≡C−H), which can be formed between the two atoms. Each line, “−”, represents a pair of shared electrons.
— It is also possible to have a single covalent bond being formed in which the pair of sharing electrons comes only from one atom, known as the donor atom. This type of bonding is known as dative covalent bond or coordinate bond.
— Some non-metallic substances such as graphite, diamond, and silicon dioxide have high melting and boiling points which means that the attractive forces binding the particles together are very strong. For melting to occur, a great amount of energy in the form of heat is required to overcome the strong covalent bonds between the atoms.
| Q | So all atoms that form covalent bonds would certainly have fulfilled the octet rule? |
A: No! There are exceptions to the octet rule, for example:
Thus, the octet rule is not a rule that an atom must obey when the atom is involved in bond formation. Rather, it is just a useful guideline to help us predict the number of valence electrons that an atom would “prefer” to have. Ultimately, what determines the number of valence electrons that an atom would “prefer” to have during the formation of a compound is still the energy change for the reaction to form that compound.
Do you know?
The strength of id–id interactions depends on:
— Number of electrons in the molecule. The greater the number of electrons (i.e., the bigger the electron cloud), the more polarizable is the electron cloud, and hence the stronger the id–id interactions.
— Surface area for contact of the molecule. The greater the surface area possible for contact between the molecules, the greater the extent of the id–id interactions.
The strength of pd–pd attractions depends on the magnitude of the molecule’s net dipole moment. The greater the magnitude of the dipole moment, the more polar the bond.
The magnitude of the dipole, in turn, depends on the magnitude of the electronegativity difference between the bonding atoms.
The greater the difference in electronegativity between the two atoms, the greater the dipole moment being created, and thus the more polar the covalent bond.
Hydrogen bonding is present between molecules which have one of the highly electronegative atoms — F, O, or N — covalently bonded to a hydrogen (H) atom.
X (F, O, or N), being more electronegative than H, attracts the bonding electrons in the H–X bond closer to itself. As a result, X is more electron-rich and gains a partial negative charge (δ–). The H atom, on the other hand, acquires a partial positive charge (δ+) is electron deficient.
Being electron deficient, this H atom is strongly attracted to the lone pair of electrons on the highly electronegative X atom (electron-rich region) in other molecules. This electrostatic attraction between the H atom and the lone pair of electrons on the highly electronegative atom, F, O, or N, is known as hydrogen bonding.
The strength of a hydrogen bond depends on:
— Dipole moment of H–X bond where X is O, F, or N
F–H- - -F–H > O–H- - -O–H > N–H- - -N–H.
— Ease of donation of the lone pair on Y where Y is O, F, or N
N–H- - -N–H > O–H- - -O–H > F–H- - -F–H.
The order of overall hydrogen bond strength is:
F–H- - -F–H > O–H- - -O–H > N–H- - -N–H.
This shows that the dipole moment is in fact the most significant factor!
It must be noted that these intermolecular attractions are also known as intermolecular bonding. They are not the same as the conventional bonds such as ionic, covalent, and metallic bonds. The attractive forces between the molecules are very much weaker than the conventional bonds. This accounts for why lesser amount of energy is needed to overcome intermolecular attractions than conventional bonds such as ionic, covalent, and metallic bonds. The lesser amount of energy needed is translated to lower boiling and melting points for such substances.
| Q | What gives rise to electronegativity? |
A: Electronegativity refers to the ability of an atom to distort shared electron clouds. Now, since valence electrons are used for sharing, a more electronegative atom would have a stronger net electrostatic attractive force acting on the valence shell than a less electronegative one!
| Q | So, does the net electrostatic attractive force acting on the valence shell also explain why the ease of donation of the lone pair of electrons on Y, where Y is O, F, or N, decreases from left to right as shown below?
N–H- - -N–H > O–H- - -O–H > F–H- - -F–H. |
A: Yes, certainly! The net electrostatic force acting on the valence shell is the strongest for a fluorine atom and weakest for a nitrogen atom. Hence, the valence electrons on the N atom are more likely to be donated than those on the O atom, which in turn is more likely to be donated as compared to those on the F atom.
| Q | Is the hydrogen bond the strongest type of intermolecular forces among the three? |
The id–id interaction for the non-polar iodine is stronger than the hydrogen bond in water. This is why iodine is a solid at room temperature while water exists as a liquid.
(c) The elements sodium, magnesium, aluminum, silicon, phosphorous, sulfur, chlorine, and argon are the elements in Period 3. Draw simple diagrams to show the structures of three elements in Period 3.
Sodium, magnesium and aluminum have metallic bonding. The main differences in their metallic lattice structure would be the charge of the metal cation and the number of delocalized electrons involved in the metallic bonding:
Silicon, phosphorous, sulfur, and chlorine form covalent bonds between its own atoms. Silicon has a macromolecular/giant covalent structure while phosphorous, sulfur, and chlorine all exist as simple discrete molecular substances:
Argon exists in the monoatomic form with weak van der Waals’ forces of the instantaneous dipole–induced dipole type between the atoms:
| Q | How would the melting points for the elements, silicon, phosphorous, sulfur, chlorine, and argon, be like? And why would they have such a trend? |
A: Silicon has the highest melting point among the four elements because we are breaking Si−Si covalent bonds when we melt silicon. The melting points of sulfur, phosphorous, and chlorine would follow the trend: S8 > P4 > Cl2 > Ar. This is because the number of electrons decreases from S8 to P4 to Cl2 to Ar. The greater the number of electrons (i.e., the bigger the electron cloud), the more polarizable the electron cloud, and thus the stronger the id–id interactions. Hence, this increases the melting point.
(d) Give the formulae of (i) two covalent compounds, and (ii) two ionic compounds, formed only from elements in Period 3.
(e) Draw the dot-and-cross diagrams of the four compounds in part (d).
(i) Covalent compounds: SiCl4, PCl3, PCl5 and AlCl3.
(ii) Ionic compounds: NaCl, MgCl2 and Mg3P2.
Do you know?
— To draw the dot-and-cross diagram of an ionic compound:
• The metal atom loses all its valence electrons easily to form a cation. Thus, the dot-and-cross diagram for the cation should show no valence electrons, but only the elemental symbol enclosed in square brackets with a superscript that denotes the charge of the cation (its value corresponding to the number of electrons lost).
• The non-metal atom “seeks” to form a stable outer octet electronic configuration. Thus, in the case of Cl, it will accept one electron. For the anion, there should be eight valence electrons depicted, and the electrons that were gained should be differentiated from the original valence electrons that the non-metal atom has. The anion would thus have a charge that corresponds to the number of electrons gained.
• The number of electrons lost by a metal atom need not necessarily be equal to that gained by a non-metal atom. To ensure electrical neutrality, you need to balance the charges by adding the required coefficients, outside the square bracket, on the left side of the symbols for the ions.
— To draw the dot-and-cross diagram of a covalent compound:
• Identify the central atom and draw the valence electrons around the central atom using either a dot or a cross symbol.
Note: The number of valence electrons that an atom has is equivalent to its (Group number -10) for elements that come from Groups 13 to 18 in the Periodic Table.
• Draw the peripheral atoms around the central atom. Take note of the number of electrons each peripheral atom requires in order to fulfill the octet rule.
• Decide on the types of bond (dative, single, double, or triple) a peripheral atom needs to form with the central atom in order to fulfill the octet rule.
| Q | Why is phosphorous able to form PCl5? Wouldn’t the phosphorous atom in PCl5 have more than eight valence electrons? |
A: Elements that cannot have more than eight electrons in the valence shell during bond formation usually come from Period 2. For these elements, the valence shell is the n = 2 electronic shell which can only “house” a maximum of only eight electrons. As for phosphorous, it comes from Period 3, which uses the n = 3 valence shell for bond formation. Since the n = 3 electronic shell can “house” a maximum of 18 electrons, phosphorous can form PCl5 which breaches the octet rule.
| Q | Why is AlCl3 a covalent compound but not an ionic compound? |
A: Imagine when Al and Cl2 react: if Al3+ and Cl− ions are formed but due to the high charge density 
of the Al3+ ion and the high polarizability (ability to be polarized) of the Cl− ion, the electron cloud of the Cl− would be distorted to the extent that when you observe AlCl3, you would find that it is a simple molecular compound, and not an ionic one! Thus, this is again a testimony of the limitations of a scientific model. Basically, when a compound is formed from its constituent elements, you do not actually know how it is formed. It is from the already formed compound that we infer how it is formed. But is this really how it is formed? We need to do more experiments to verify it! This is also a very good example that defies the maxim, “a metal reacts with a nonmetal to give an ionic compound!”
| Q | How would the trend of the melting point of NaCl, MgCl2, and Mg3P2 be like? And why would there be such a trend? |
A: NaCl will have a lower melting point than MgCl2 because the ionic bond in MgCl2 is between a doubly positively charged Mg2+ cation and a Cl− anion, while that in NaCl is between a singly positively charged Na+ and a Cl− anion. As for Mg3P2, it has a higher melting point than MgCl2 as the ionic bond in Mg3P2 is between a Mg2+ cation and a triply negatively charged P3+ anion, while that in MgCl2 is between a doubly positively charged Mg2+ cation and a singly negatively charged Cl− anion. Hence, the ionic bond in Mg3P2 is stronger than in MgCl2, while the ionic bond in MgCl2 is stronger than that in NaCl.
| Q | So, we cannot say that “since MgCl2 has two Cl− anions, its melting point is higher than NaCl because each Mg2+ is attracted to two Cl− anions while each Na+ is only attracted to one Cl− anion?” |
A: You cannot say that. An ionic bond is always between a cation and an anion. A cation can form more than one ionic bond in the ionic lattice, and vice versa for an anion. The number of ionic bonds that a cation or an anion can form in an ionic lattice is not determined by the chemical formula of the ionic compound. Take for instance, each Na+ in NaCl is surrounded by six Cl− anions, and vice versa for each Cl− anion. This number of cations surrounding the anion is known as the coordination number. Thus, each Na+ cation in fact forms six ionic bonds, and similarly for each Cl− anion.
(a) Give the number of protons, neutrons, and electrons for isotopes, 34S and 35S.
(b) A sample of sulfur was found to contain 15% of 34S and 85% of 35S. Suggest how could this information be found experimentally and calculate the relative atomic mass of this sample of sulfur.
The isotopic composition can be determined by a method known as mass spectrometry (refer to Understanding Advanced Organic and Analytical Chemistry by K.S. Chan and J. Tan).
The relative atomic mass is a weighted average of the relative isotopic masses of the different isotopes:
Did you notice that the relative atomic mass is closer to the relative isotopic mass of the isotope that has a higher proportion in the mixture of isotopes?
| Q | Why is the relative isotopic mass equivalent to the sum of the relative mass of the nucleons? |
A: The bulk mass of an atom comes from the nucleus, which consists of protons and neutrons, as the mass of an electron is negligible as compared to the mass of a nucleon.
(c) The eruption of volcano releases hydrogen sulfide (H2S), a toxic gas. Give the dot-and-cross diagram of hydrogen sulfide, showing the valence electrons only.
(a) Describe the bonding present in solid aluminum. Explain why aluminum is a conductor of electricity.
The aluminum atoms are being held together by metallic bonds. The loosely bound valence electrons of each aluminum atom are delocalized. A metallic bond is the electrostatic attraction between the positive ions and the delocalized valence electrons.
As the valence electrons are delocalized, they can serve as charge carriers when an electrical potential difference is applied across a piece of metal. This thus gives rise to the electrical conductivity nature of aluminum.
| Q | So, when an electrical potential difference is applied across a piece of metal, the delocalized electrons actually move from one end of the metal to the other? |
A: No! Although the valence electrons are delocalized, they are not so mobile that they can actually move from one end of the metal to the other. In fact, when a potential difference is applied, the electrons at one end of the metal move out of the metal first, thus creating “holes.” Subsequently, nearby electrons then “hop” into these holes. This process repeated itself till the other end of the metal piece. That is really how metals conduct electricity.
In addition, metals are also good thermal conductors due to their low mass and mobile delocalized valence electrons. When heat is applied to one end of the metal, the vibrational kinetic energy of the electrons and cations increases. Being low in mass, the kinetic energy of the mobile electrons will rapidly be transferred down the solid lattice structure.
(b) Give one chemical property of aluminum and support your answer with a chemical equation.
Being a metal, aluminum can react with acid to give hydrogen gas:
2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g).
(i) Suggest, using the information above, the name of the bond type present in both aluminum chloride and aluminum fluoride. Draw the dot-and-cross diagrams for these two compounds.
The useful guideline to use is, “a metal reacts with a non-metal to give an ionic compound.” Based on the higher melting point of aluminum fluoride than aluminum chloride, aluminum fluoride is more likely to be an ionic compound with ionic bonding between the Al3+ and F− ions. In addition, AlF3 is insoluble in non-polar solvent, an indication of its ionic nature. In contrast, aluminum chloride is a covalent compound, with covalent bonds between the Al and the three Cl atoms. The fact that AlCl3 is soluble in non-polar solvent is a good evidence of its molecular nature.
| Q | Why are ionic compounds soluble in water but not in non-polar solvents such as hexane? |
A: Polar molecules such as water molecules can interact with the ions as shown below:
The interaction releases energy since such an interaction is actually a type of bond formation and when bonds form, energy is evolved. The energy that is released can be channeled to break the ionic bond. Non-polar solvent molecules cannot form sufficiently strong interactions with the ions. Hence, the ionic compound is insoluble in non-polar solvents as insufficient energy is released from the interaction of the ions and the solvent molecules to compensate for the energy that is needed to overcome the ionic bond. The above is the explanation behind the maxim. “like dissolves in like.”
(ii) Explain why the bonding in aluminum fluoride leads to a high melting temperature, while the bonding in aluminum chloride results in a compound which sublimes.
The higher melting point of aluminum fluoride is due to the stronger ionic bond between the Al3+ and F− ions. The lower sublimation temperature of aluminum chloride is due to the weak intermolecular forces between the AlCl3 molecules.
| Q | So, the covalent bonds in AlCl3 are still intact when AlCl3 sublimes? |
A: Absolutely right. When you melt or boil simple molecular compounds such as water, AlCl3, etc., you are breaking intermolecular forces and not the intramolecular covalent bonds.
| Q | So, does it mean that there is still ionic bonding in a molten ionic compound? |
A: Yes, of course. Similarly, there is still metallic bonding in molten metal.
| Q | So, the ability of a molten metal to conduct electricity is still due to the sea of delocalized electrons? |
A: Yes!
(d) Aluminum is a preferred choice for high-voltage electric cables relative to copper. Suggest a reason for this.
Although copper is a better electrical conductor than aluminum due to the greater number of delocalized electrons as charge carriers, aluminum is lighter than copper. This makes aluminum able to provide a better conductivity: weight ratio than copper.
(b) Dichloromethane, CH2Cl2, is a covalent compound with carbon as the central atom and single bonds formed between it and the other atoms. Draw a dot-and-cross diagram for the molecule.
(i) silicon and chlorine;
Using the octet rule as a guideline, the chemical formula for silicon tetrachloride is SiCl4.
(ii) carbon and oxygen;
Carbon and oxygen can form both CO and CO2, in which all the atoms in both compounds fulfill the octet rule:
There is a triple bond in CO in which one of the bonds is a dative covalent bond, where the electrons are donated by the oxygen atom.
(iii) hydrogen and phosphorous;
Hydrogen and phosphorous can form PH3.
(iv) carbon and chlorine;
Using the octet rule as a guideline, the chemical formula for carbon tetrachloride is CCl4.
| Q | Which compound, CCl4 or SiCl4, would have a higher boiling point? |
A: Since both CCl4 and SiCl4 are non-polar molecules, the intermolecular forces that hold the particles together are id−id interactions. As SiCl4 has more electrons than CCl4, its electron cloud is more polarizable. Hence, the id−id interactions between SiCl4 molecules are stronger than those between the CCl4 molecules. Thus, SiCl4 would have a higher boiling point than CCl4.
(v) carbon and sulfur; and
Since both sulfur and oxygen belongs to the same group, based on the octet rule, the chemical formula for carbon disulfide is CS2.
(vi) nitrogen and fluorine.
Using the octet rule as a guideline, the chemical formula for nitrogen trifluoride is NF3.
| Q | Does NF5 exist? |
A: No. A nitrogen atom cannot have more than eight electrons in its valence shell. This is because its valence shell is the n = 2 electronic shell, which can only accommodate a maximum of eight electrons.
The dot and cross diagram of H2CO3 is:
| Q | Based on the dot-and-cross diagram, can H2CO3 have hydrogen bonding? |
A: Yes. Since there is an H atom bonded to an O atom and there are lone pair of electrons on the various O atoms, hydrogen bonding can form between the H2CO3 molecules.
| Q | Is carbonic acid acidic because the O−H bonds can break to release H3O+ ions? |
A: Absolutely right. The H2CO3 can dissociate as follows:
H2CO3(aq) + 2H2O(l) ⇌ CO32−(aq) + 2H3O+(aq).
(b) Iodine can form either covalent or ionic bonds. Draw dot-and-cross diagrams to show the bonding in potassium iodide (KI) and iodine (I2).
(c) Explain why potassium iodide has a higher melting point than iodine.
The strong ionic bond in potassium iodide causes it to have a higher melting point than the intermolecular forces of the id−id type between the iodine molecules.
Do you know?
— When we melt or boil simple molecular compounds, we are breaking the intermolecular forces between the molecules and not the intramolecular covalent bonds.
(d) Explain why iodine and potassium iodide do not conduct electricity when in the solid state, but when in the molten and aqueous states, only potassium iodide can conduct electricity.
In the solid state, the ions in potassium iodide are rigidly held in the lattice structure. Thus, they cannot move when an electrical potential difference is applied. But when melted or in the aqueous state, both the cations and anions are mobile. Hence, these ions can act as charge carriers. As for iodine, the molecules are overall electrically neutral. Therefore, the molecules cannot move under the influence of an electrical potential difference.
(i) Explain the meaning of the term isoelectronic.
When two particles have the same number of electrons, they are termed isoelectronic.
| Q | What is the significance of knowing the concept of being isoelectronic? |
A: If the particles are isoelectronic, where they have the same number of electrons, the inter-electronic repulsion within each of the particles are the same.
(ii) Suggest the type of bonding which is present within the layers.
Since the structure of boron nitride is similar to that of graphite, the B and N atoms must be held together by B−N covalent bonds.
(iii) Suggest the type of interaction between the layers.
From the outset, a layer of atoms is none other than a mass of electron cloud. Since there is no covalent bond between the layers, what holds the different layers together would be intermolecular forces of the id−id type.
Do you know?
— Since the layers are being held together by intermolecular forces of the id−id dipole type, which are not strong in nature, the different layers can slide over each other easily. This allows boron nitride to be used as a lubricant, like graphite.
(iv) Suggest a possible use in which this compound would behave similarly to the corresponding carbon compound.
Boron nitride can replace graphite as a high-temperature lubricant.
Do you know?
— Since the atoms in boron nitride are bonded by strong B−N covalent bonds which will only break at high temperatures, it can be used as a lubricant at high temperatures.
(v) When heated under high pressure, this form of boron nitride is converted into another form which is an extremely hard solid. Suggest the type of structure adopted by this new material.
In this new material, each of the boron and nitrogen would adopt a tetrahedral shape. Among the four covalent bonds that an N atom forms with four B atoms, one of the N→B bond is a dative covalent bond. The overall structure of this new material is very similar to the structure of diamond.
| Q | If we have not come across the chemistry of boron nitride before, how would we know how to answer this question? |
A: In this chapter on Chemical Bonding, you need to know the structures and bonding of some basic substances such as diamond, graphite, sodium chloride, water, silicon dioxide, etc. In the very beginning, the question already tells you that boron nitride is similar to graphite. So, you need to use the physical and chemical properties of graphite here. Now, since diamond is related to graphite in terms of allotropy, we thus can infer that the planar layered structure of boron nitride can be transformed into the tetrahedral form. This is applying what we have learned into a new context.
| Q | What is the meaning of allotropy? |
A: Allotropy refers to the phenomenon in which atoms of the same element are bonded differently together. For example, both graphite and diamond consist of carbon atoms. But in graphite, each carbon atom is bonded to three other carbon atoms. But each carbon atom is bonded to four other carbon atoms in diamond. The term allotropy is applicable only to elements but not compounds!
| Substance | Boiling point/K | Molar mass/g mol–1 |
| CH4 | 109 | 16 |
| NH3 | 240 | 17 |
| H2O | 373 | 18 |
(i) Suggest reasons for the difference in boiling points between NH3 and CH4 in terms of the type of molecules involved and the nature of the forces between them.
Both NH3 and CH4 exist as simple discrete molecular compounds held together by weak intermolecular forces. The intermolecular forces between the non-polar CH4 molecules are of the id−id type. This is much weaker than the hydrogen bonds between the NH3 molecules. Hence, CH4 has a lower boiling point than NH3.
(ii) Why does H2O have a higher boiling point than NH3?
Both NH3 and H2O exist as simple discrete molecular compounds held together by hydrogen bonds. The hydrogen bonds between the H2O molecules are more extensive than those between NH3 molecules. This is because a H2O molecule can form, on average, two hydrogen bonds per molecule, whereas a NH3 molecule can only form one hydrogen bond per molecule. Hence, the boiling point of water is higher than that of ammonia.
| Q | So, a HF molecule can also form, on average, one hydrogen bond per HF molecule? |
A: Yes! This thus explains why although the hydrogen bond between two HF molecules is stronger than that between water molecules, water has a higher boiling point than HF because there are more hydrogen bonds being formed per water molecule than per HF molecule.