CHAPTER 1: Question 3

Answers with 3D Explanations

a) A carbon atom has four electrons in its outer shell. In CCl₄, each of these is combined with one of the seven electrons in the outer shell of a chlorine atom, to give the basic structure shown below in which both the carbon and chlorine atoms are surrounded by eight electrons.

Thus the carbon atom is surrounded by four identical bond pairs of electrons. Each bond pair repels the other three bond pairs of electrons, and the geometry which allows the four pairs of electrons to be as far from one another as possible is a tetrahedron. Thus both the bond and electron structures of CCl₄ are tetrahedral. A wedge / hash structure and a rotating 3D structure are shown below.

b) A phosphorus atom has five electrons in its outer shell. In PMe₃, three of these are combined with one of the four electrons in the outer shell of a carbon atom, whilst the remaining two electrons are held as a lone pair of electrons associated with the phosphorus atom. This gives the basic structure shown below in which both the phosphorus and carbon atoms are surrounded by eight electrons.

Thus the phosphorus atom is surrounded by four pairs of electrons (one lone pair and three bond pairs). Hence, the basic electron structure will be tetrahedral as discussed in part a above. However, the lone pair of electrons is held closer to the phosphorus atom than the bond pairs, so that the repulsion between the lone pair and the bond pairs of electrons is greater than the repulsion between the bond pairs. This gives a distorted tetrahedral structure in which the C-P-C bond angles are less than 109o. The bond structure is described as a trigonal pyramid. A wedge / hash structure and a rotating 3D are shown below.

c) A beryllium atom has two electrons in its outer shell. In Be(OH)₂, each of these is combined with one of the six electrons around an oxygen atom to give the structure shown below in which the beryllium atom is surround by just two bond pairs of electrons. The furthest apart that these two electron pairs can be from one
another is 180^o, hence both the bond and electron structures around the beryllium atom are linear.

An oxygen atom has six electrons in its outer shell, one of these can be combined with an electron from the beryllium to form the Be-O bond, and another can be combined with the electron from the hydrogen atom to form the O-H bond. This leaves four unshared electrons around the oxygen atom which form two lone pairs.
Hence, the oxygen atom is surrounded by four electron pairs, two lone pairs and two bond pairs. The furthest apart that these four electron pairs can orientate themselves is at the vertices of a tetrahedron, however, the lone pairs of electrons are held closer to the oxygen atom than the bond pairs, so that the repulsion between the lone pairs is greater than the repulsion between a lone pair and a bond pair which is greater than the repulsion between the bond pairs. Thus the electron structure is a distorted tetrahedron in which the Be-O-H bond angle is less than 109^o. The bond structure is described as bent. The resulting structure is
shown below.

d) A sulphur atom has six electrons in its outer shell. In H₂S two of these are combined with electrons from the hydrogen atoms to form the S-H bonds. This leaves four unshared electrons around the sulphur atom which form two lone pairs. Hence, the sulphur atom is surrounded by four electron pairs, two lone pairs and two bond pairs. The furthest apart that these four electron pairs can orientate themselves is at the vertices of a tetrahedron, however, the lone pairs of electrons are held closer to the sulphur atom than the bond pairs, so that the repulsion between the lone pairs is greater than the repulsion between a lone pair and a bond pair which is greater than the repulsion between the bond pairs. Thus the electron structure is a distorted tetrahedron in which the H-S-H bond angle is less than 109^o. The bond structure is described as bent. The resulting structure is shown below. It is worth noting that this is the same is the geometry around the oxygen atom in part c. This is as expected, since oxygen and sulphur are in the same group of the periodic table and in both cases two covalent bonds are formed to the oxygen / sulphur atom.

e) The carbon atom is surrounded by four bond pairs of electrons which will orientate themselves so as to produce a tetrahedral bond and electron structure (cf part a). In this case however, the four atoms bound to the carbon are not all identical, an oxygen atom is larger than hydrogen and so requires slightly more space. Thus
both the bond and electron structures will be slightly distorted tetrahedra in which the O-C-H bond angle is greater than the H-C-H bond angle. The oxygen atom is surrounded by two lone pairs and two bond pairs, hence the electron structure is a distorted tetrahedron and the bond structure is bent (cf parts c and d). The resulting structure is shown below.

f) A sulphur atom has six electrons in its outer shell. In SOCl₂, two of these are each combined with an electron from one of the chlorine atoms to form the S-Cl bonds. A further two of the sulphur electrons are combined with two electrons from the oxygen atom to form the S-O double bond. This leave two electrons which are held as a lone pair associated with the sulphur atom. Hence, the sulphur atom is surrounded by three bond pairs (remember that the p-component of the double bond does not count) and a lone pair of electrons. This is similar to part b, thus the electron geometry will be tetrahedral and the bond geometry a trigonal pyramid.
In this case however, both the bond and electron structures will be distorted since three different types of electrons need to be considered: the lone pair; the four electrons in the double bonds; and the two electrons in each of the S-Cl bonds. The largest repulsion will be between the lone pair and the bond pairs, however the four electrons in the S-O double bond will also repel the two electrons in each of the S-Cl bonds more than these electrons repel one another. Hence, the O-S-Cl bond angle will be greater than the Cl-S-Cl bond angle. The resulting structure is shown below.

g) Of the six electrons in the sulphur atom's outer shell, two are used to form the two S-Cl bonds. Each S-O double bond requires two electrons from the sulphur atom (one for the s-bond and one for the p-bond). Thus all six electrons are involved in bond pairs of electrons, and the sulphur atom is surrounded by four bond pairs (remember that the p-component of the double bond does not count) of electrons. This is the same as in part a, so both the electronic and bond structures will be tetrahedral. In this case however, both the bond and electron structures will be distorted since two different types of bonding electrons are present; the four electrons in each S-O double bond and the two electrons in each S-Cl bond. The largest repulsion will be between the two S-O double bonds, so the O-S-O bond angle will be greater than the O-S-Cl bond angle which will be greater than the Cl-S-Cl bond angle. This answer should be compared to that for part f. The resulting structure is shown below.

h) Of the six electrons in the sulphur atom's outer shell, two are used to form each S-O double bond. This leaves two electrons which are held as a lone pair associated with the sulphur atom. Thus the sulphur atom is surrounded by three pairs of electrons (remember that the p-component of the double bonds do not count), so the electron structure will be distorted trigonal planar in which the angle between the lone pair and the S-O bonds is greater than the O-S-O bond angle. The bond structure is bent, and the resulting structure is shown below.

i) Each C-O double bond requires two electrons from the carbon atom, and this accounts for all four electrons in the outer shell of the carbon atom. Thus the carbon atom is surrounded by two pairs of electrons (remember that the p-component of the double bonds do not count), and both the electron and bond structures will be linear. It is instructive to compare this structure with that obtained for an apparently similar situation in part h. The resulting structure is shown below.

j) Each S-O double bond requires two electrons from the sulphur atom, and this accounts for all six electrons in the outer shell of the sulphur atom. Thus the sulphur atom is surrounded by three identical pairs of electrons (remember that the p-component of the double bonds do not count), so both the electron and bond structures will be trigonal planar with O-S-O bond angles of exactly 120^o. It is instructive to compare this with the structures obtained for similar molecules in parts h and i. The resulting structure is shown below.

k) An oxygen atom has six electrons in its outer shell. Three of these electrons are used to form the covalent bonds to the carbon atoms, and one is removed to leave a positive charge on the oxygen. This leaves two electrons which are held as a lone pair associated with the oxygen atom. Thus the oxygen is surrounded by three bond pairs and one lone pair, resulting in a distorted tetrahedral electron geometry and a trigonal pyramidal bond geometry. It is instructive to compare this example with case b), since the removal of an electron from the oxygen atom makes its bonding equivalent to a group five element. The structure is shown below.

l) The two nitrogen atoms are equivalent and so will have the same electron and bond geometries. Of the five electrons in the outer shell of a nitrogen atom, one is used to form the bond to fluorine, and two (from each nitrogen atom) are needed to form the nitrogen-nitrogen double bond. This leaves two electrons on each
nitrogen atom which will form a lone pair. Thus each nitrogen is surrounded by two bond pairs (remember that the p-component of the double bond does not count), and a lone pair. Hence, the electron distribution will be a distorted trigonal planar structure in which the angle between the lone pair and a bond pair is greater than the F-N-N bond angle. The bond geometry will be bent with a F-N-N angle of less than 120^o. It is also possible to predict that the whole molecule will be planar because of the prevention of rotation around the p-component of the nitrogen-nitrogen bond. The structure is shown below.

m) This example is exactly analogous to the oxygen or sulphur atoms in examples c-e. Thus the electron distribution is a distorted tetrahedron, and the bond structure is bent. It is instructive to compare this with the related compound in part k. The structure is shown below.

n) The geometry around both carbon atoms is essentially the same, only the H₂C= group is discussed here. A carbon atom has four electrons in its outer shell. Two of these are used to form the carbon-carbon double bond, and the remaining two are used to form the two carbon-hydrogen bonds. Thus the carbon atom is surrounded by three bond pairs and both the electron and bond geometries will be distorted trigonal planar. The distortion is caused by the four electrons in the double bond exerting a greater repulsive effect than the electrons in the carbon-hydrogen bonds. Thus the C-C-H bond angle will be greater than the H-C-C bond angle. All of the carbon atoms and the two hydrogens attached to the doubly bonded carbon will be coplanar because of the lack of rotation around the p-component of the carbon-carbon double bond. The structure is shown below.

o) The geometry around the carbon atom is identical to that of the carbon atom in example e. Thus both the bond and electron geometries are slightly distorted tetrahedra. A nitrogen atom has five electrons in its outer shell. Three of these are needed to form the three covalent bonds, leaving two electrons to form a lone pair of electrons. Thus the nitrogen is surrounded by three bond pairs and one lone pair, which will give a distorted tetrahedral electron geometry in which the angle between the lone pair and any of the bond pairs is greater than the angle between any of the bond pairs. The bond geometry is a trigonal bipyramid which will be slightly distorted since there are two different types of bond pair. It is instructive to compare this example with that of Me₃P (part b), since both nitrogen and phosphorus are in group five of the periodic table and so have the same number of electrons in their outer shell. The structure is shown below.

p) This case is exactly the same as part k since both sulphur and oxygen are in group six of the periodic table and so have six electrons in their outer shell. Thus the electron distribution is a distorted tetrahedron and the bond geometry is a trigonal pyramid. The structure is shown below.

q) An iodine atom has seventeen electrons in its outer shell, however the ten electrons in the filled d-orbitals can be ignored leaving seven electrons. The negative charge in this molecule is located on the central atom, so this iodine atom is surrounded by eight electrons. Two of these electrons are needed to form the two I-I bonds, which leaves six electrons to be held as three lone pairs. Thus the central iodine atom is surrounded by five pairs of electrons: two bond pairs and three lone pairs. The electron distribution will thus be based on a trigonal bipyramid, and in this geometry there are two different types of position: axial and equatorial. At least one of the lone pairs must occupy an equatorial position since there are only two axial positions and three lone pairs. If either of the other lone pairs was to occupy an axial position, it would be located at just 90^o to the equatorial lone pair, whereas if all three lone pairs occupy equatorial positions, then they will be at 120^o to one another. The latter is clearly the more favourable since the repulsion between electron pairs decreases as the angle between them increases. Thus the three lone pairs occupy the three equatorial positions and the two bond pairs occupy the two axial positions. This gives a trigonal bipyramidal electron structure, and a linear bond structure as illustrated below. In the diagram shown below, the three bonds with no atoms at their end represent the three lone pairs of electrons.

r) Consider the cation first. A phosphorus atom has five electrons in its outer shell. One of these is removed to form the cation, and the remaining four are needed to form the four P-F bonds. Thus the phosphorus atom is surrounded by four identical bond pairs of electrons, and both the electron and bond structures will be
tetrahedrons (cf part a). For the anion, the phosphorus atom gains an electron giving six electrons in the outer shell. These are all needed to form the six P-F bonds, so the phosphorus atom is surrounded by six identical bond pairs. The furthest apart that these electron pairs can get is if they form an octahedron, so both the electron and bond structures will be octahedral. Diagrams of both the cation and anion are shown below.

s) Iodine has seven electrons in its outer shell (ignoring the d-electrons, see part q). Five of these electrons are needed to form the five I-F bonds, leaving two electrons to form a lone pair. Thus the iodine atom is surrounded by six electron pairs, and the electron geometry will be based on an octahedron. However, the lone pair will repel the four fluorine atoms at 90^o to it more than they will be repelled by the fluorine at 180^o to the lone pair. Thus, the electron structure will be a distorted octahedron and the bond structure will be a distorted square based pyramid as shown in the diagram below.

t) Iodine has seven electrons in its outer shell (ignoring the d-electrons, see part q), and the negative charge makes this eight electrons around the iodine atom. Four of these electrons are needed to form the I-Cl bonds, leaving four electrons which will form two lone pairs. Thus the iodine is surrounded by six pairs of electrons and the electron geometry will be based on an octahedron. The most important interaction will be between the two lone pairs, and the furthest apart that they can get is 180^o if they occupy positions trans to one another. This the electron structure will be octahedral whilst the bond structure will be square planar as shown below.

u) Xenon has eight electrons in its outer shell (ignoring the d-electrons, see part q). Two these are required to form the Xe-F bonds, leaving six electrons to form three lone pairs. Hence, the xenon atom is surrounded by three lone pairs and two bond pairs which will give a trigonal bipyramidal electron structure and a linear bond structure for the reasons discussed in part q. The structure is shown below, with the bonds with no atoms on their ends representing the lone pairs.

v) To answer this problem, it is necessary to first draw a single resonance form of the sulphate anion as shown below. Sulphur has six electrons in its outer shell, and all six of these are needed to form the bonds to oxygen atoms (one for each of the single bonds in the resonance structure and two for each of the double bonds). Note that the two negative charges are located on the oxygen atoms rather than on the sulphur, and so do not affect the number of electrons around the sulphur.

Hence the sulphur is surrounded by four bond pairs of electrons (the p-bonds do not count), giving a tetrahedral electron and bond structure. The resonance structure would suggest that this should be a distorted tetrahedral structure, since two of the S-O bonds are double bonds whilst the other two are single bonds. However, this is only one of six possible resonance structures for the sulphate anion, and each of the S-O bonds is actually identical (being half way between a single and a double bond), so both the bond and electron structures are regular tetrahedra as illustrated below.

w) Nitrogen has five electrons in its outer shell. Four of these are needed to form the two nitrogen-oxygen double bonds, leaving a single electron (a radical) associated only with the nitrogen atom. Note that this is a very unusual case where a stable molecule possesses an unpaired electron. This unpaired electron can be treated as if it was a lone pair, but it will repel electron pairs much less than a lone pair would. Thus the electron structure will be a distorted trigonal planar geometry, and the bond structure will be bent with an O-N-O bond angle greater than 120^o as shown below.

x) This case is almost identical to part w, except that the positive charge means that one fewer electrons is present around the nitrogen atom. Thus the unpaired electron no longer exists, and there are now just two electron pairs (neglecting the p-bonds) around the nitrogen atom. Thus both the electron and bond geometries
will be linear. as illustrated below.

y) This case is also very similar to part w, except that an extra electron is added to the nitrogen atom so that the unpaired electron in part w becomes a lone pair in the present case. Thus there are again three pairs of electrons (ignoring the pi-bonds) around the nitrogen atom, giving a distorted trigonal planar electron structure
and a bent bond structure. In this case however, the O-N-O bond angle will be less than 120^o since the lone pair will exert a greater repulsive effect than the bond pairs. The structure is shown below, and it is important to compare this answer with those to parts w and x and to appreciate the reasons for the differences.

z) To answer this problem, it is necessary to first draw a single resonance form of the nitrate anion as shown below. Nitrogen has five electrons in its outer shell, and all five of these are needed to form the bonds to oxygen atoms (one for the single bond in the resonance structure and two for each of the double bonds). Note that the negative charge is located on the oxygen atom rather than on nitrogen, and so does not affect the number of electrons around the nitrogen.

Hence the nitrogen is surrounded by three bond pairs of electrons (the pi-bonds do not count), giving a trigonal planar electron and bond structure. The resonance structure would suggest that this should be a distorted trigonal planar structure, since two of the N-O bonds are double bonds whilst the other is a single bond.
However, this is only one of three possible resonance structures for the nitrate anion, and each of the N-O bonds is actually identical (being part way between a single and a double bond), so both the bond and electron structures are exactly trigonal planar as illustrated below.

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