Other electrons in the valence shell of the atoms, not involved in bonding,
are either simply ignored, or shown as dots on the atoms. Two such dots can also
be taken together as a line, representing a so-called free electron (f.e.) pair.
In so-called Lewis structures, we try to account for the number of electrons
involved in the formation of bonds. A line between two atoms represents the two
electrons that form a covalent bond between the two.
Other electrons in the valence shell of the atoms, not involved in bonding,
are either simply ignored, or shown as dots on the atoms. Two such dots can also
be taken together as a line, representing a so-called free electron (f.e.) pair.
Before we start drawing lines, we have to know how many electrons we have to
account for, how many electrons the common atoms (C, N, O, Cl, S,...) have in their
outer shell. Which in turn is related to their position in the periodic system.
To summerize:
Hydrogen has 1 electron in its 1s shell;
B, C, N, O, and F have their 2s2p shells partially filled with 3, 4, 5, 6, and 7 electrons
respectively.
With this information, we can start constructing simple molecules.
In H-H the two electrons form a bond, only one line is needed, finished.
In CH4 we have 4 (C) + 4 (4x1 H) = 8 electrons, so with four lines for the C - H bonds all electrons are accounted for.
NH3: N has 5, 3 form a bond, 2 remaining in free electron pair,
4 pairs in total around N (see figure above).
Ammonium ion: still 8 electrons, while N started with 5 and the four H's with one each,
so there is one missing; hence overall charge +1.
Water you can now do for yourself. And H3O+.
Octet rule: first row atoms (Be B C N O F Ne) prefer to accommodate 8 electrons, 4 pairs,
in the '2s2p' shell. Less is undesirable, more is not possible.
Second row atoms (P, S, Cl...) can have more, as we will see later.
Some interactive exercises on more complicated molecules (in Dutch)
At this point we jump from the 2D drawing, well described in the book, to the 3D spatial structure:
Second rule: from the point of view of an atom, there is not much difference between
a bonding electron pair and a free electron pair. They occupy the same type of orbitals,
and, important for understanding 3D structures: they all repel eachother. They tend to
be as far apart as possible.
| This repulsion results in a tetrahedral structure, i.e. the hydrogens in methane occupy the corners of a tetraeder, and all bond angles are 109.5°. | |
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In ammonia, NH3, the three bonding pairs and the free electron pair minimize their
mutual repulsion by adopting a tetrahedral environment. Just the same as the
four bonds do in methane. The bond angles are not exactly the same as in methane, but fairly close: 106.7°. |
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The same rule applies to water: two bonds and two f.e. pairs in tetraeder, H-O-H angle 104.5° From which we can conclude the f.e.- bond repulsion is
somewhat larger than O-H bond-bond repulsion. Keep this in mind for later.
Like in ammonia, one f.e. pair in water can easily be protonated, to form H3O+. From this point of view it could be protonated twice, were it not that coulombic repulsion makes this less probable. |
So we can have ('quasi') tetrahedral compounds of several types:
The ammonia picture shows equivalence of the four molecular orbitals: three that
describe the three N-H bonds in ammonia and the one occupied by the lone pair.
We say that the 2s and 3 2p atomic
orbitals of nitrogen hybridize to form four equivalent sp3
orbitals.
This is shown by experiment,
it is not 'trivial' from a theoretical point of view. It could have
been sp2 hybridization,
with the three N-H bonds in a plane. and the lone pair occupying a p orbital
perpendicular to this plane.
In fact, this is what happens if this p
orbital is 'stabilized' by overlap with some other p orbital , like in amides, or in aniline.
In a future chapter we will discuss orbitals and hybridization, double bonds and resonance.
As stated above, less than four electrons pairs is not impossible, it
is just undesirable. Compounds with three electron pairs are reactive
(electrophilic), in
their attempts to fill the gap.
Good examples are the boron compounds, like BF3. The 3 electrons
of boron form bonds with fluorine (which has 7 electrons). This results
in three bonding pairs around boron, and three f.e. pairs at each fluorine.
The three bonds, staying as far apart as possible, lie in one plane with bond
angles of 120°. The general type is AX3.
Trifluoroborane will react rapidly with compounds that have a f.e. pair available for bonding, e.g. a fluoride anion, with which a tetrahedral octet arrangement can be formed.
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Note the change in geometry around the boron atom; measure the bond angles
in the first and the last structure.
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Questions: (Answers on the quiz page of this chapter.)
We continue with other types of spatial arrangements in second row elements, consisting of more than four bonds and f.e. pairs.
