Chapter 7:  Covalent Bonds & Molecular Structure

Along with what is presented here, there is a very long writeup in the "Supplementary Notes" section of these web pages.  Chapters 7-10 of those notes present a lot of this same material from a different point of view and in considerable depth.  Also, there is an alternate set of excellent problems.  Because that section is so thorough, this one will just hit the highlights (it won't be short but will have a different emphasis).

Assigned Problems:

Problem 7.36 (p. 292):

Vitamin C (ascorbic acid) has the following connections among atoms.  Complete the Lewis electron-dot structure for vitamin C and identify any multiple bonds.

This is relatively straightforward if you remember the following:

Upon a quick examination, we see two carbons with only three bonds.  These need double bonds to complete their octets.  After that is done, we add extra electrons to the oxygens where needed.  There happen to be 68 valence electrons in this molecule (C6H8O6) and you can check your final answer against the diagram below.  Extra bonds and electrons are drawn in red.

Problem 7.37 (p. 292):
The following ball-and-stick molecular model is a representation of thalidomide, a drug that causes birth defects when taken by expectant mothers but that is valuable for its use against leprosy.  The lines indicate only the connections between atoms, not whether the bonds are single, double, or triple (red = O, gray = C, blue = N, ivory = H).

(a)  What is the molecular formula of thalidomide?
(b)  Indicate the positions of the multiple bonds in thalidomide?
(c)  What is the geometry around each carbon and each nitrogen?

This is a little tedious but, nevertheless, not really out of control complicated.  We answer the individual parts of the question now.

(a)  Just counting the atoms by color, we see that the formula is C13H10N2O4.

Now comes the fun part.  We need to complete the Lewis dot structure.  I have done this by noting that each C, N, and O must have 8 electrons and arranging the double bonds accordingly.  This is shown (additions in green) by mangling the authors' pretty structure:

(b)  You can see where the double bonds are (I drew them in green).

(c)  Here are the geometries around each C and N:

In the ring to the left (an aromatic ring), the structures are trigonal planar.  In the next ring, both carbons have double bonds with their oxygens and, hence, are also trigonal planar.  In the last (rightmost) ring, the two carbons bonded to oxygen are also trigonal planar.  The other 3 carbons have tetrahedral structures.  Finally, both nitrogens are surrounded by four "objects" (three single bonds and one lone pair) and, thus, also have tetrahedral structure.  The molecule has 96 valence electrons.  20 of these are in lone pairs, 14 are in the added bonds (green), and 62 are in the "sticks."  Count these yourself to be sure!
Problem 7.38 (p. 292):
What general trends in electronegativity occur in the periodic table?

This should be easy.  EN increases from left to right and from bottom to top.  Or, to put it another, "up and to the right" means increasing and "down and to the left" means decreasing.  The following picture should make this clear:

Note, however, occasional blips such as those from Cu to Zn, Ag to Cd, and Au to Hg.  Why do you think these occur?
Problem 7.48 (p. 293):
Draw electron-dot [Lewis] structures for the following molecules or ions:
 
(a) CBr4  (b) NCl3  (c) C2H5Cl 
(d) BF4- (e) O22- (f ) NO+
These are all fairly routine Lewis dot structures.  All we need do is count the valence electrons and proceed from there.  The numbers of valence electrons for (a)-(f ), respectively, are 32, 26, 20, 32, 14, and 10.  These are now given as free-hand drawings:

Remember that the number of valence electrons is just the sum of the column numbers of the various atoms minus any ionic charge.  For instance, for O22-, we have (6 x 2) - (-2) = 14.  With Lewis structures of this type, the process is relatively easy and "painless."
Problem 7.50 (p. 293):
Draw as many resonance structures as you can for each of the following molecules or ions:

(a)  HN3        (b)  SO3        (c)  SCN-

This is only moderately painful.  First, we look at case (a).  This has three obvious resonance structures (16 valence electrons to be shared to form octets on the nitrogens and a pair on the hydrogen):

The above three forms are nonequivalent because of the presence of the hydrogne.  Note that, if the H were removed to give the anion, the outer two forms would be equivalent!

For part (b), we have some fun with the sulfur.  We have 24 valence electrons since we have 4 group 6A elements.  If sulfur is restricted to obeying the octet rule here, then we get the following three equivalent forms (these are drawn by hand since the math word processor has a nervous breakdown with these!):

The top three forms obey the octet rule for S as advertized.  However, what about the one on the bottom?  S does have some vacant d-orbitals and can accomodate these extra double bonds.  Why draw this one?  Well, please note that the formal charges on all atoms are zero!  That is the reason!  In the three resonance forms, the formal charges on the S are all +2 with -1 charges on the singly-bonded oxygens.  I'll let you think about the implications of this!

With the SCN- anion (c), we can go back to the math processor and spare you the agony of my poor art work.  In this case, there are three structures which we now draw:

Here, the 16 valence electrons are accomodated quite well.  Looking at formal charges, probably the middle structure is the best.

Problem 7.62 (p. 294):
Assign formal charges to the atoms in the following resonance forms of ClO2-:

Just following the rules, getting the charges is easy.  Here are the answers:

In the structure on the left, each O has 7 electrons and 6 - 7 = -1 whereas the Cl has 6 electrons (net) and 7 - 6 = +1.  On the right, the rightmost oxygen "has title" to 6 electrons and 6 - 6 = 0; similarly the chlorine has 7 - 7 = 0 electrons and we are left with a charge of -1 for the leftmost oxygen.  Of the two structures, the one on the right is probably the better since the differences in the formal charges are minimized.  Also, the structure on the right has two equivalent resonance forms; this really "seals the deal"!

Problem 7.64 (p. 294):
Assign formal charges to the atoms in the following structures.  Which of the two do yo think is the more important contributor to the resonance hybrid?

The authors must have been smoking something illegal when they did this one!  The molecule in question has 16 valence electrons and (a) is certainly an acceptable structure since the octet rule is satisfied for the C atom and the two N atoms.  This is definitely not the case with structure (b).  This is not an acceptable Lewis structure since the carbon has only 6 electrons or, in other words, the octet rule is violated!  Parenthetically, we do note that structures such as this do enter into valence bond calculations.  However, their contributions are minor and only serve to slightly improve a final wave function.  In any event, (b) is certainly not the type of resonance structure we have been talking about in our discussions!
Problem 7.66 (p. 294):
What geometric arrangement of charge clouds do you expect for atoms that have the following number of charge clouds?

(a)  3        (b)  5        (c)  2        (d)  6

For these, I would immediately say "triangular," "trigonal bipyramidal," "linear," and "octahedral."  Of course, I have been at this for several hundred years!  The following picture suffices to give all shapes for charge clouds vs. bonds in scorching detail:


Please remember the logic that I employ:

The above drawings pretty much tell the whole story up to SIX.  There are also structures for things surrounded by 7 or 8 objects--but we leave that to an advanced inorganic chemistry class!
Problem 7.68 (p. 294):
How many charge clouds are there around the central atom in molecules that have the following geometry?
 
(a) Tetrahedral (b) Octahedral
(c) Bent (d) Linear
(e) Square pyramidal (f ) Trigonal pyramidal

Here, the emphasis is on molecular geometry which stresses where the atoms are and basically ignores the charge clouds.  However, the latter must be taken into account if a full understanding of the chemistries of the various species is required.  In light of this, we shall give rather full answers for each of the above.  Refer also to the huge figure shown in the answer to the previous question (7.66).
 
(a)  With a tetrahedral molecular geometry, this can only be 4.  There is no way in which extra "objects" can be accomodated while keeping the geometry tetrahedral.
(b) Here, the number of objects surrounding the central atom must be 6.  (These can be either bonds or lone pairs.)  In any event, once you have 6 objects present--with an octahedral molecular geometry these would all be bonds--you are "stuck" with an octahedral geometry.
(c) Here, there are more possibilities.  A "bent" structure simply means that you have lone pairs along with the atoms bound.  For the cases shown in the figure of the previous section, you see that the number of clouds would be either 3 or 4.  Note that this assumes that the atoms would be arranged as "A-B-C" and that we are considering just three atoms here!
(d) A linear molecular structure means that 3 atoms are in a straight line.  For the cases cited here, there would be either 2 charge clouds (objects) or 5 charge clouds about the central atom.
(e) The only way a square pyramid can occur is with 6 charge clouds; 5 of these go to other atoms and the last to a lone pair.  This is shown in the table given with the previous problem.
(f ) A trigonal bipyramid requires 5 electrons clouds--and that is the only possibility.  (Note that the answer in the back of the book is wrong!  They said "6"!  Obviously, they were smoking or drinking something!)


Problem 7.72 (p. 294):

What shape do you expect for each of the following molecules or ions?

(a)  SbF5        (b)  IF4+        (c)  SeO32-        (d)  CrO42-

Hey, boys and girls, I expect great shapes!!!  (Now we get serious!)

(a) involves a 5A element and 5 atoms of a 7A element.  This involves 40 electrons.  The main idea here is that a central atom is surrounded by 5 atoms with 8 electrons each (for a total of 40).  Since all electrons are consumed here and we have 5 objects (electron clouds) surrounding the central Sb, this means that we have a trigonal bipyramidal structure.  If it helps, this is exactly the same as if we had had PF5.

(b) is much more fun.  Here we have 5 group 7A atoms with a net positive charge to give us a total of 7 x 5 - 1 = 34 electrons.  The 4 F's take up 32 electrons; this leaves us with one extra lone pair on the I.  Thus, we have a "see-saw" structure (cf. the table in problem 7.66).

(c) involves 4 type 6A elements for 24 electrons + 2 for the anion's charge.  Thus, we have 26 electrons here.  The three oxygens take 8 electrons each for 24 and this leaves one lone pair.  Thus, Se is surrounded by 4 electron clouds and we get a trigonal pyramid structure (much like that of ammonia).  We have the three oxygens at the base of a three-sided pyramid and a single lone pair sticking out of the top of this magnificant ediface.

(d) is a simple tetrahedron.  There are no other charge clouds involved other than those with the oxygen atoms.  Hence, a simple tetrahedral structure.  There is really nothing much we can say beyond this point since the oxygens form all 4 of the bonds in chromate.
 

Problem 7.76 (p. 295):
What bond angles do you expect fo reach of the following?

(a)  The F-S-F angle in SF2.
(b)  The H-N-N angle in N2H2.
(c)  The F-Kr-F angle in KrF4.
(d)  the Cl-N-O angle in NOCl.

Again, just a little thinking is all that is necessary.

(a) has 20 valence electrons.  Each F has 8 and, thus, the sulfur has two extra lone pairs.  Thus the bond angle here is about 109-degrees (tetrahedral).  The Lewis structure would look like this:

As you can see, this is a bent structure (not much different than that of water!).

(b) has 12 valence electrons.  If you draw the Lewis structure for this, here is what you get:

From this ISBOT the bond angle is 120-degrees, approximately, at least!

(c) is a rather obscure compound.  However, let us do a little thinking.  We see that we have 8 + (4 x 7) = 36 electrons.  The four F's take up 32 and this leaves 4 more electrons which obviously fall into two lone pairs.  Hence, KrF4 has a square-planar configuration and the bond angle is thus 90o.

(d) ClNO has 18 (7 + 5 + 6) valence electrons.  We now show what should be a reasonable Lewis dot structure for this molecule:

Since there are three "objects" (electron clouds) around the central N atom, the bond angle should be of the order of 120o.
 

Problem 7.84 (p. 295):
What hybridization do you expect for atoms that have the following numbers of charge clouds?

(a)  2        (b)  5        (c)  6        (d)  4

This is most easily discussed by just looking at the following table (again, pour moi, this is a very easy piece of cake!):

Thus we can answer immediately:

(a)  2 means sp, bagel breath!
(b)  5 means sp3d, pizza face!
(c)  6 can be nothing other than sp3d2 (or your mother drives a beer truck)!
(d)  Finally, 4 means "tetrahedral!"  In turn, this means sp3!  (If you can't see this, may a camel sit on your face!)
 

Problem 7.88 (p. 295):
What hybridization would you expect for the indicated atom in each of the following?

(a)  H2C=O          (b)  BH4-          (c)  XeOF4          (d)  SO3

Fairly straightforward!

(a):  The carbon atom has three bonds.  This means sp2 with C automatically!

(b):  The structure here is tetrahedral.  Obviously, sp3.

(c):  Here, you might want to draw a Lewis structure first.  If you do, you first note that the molecule has 8 + 6 + (4 x 7) = 42 valence electrons.  If you let the F and O atoms surround the Xe, you see that these take up 40 electrons to give a square pyramid structure.  There are two electrons left over, however, and this means that there are 6 electron clouds surrounding the Xe or in other words, the hybridization is sp3d2.

(d):  Finally, we get to sulfur trioxide.  In this case, if the S obeys the octet rule (cf. problem 7.50b above), we have three charge clouds around the central atom and the hybridization is sp2.


Problem 7.92 (p. 296):

Use the MO diagram in Figure 7.18b to describe the bonding on O2+, O2, and O2-.  Which of the three should be stable?  What is the bond order of each?  Which contain unpaired electrons?

Before we really get going, here is the diagram in question:

This probably looks like a mess but it does inform us that the order of filling of MO's changes as we go from nitrogen to oxygen (and fluorine).  To make this point really dramatic, we put in the electrons into the O2 diagram for the three species mentioned in this problem.

First, we look at O2+.  This has (6 x 2) - 1 = 11 electrons.  We can load these electrons into the bonding and antibonding MO's now (in the order of energy and following Hund's rule just as we did with atoms in an earlier chapter).  To do this I "stole" the figure from the book and put my own arrows on it to give the following presentation for O2+:

(This is probably not as pretty as it should be, but it does work!)  This is a stable species (although it would react rapidy in air to form even more stable species).  How do we know this?  The answer is in the bond order.  This is simply

(This may be overkill in that the s2sand s*2s electrons were included here but--what the heck--let's be sure all the valence electrons are counted!)  This species has one unpaired electron and, hence, is paramanetic.

The next species, O2, is certainly stable!  (After all, you are breathing it now and it is about 20% of the earth's atmosphere.)  In any event, it has two unpaired electrons and is quite paramagnetic (in fact, liquid oxygen can be held between the poles of a magnet).  In any event, here is its electronic structure.  This molecule has (6 x 2) = 12 valence electrons and we now see them in all their glory:

Following what we did above, we see that the bond order is just

This is a bond order of two--but can we call this a double bond?  The Lewis structure,

is certainly not correct!  Rather, it should be noted that oxygen gas is a free radical with a structure something like this:

(The dot at the far right is the period at the end of the sentence!)

This particular prediction is a case where MO theory does better than VB theory at the elementary level.  (Remember that both do converge to the same result if carried far enough!)  Please note that this is a very stable free radical.  Reactions such as combustion need a "kick" before they can start (otherwise, wood houses would not be a good idea!).

The final species is also stable (by itself--if other things are present, it might also combine).  We show the MO diagram for the O2- anion.  This anion has a name:  "superoxide."  Some superoxides actually exist (such as KO2).  The anion is stable as long as there is nothing nearby it likes to react with.  It is easy to see that the bond order of this species is 1.5 ( = [8 - 5]/2) and this can be easily deduced from the following diagram:

This has one unpaired electron and is definitely paramagnetic.

We did not carry this analysis all the way to the peroxide anion (O22-).  However, it should be easy to see that this ion is diamagnetic (with no unpaired electrons) and has a bond order of 1 (one).  Peroxides, being quite commonly observed compounds, are obviously also stable.  (But not stable compared to O2--they eventually decompose to give that!)

What have we shown here?

Covalent bonds are
pHun!!!