Covalent Molecules

I. The Covalent Bond

A. Electrons shared between 2 or more nonmetals

B. Using 2 H atoms as the example:  as two atoms approach each other at a certain distance they begin to “feel” each other.  The proton(s) in nucleus of the first atom attract the electrons of the second atoms and vice versa.  This attraction lowers the energy of the 2 separate atoms.  Eventually an optimum distance between the atoms is reached where the energy is lowest = bond length.  If the atoms get closer, the protons in the nuclei repel each other – they are getting too close – you can’t smush two nuclei together.  Thus the energy sky rockets.

C. You should understand this graph.  Watch the short video, click on Activities and then on H₂ Bond Movie.

II. Covalent Bond Strength

A. The H₂ molecule is lower in energy than 2 separate H atoms.  This is why hydrogen is diatomic.  Similar for the other diatomic elements:  N₂, O₂, F₂, Cl₂, Br₂, I₂.  IMPORTANT

B. Forming H₂ lowers energy, DE is -436 kJ (negative since the E went down).  Exothermic (heat exits) = energy is released.

C. Breaking apart H₂ raises energy, DE is +436 kJ (positive since E went up) Endothermic (heat enters) = heat is required

D. Bonds only form if the DE will be negative, otherwise there is no reason for the bond to form.  Remember the goal is always to lower energy.

E. lists various bond energies.  Notice multiple bonds have higher E than single bonds. 

III. Ionic vs Covalent Compounds

A. Ionic compounds

1. metals + nonmetals atoms (electrons transferred)

2. solids

3. high mp and bp

4. endless 3D pattern (crystal lattice structure)

5. whole sample is held by ionic bonds – can’t really tell which Na goes with which Cl in salt for example. The ions just continue in all 3 directions held by only ionic bonds throughout the sample.

6. Because all the ions are held by strong ionic bonds it is hard to melt and boil ionic solids.

B. Covalent Compounds

1. 2 or more nonmetals sharing electrons

2. gases, liquids and some solids

3. low mp and bp usually

4. Can see distinct molecules – not endless like ionic solids.  Can see distinct water molecules for example – can always tell which H and which O go together. 

5. Only within one molecule do the covalent bonds exist holding the atoms together

6. Between molecules forces hold them together, not covalent bonds.  Why one water molecule sticks with another water molecule is NOT covalent bonds, but a force. 

7. These forces are much, much weaker than covalent bonds, which is why covalent compounds are much easier in general to melt and boil.  IMPORTANT

C. Nice movie – click on Ionic vs Covalent Bonding

IV. Electronegativity (EN)

A. The ability of an atom to pull the shared electrons in a bond closer.

B. In general EN increases g  and h except the noble gases. Why?  Well because the Halogens want to have electrons near them to get their octet.  So they pull them really close.  The metals want to lose an electron so don’t pull them close.  What about the Noble gases?  Do they have a low or high EN?  Trick question – they don’t even have EN since they don’t bond.  Remember the definition of EN – pulling electrons close in a bond.

C. Polar = electrons shared unequally between atoms with different EN so that the more EN atom pulls electrons closer and is thus a little bit negative d- since the negative electrons are closer.  The less EN atom is deficient in electrons so is a little bit positive d +.  It is like a tug-o-war between the two atoms with the more EN atom winning. A polar bond is kinda like a little magnet. It has “poles” like our planet has North and South poles.  Remember poles is polar. IMPORTANT  

D. Nonpolar = electrons shared equally between atoms with the same EN so that neither atom can pull them closer (tug-o-war tied).  So no “poles”

E. Compare H₂ with HCl.  H₂ is H : H .  The two electrons are exactly between the two H’s.  Equally shared.  Compare to  H :Cl  where the shared electrons are much closer to the more EN Cl atom.  Thus we can write ^d+H     :Cl^d- indicating which has the partial negative charge and which has the partial positive charge.  You can also draw an arrow pointing from the H to the Cl indicating the Cl is more EN.

F. Imagine a continuum: 100% ionic ------------polar-------------100% nonpolar. NaCl which is really Na⁺Cl^- is near the ionic end, HCl is in the middle in the polar area, and H₂ is at the nonpolar end.

G. Pretty much all covalent bonds are polar to some degree unless an atom bonds with itself (tug-o-war is tied) and the C-H bond is nonpolar since their EN values are so close.

H. Polar, ionic, or nonpolar bonds?  O=O,     C-Cl,     N-H,     C-H,    Br-Br,     P-F, S-F     IMPORTANT

I. Answers:  All are polar except O=O, C-H, Br-Br

J.

V. Lewis Dot Structures

A. Used for covalent compounds.  Practice makes perfect.  Here are some general guidelines:   IMPORTANT

1. Add up the total number of valence electrons for all the atoms in the molecule

2. The least EN atoms usually in the center

3. Arrange the electrons so that everyone has 8, except H has 2 of course, and start with single bonds first.  Only if it can’t work try double then triple bonds. Note than only transition metals can make quadruple bonds.

4. Remember that the electrons shared between two atoms “count” for both atoms when reaching their octet.

5. Check and make sure your drawing has the same number of electrons as in step one.  You can’t create or destroy electrons.

B. We call the shared electrons bonded electrons and one line – represents two electrons. A double bond = would be 4 electrons. A triple bond = would be 6 electrons. The unshared electrons are almost always paired up  :  and we call them lone pairs and always draw them as dots.  Quadruple bonds exist, but only with transition metals.

C. Try drawing Lewis dot structures for NH₃, CO, HOCl, I₂

D. Answers:  NH₃ is ammonia and has N in the middle with 3 single bonds to the H’s and one lone pair on N.  8 electrons total.   CO is carbon monoxide and has C triple bonded to O, each has a lone pair.  10 electrons total.    HOCl has H single bonded to O single bonded to Cl.  O has 2 lone pairs and Cl has 3 lone pairs.  14 electrons total.  Finally I₂ has I single bonded to I and each has 3 lone pairs.  14 electrons total.

E. Cheat Sheet:  IMPORTANT

1. Note that C almost always has 4 bonds (4 singles, 1 double + 2 singles, triple + single, or 2 doubles).  This is because C atom has 4 valence electrons (2s²2p²) and thus needs 4 more to make the octet.  So it needs to share 4 more = 4 bonds.  Carbon rarely has a triple bond with 1 lone pair. Atoms in C column are similar.

2. Note that N almost always has 3 bonds and 1 lone pair (3 singles, 1 double + 1 single, triple).  This is because N atom has 5 valence electrons (2s²2p³) and thus needs 3 more to make the octet.  So it needs to share 3 more = 3 bonds.  Atoms in N column are similar.

3. Note that O almost always has 2 bonds and 2 lone pairs (2 singles or 1 double).  This is because O atom has 6 valence electrons (2s²2p⁴) and thus needs 2 more to make the octet.  So it needs to share 2 more = 2 bonds.  Note that O with a triple bond and 1 lone pair can also exist, but is rarer. Atoms in Chalcogen column are similar.

4. H can only have one single bond and no lone pairs.  This is because it only has the first shell to fill 1s² and it is done. For this reason H is NEVER in the middle of a structure.

5. F is always one single bond with 3 lone pairs. It has 7 valence electrons (2s²2p⁵) so only needs one more = 1 bond. The other halogens are usually this way also unless they are bonded to something more EN than they are – like if they bond to F.

VI. Exceptions

A. B often makes only 3 bonds since it starts with just 3 electrons to share (2s²2p¹).  For example BH₃ is just B in the middle with 3 single bonds to the H’s.  No lone pair as we only have 6 electrons total.  Now this molecule is not very stable since it doesn’t have an octet.

B. Atoms in rows 3, 4, 5, 6, 7, can all have more than an octet.  Why???  With the s and p subshells full that is 8 electrons?  Where could more go???  Well remember starting with the 3^rd shell there is d subshells, so electrons can go into them.  The 3^rd shell can hold up to 18 electrons.  So these atoms can have 10, 12 or even 14 electrons in them.  Now 8 is still the best, but we can “expand the octet” if there is absolutely no other option. IMPORTANT

C. Look at example 7.2, 7.3, 7.4, 7.5, 7.6, 7.7, 7.8 and try problems 7.6, 7.7, 7.8, 7.9, 7.10, 7.11

D. Practice here, click on Activities then on Electron dot structures Activities.  Lots of practice here.

VII. Resonance

A. Draw the Lewis structure for ozone, O₃.  (16 e total) You should get an O with two lone pairs doubly bonded to the center O with 1 lone pair singly bonded to an O with 3 lone pairs.  Without showing the lone pairs you can draw either O=O-O   OR   O-O=O.  There are 2 pictures – this is called resonance.  They are resonance structures.  Both contribute to the real existence of ozone.  Actualy data tells us that the bonds are not really single and double, but they are same, kinda like 1.5 each.  Reality is an average of the resonance structures.

B. Here is a tutorial.  Scroll to Resonance.

C.

IX. VSEPR (Valence Shell Electron Pair Repulsion Theory)

A. Lewis dot structures are 2D.  Reality is 3D.  We need to know the actual shapes of molecules. 

B. All electrons take up space and repel each other since they are negative.  So electrons arrange themselves around the center atom as far apart as possible.  Awesome short movie – click on Activities then on VSEPR movie.  Another great movie with lots of practice problems. Try them first, then watch the answers.

C. To get the shape (also called molecular geometry) you must count up all the attachments (electron densities or clouds) around the central atom.  Note than atoms and lone pairs are attachments, not bonds.  An atom with a single bond, double or triple is still one attachment. The number of attachments determines the arrangement around the central atom. Note you have to get the correct Lewis dot structure in order to get the correct shape!!!

D. Here is the summary table. Go over it really well.  IMPORTANT

E. Examples:  for each draw the Lewis dog structure, give the ABE pattern, give the shape, give the bond angle(s) and finally give the electron geometry (when you include the lone pairs)  O₃, BH₃, SeCl₄, CH₂Cl₂, SeF₅^-.  IMPORTANT

1. Ozone is already described above.  Pattern is AB₂E.  Shape is bent. Angles are 120, e geometry is trigonal planar.

2. BH₃ is already described above.  Pattern is AB₃.  Shape is trigonal planar.  Angles are 120, e geometry is trigonal planar.

3. SeCl₄ has Se in the middle with 1 lone pair and with 4 single bonds to Cl’s which have 3 lone pairs each. (34 e total)  Pattern is AB₄E.  Shape is see saw.  Angles are 90 and 120, e geometry is trigonal bipyramid.

4. CH₂Cl₂ (20 e total) has C in the center with 2 single bonds to H’s and 2 single bonds to the Cl’s which have 3 lone pairs each. Pattern is AB₄.  Shape is tetrahedral.  Angles are 109.5, e geometry is tetrahedral.

5. SeF₅^-.  (42 e total)  Se in the middle with one lone pair and 5 single bonds to the F’s which have 3 lone pairs each. Pattern is AB₅E.  Shape is square pyramic.  Angles are 90, e geometry is octahedral.

G. Polar bonds vs polar molecules (covered more in section 10.1)

1. Animation tutorial

2. If a molecule has polar bonds, it may or may not be polar overall.  If the dipoles cancel out (the arrows cancel out) then the molecule overall is nonpolar.

3. CF₄ for example, a tetrahedral molecule has 4 very polar C-F bonds.  However they all cancel out so the overall molecule is nonpolar.

4. H₂O has two polar bonds and is also tetrahedral in shape, but the bonds don’t cancel out so the overall molecule is polar.

X. Valence Bond Theory

A. Sharing electrons requires the orbitals to overlap.  Usually two atomic orbitals with 1 electron each overlap (covalent bond) so that there are now 2 electrons being shared between the two orbitals. 

B. The orbitals are not really atomic orbitals anymore as we will see next.

XI. Hybridization of C

A. Check out this short animation. IMPORTANT

B. I highly recommend this tutorial.  Scroll to the hybridizations tutorial and click on the icon on the right side.

C. One more movie, click on Activities then hybridization movie.

D. So Carbon which is 2s²2p² can take these 4 atomic orbitals and rearrange them to get 4 hybrid orbitals called sp³ for where they came from.  The resulting 4 hybrid orbitals are equal and have one electron each so we can make 4 equal single bonds = tetrahedral.  Methane CH₄ is an example with sp³ hybridization.

XII. More Hybridization

A. What about when C makes a double and 2 single bonds?  Carbon which is 2s²2p² can take these 4 atomic orbitals and rearrange them to get 3 hybrid orbitals called sp² for where they came from and one untouched p orbital.  The resulting 3 hybrid orbitals are equal and have one electron each so we can make 3 equal single bonds = trigonal planar.  Formaldehyde CH₂O is an example with sp² hybridization. What about the untouched p orbital – it has one electron in it also and makes the double bond! 

B. What about when C makes 2 double bonds or a triple and single???  Carbon which is 2s²2p² can take these 4 atomic orbitals and rearrange them to get 2 hybrid orbitals called sp for where they came from and two untouched p orbitals.  The resulting 2 hybrid orbitals are equal and have one electron each so we can make 2 equal single bonds = linear.  Carbon dioxide CO₂ is an example with sp hybridization. What about the untouched p orbitals – they have one electron each and make the 2double bonds (or in another molecule could make a triple bond)! 

C. What about 5 bonds in trigonal bipyramids (not C of course)  Well we need 5 hybrid orbitals – where do they come from?  There are only 4 orbitals in the s and p subshells???  Hum!  Oh yeah – we have d orbitals for atoms in rows 3 – 7.  Which is why only atoms in these rows can make trigonal bipyramids in the first place!  When we have 5 attachments we have sp³d hybridization. 

D. What about 6 bonds in octahedrals (not C of course)  Well we need 6 hybrid orbitals. When we have 6 attachments we have sp³d² hybridization. Remember only atoms in row 3-7 can make octahedrals. See Figure 7.13

F. Assign hybridization to all our examples so far!  IMPORTANT

1. NH₃ the N is sp³,  CO the C and O are sp, HOCl the O is sp³, I₂ the I’s are sp³

2. O₃ the center O is sp², BH₃ the B is sp², SeCl₄ the Se is sp³d, CH₂Cl₂ the C is sp³, SeF₅^- the Se is sp³d². 

G. Look at example 7.12 and try problems 7.21 to 7.25

H. Wondering about those untouched p orbitals in sp and sp²? How do double and triple bonds form?  Well watch this, especially if you are going to take organic chemistry.

Study hard. This is an important chapter.