Hydrogen Bonding

When a hydrogen atom forms a covalent bond with a highly electronegative atom (such as oxygen or nitrogen) in a molecule, a polar bond is formed between them.

The highly electronegative atom draws the shared electron pair to itself most of the time and thus attains a partially negative charge.

Consequently, the hydrogen atom attains a partially positive charge, being virtually electron deficient most of the time.

As shown from the figure above, the electron cloud is asymetrically located between the two bonding atoms. The arrow simply indicates the net dipole moment of the molecule and the direction of increasing electronegativity.

This frequent uneven sharing of electrons between the two atoms imparts a polarity to the molecule containing these bonded atoms.

The resulting polarity creates a dipole-dipole attraction between molecules called hydrogen bonding.

Hydrogen bonding accounts for the following phenomena:

• associated compounds have higher boiling point than non-associated compounds with nearly the same molecular weights;


• greater solubility of associated compounds in water;


• variations in the boiling points and solubility in water of the isomers of certain phenolic compounds.

Hydrogen bonding can occur between:

• the same molecules;


• different molecules (intermolecular hydrogen bonding);


• within the molecule (intramolecular hydrogen bonding).

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Intermolecular Hydrogen Bonding

In the following figures below, the conceptualized intermolecular hydrogen bonding of several organic molecules with themselves and with water molecules is shown.







The hydrogen bonding of the ethyl alcohol molecules above explains its higher boiling point than those of propane's and dimethyl ether's.







The wide difference in the boiling points of the primary, secondary and tertiary amines can be attibuted to hydrogen bonding. Primary amines, like methyl amine molecules shown in hydrogen bonding above, and secondary amines form hydrogen bonds while tertiary amines can not do so due to lack of hydrogen atom bonded to nitrogen atom.







Phenol has higher boiling point than toluene because of hydrogen bonding of phenol molecules as illustrated above.







The association of ethyl alcohol molecules (above) and phenol molecules (below) with water molecules through hydrogen bonding explains their relatively greater solubility in water as compared to those compounds with nearly the same molecular weights.

Intramolecular Hydrogen Bonding

The isomers of some phenolic compounds exhibit great difference in their boiling points and water solubility because of intramolecular hydrogen bonding (that is, association within the molecules) that occurs in the ortho isomers. This intramolecular hydrogen bonding reduces the chances of intermolecular hydrogen bonding of the said isomers resulting to their relatively lower boiling points and lower water solubility.

Lewis Structures of Some Common Compounds and Ions

It's been quite a while since my last post....

...and there's so much catching up to do!

And so without further ado, Chemistry Partner resumes to make your life in chemistry a little bit easier.

This current update serves as a supplement to my October 23, 2009 post, Drawing Lewis Electron Dot Structure or Formula.

It consists of illustrations of Lewis structures showing the covalent bonding of the following:

1. Inorganic Compounds




• NH₃ (ammonia)


• HNO₃ (nitric acid)


• SO₃ (sulfur trioxide)


• H₂S (hydrogen sulfide)


• N₂ (nitrogen gas)


• O₃ (ozone)


• H₂SO₄ (sulfuric acid)


• H₂O (water)







2. Organic Compounds




• HNCO (isocyanic acid)


• CO (carbon monoxide)


• HCN (hydrogen cyanide)


• CH₃COOH (acetic acid)


• H₂CO₃ (hydrogen carbonate or carbonic acid)







3. Polyatomic Ions




• SO₄^2- (sulfate ion)


• NH₄⁺ (ammonium ion)


• PO₄^3- (phosphate ion)


• C₂O₄^2- (oxalate ion)


• NO₃^- (nitrate ion)


• CO₄^2- (carbonate ion)

In order to identify each atom's respective valence electrons easily, different colors for atoms were used.

Also, red and grey shades were used to highlight the atoms with negative and positive formal charges.

Lewis Structures of Inorganic Compounds

Where necessary, I used different colors for the same kind of atom. Here in the example of O₃ (ozone) below, I used three different colors to distinguish the three different oxygen atoms.

For the structures of HNO₃ (nitric acid) and SO₃ (sulfur trioxide) above, the use of one color for the same kind of atom is enough to distinguish easily the individual atom's valence electrons.

For the rest of the illustrations below, you'll find the same methods of color assignment for the atoms.

When more than one Lewis structures for a species are possible, the resonance structures are given as in the case of ozone, isocyanic acid, nitrate and carbonate ions.


For determining the most stable resonance structure(s) of a species, click here to see my earlier post about it.

Lewis Structures of Organic Compounds

With the exception of hydrogen, you can see that an atom can share one or more of its valence electrons, with a maximum of six electrons shared between any two atoms.

Lewis Structures of Ions

Here, the charge on a species is determined by either of the following:

• the sum of all of the valence electrons of the atoms of the species minus the total number of electrons in the Lewis electron-dot structure of the species


• the algebraic sum of all the formal charges of the species

For a discussion of the determination of the formal charges of atoms of a species, follow the link given above.









For some ions, just as the case is with the ammonium ion, the total number of electrons in its electron-dot structure is less than the total number of valence electrons of its atoms.

Referring to the ammonium ion above, it should have a total of 9 valence electrons (5 from N + 4x1 from H), but its electron-dot structure has only eight electrons.

Ammonia acts as Brönsted base, though a weak one, by accepting a proton thus forming ammonium ion. (See my article Solving Weak Acid/Base Dissociation Problems for a discussion of Brönsted acid-base reaction.)

And for some ions, the total number of electrons in its electron-dot structure is more than the total number of valence electrons of its atoms.

Observe that some oxygen atoms of the phosphate, oxalate, nitrate and carbonate ions have acquired electrons from atoms other than those its own atoms.

Solving Weak Acid/Base Dissociation Problems

Dissociation problems are about calculations of hydronium ion concentration, [H₃O⁺], and/or pH of aqueous solutions containing any of the following:

a weak acid or a weak base

a strong acid or a strong base

a salt of a weak acid or a salt of a weak base

a weak acid and its salt (weak acid/conjugate base composition) or a weak base and its salt (weak base/conjugate acid composition)

Weak acid and weak base dissociate (or ionize) partially in aqueous solution. Their dissociation products combine to form again the initial weak acid and base.These two reactions occur simultaneously and form an equilibrium. This is in contrast to the strong acids and bases that undergo complete dissociation or ionization in aqueous solution.

Basically, these reactions are about giving up and gaining protons. Those that donate protons are called Bronsted-Lowry acids; and those that accept protons are called Bronsted-Lowry bases.

Their dissociation equilibria given below show that a weak acid loses a proton and forms conjugate base while a weak base gains a proton and forms conjugate acid.

The extent of dissociation (or ionization) is determined by their so called dissociation constants ( K_a for acid and K_b for base ), measured at 25° C.

The dissociation constant is a measure of the tendency of a weak acid to give up a proton and tendency of a weak base to gain a proton. A very small dissociation constant value indicates a very low degree of dissociation.

Values in [ ] denote concentration in moles/liter.

Since the [H₂O] in each case is considered constant, it is incorporated into K_a and K_b.

Take note here that water acts as a weak base in weak acid dissociation and acts as a weak acid in weak base dissociation. A solvent that can act either as an acid or as a base depending upon the solute is called amphiprotic ( or amphoteric ) solvent.

The following discussion is about the calculations of hydronium ion concentration of the aqueous solutions containing (a) a weak acid or a weak base, (b) a strong acid or a strong base, (c) a salt of a weak acid or a salt of a weak base, and (d) a weak acid and its salt or a weak base and its salt.

A. Aqueous Solution of a Weak Acid or a Weak Base

The [H₃O⁺] and/or pH of this solution can be calculated using equations 1 and 2.

Since the amount (or concentration) of dissociated acid or base is equal to the amount (or concentration) of dissociation products:

[dissociated acid] = [H₃O⁺] = [A^-] = x, and

[ionized base] = [BH⁺] = [OH^-] = x, then

we can set equations 1 and 2 to one unknown:

Equations 3 and 4 have the general form of ax² + bx + c = 0 which can be solved by using the quadratic equation:

If the dissociation constant value is very low such that x is negligible, the following approximation can be made:



If the dissociation constant value or % error is high, equations 3 and 4 should then be used for calculation.

B. Aqueous Solution of a Strong Acid or a Strong Base

At 25° C, water has the following dissociation equilibrium:

Since [H₂O] is considered constant, it is incorporated into K_w.

K_w is called the ion-product constant of water at 25° C. This means that pure water has this much amount dissociated at this temperature, that is, [H₃O⁺] = [OH^-] = 1 x 10^-7 M. The [H₃O⁺] or [OH^-] of an aqueous solution can be calculated when one of them is known.

Addition of a strong acid or base in water will shift the equilibrium to the left, as predicted by Le Chatelier principle, lowering the [H₃O⁺] and [OH^-] by depressing the dissociation of water.

Since a strong acid or base dissociates completely in water, then:

[acid] = [H₃O⁺]

[base] = [OH^-]

So,

Setting equation 7 to one unknown:

[dissociated water] = [H₃O⁺] = [OH^-] = x

1 x 10^-14 = ( [acid or base] + x )( x )

which can be solved by using quadratic equation.

But x is very small, we can assume that:

The sample problem below shows the validity of our assumption that x is negligible.

C. Aqueous Solution of a Salt of a Weak Acid or a Salt of a Weak Base

Salt of a weak acid dissociates completely in water as Na⁺ and A^-.

A^-, as a conjugate base of HA, reacts with water to form the following equilibrium:

Multiplying K_a (equation 1) by K_b, we get:



Equation 8 indicates the relationship that the strength of a weak acid or base tends to increase as the strength of its conjugate pair decreases, or vice versa.

This equation enables us to compute the dissociation constant value of the conjugate base of a weak acid and the conjugate acid of a weak base, as well as the [H₃O⁺] of this solution.

D. Aqueous Solution of a Weak Acid And Its Salt or a Weak Base And Its Salt

An aqueous solution containing a weak acid and its salt (or a weak base and its salt) is called a buffered solution or simply buffer solution. This is so because a buffer solution's hydronium ion concentration changes very little upon its dilution or upon addition of a strong acid/base.
The changes in pH value are so small such that the pH of the solution remains practically constant.

Calculation of the buffer solution's pH involves the use of equation 1 or 2:

By taking the negative logarithm of each term of the rearranged equation 1, we get equation 9 which is known as the Henderson-Hasselbach equation.



This equation is commonly used for pH calculation in biochemistry problems.

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Problems

The best way to improve your problem solving skill in chemistry is to practice solving as many chemistry problems as possible. Try solving the following problems.

Problems are solved using approximation and the quadratic equation. Values in parenthesis are obtained using the quadratic equation.

For more problems on this topic and other chemistry topics, go to www.tutorpartner.blogspot.com.

1.	Calculate the dissociation constant of the conjugate base of HC2H2ClO2. Ka = 1.36 x 10-3.
	Answer: 7.35 x 10-12
2.	A one liter aqueous solution contains 9.54 g KOH. What is the pH of this solution? The dissociation constant for water, Kw, is 1 x 10-14. ( KOH = 56.1056 g )
	Answer: pH = 13.23
3.	An aqueous solution has a strength of 0.05 M HOCN and 0.07 M NaOCN. Calculate the pH of this solution. ( Ka = 3.3 x 10-4)
	Answer: pH = 3.63 ( 3.63 )
4.	Calculate the pH of a 0.3 M NaC6H5O solution. (Ka for phenol acid is 1.05 x 10-10; Kw = 1 x 10-14)
	Answer: pH = 11.73 ( 11.72 )
5.	110 mL of 0.01 M NaOH solution is added to 110 mL of 0.11 M HNO2 solution. Find the pH of the resulting solution. Given: Ka for nitrous acid is 4.5 x 10-4.
	Answer: pH = 2.35 ( 2.56)
6.	A one liter aqueous solution contains 9.2 g NaOH. What is the pH of this solution? The dissociation constant for water, Kw, is 1 x 10-14. ( NaOH = 39.99707 g )
	Answer: pH = 13.36
7.	What is the pH of an aqueous solution having a concentration of 0.07 M HOCN and 0.02 M NaOCN? ( Ka = 3.3 x 10-4)
	Answer: pH = 2.94 ( 2.97 )
8.	If a solution of HOCN has a concentration of 0.9 M, what is its pH? Ka for cyanic acid is 3.3 x 10-4.
	Answer: pH = 1.76 ( 1.77 )
9.	Find the pH of a 0.86 M NaC6H5COO solution. (Ka for benzoic acid is 6.3 x 10-5; Kw = 1 x 10-14)
	Answer: pH = 9.07 ( 9.07 )
10.	An aqueous solution of 0.12 M CH3COOH and 0.12 M NaCH3COO has a volume of 340 mL. If 1.6 g of NaOH is added to this solution, find the change in pH. Assume no change in the volume of solution. ( Ka = 1.8 x 10-5; NaOH = 39.99707 g )
	Answer: pH = 6.75 ( 6.75)
11.	What is the pH of a solution of 0.28 M NaCH3COO? Calculate the percent hydrolysis. (Ka = 1.8 x 10-5; Kw = 1 x 10-14)
	Answer: pH = 9.1; 4.45 x 10-3 % ( 9.1; 4.45 x 10-3 % )
12.	What is the pH of an aqueous solution of 0.09 M C6H5NH3Cl? ( Given: Kb = 3.94 x 10-10; Kw = 1 x 10-14)
	Answer: pH = 2.82 ( 2.82 )
13.	If 3 L of 0.01 M HOC2H4NH2 solution has a pH of 10.37, find the weight of HOC2H4NH3Cl dissolved in the solution. ( Kb for ethanolamine is 3.18 x 10-5; mol. wt. = 97.5443 )
	Answer: 0.4 g
14.	In a buffer solution of 0.07 M HC2H2ClO2 and 0.07 M NaC2H2ClO2, HCl is added to it such that its calculated concentration in the solution is 0.02 M. Assuming a constant volume of solution, find the change in pH of the solution. ( Ka = 1.36 x 10-3)
	Answer: pH = 2.61 ( 2.64 )
15.	Find the [H+] and [OH-] of a 0.24 M of HF solution. Ka = 6.7 x 10-4; Kw = 1 x 10-14.
	Answer: [H+] = 1.27 x 10-2 M; [OH-] = 7.89 x 10-13 M ( 1.24 x 10-2; 8.1 x 10-13 )
16.	What is the dissociation constant of the conjugate acid of HOC2H4NH2. Kb = 3.18 x 10-5.
	Answer: 3.14 x 10-10
17.	What is the pH of a 0.38 M CH3COOH solution. ( Ka = 1.8 x 10-5 )
	Answer: pH = 2.58; ( 2.58)
18.	What is the pH of an aqueous solution of NH2C2H4NH2 having a concentration of 0.01 M? The dissociation constant of ethylenediamine is 8.5 x 10-5.
	Answer: pH = 10.96 ( 10.94)

Drawing Lewis Electron Dot Structure or Formula

A Lewis electron dot structure shows how the atoms of a molecule or an ion share their outermost electrons or valence electrons to form covalent bonds with each other.

It indicates the number of shared and unshared valence electrons.

This sharing of electrons results in their outermost shells being filled up thus attaining electronic configuration similar to those of the noble gases'.

Observe the electronic configurations of the noble gases below.

He	1s2
Ne	1s22s22p6
Ar	1s22s22p63s23p6
Kr	1s22s22p63s23p64s23d104p6
Xe	1s22s22p63s23p64s23d104p65s24d105p6
Rn	1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p6

You can see that their outermost shells are filled up which account for their being highly stable.

Now, compare them to the electronic configurations of several atoms that commonly form covalent bonds with other atoms.

H	1s1
C	1s22s22p2
O	1s22s22p4
N	1s22s22p3
S	1s22s22p63s23p4
B	1s22s22p1
P	1s22s22p63s23p3
Si	1s22s22p63s23p2
F	1s22s22p5

With the exception of hydrogen, they share their valence electrons with other atoms to fill up their outermost shells which contain a maximum of eight electrons. Hence, a Lewis electron dot formula must follow the octet (eight) rule.

Take a look at the following examples.

For our purposes, we're going to use color codes in our illustrations to distinguish the individual atoms with their respective valence electrons.

Lewis Electron Dot Structure of Ethane, C₂H₆

The electronic configuration of carbon atom shows that it needs four more electrons to fill up its outer shell; the hydrogen atom, on the other hand, needs just one more electron for its lone electron shell.

Each carbon atom satisfies the octet rule by sharing 3 valence electrons with 3 hydrogen atoms and one with another carbon atom. The hydrogen atom fills up its shell by sharing one electron with carbon atom.

Take note that each pair of shared electrons constitue a single bond.

Lewis Electron Dot Structure of Ethylene (Ethene), C₂H₄

This time, the carbon atoms share 2 electrons with each other and 2 electrons with 2 hydrogen atoms.

Sharing of two pairs of electrons constitute double bond.

Lewis Electron Dot Structure of Acetylene (Ethyne), C₂H₂

Here the carbon atoms share 3 valence electrons with each other and 1 electron with 1 hydrogen atom.

Sharing of three pairs of electrons constitute triple bond.

So far, we've been looking at the Lewis electron dot structures of molecules. We're going to look at the electron dot structures of some ions.

Lewis Electron Dot Structure of Hydroxide Ion

From the electronic configurations given above, we can see that oxygen atom has 6 valence electrons; it needs 2 more electrons to fill up its outer electron shell.

In order to follow the octet rule, oxygen forms the hydroxide ion with the hydrogen atom by sharing one electron with one hydrogen atom and acquiring one more electron .

The gain of one electron results to the ion having a negative net charge of one.

The structure has one electron more than the total valence electrons of the ion (6 from O + 1 from H = 7 valence electrons).

Lewis Electron Dot Structure of Ammonium Ion

The nitrogen atom has 5 valence electrons as indicated by its electronic configuration; it needs three more electrons to fill up its last shell.

But the nitrogen atom in the ammonium ion satisfies the octet rule by sharing three of its valence electrons with each of the three hydrogen atoms.

The structure has one electron less than the total valence electrons of the ion (5 from N + 4 from 4 H = 9 valence electrons).

Determining The Net Charge of an Electron Dot Structure

From these examples, we can say that the:

positive net charge of an ion is equal to the number of electrons deficient in its total valence electrons;

negave net charge of an ion is equal to the number of electrons in excess of its total valence electrons.

The net charge of an ion is also determined by the algebraic sum of the formal charge of all the atoms of the ion.

The formal charge of an atom is the encircled + or - sign near the atom. It is calculated as follows:

formal charge of an atom = valence electrons of the atom - (number of unshared electrons of the atom + number of pairs of shared electrons by the atom)

Formal charge of hydroxide ion's atoms:

O = 6 - (6 + 1) = -1

H = 1 - (0 + 1) =   0

net charge of hydroxide ion = -1 + 0 = -1

Formal charge of ammonium ion's atoms:

N = 5 - (0 + 4) = +1

H = 1 - (0 + 1) =   0

H = 1 - (0 + 1) =   0

H = 1 - (0 + 1) =   0

H = 1 - (0 + 1) =   0

net charge of ammonium ion = +1 + (4 x 0) = +1

Steps To Be Followed In Drawing A Lewis Electron Dot Structure

To summarize, the following steps may be followed to draw the Lewis electron dot structure of a chemical species.

1. Draw all possible structures by using single, double or triple pairs of electrons for atom-to-atom bonds.

2. Make sure that the structures contain the total valence electrons of all the atoms of the species.

3. If the species has a net charge, subtract electrons from the structure as indicated by the + charge or add electrons to the structure as indicated by the - charge.

4. Compute the formal charge of the atoms.

5. The structure that satisfies the octet rule for all atoms must be the correct Lewis structure.

Following the above steps, we're going to draw the electron dot structures of oxygen molecule (O₂) and cyanide ion (CN^-)

Drawing the Lewis Structure of a Molecule

Here are the 3 possible Lewis structures of N₂ molecule.

Structures I and II are incorrect because they violate the octet rule:

structure I has nitrogen atoms each containing only 6 electrons;

structure II has nitrogen atoms each containing only 5 electrons.

Structure III is the correct Lewis electron dot structure of N₂ molecule.

Drawing the Lewis Structure of an Ion

For the structures of cyanide ion as illustrated below, we added one electron to the total valence electrons of 9 (4 from C + 5 from N = 9) since it has a net charge of -1.

We rule out structures I and II because their atoms are electron deficient.

Structure III is the correct Lewis electron dot structure of cyanide ion.

The Resonance Theory

Sometimes, the step by step procedure we used above does not suffice because there are chemical species that have more than one possible Lewis electron dot structures.

One such species is the benzene molecule.

The Lewis structures shown below are equivalent but neither of them represents the actual structure of benzene molecule.

(We used single and double bonds here for easier viewing of the illustrations.)

Instead, benzene is represented by this structure:



This representation of a chemical species by a structure which is intermediate of 2 or more equivalent Lewis structures is called resonance.

The above structure is called resonance hybrid and structures I and II above are called resonance structures.

It should be noted here that resonance structures retain the same atomic arrangement; the only difference among these resonance structures is the arrangement of their electrons.

The Resonance Rules

There are several resonance rules that are used to determine the stability of resonance structures.

The most stable structures are the most important ones.

According to the resonance rules, the most stable structure is the one which has:

1. the greatest number of covalent bonds;

2. the least number of formal charges;

3. the - sign on the more electronegative atom and the + sign on the more electropositive atom, if formal charges are present.

We will use these rules to help us determine the most important Lewis structures of compounds having several resonance structures.

Determining The Most Stable Resonance Structure

Given below are the possible electron dot structures of CO₂.



We strike off structures I, II and III because of their electron deficient atoms.

Our choices then boil down to structures IV - VI.

We can see that structures V and VI are equivalent; we can use any of the two to compare with structure IV.

Being equivalent in the number of covalent bonds, we compare their number of formal charge.

Structure IV is more stable than structure V or VI because it has no formal charge. Hence, structure IV is the Lewis electron dot structure of CO₂.

Let's have another example: an ion this time.

The possible structures of thiocyanate ion (SCN^-) are given below.

How did we determine that C atom is the central atom?

We take the atom with the lowest valence electrons (except H) as the central atom because it has the maximum number of unpaired electrons that can be used to form covalent bonds with other atoms.

We exclude structures I - III from our choices because they don't follow the octet rule.

Since the above 3 structures have all equal number of covalent bonds, we will evaluate their stability based on the number of formal charge.

Structures IV and V are equivalent and are more stable than structure VI because they have less number of formal charge.

We choose structure IV as the Lewis structure of thiocyanate ion because the - sign is on the more electronegative atom.

Conclusion

By using the procedure and rules given above, you are now ready to draw the Lewis electron dot formula of most compounds having covalent bonds. Try the problems below.

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Problems

The best way to improve your problem solving skill in chemistry is to practice solving as many chemistry problems as possible. Here are some common ions and molecules. Draw their Lewis electron dot structures.

For more problems on other chemistry topics, go to www.tutorpartner.blogspot.com.

1.	C2H5NH3+
2.	CH3COO-
3.	H2O2
4.	HSO3-
5.	HCO3-
6.	CH3CN
7.	C2O4-2
8.	HNO2
9.	HCOCl
10.	N2
11.	CH3CONH2
12.	SO3-2
13.	C4H6
14.	NO3-
15.	NO2-