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October 23, 2009

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.


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-

October 6, 2009

How to Derive a Gas Law From the Ideal Gas Law Equation

There are several equations that relate the properties of a gas. These relatioships of the pressure, the volume and the temperature for a fixed amount of a gas are given below.

The variables with subscript 1 indicate initial values; and those with subscript 2 indicate final values.






For one who is taking up chemistry for the first time, remembering all these equations is a little bit tricky.


Using the Ideal Gas Law equation, the above gas laws can be derived easily.


This way, you need to remember fewer equations.





The Ideal Gas Law Equation


PV = nRT



where:



  • P is the pressure in atmosphere

  • V is the volume in liters

  • n is the amount of gas in moles

  • R is the universal gas constant in L-atm/K-mol

  • T is the temperature in Kelvin


This law relates the four variables of an ideal gas (pressure, volume, amount, and temperature) such that the value of PV/nT is always 0.082057.

It means that a change in the value of any of the four variables will bring about corresponding changes in the values of the other three variables so that the value of R remains constant.


Mathematically, it is equivalent to:



Deriving a Gas Law Equation


As mentioned earlier, we are dealing here with a fixed amount of gas.

It means the amount of gas remains the same from the initial state to the final state. So,



The value of nR must be constant.


Since the values of P1V1/T1 and P2V2/T2 are both equal to nR, then

Thus, we have derived the Combined Gas Law.

Using the above method, we can easily derive the other gas laws.


Suppose a problem states that the temperature must remain constant.


Using the Ideal Gas Law equation for the initial values and final values,



The value of nRT must be constant. Hence,



This is the Boyle's Law.

Now, you can derive the rest of the gas laws using the method we used here.




Problems


Improve your problem solving skill in chemistry by solving as many chemistry problems as possible. Here are some problems dealing with gas laws. The answer is provided for each problem.

For more chemistry problems on gases and other chemistry topics, you can go to www.tutorpartner.blogspot.com.


Boyle's Law Problems




1.Calculate the pressure required to reduce the volume of 18.0 L of gas at a pressure of 0.333 atm to 6.0 L with the temperature remaining constant.
Answer: 0.999 atm
2.Under constant temperature, 7000 mL of a sample gas was allowed to expand to a final volume of 23000 mL. If the initial pressure of the gas was 1085 torr, what was the final pressure of the gas after its expansion?
Answer: 330 torr
3.A sample of a gas which occupies 0.93 L has a pressure of 3 atm. Determine the resulting volume of the gas if the pressure is changed to 9 atm and the temperature is maintained constant.
Answer: 0.31 L
4.Calculate the pressure required to reduce the volume of 3000 mL of gas at a pressure of 1260 torr to 1000 mL with the temperature remaining constant.
Answer: 3780 torr
5.Calculate the pressure required to reduce the volume of 24.0 L of gas at a pressure of 0.25 atm to 13.0 L with the temperature remaining constant.
Answer: 0.462 atm
6.At a pressure of 7 atm, a sample of gas has a volume of 8.634 L. If the pressure is reduced to 2.75 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
Answer: 21.977 L
7.Under constant temperature, 10000 mL of a sample gas was allowed to expand to a final volume of 17000 mL. If the initial pressure of the gas was 21489 torr, what was the final pressure of the gas after its expansion?
Answer: 12641 torr
8.At a pressure of 10 atm, a sample of gas has a volume of 0.497 L. If the pressure is reduced to 5 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
Answer: 0.994 L
9.A sample of a gas which occupies 2.8 L has a pressure of 1 atm. Determine the resulting volume of the gas if the pressure is changed to 8 atm and the temperature is maintained constant.
Answer: 0.35 L
10.Under constant temperature, 5.0 L of a sample gas was allowed to expand to a final volume of 19.0 L. If the initial pressure of the gas was 12.088 atm, what was the final pressure of the gas after its expansion?
Answer: 3.181 atm
11.Calculate the pressure required to reduce the volume of 14000 mL of gas at a pressure of 3281 torr to 3000 mL with the temperature remaining constant.
Answer: 15311 torr
12.At a pressure of 10.425 atm, a sample of gas has a volume of 0.214 L. If the pressure is reduced to 1.882 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
Answer: 1.185 L
13.Under constant temperature, 8.0 L of a sample gas was allowed to expand to a final volume of 25.0 L. If the initial pressure of the gas was 0.342 atm, what was the final pressure of the gas after its expansion?
Answer: 0.109 atm
14.At a pressure of 9.679 atm, a sample of gas has a volume of 0.397 L. If the pressure is reduced to 5.373 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
Answer: 0.715 L
15.At a pressure of 8.537 atm, a sample of gas has a volume of 0.45 L. If the pressure is reduced to 2.531 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
Answer: 1.518 L

Charles' Law Problems







1.Calculate the final volume of a 3.65 L gas at 92° C when the temperature is brought down to 9° C as the pressure remains unchanged.
Answer: 2.82 L
2.Calculate the final volume of a 7.39 L gas at 43° C when the temperature is brought down to 13° C as the pressure remains unchanged.
Answer: 6.69 L
3.A sample of gas with a volume of 6.0 L at -17° C was allowed to expand to 21.0 L while maintaining a constant pressure. What was the new temperature of the gas?
Answer: 623° C
4.18 L of gas at 294° C was compressed to 2 L at constant pressure. Determine the resulting temperature of the gas.
Answer: -210° C
5.Maintaining a constant pressure, a sample of gas having a volume of 5.0 L at -127° C is heated to 369° C. What is the final volume of the gas?
Answer: 22.0 L
6.20 L of gas at 582° C was compressed to 4 L at constant pressure. Determine the resulting temperature of the gas.
Answer: -102° C
7.A sample of gas with a volume of 3.0 L at -145° C was allowed to expand to 17.0 L while maintaining a constant pressure. What was the new temperature of the gas?
Answer: 452° C
8.Maintaining a constant pressure, a sample of gas having a volume of 9.0 L at -210° C is heated to -7° C. What is the final volume of the gas?
Answer: 38.0 L

Gay-Lussac's Law Problems




1.A sample of a gas sealed in a vessel was determined to have a pressure of 0.795 atm at 9° C. What would be the pressure of the gas if the vessel was heated to increase the temperature by 88° C?
Answer: 1.043 atm
2.A gas cylinder containing a certain gas is found to have a pressure of 18.053 atm at 57° C. If the cylinder is immersed in a cooling solution so that the temperature is lowered by 0° C, calculate the resulting pressure inside the cylinder.
Answer: 18.053 atm
3.A sample of a gas has a pressure of 10 atm and a temperature of 32° C. Find the corresponding change in the temperature if the pressure is (a) reduced to 5 atm, (b) tripled. Assume the volume of the gas is constant.
Answer: (a) -120° C, (b) 642° C
4.A sample of a gas sealed in a vessel was determined to have a pressure of 1.007 atm at 2° C. What would be the pressure of the gas if the vessel was heated to increase the temperature by 81° C?
Answer: 1.304 atm
5.A gas cylinder containing a certain gas is found to have a pressure of 147.177 atm at 50° C. If the cylinder is immersed in a cooling solution so that the temperature is lowered by 1° C, calculate the resulting pressure inside the cylinder.
Answer: 146.721 atm
6.A sample of a gas sealed in a vessel was determined to have a pressure of 0.906 atm at 3° C. What would be the pressure of the gas if the vessel was heated to increase the temperature by 84° C?
Answer: 1.182 atm

Combined Gas Law Problems




1.What will be the final temperature of a 51.413 L gas at 27° C if its volume is changed to 18.283 L and its pressure changed from 0.974 atm to 4.063 atm?
Answer: 172° C
2.784.96 L of a sample of gas has a pressure of 0.23 atm and a temperature of 19° C. After compressing the volume down to 1.37 L and increasing the temperature to 98° C, calculate the change in the pressure of the gas.
Answer: 167.435 atm
3.69.73 L of a sample of gas has a pressure of 0.98 atm and a temperature of 29° C. After compressing the volume down to 7.47 L and increasing the temperature to 72° C, calculate the change in the pressure of the gas.
Answer: 10.451 atm
4.An expandable air balloon containing 88.2 L of a sample of a gas has a pressure of 1.039 atm at 47° C. What will be the resulting volume of the gas if the pressure is changed to 7 atm and the temperature is increased by 15° C?
Answer: 13.7 L
5.What will be the final temperature of a 161.519 L gas at 12° C if its volume is changed to 24.129 L and its pressure changed from 0.225 atm to 11.314 atm?
Answer: 1866° C
6.An expandable air balloon containing 8.3 L of a sample of a gas has a pressure of 31.941 atm at 42° C. What will be the resulting volume of the gas if the pressure is changed to 9 atm and the temperature is increased by 14° C?
Answer: 30.8 L

Ideal Gas Law Problems




1.A 29.135 L tank containing a sample of gas has a pressure of 0.968 atm and a temperature of 70° C. Find the moles of gas contained inside the tank.
Answer: 1.002 mol
2.Calculate the volume of 9.385 mol of a gas confined in a sealed vessel with a pressure of 39.66 atm and a temperature of 36° C.
Answer: 6 L
3.Given a volume of 25.07 L of the following gases at STP : (a) SO2, (b) N2, (c) H2S, find the weight in grams of each gas. (mol. wt.: SO2=64.0588, N2=28.0134, H2S=34.0758)
Answer: (a) 71.69 g, (b) 31.35 g, (c) 38.14 g
4.Given a weight of 8.8 g of each of the following gases: (a) NH3, (b) He, (c) NH3, calculate the volume of each gas at STP. (mol. wt.: NH3=17.0304, He=4.0026, NH3=17.0304)
Answer: (a) 11.57 L, (b) 49.25 L, (c) 11.57 L