Bonding : Bonds and Electronegativity

Types of Bonds

- There are three types of bonds :

• Ionic (metal - non metal) - e- and transferred from metal to non metal
• Covalent (non metal - non metal) e- shared between non metals
• Metallic (metal) - Holds pure metal together by electrostatic attraction

Electronegativity

- Electronegativity (en) is a measure of an atom's attraction for electrons in a bond

- Fluorine = 4.0

- Chlorine = 3.0

- Cesium = 0.9

- Atoms with greater en attract e- more

- Polar covalent bonds form from an unequal sharing of e-

- Non-polar covalent bonds form from equal sharing

Bonds (no.. Not James Bond)

- The type of bond formed can be predicted by looking at the difference in electronegativity of the element

en > 1.7 = Ionic bond

en < 1.7 = Polar covalent bond

en = 0 is non-polar covalent bond

Examples

- Predict the type of bond formed

O - O

Since the en charge of Oxygen is 3.44, we would sub it in like math.

3.44 - 3.44 = 0

Since it equals zero, you have to look back at the Bonds, and since it equals 0, it's a non-polar covalent bond

- Identify the positive and negative side of the Polar Bonds

Now you would have to figure out the positive the negative of each chemical in the electronegativity.

You need to know a symbol that means partially. ( It's suppose to look like a really messed up six, you're pretty much drawing it backwards, but for now we'll use this, '[')

Example

N - O
3.04 - 3.44 = 0.40 - Polar covalent

You would think that would be the answer, but it's not.
You would need to state which is positive and which is negative.
So you're final answer would be :

[+N - O[-
That means, Nitrogen is partially positive and Oxygen is partially negative.
And don't worry about if the number you subtract becomes a negative number. You would always want to imply the Absolute value. (Example)

|N - O = ?|

|3.04 - 3.44 = ?|
= |-0.40|
=0.40

Here's a video example of what I was trying to explain.

http://www.youtube.com/watch?v=Kj3o0XvhVqQ

-Meldrick Mendoza

Dilution and Creating Solutions

In this lab, we are trying to find out which test tube is the right when when we are given 0.2M.

Materials :
50g of Copper Chlorine
Scale
Wax paper
30mL of water
Graduated Cylinder

Procedure :

1. We weighed 0.807g of Copper Chloride (Plus the weight of the wax paper, which was 0.78g)
2. We diluted the substance in 30mL of water after accurately measuring the water in a graduated cylinder using the minuscus.
3. Then you repeat the first two steps to get an accurate answer.

-Meldrick Mendoza

Acid - Base Reactions Strong Acids and Strong Bases

-Strong Acids (SA) dissociate to produce H+ ions
     -HCl, H2SO4, HClO4
-Strong Bases (SB) dissociate to produce OH- ions
     -NaOH, Ba(OH)2, LiOH
-When a SA and SB mix they form water and an ionic salt
-Total volume changes

pH and pOH

-pH is a measure of the hydrogen ions present in a solution

     pH = -log[H+]

-pOH is a measure of the hydroxide ions present in a solution 

     pOH = -log[OH-]

Example

-100mL of 0.250M HCl is added to 100mL of 0.250M LiOH

-Determine the mass of water produced.

 HCl + LiOH ----> HOH + LiCl

 0.250mol/L x .100L x 1/1 x 18g/1mol = 0.45g

-Determine the mass of the salt produced

 .45g x 1mol/18g x 1/1 x 42.4g/1mol =1.06g

Example

-100mL of .250M HCl is added to 200mL of .150M LiOH

-Determine the L.R.

 0.250mol/L x .100L = .0250mol

.150mol/L x .200L x 1/1 = .0300mol

L.R. HCl

-How much LiOH is left over when the reaction is complete?

0.206L x 0.150mol/L = .0300mol LiOH started with

0.0250mol x 1/1 = .0250mol LiOH used 

.0300mol - .0250mol = .0050mol LiOH left

-Determine [LiOH] after the reaction is complete

 0.0050mol x 1/.300L = 0.0167 M = [LiOH]

-Find the pOH of the final solution

 LiOH ----> Li+ + OH-

.0167M   .0167M  .0167M

pOH = -log[OH-]

pOH= -log[0.0167M]

       = 1.78

-T.H

Ion Concentration

Dissociation

- Ionic compounds are made up of two parts

- Cation: positively charged particle

- Anion- Negatively charged particle

- When ionic compounds are dissolved in water the cation and anion separate fro m each other

- This process is called dissociation (splitting apart the ions)

When writing dissociation equation, the atoms and charges must balance

The Dissociation of Sodium Chloride is:

NaCl ----> Na (+1) +  Cl (-1)

Complete the following dissociation equation:

Fe(OH)2  ---->   Fe (+2) + OH (-1)

Balanced ----->  Fe (+2) + 2OH (-1)

If the volume does not change then the concentration of individual ion depends on the balanced

coefficients in the dissociation equation.

Determine the [NA(+1)] and [PO4 (-3)] in a 1.5 M solution of Na3PO4:

Na3PO4 ---> 3Na (+1) + PO4 (-3)

1.5 M        (1.5x3) 4.5 M       1.5 M

If the [Cl (-1)] = 0.752 M in a solution of ZnCl2 determine the solution's concentration:

ZnCl2 -----> Zn (+2) + 2Cl (-1)

-0.752M = 0.376 M

-If two solutions are added together the volume changes.

-The ions are diluted (C1V1 = C2V2)

http://www.youtube.com/watch?v=EBfGcTAJF4o
Dissociation of Salt

Jomar Delos Santos

DILUTIONS

-When two solutions are mixed the concentration changes
-Dilution is the process of decreasing the concentration by adding a solvent ( usually water)
-The amount of solute does not change n1=n2
-Because concentration is mol/L we can write
C=m/v
n-cv so c1v1=c2v2
Example
-How much water must be added to 10.mL of 10.0M Na2So4 to give a solution with a concentration of 0.50M?

c1v1=c2v2
(0.01L)(10.0M)=(0.50M) v2
v2=0.200L
200mL-10mL=190mL

KP

Diluting Solutions

- When two solutions are mixed, the concentration changes

- Dilution is the process of decreasing the concentration by adding a solvent (usually water)

- This amount of solute does not change

n1= n2

- Because the concentration is mol/L, we can write:

C = n/v  C = mol/v

and n (of mol) = CV

c1v1 = c2v2

Example: Determine the “C” when 100 ml of 0.10 M of HCl is diluted t a final volume of 400 ml

c1v1 = c2v

c2 = c1v1/v2= (0.1L)(0.1M)/(0.4L) = 0.025 mol/L

*Remember ‘[ ]’ means concentration

100 mL of 0.50 M Sodium Nitrate is mixed with 200mL of 100M Sodium Nitrate

- Determine the # of moles of Sodium Nitrate in the resulting solutions

n1 + n2 = (0.100L) x 250mol/ L + (0.200L) x 0.1 mol/L

             =  0.025 mol + 0.02 mol

             =  0.0450 mol

Determine the final NaNo3 --> (0.0450mol)/(0.300L) = 0.150 M

Jomar Delos Santos

Chemistry Lab: Titration of Vinegar

In class we were looking for the Molar Concentration of Acetic Acid in house hold vinegar.

Balanced equation: NaOH + CH3COOH ----> NaCH3COOH + H2O

Lab:

Trial                                        1                                2                                  3
Initial Volume                       33.2                           41.5                             49.9
Final Volume                        24.7                           33.2                             41.5
NaOH added                        8.5                             8.3                               8.4

Determine the average volume of NaOH added:

8.5 + 8.4 + 8.3 = 8.4mL -----> .0084L
           3

Determine the concentration of Acetic Acid.

1.00 mol x 0.0084L x 1 x 1 mol = .84 mol/L
      L                             1    .010 L

The actual value was .85 M.
My percent error was 1.2%

-Thomas H