POLAR BONDS AND MOLECULES Ms. Withrow November 10, 2008.
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Transcript of POLAR BONDS AND MOLECULES Ms. Withrow November 10, 2008.
POLAR BONDS AND MOLECULES
Ms. WithrowNovember 10, 2008
Polar Bonds
When involved in a bond, atoms of some elements attract the shared electrons to a greater extent than atoms of other elements – This property is called Electronegativity (EN)
The following chart is used to determine the electronegativities of each atom
Based on the difference in electronegativities of atoms we can predict the type of bond that will form
Formula: ∆EN = ENA – ENB
Chart:
Examples
Potassium Fluoride KF ∆EN = ENF – ENK = 3.98 – 0.82 = 3.16 IONIC BOND
Two Oxygen Atoms O2 ∆EN = ENO – ENO = 3.44 – 3.44 = 0 NON-POLAR COVALENT
Carbon Tetrachloride CCl4 ∆EN = ENCl – ENC = 3.16 – 2.55 = 0.61 POLAR COVALENT
With respect to polar covalent bonds, the differences in electronegativity tell us about the sharing of electrons
Example: Carbon Tetrachloride (CCl4) Cl has EN = 3.16 C has EN = 2.55 From this, we say that chlorine has stronger
attraction for electrons than carbon Thus, electrons will spend more time
around the Cl than C
This results in a slight separation of positive and negative charges which we call “partial charges” and represent them as δ+ or δ-
Example: CCl4 Chlorine with greater EN will have greater
attraction of e- and thus will have partial negative charge δ-
Carbon with lower EN will have less attraction of e- and thus will have partial positive charge δ+
Shown as δ+C-Cl δ-
When the bond is separated into partial positive and negative charges we call this bond a dipole bond
We represent dipole bonds with a vector arrow that points to the more electronegative atom
Example CCl4
δ+C-Cl δ-
Examples
Remember to Determine the bond type (by finding ∆EN) Assign the partial charges Place the dipole moment
Copper and Oxygenδ+C-O δ-
Carbon and Fluorine
δ+C-F δ-
Polar Molecules
We use our information on polar bonds to predict whether molecules will be polar or non-polar
We also must know our VSEPR shapes in order to do this!!
Water H20
Determine bond type ∆EN = ENO – ENH = 3.44 – 2.20 = 1.24 Thus is POLAR COVALENT
Determine partial chargesO has higher EN and H has lower ENOur partial charges are:
If we include the dipoles
Bent shape according to VSEPR
This is where VSEPR is important! -- You must know the shape of the molecule in order to determine it’s polarity
Water has two partially positive ends and one partially negative end The two dipole arrows point in the same direction. If we add these together we can see the molecule will have an overall net dipole
Because the dipoles do not cancel each other a net dipole is produced and we say that the molecule is POLAR
Carbon Dioxide CO2
Determine bond type ∆EN = ENO – ENC = 3.44 – 2.55 = 0.89 Thus is POLAR COVALENT
Determine partial chargesO has greater EN than COur partial charges are:
If we include the dipoles
Linear shape according to VSEPR
The dipoles created in this molecule are pointing in opposite directions and thus will cancel each other
This molecule has no net dipole and therefore is said to be NON-POLAR
Determine bond type ∆EN = ENN – ENC = 3.04 – 2.55 = 0.49 Thus is slightly POLAR COVALENT ∆EN = ENC – ENH = 2.55 – 2.20 = 0.35 Is also slightly POLAR COVALENT
Determine partial chargesN has greater EN than C – N will have δ-
C has greater EN than H – C will have δ-
Hydrogen Cyanide HCN
When we assign the dipoles
We see that they are both pointing the same direction
Thus they will not cancel, but will result in an overall net dipole
This molecule is said to be POLAR
Note the Difference!
When we had a linear molecule with the same atoms attached to the central atom the molecule was non-polar ex. CO2
When we had a linear molecule with two different atoms attached to the central atom, the molecule was polar Ex. HCN
It is very important to look at the electronegativities associated with the atoms and not just the VSEPR shape
Sulfur Trioxide SO3
Determine bond type ∆EN = ENO – ENS = 3.44 – 2.58 = 0.86 Thus is POLAR COVALENT
Determine partial chargesO has greater EN than SOur partial charges are:
Trigonal Planar shape according to VSEPR
When we assign dipole arrows All the dipoles are pulling
away from the central atom
You may think that because there are three dipoles they will not cancel and will result in a polar molecule
This is not correct however!!
Look at the horizontal and vertical components of the vectors (red and green arrows)
The red arrows will cancel The green arrows can add
together This green arrow will cancel
with the blue vector created by the top O
Therefore all dipole vectors will cancel in this molecule creating no net dipole and therefore the molecule is NON-POLAR
Similar to our linear molecule, difference will occur when the atoms attached to the central atom are different
We must be sure to look at the electronegativities of each atom when comparing the dipole vectors
Ex. CCl2O O has higher EN than Cl and will therefore have a greater dipole
The two dipoles from Cl will add together but they will still be less than that of O
Overall net dipole will result and thus molecule is POLAR
Ammonia NH3
Determine bond type ∆EN = ENN – ENH = 304 – 2.20 = 0.84 Thus is POLAR COVALENT
Determine partial chargesN has greater EN than HOur partial charges are:
Pyramidal shape according to VSEPR
Assign dipole vectors
The three vectors will add together to create an overall net dipole
This will result in a POLAR molecule
Carbon Tetrachloride CCl4
Determine bond type ∆EN = ENCl – ENC = 3.16 – 2.55 = 0.61 Thus is POLAR COVALENT
Determine partial chargesCl has greater EN than COur partial charges are:
Tetrahedral shape according to VSEPR
When we assign dipoles We can see that all the dipoles are
of the same magnitude because the EN differences are all the same
There are equal amounts of dipoles in opposite directions and thus they will all cancel
This results in no net dipole and therefore the molecule is NON-POLAR
Chloroform CHCl3
Determine bond type ∆EN = ENCl – ENC = 3.16 – 2.55 = 0.61 Thus is POLAR COVALENT ∆EN = ENC – ENH = 2.55 – 2.20 = 0.35 Thus is slightly POLAR COVALENT
Determine partial charges Cl has greater EN than C C has greater EN than H Our partial charges are:
Tetrahedral shape according to VSEPR
Assign dipoles (blue arrows) We can see that the dipoles to
Cl will all add up to create the larger green dipole vector
This is opposite to the dipole vector created by H-C but does not have the same magnitude
Thus, it will not cancel and result in a net dipole
This molecule is POLAR
Summary of Polarity of Molecules
Linear: When the two atoms attached to central
atom are the same the dipoles will cancel, leaving no net dipole, and the molecule will be Non-Polar
When the two atoms are different the dipoles will not cancel, resulting in a net dipole, and the molecule will be Polar
Bent: The dipoles created from this
molecule will not cancel creating a net dipole and the molecule will be Polar
Pyramidal: The dipoles created from this
molecule will not cancel creating a net dipole and the molecule will be Polar
Summary of Polarity of Molecules
Trigonal Planar: When the three atoms attached
to central atom are the same the dipoles will cancel, leaving no net dipole, and the molecule will be Non-Polar
When the three atoms are different the dipoles will not cancel, resulting in a net dipole, and the molecule will be Polar
Summary of Polarity of Molecules
Tetrahedral: When the four atoms attached to the
central atom are the same, the dipoles will cancel, leaving no net dipole, and the molecule will be Non-Polar
When the four atoms are different, the dipoles will not cancel, resulting in a net dipole, and the molecule will be Polar
Summary of Polarity of Molecules
Examples to Try
Determine whether the following molecules will be polar or non-polar SI2 CH3F AsI3 H2O2