Polar bonds and molecules
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Polar Bonds and Molecules. Electronegativity. Why does ice float?. 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)

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Polar Bonds and Molecules

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Polar Bonds and Molecules

Electronegativity


Why does ice float?


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


  • In general, electronegativity increases from left to right and from the bottom up

  • As atomic radius increases, electronegativity decreases.


Difference in Electronegativity = ΔEN

  • 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

  • Oxygen, O2

    • ∆EN = ENO – ENO = 3.44 – 3.44 = 0

    • NON-POLAR COVALENT

  • Carbon Tetrachloride, CCl4

    (look at the ∆EN for one of the C-Cl bonds)

    • ∆EN = ENCl – ENC = 3.0 – 2.5 = 0.5

    • 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 = 3.0

    • C has EN = 2.5

    • 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+ orδ-

  • 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-Clδ-


Equal sharing of electrons

ΔEN = 0

Unequal sharing of electrons

ΔEN > 0


Dipoles in Molecules


  • When the bond is separated into partial positive and negative charges we call this bond a polar 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

  • Carbon and Oxygen

    δ+C-O δ-

  • Carbon and Fluorine

    δ+C-F δ-


With your group, build the following molecules…

  • CH4

  • NH3

  • CO2

  • HCN

  • H2O

  • COCl2


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 charges

    • O has higher EN and H has lower EN

    • Our partial charges are:

  • If we include the dipoles

Bent shape according to VSEPR


Back to the question:Why does ice float?

  • Density of water=1g/mL Density of ice = 0.92g/mL


What causes this difference in density?

  • The polar bonds in water.


VSEPR Theory

  • Valence Shell Electron Pair Repulsion Theory

  • This theory predicts the shapes of molecules based on the number of areas of electron density around the central atom

  • Electron density can be a lone pair or a bonding pair of electrons

  • The areas of electron density want to be as far apart as possible and as such form predictable molecular shapes


VSEPR Shapes of Molecules


  • 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 charges

    • O has greater EN than C

    • Our 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


Hydrogen Cyanide HCN

  • 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 charges

    • N has greater EN than C – N will have δ-

    • C has greater EN than H – C will have δ-


  • 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 charges

    • O has greater EN than S

    • Our 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 charges

    • N has greater EN than H

    • Our 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 charges

    • Cl has greater EN than C

    • Our 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


With your group,

  • Read through the tutorial on pg 106-107 and answer question 1 on pg 107

    Homework

  • Read pg 102-108Questions pg 108 # 1, 2, 5


Examples to Try

  • Determine whether the following molecules will be polar or non-polar

    • SI2

    • CH3F

    • AsI3

    • H2O2


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