Unit 9 - VSEPR

Question 1

VSEPR (Valence Shell Electron Pair Repulsion)

Answer 1

Repulsion Order Rules:

LP - LP > LP - BP > BP - BP

Question 2

Bent’s Rule

Answer 2

more electronegative (smaller) atoms prefer axial positions and less electronegative (larger) atoms and LPs prefer equatorial positions in tbp geometries

Question 3

Steric Number

Answer 3

total number of bonded atoms and lone electron pairs around a central atom

Question 4

What steric number (SN) corresponds with what basic geometry?

Answer 4

SN = 2: linear
SN = 3: trigonal planar
SN = 4: tetrahedral (Td)
SN = 5: trigonal bipyramidal (tbp)
SN = 6: Octahedral

Question 5

Coordination Number (CN)

Answer 5

total number of bonded atoms around a central atom

Question 6

SN = 2

Answer 6

• Basic geometry: Linear
• linear and no LPs therefore the bond angle is 180˚

Question 7

SN = 3

Answer 7

• Basic geometry: trigonal planar
• tp and no LPs therefore the bond angle is 120˚

Question 8

SN = 3 with 1 LP

Answer 8

• CN = 2
• Basic geometry: trigonal planar
• Molecular geometry: bent or angular
• tp and 1 LP so the bond angle is >120˚ between the LPs and bonds, and <120˚ between the two bonds bc the LP pushes the bonds away more (LP-LP repulsion, around 118˚)

Question 9

SN = 4

Answer 9

• CN = 4
• Basic geometry: tetrahedral
• tetrahedral and 1 LP so the bond angle is 109.5˚
• one out of page and one into page

Question 10

SN = 4 with 1 LP

Answer 10

• CN = 3
• Basic geometry: tetrahedral (td)
• Molecular geometry: trigonal pyramidal
• LP pushes down 3 BPs
• > 109.5 for LP and bonds
• <109.5 for bond and bonds

Question 11

SN = 4 with 2 LPs

Answer 11

• CN = 2
• Basic geometry: tetrahedral (td)
• Molecular geometry: bent angular
• «109.5 between BP - BP

Question 12

SN = 5

Answer 12

• CN = 5
• apparet octet expansion
• Basic geometry: trigonal biprymidal (tbp)
• axial positions are directly above and below the
• equatorial positions are left, right, and forward/backward

Question 13

SN = 5 with 1 LP

Answer 13

• CN = 4
• Basic geometry: trigonal biprymidal (tbp)
• Molecular geometry: sawhorse or seesaw
• to minimize repulsion the pairs are arranged with angles of 120˚between equatorial bonds and 90˚between equatorial and axial bonds

Question 14

SN = 5 with 2 LP

Answer 14

• CN = 3
• Basic geometry: trigonal biprymidal (tbp)
• Molecular geometry: T - shape
• <90˚between the axial and equatorial

Question 15

SN = 5 with 3 LPs

Answer 15

• CN = 2
  Basic geometry: trigonal biprymidal (tbp)
• Molecular geometry: linear
• 180˚because the axial atoms (BP-BP)

Question 16

SN = 6

Answer 16

• CN = 6
• also apparent octet expansion
• cannot separate as axial or equatorial because they don’t have different angles
• Basic geometry: octahedral (Oh)
• 90˚between all the BP-BP

Question 17

SN = 6 with 1 LP

Answer 17

• CN = 5
• also apparent octet expansion
• Basic geometry: octahedral (Oh)
• Molecular geometry: square pyramidal
• all 6 locations for LP is equivalent
• <90˚between all the BP-BP

Question 18

SN = 6 with 2 LPs

Answer 18

• CN = 4
• Basic geometry: octahedral (Oh)
• Molecular geometry: square planar
• repulsion cancels out so 90˚angles

Question 19

SN = 6 with 3 LPs

Answer 19

• CN = 3
• Basic geometry: octahedral (Oh)
• Molecular geometry: T - shape
• <90˚between the atoms (BP-BP)

Question 20

SN = 6 with 4 LPs

Answer 20

• CN = 2
• Basic geometry: octahedral (Oh)
• Molecular geometry: linear