Orbitals, Shapes and Polarity Quiz /17

Name:

Knowledge. Answer the following questions on foolscap.

/2

1. Explain why the p sub-level can appear to be spherical like the s sub-level?

/2 /2

2.a) What is the maximum number of pi bonds that a p sub-level can form? Explain. b) A bond forms between a sp2 hybridized orbital from one atom and a sp2 hybridized orbital from another atom. Is this bond likely to be a sigma or pi bond? Explain your reasoning.

/7

3. With the aid of diagrams, predict the most stable shape of a H2SO molecule.

/2

4. Arrange the following molecules/ions in order of decreasing angle between the marked (curved line) bonded atoms?

/2

5. Explain what effect losing 1 electron from N in NH3 would have on its shape.

/20

Communication. Answer the following questions on foolscap.

/2

6. Draw the orbital diagram to represent 3p4.

/5 /5

7.a) Draw orbital diagrams to show the bonding in the most stable predicted HNCH2 molecule. b) Label each bond and orbital in your diagram.

/8

8. For each of the following combinations of molecules:

i)

IO3

ii) XeOF4 (Xe – electronegativity = 2.6)

a) draw and name the geometric shape of the molecule. b) illustrate the bond polarity and state whether or not the molecule is polar.

Orbitals, Shapes and Polarity Quiz /17 /2

Knowledge. Answer the following questions on foolscap.

1. Explain why the p sub-level can appear to be spherical like the s sub-level? !

/2

electrons mostly found at the tips of the orbitals but the orbitals are constantly moving around the nucleus creating the appearance of spherical shape (like a spinning triangle looks like a sphere)

2.a) What is the maximum number of pi bonds that a p sub-level can form? Explain. ! !

/2

Name:

2 there are 3 orbitals and 1 must form a sigma bond to allow any pi bonding leaving only the remaining 2 available for pi bonding

b) A bond forms between a sp2 hybridized orbital from one atom and a sp2 hybridized orbital from another atom.

Is this bond likely to be a sigma or pi bond? Explain your reasoning. !

/7

sigma bond because a pi bond involving sideways overlapping and the bond forming between sp2 orbitals constrains the shared electrons into the one significant sp2 lobe from each atom would require too much energy compared to the direct overlap of the sigma bond that could form

3. With the aid of diagrams, predict the most stable shape of a H2SO molecule.

O !

H

S

H ! !

sulfur could single bond to an H and the O and then the other H could single bond to the O (recall H easily bonds to very electronegative atoms) this creates a line of bent shapes as the S and O both have 4 electron pairs with 2 bonding pairs and 2 non-bonding pairs each

!

from left to right F.C:

!

works since there is no electron distribution issues and all atoms have stable electron configurations

H: S: O: H:

1-0-1=0 6-4-2=0 6-4-2=0 1-0-1=0

OR

O !

H

S

H ! !

sulfur could single bond to the H atoms and then hybridize a pair of electrons, single bond to the O and then redistribute the lone remaining electron to the O creates a trigonal pyramid shape (results in 4 electron pairs around S, 3 sigma bonds and 1 lone-pair)

!

from top to bottom F.C:

!

even though all atoms have stable electron configurations, this is not as likely due to an electron deficit on S and electron loading on O

O: H: S: H:

6 - 6 - 1 = -1 1-0-1=0 6-2-3=1 1-0-1=0

OR

O !

H

S

H

/2

! !

sulfur single bonds to the H atoms and then hybridize a pair of electrons and double bond to the O this creates a trigonal pyramid shape (5 electrons around S with 3 sigma bonds, 1 lone pair and 1 pi bond–the 3 sigma and 1 lone pair create a tetrahedral shape (4,1), but only 3 pairs are actual bonds)

!

from top to bottom F.C:

!

works since there is no electron distribution issues and all atoms have stable electron configurations (well sort of, the pi bond pulls one pair of electrons further away from the other 4 pairs making them less repulsive than if there were just 5 pairs of electrons in sigma positions meaning sulfur can tolerate 10 outer electrons)–so you would not likely predict this one as being a stable possibility but if told it did work, you could now explain it

!

so, in the end, two possible configurations but you would pick the first one as being most stable

6-4-2=0 1-0-1=0 6-2-4=0 1-0-1=0

4. Arrange the following molecules/ions in order of decreasing angle between the marked (curved line) bonded atoms?

! !

! !

/2

O: H: S: H:

b, a, c in b, repulsions on bonding pairs but the F draws its shared electron pair away from P so the repulsion between F-Cl atoms is less allowing the repulsion between the Cls to be felt more and they push away from each other creating an angle bigger than 120 in a, all the repulsions are equal and the angle is 120 in c, the lone pair of electrons are most repulsive since they are very close to P making the lone pairCl repulsion significant pushing the Cl atoms together making the angle less than 120

5. Explain what effect losing 1 electron from N in NH3 would have on its shape. ! !

there will be a lone electron on N which means it will be less repulsive than when there was a pair of electrons thus, the angle between the H atoms will relax somewhat and become slightly larger, but not 109.5 as in the tetrahedral

/20

/2

Communication. Answer the following questions on foolscap.

6. Draw the orbital diagram to represent 3p4. !

/5 /5

7.a) Draw orbital diagrams to show the bonding in the most stable predicted HNCH2 molecule. b) Label each bond and orbital in your diagram.

! p



p

p

p

sp

2

 s

/8

 sp2 sp2



8. For each of the following combinations of molecules:

s

 s

i)

IO3

ii) XeOF4 (Xe – electronegativity = 2.6)

a) draw and name the geometric shape of the molecule. b) illustrate the bond polarity and state whether or not the molecule is polar. i)

IO3

3 possibilities: ! ! ! ! ! !

!

I hybridizes twice, all O atoms single bond with electrons shifting from I to O but F.C. poor I hybridizes twice, one O atom double bonds and two O atoms single bond with an electron shifting from I to one of the O atoms but F.C. poor I hybridizes twice, two O atoms double bond and the other O atom single bonds and F.C. is best so I has 6 electron pairs, but only 3 sigma and one non-bonding pair so I has an architecture of (4,1) trigonal pyramidal

vectors create a net polarity so the molecule is polar

ii) XeOF4 ! ! ! ! !

2 possibilities Xe hybridizes three times, all atoms single bond and twice, with an electron shifting from Xe to the O but F.C. poor Xe hybridizes three times, all F atoms single bond and O atom double bonds–F.C. is best Xe has 7 electron pairs, but only 5 sigma and 1 non-bonding pair so Xe has an architecture of (6,1) square pyramidal

!

vectors create a net polarity so the molecule is polar (F has more pull and unsymmetrical shape)

! !

putting O is unlikely since it is much more electronegative than Xe attaching a F to O is possible but then the shape would not have a name so it is implied that a better, more likely answer should be used

! !

note that there is often more than one way to do these and sometimes this creates different answers the most stable version is the most likely, but this was not part of the question (so you don’t actually have to do all the FC work–just pick one and go from there)

Prepared by K. Zuber