ch3

Question 1

Do core electrons contribute to bonding? (Yes/No)

Answer 1

No.

Question 2

Which elements are not limited to the octet rule?

Answer 2

Row 3 and below, because (n+1)s and nd orbitals are not exceedingly higher in their energy from the atoms regular valence orbitals.

Question 3

Which orbitals can form an excited valence state?

Answer 3

Empty nd orbitals, since they can house excess valence e-s.

Question 4

What is another term for an excited valence state?

Answer 4

hypervalent or expanded shell structures.

Question 5

What are the steps in drawing Lewis structures?

Answer 5

1. count total # valence e-s.
2. identify central atom: usually lowest in electronegativity.
3. make single bonds to connect peripheral atoms to central atom: keep track of e-s used.
4. use remaining electrons in adding lone pairs to achieve octet: start with atoms in higher electronegativity because they pull harder on e-s and less likely to share.
5. if central atom has less than 8 e-s form multiple bonds by converting lone pairs of least electronegative peripheral atoms to bonds, because they pull less hard on e-s and more likely to share.
6. assign formal charges: repeat step 5 to minimize FCs.

Question 6

How do you calculate formal charge of an atom?

Answer 6

valence e-s - (non-bonding e-s + 1 e- from each bond)

Question 7

Define a canonical structure.

Answer 7

Canonical structures is a way of describing the delocalized e-s within molecules or polyatomic ions where the bonding cannot be expressed in terms of single lewis structure. the true structure is an average of 2 (or more) Lewis structures.

Question 8

Which term is used for “true structure”.

Answer 8

resonance hybrid.

Question 9

Why does a single Lewis structure fail to represent the true structure of a molecule (provide example)?

Answer 9

Because Lewis structures predicts a difference in bond lengths between identical pairs of atoms. For example, it predicts one double and one single bond in O3.

Question 10

When does resonance occur? Energy of molecule vs contributing forms?

Answer 10

It occurs when e-s are delocalized over several overlapping orbitals. Also the energy of the molecule or ion ends up lower than the contributing forms.

Question 11

Why does resonance increase stability?

Answer 11

Because resonance allows for delocalization of e-s, the overall energy of a molecule is lowered since its e-s occupy a greater volume.

Question 12

What are three factors of Pauling’s principle of electroneutrality? Where does his theory apply?

Answer 12

1) FC’s as close to zero are preferred.
2) Any negative FC is most likely on the most electronegative atom(s).
3) Any positive FC is most likely on the least electronegative atom(s).
* *It applies when identifying the resonance hybrid.

Question 13

Which structure(s) contribute most to the resonance hybrid?

Answer 13

The most stable canonical structure(s).

Question 14

What does the most realistic Lewis structure have (contributes most to the resonance hybrid)?

Answer 14

• all atoms with full valence shell.
• minimum charge separation.
• formal charges agreeing with atoms electronegativity.

Question 15

What is the proceeding step from establishing the most stable structure?

Answer 15

To deduce how many equivalent most-stable canonical structures there are (resonance contributors).

Question 16

Do inequivalent resonance structures contribute to the resonance hybrid? (Yes, No)

Answer 16

Yes, but not as much.

Question 17

How do you calculate average bond order?

Answer 17

(sum of individual bond orders) / (# of canonical structures)

Question 18

What is the difference between resonance structures and isomerism?

Answer 18

Resonance structures have the same atoms in the same arrangement, however e-s can move around the molecule. Isomerism is the possibility of differing compounds having the same atoms, but in a different arrangement and e- arrangement.

Question 19

Explain why some elements have more than the octet.

Answer 19

In row 3, empty d orbitals can house extra e-s , allowing elements like silicon, phosphorus, sulfur, chlorine, to have more than the octet.

Question 20

Explain why some elements have less than the octet.

Answer 20

This exception applies to Be, B and Al which (remain “open-shell”). Although they’re stable, they’ll try to form a fourth bond to get eight electrons.

Question 21

What is a unique feature of elements that have less than the octet?

Answer 21

They’re good Lewis acids because they can accept e-s.

Question 22

For the molecule BF3, why is B-F(obsd)=131 pm and B-F(calc)=152 pm?

Answer 22

BF3 has 2 inequivalent canonical structures, 3 for B=F and 1 B-F. B=O has a full valence shell, but highly electronegative F pulls strongly on e-s, which is less likely to share e-s as a double bond. The overall B-F bond is less than predicted, because it shows partial double bond character.

Question 23

Which is structure expected to be a Lewis acid for BF3, B=F or B-F?

Answer 23

B-F because it only has 6 e-s around B, can achieve 8 by accepting e-s which makes it a good Lewis acid.

Question 24

Define oxidizing agent and rationalize the oxidizing ability of K2CrO7.

Answer 24

An oxidizing agent gains e-s and is reduced in the process. Since O is more electronegative than Cr it has a stronger pull on e-s, and is less likely to share. Cr hold 6 valence e-s and its oxidation state is very high because it is bonded to so many strong pulling atoms, so in reaction with reducing agent, Cr could be converted into a form in which it is more e- rich.

Question 25

How is the geometry of molecules predicted?

Answer 25

Predicted by arranging regions of enhanced e- density most efficiently.

Question 26

Define a steric number/electron grouping?

Answer 26

The number of atoms bonded to central atom.

Question 27

Types of electron groupings.

Answer 27

1. Bonds (single, double, triple).
2. Stereochemically inactive lone pairs.
3. One lone electron.

Question 28

Provide steric number, basic geometry, molecular geometry, and predicted bond angles (up to 6).

Answer 28

2, linear, 180.

3, trigonal planar,
bent or angular (1 lone e- pair), 120.

4, tetrahedral,
trigonal pyramidal (1 lone e- pair), bent/angular (2 lone e-pairs), 109.5.

5, trigonal bipyramidal,
seesaw (1 lone e- pairs), T-shape (2 lone e- pairs), Linear (3 lone e- pairs), 120&90.

6, octahedral,
square pyramidal (1 lone e- pair), square-planar (2 lone e- pair), 90.

Question 29

What is the geometry and predicted bond angle for steric numbers 7 & 8?

Answer 29

7, pentagonal bipyramidal, 72/90.

8, square antiprismatic, 70.5/99.6/109.5

Question 30

Factors affecting bond length.

Answer 30

• Bond polarity, thus electronegativity.

- size.

Question 31

Explain the effect(s) of bond polarity (electronegativity) on bond angles.

Answer 31

If peripheral atoms are more electronegative than central atom, there’s less e- density around central atom, lone pair has room to expand pushing bonds closer together.

Question 32

How do electronegativity and size work against each other? Compare N(CH3)3 & N(CF3)3. Which dominates size or bond polarity?

Answer 32

CF3 has a bigger bond angle than CH3. Since CF3 is more electronegative than CH3, you expect less e- density near the central atom resulting in smaller bond angle. CF3 has a bigger radius than CH3, therefore you expect more steric repulsion between F atoms, thus bonds cannot squeeze closer together, lone pair must act smaller.

• Data shows that size dominates in this particular case.

Question 33

Bond angle trends with change in peripheral atoms.

Answer 33

• > bond angle as electronegativity of group decreases.

- >bond angle as size of group increases.

Question 34

Bond angle trends with change in central atom.

Answer 34

-

Question 35

Which position do lone pairs prefer, equatorial or axial?

Answer 35

Lone pairs prefer equatorial position, because there’s less neighboring bonding pairs 90 degrees away and more e- repulsive forces than bonding pairs, therefore prefer their space.

Question 36

Which information is provided and missing with the Lewis structure?

Answer 36

Provides # of lone pairs and types of bonds pbetween the atoms. Missing information about the actual arrangement of atoms in space.

Question 37

Which information is provided and missing with the VSEPR Theory?

Answer 37

Provides the molecular geometry of molecules. Missing information about bond lengths.

Question 38

A combination of VSEPR and a bonding model, such as Lewis electron structures, is necessary to understand the presence of multiple bonds (T/F).

Answer 38

True.

Question 39

Limitations of VSEPR Theory.

Answer 39

Heavy p-block elements in their lower valence state have geometries that can’t be predicted by VSEPR, because they prefer to hold their lone pairs in an unhybridized s orbital.

Question 40

What does LCP stand for?

Answer 40

Ligand Close-Packing, to predict bond lengths.

Question 41

What bond lengths do we expect as bond angles get smaller?

Answer 41

Longer bond lengths.