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CHEM 241 > ch3 > Flashcards

ch3 Flashcards

1
Q

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

A

No.

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2
Q

Which elements are not limited to the octet rule?

A

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

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3
Q

Which orbitals can form an excited valence state?

A

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

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4
Q

What is another term for an excited valence state?

A

hypervalent or expanded shell structures.

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5
Q

What are the steps in drawing Lewis structures?

A
  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.
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6
Q

How do you calculate formal charge of an atom?

A

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

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7
Q

Define a canonical structure.

A

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.

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8
Q

Which term is used for “true structure”.

A

resonance hybrid.

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9
Q

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

A

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.

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10
Q

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

A

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.

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11
Q

Why does resonance increase stability?

A

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

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12
Q

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

A

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.

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13
Q

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

A

The most stable canonical structure(s).

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14
Q

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

A
  • all atoms with full valence shell.
  • minimum charge separation.
  • formal charges agreeing with atoms electronegativity.
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15
Q

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

A

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

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16
Q

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

A

Yes, but not as much.

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17
Q

How do you calculate average bond order?

A

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

18
Q

What is the difference between resonance structures and isomerism?

A

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.

19
Q

Explain why some elements have more than the octet.

A

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

20
Q

Explain why some elements have less than the octet.

A

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.

21
Q

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

A

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

22
Q

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

A

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.

23
Q

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

A

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.

24
Q

Define oxidizing agent and rationalize the oxidizing ability of K2CrO7.

A

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.

25
Q

How is the geometry of molecules predicted?

A

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

26
Q

Define a steric number/electron grouping?

A

The number of atoms bonded to central atom.

27
Q

Types of electron groupings.

A
  1. Bonds (single, double, triple).
  2. Stereochemically inactive lone pairs.
  3. One lone electron.
28
Q

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

A

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.
29
Q

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

A

7, pentagonal bipyramidal, 72/90.

8, square antiprismatic, 70.5/99.6/109.5

30
Q

Factors affecting bond length.

A
  • Bond polarity, thus electronegativity.

- size.

31
Q

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

A

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.

32
Q

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

A

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.
33
Q

Bond angle trends with change in peripheral atoms.

A
  • > bond angle as electronegativity of group decreases.

- >bond angle as size of group increases.

34
Q

Bond angle trends with change in central atom.

A

-

35
Q

Which position do lone pairs prefer, equatorial or axial?

A

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.

36
Q

Which information is provided and missing with the Lewis structure?

A

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

37
Q

Which information is provided and missing with the VSEPR Theory?

A

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

38
Q

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

A

True.

39
Q

Limitations of VSEPR Theory.

A

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.

40
Q

What does LCP stand for?

A

Ligand Close-Packing, to predict bond lengths.

41
Q

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

A

Longer bond lengths.

CHEM 241 (6 decks)

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