“I Pledge”: ( Initial DON’T FORGET WHAT THIS … · What does VSEPR stand for and what it is...
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4.2.A – ReIntro to Bonding “I Pledge”: _____ ( Initial ) DON’T FORGET WHAT THIS REPRESENTS…. Instructions: Provide a response for each question that is well thought out, satisfies the prompt, is clearly explained, and LEGIBLE. 1. Why do atoms and ions form bonds? Atoms / Ions donate / accept / share e ─ in order to form a low energy stable state (communally isoelectric to noble gases) _ 2. What is the difference between an intramolecular force and an intermolecular force (VERY IMPORTANT, differences must take into account EN and polarity)? Intra- = Strong = Chemical = determined by ∆EN resulting in polar and nonpolar molecules = between atoms / ions forming a molecule / FU (determines chemical . properties) _ Inter- = Weak = Physical = determined by polarity and e ─ movement = between molecules (determines physical properties ) . 3. What role do e - play in ionic bonding and why? e ─ are transferred from low EN atoms ( metals ) to a high EN atom ( nonmetals ) due to an ∆EN ≥ 1.7 . Difference . is enough that the more EN atom is able to rip away, or “bully” the high energy e ─ of the low “whimpy” EN atom resulting in a “permanent” Coulombic attraction . 4. What role do e - play in covalent bonding and why? e ─ are shared between high EN atoms ( nonmetals ) due to an ∆EN < 1.7 . Difference between the two high EN . “bullies” is not enough that one atom is able rip away the high e ─ of the other so some form of imperfect sharing occurs . 5. What role do e - play in metallic bonding and why? e ─ are shared between uniformly between low EN atoms ( metals ) due to an ∆EN < 1.7 . Difference between the two low EN “wimps” is not enough that ANY atom in the bond is able rip away the high e ─ of the others so perfectly uniform sharing occurs = e ─ sea . Instructions: Determine the electronegativity difference (ΔEN) for the following chemical bonds in the indicated blank. No work need be shown. Then using the ΔEN determine if the bond is considered a nonpolar covalent bond [NC], polar covalent [PC], ionic bond [I], or metallic bond [M]. Remember: ∆EN = │EN1 – EN2 │, is NEVER a ─# ΔEN Bond Type ΔEN Bond Type ΔEN Bond Type 6. Li – H 1.22 [PC] . 19. Cu – Zn 0.25 [M] . 32. C – H 0.35 [NC] . 0.98 2.20 ∆ EN ≤1.7 but ≥ 0.5 1.90 1.65 ∆ EN < 0.5 & 2M’s 2.55 2.20 ∆ EN < 0.5 & 2NM’s 7. Li – Cl 2.18 [I] . 20. Ca – Cl 2.16 [I] . 33. Si – H 0.30 [NC] . 0.98 3.16 ∆ EN ≥1.7 1.00 3.16 ∆ EN ≥1.7 1.90 2.20 ∆ EN < 0.5 & 2NM’s 8. Na – H 1.27 [PC] . 21. Ca – S 1.58 [PC] . 34. C – O 0.89 [PC] . 0.93 2.20 ∆ EN ≤1.7 but ≥ 0.5 1.00 2.58 ∆ EN ≤1.7 but ≥ 0.5 2.55 3.44 ∆ EN ≤1.7 but ≥ 0.5 9. Na – Cl 2.23 [I] . 22. Sr – F 3.05 [I] . 35. Al – Cl 0.96 [PC] . 0.93 3.16 ∆ EN ≤1.7 but ≥ 0.5 0.95 4.00 ∆ EN ≥1.7 2.20 3.16 ∆ EN ≤1.7 but ≥ 0.5 10. K – H 1.38 [PC] . 23. Ba – F 3.11 [I] . 36. C – Cl 0.61 [PC] . 0.82 2.20 ∆ EN ≤1.7 but ≥ 0.5 0.89 4.00 ∆ EN ≥1.7 2.55 3.16 ∆ EN ≤1.7 but ≥ 0.5 11. K – Cl 2.34 [I] . 24. Fr – F 3.30 [I] . 37. H – H 0.00 [NC] . 0.82 3.16 ∆ EN ≥1.7 0.70 4.00 ∆ EN ≥1.7 2.55 2.55 ∆ EN < 0.5 & 2NM’s 12. Fr – N 2.34 [I] . 25. H – O 1.24 [PC] . 38. S – O 0.86 [PC] . 0.70 3.04 ∆ EN ≥1.7 2.20 3.44 ∆ EN ≤1.7 but ≥ 0.5 2.58 3.44 ∆ EN ≤1.7 but ≥ 0.5 13. Fr – Cl 2.46 [I] . 26. H – S 0.38 [NC] . 39. P – O 1.25 [PC] . 0.70 3.16 ∆ EN ≥1.7 2.20 2.58 ∆ EN < 0.5 & 2NM’s 2.19 3.44 ∆ EN ≤1.7 but ≥ 0.5 14. Fe – Fe 0.00 [M] . 27. H – Se 0.35 [NC] . 40. Cs – F 3.21 [I] . 1.83 1.83 ∆ EN < 0.5 & 2M’s 2.20 2.55 ∆ EN < 0.5 & 2NM’s 0.79 4.00 ∆ EN ≥1.7 15. Mg – F 2.69 [I] . 28. H – Br 0.76 [PC] . 41. O – O 0.00 [NC] . 1.31 4.00 ∆ EN ≥1.7 2.20 2.96 ∆ EN ≤1.7 but ≥ 0.5 3.44 3.44 ∆ EN < 0.5 & 2NM’s 16. Mg – O 2.13 [I] . 29. H – I 0.46 [NC] . 42. N – O 0.40 [NC] . 1.31 3.44 ∆ EN ≥1.7 2.20 2.66 ∆ EN < 0.5 & 2NM’s 3.04 3.44 ∆ EN < 0.5 & 2NM’s 17. Cu – Cu 0.00 [M] . 30. N – H 0.84 [PC] . 43. Fe – Co 0.05 [M] . 1.90 1.90 ∆ EN < 0.5 & 2M’s 3.04 2.20 ∆ EN ≤1.7 but ≥ 0.5 1.83 1.88 ∆ EN < 0.5 & 2M’s 18. Ca – O 2.44 [I] . 31. P – H 0.01 [NC] . 44. I – Br 0.30 [NC] . 1.00 3.44 ∆ EN ≥1.7 2.19 2.20 ∆ EN < 0.5 & 2NM’s 2.66 2.96 ∆ EN < 0.5 & 2NM’s Remember: ∆EN = │EN 1 – EN 2 │, is NEVER a – #, PLEASE GO BACK AND ENSURE NO ∆EN IS A “– #” Regarding Electronegativity Difference: It is somewhat artificial to classify bonds based on the differences in the electronegativities (ΔEN) of the two atoms. However, we will use these ranges to do so: Ionic ΔEN > 1.7 (symbolized as A + and Z – ) Metallic ΔEN < 0.5 (symbolized as M1 δ+ and M2 δ+ ) Polar Covalent 1.7 ≥ ΔEN ≤ 0.5 (symbolized as A δ+ and Z δ– ) Non-Polar Covalent ΔEN < 0.5 (symbolized as A 0 and Z 0 ) YOU NEED TO KNOW THESE!!!!
Transcript of “I Pledge”: ( Initial DON’T FORGET WHAT THIS … · What does VSEPR stand for and what it is...
4.2.A – ReIntro to Bonding “I Pledge”: _____ ( Initial ) DON’T FORGET WHAT THIS REPRESENTS…. Instructions: Provide a response for each question that is well thought out, satisfies the prompt, is clearly explained, and LEGIBLE.
1. Why do atoms and ions form bonds? Atoms / Ions donate / accept / share e─ in order to form a low energy stable state (communally isoelectric to noble gases) _
2. What is the difference between an intramolecular force and an intermolecular force (VERY IMPORTANT, differences must take into account EN and polarity)?
Intra- = Strong = Chemical = determined by ∆EN resulting in polar and nonpolar molecules = between atoms / ions forming a molecule / FU (determines chemical .
properties) _
Inter- = Weak = Physical = determined by polarity and e─ movement = between molecules (determines physical properties ) .
3. What role do e- play in ionic bonding and why? e─ are transferred from low EN atoms ( metals ) to a high EN atom ( nonmetals ) due to an ∆EN ≥ 1.7 . Difference .
is enough that the more EN atom is able to rip away, or “bully” the high energy e─ of the low “whimpy” EN atom resulting in a “permanent” Coulombic attraction .
4. What role do e- play in covalent bonding and why? e─ are shared between high EN atoms ( nonmetals ) due to an ∆EN < 1.7 . Difference between the two high EN .
“bullies” is not enough that one atom is able rip away the high e─ of the other so some form of imperfect sharing occurs .
5. What role do e- play in metallic bonding and why? e─ are shared between uniformly between low EN atoms ( metals ) due to an ∆EN < 1.7 . Difference between the two
low EN “wimps” is not enough that ANY atom in the bond is able rip away the high e─ of the others so perfectly uniform sharing occurs = e─ sea .
Instructions: Determine the electronegativity difference (ΔEN) for the following chemical bonds in the indicated blank. No work need be shown. Then using the ΔEN determine if the bond is considered a nonpolar covalent bond [NC], polar covalent [PC], ionic bond [I], or metallic bond [M]. Remember: ∆EN = │EN1 – EN2 │, is NEVER a ─# ΔEN Bond Type ΔEN Bond Type ΔEN Bond Type 6. Li – H 1.22 [PC] . 19. Cu – Zn 0.25 [M] . 32. C – H 0.35 [NC] . 0.98 2.20 ∆EN ≤1.7 but ≥ 0.5 1.90 1.65 ∆EN < 0.5 & 2M’s 2.55 2.20 ∆EN < 0.5 & 2NM’s
7. Li – Cl 2.18 [I] . 20. Ca – Cl 2.16 [I] . 33. Si – H 0.30 [NC] . 0.98 3.16 ∆EN ≥1.7 1.00 3.16 ∆EN ≥1.7 1.90 2.20 ∆EN < 0.5 & 2NM’s
8. Na – H 1.27 [PC] . 21. Ca – S 1.58 [PC] . 34. C – O 0.89 [PC] . 0.93 2.20 ∆EN ≤1.7 but ≥ 0.5 1.00 2.58 ∆EN ≤1.7 but ≥ 0.5 2.55 3.44 ∆EN ≤1.7 but ≥ 0.5
9. Na – Cl 2.23 [I] . 22. Sr – F 3.05 [I] . 35. Al – Cl 0.96 [PC] . 0.93 3.16 ∆EN ≤1.7 but ≥ 0.5 0.95 4.00 ∆EN ≥1.7 2.20 3.16 ∆EN ≤1.7 but ≥ 0.5
10. K – H 1.38 [PC] . 23. Ba – F 3.11 [I] . 36. C – Cl 0.61 [PC] . 0.82 2.20 ∆EN ≤1.7 but ≥ 0.5 0.89 4.00 ∆EN ≥1.7 2.55 3.16 ∆EN ≤1.7 but ≥ 0.5
11. K – Cl 2.34 [I] . 24. Fr – F 3.30 [I] . 37. H – H 0.00 [NC] . 0.82 3.16 ∆EN ≥1.7 0.70 4.00 ∆EN ≥1.7 2.55 2.55 ∆EN < 0.5 & 2NM’s
12. Fr – N 2.34 [I] . 25. H – O 1.24 [PC] . 38. S – O 0.86 [PC] . 0.70 3.04 ∆EN ≥1.7 2.20 3.44 ∆EN ≤1.7 but ≥ 0.5 2.58 3.44 ∆EN ≤1.7 but ≥ 0.5
13. Fr – Cl 2.46 [I] . 26. H – S 0.38 [NC] . 39. P – O 1.25 [PC] . 0.70 3.16 ∆EN ≥1.7 2.20 2.58 ∆EN < 0.5 & 2NM’s 2.19 3.44 ∆EN ≤1.7 but ≥ 0.5
14. Fe – Fe 0.00 [M] . 27. H – Se 0.35 [NC] . 40. Cs – F 3.21 [I] . 1.83 1.83 ∆EN < 0.5 & 2M’s 2.20 2.55 ∆EN < 0.5 & 2NM’s 0.79 4.00 ∆EN ≥1.7
15. Mg – F 2.69 [I] . 28. H – Br 0.76 [PC] . 41. O – O 0.00 [NC] . 1.31 4.00 ∆EN ≥1.7 2.20 2.96 ∆EN ≤1.7 but ≥ 0.5 3.44 3.44 ∆EN < 0.5 & 2NM’s
16. Mg – O 2.13 [I] . 29. H – I 0.46 [NC] . 42. N – O 0.40 [NC] . 1.31 3.44 ∆EN ≥1.7 2.20 2.66 ∆EN < 0.5 & 2NM’s 3.04 3.44 ∆EN < 0.5 & 2NM’s
17. Cu – Cu 0.00 [M] . 30. N – H 0.84 [PC] . 43. Fe – Co 0.05 [M] . 1.90 1.90 ∆EN < 0.5 & 2M’s 3.04 2.20 ∆EN ≤1.7 but ≥ 0.5 1.83 1.88 ∆EN < 0.5 & 2M’s
18. Ca – O 2.44 [I] . 31. P – H 0.01 [NC] . 44. I – Br 0.30 [NC] . 1.00 3.44 ∆EN ≥1.7 2.19 2.20 ∆EN < 0.5 & 2NM’s 2.66 2.96 ∆EN < 0.5 & 2NM’s
Remember: ∆EN = │EN1 – EN2 │, is NEVER a – #, PLEASE GO BACK AND ENSURE NO ∆EN IS A “– #”
Regarding Electronegativity Difference: It is somewhat artificial to classify bonds based on the differences in the electronegativities (ΔEN) of the two atoms. However, we will use these ranges to do so:
Ionic ΔEN > 1.7 (symbolized as A+ and Z–) Metallic ΔEN < 0.5 (symbolized as M1δ+ and M2δ+)
Polar Covalent 1.7 ≥ ΔEN ≤ 0.5 (symbolized as Aδ+ and Zδ–) Non-Polar Covalent ΔEN < 0.5 (symbolized as A0 and Z0)
YOU NEED TO KNOW THESE!!!!
4.2.B – Lewis Structures , “I Pledge”: _____ ( Initial ) DON’T FORGET WHAT THIS REPRESENTS…. Instructions: Draw Lewis Structures for each of the following atoms and then draw the Lewis Structures for its most common ion identifying how many valence electrons are present.
Sodium Atom
# Val e-
1 . Sodium Ion # Val e-
0 . Calcium
Atom # Val e-
2 . Calcium
Ion # Val e-
0 . Aluminum
Atom # Val e-
3 . Aluminum
Ion # Val e-
0 .
Lead Atom # Val e-
4 . Lead (IV)
Ion # Val e-
0 . Nitrogen
Atom # Val e-
5 . Nitride Ion # Val e-
8 . Sulfur Atom # Val e-
6 . Sulfide Ion # Val e-
8 .
Strontium Atom
# Val e-
2 . Strontium
Ion # Val e-
0 . Selenium
Atom # Val e-
6 . Selenide
Ion # Val e-
8 . Bromine
Atom # Val e-
7 . Bromide
Ion # Val e-
8 .
Instructions: For problems 19 – 30, given the following Lewis structures, provide the formula they represent. REMEMBER, the more EN atom is written last.
Formula HF . Formula N2 . Formula NH3 . Formula CH4 . Formula CF4 . Formula NO+ .
Formula H4C3O3 . Formula H6C2O . Formula H6C6 . Formula PF5 . Formula NH4+ . Formula ClO2─ .
Instructions: For problems 31 – 46, for each formula given, write the correct Lewis electron-dot structure for the compound ON A SEPARATE SHEE OF PAPER. All 3 steps
must be shown for CR. MUST SHOW ALL 3 “Steps” and circle final step to receive credit31. CH4
32. C2H4
33. PO33 ―
34. NO2 ―
35. NF3
36. Al H3
37. H2O
38. C2H2
39. O2
40. N2
41. CH3COO ― Use hint of how the formula is written
42. CN ―
43. C2F2O2
44. SO22-
45. C6H12 (Hint: Ring)
46. C6H6 (Hint: Ring)
4.2.C – VSEPR , “I Pledge”: _____ ( Initial ) DON’T FORGET WHAT THIS REPRESENTS…. Instructions: Provide a response for each question that is well thought out, satisfies the prompt, is clearly explained, and LEGIBLE.
1. What does VSEPR stand for and what it is it significance in determining chemical structures? Valence Shell Electron Pair Repulsion .
The 3-D arrangement of a molecule is determined by the repulsion of it e─ pairs ( both bonding and non-bonding ). Note, lone pairs have a greater overall effect on .
the structure due to being in a less defined space .
Instructions: Given the following Lewis Structures, complete the table as indicated. Lewis Structure Chemical
Formula Central Atom
Steric Number
AXE Notation
VSEPR Geometry Lewis Structure Chemical Formula
Central Atom
Steric Number
AXE Notation
VSEPR Geometry
BeCl2 Be 2 AX2 Linear
C2H2 C 2 AX2 Linear
BH3 B 3 AX3 Trigonal Planar
C2H4 C 3 AX3 Trigonal Planar
COH2 C 3 AX3 Trigonal Planar
C2H4 C 3 AX3 Trigonal Planar
CO2 C 2 AX2 Linear
HF F 1 AXE3 Linear
Instructions: For problems 10 - 24, identify what hybridization is required at the central atom of the following molecules or ions by sketching the Lewis electron-dot diagram;
state the Steric Number (SN), then state the AXE notation, the molecular geometry, and whether the molecule is polar or nonpolar.
10. AlCl3 3 3 Bonded + 0 Lone Pairs AX3
trigonal planar nonpolar symmetric charge distribution on central.
11. BCl3 3 3 Bonded + 0 Lone Pairs AX3
trigonal planar nonpolar symmetric charge distribution on central.
12. SCl2 4 2 Bonded + 2 Lone Pairs AX2E2
bent polar asymmetric charge distribution on central.
13. NO2+
2 2 Bonded + 0 Lone Pairs AX2
linear polar is an ion…
14. ClO4-
4 4 Bonded + 0 Lone Pairs AX4
tetrahedral polar is an ion…
15. PCl3 4 3 Bonded + 1 Lone Pairs AX3E
trigonal pyramidal polar asymmetric charge distribution on central.
16. BF3 3 3 Bonded + 0 Lone Pairs AX3
trigonal planar nonpolar symmetric charge distribution on central.
17. CCl4 4 4 Bonded + 0 Lone Pairs AX4
tetrahedral nonpolar symmetric charge distribution on central.
18. GaH3 3 3 Bonded + 0 Lone Pairs AX3
trigonal planar nonpolar symmetric charge distribution on central.
19. BeCl2 2 2 Bonded + 0 Lone Pairs AX2
linear polar asymmetric charge distribution on central.
20. PCl5 5 5 Bonded + 0 Lone Pairs AX5
Trigonal bipyramidal nonpolar symmetric charge distribution on central.
21. SiCl4 4 4 Bonded + 0 Lone Pairs AX4
tetrahedral nonpolar symmetric charge distribution on central.
22. NCl3 4 3 Bonded + 1 Lone Pairs AX3E
trigonal pyramidal polar asymmetric charge distribution on central.
23. OCS 2 2 Bonded + 0 Lone Pairs AX2
linear polar asymmetric charge distribution due to differences in O / S.
24. ClF2+
4 2 Bonded + 2 Lone Pairs AX2E2
linear polar is an ion…
4.2.D – Compounds: Polarity and Lattices “I Pledge”: _____ ( Initial ) DON’T FORGET WHAT THIS REPRESENTS…. Instructions: For each of the bonds listed below, classify each bond as either nonpolar covalent bond [NC], polar covalent [PC], ionic bond [I], or metallic bond [M] based on ΔEN and indicate full ( + / ― ), partial charges ( δ+ / δ― ) , or no charge ( 0 ).
1. I Na—Cl 3. PC Mg—H 5. PC C—O 7. I Cs—F 0.93 3.16 1.31 2.20 2.55 3.44 0.79 4.00
2. PC Cu—O 4. NC Cl—Cl 6. NC C—H 8. PC Al—C 1.90 3.44 3.16 3.16 2.55 2.20 2.20 2.55
Instructions: For each of the bonds listed below, indicate the direction of bond polarity ( ), or if no polarity ( )
9. Na—Cl 11. Mg—H 13. C—O 15. Cs—F 0.93 3.16 1.31 2.20 2.55 3.44 0.79 4.00 10. Cu—O 12. Cl—Cl 14. C—H 16. Al—Cl
1.90 3.44 3.16 3.16 2.55 2.20 2.20 2.55 Instructions: For problems 17 – 30, indicate the strongest IMF (not a trick question) holding together crystals of the following:
Molecular Crystals Metal Ionic Crystals
Network Solid
Molecular Crystals Metal Ionic Crystals
Network Solid
Dispersion Forces
Dipole-Dipole
Hydrogen Bonds
Metallic Bonds
Ionic Bonds Covalent Bonds
Dispersion Forces
Dipole-Dipole
Hydrogen Bonds
Metallic Bonds
Ionic Bonds Covalent Bonds
17. NH3 X 32. N2 X No metals involved, bond is polar and is N w/ H. Remember that H w/ N, O or F is hydrogen bonding No metals involved, no polar bond. As this is an element bonded to itself, no permanent IMF’s exist
18. Kr X 33. CH3OH X No metals involved, no polar bond. Also this is a noble gas, so no permanent IMF’s exist No metals involved, is A bond polar and is O w/ H. Remember that H w/ N, O or F is hydrogen bonding
19. HCl X 34. Ag X No metals involved, bond is polar but is not H w/ N, O or F Metals involved, and just metals.
20. F2 X 35. NaOH X No metals involved, no polar bond. As this is an element bonded to itself, no permanent IMF’s exist Metal involved with a polar ionic bond present.
21. KMnO4 X 36. Al X Metal involved with a polar ionic bond present. Metals involved, and just metals.
22. NaCl X 37. PCl3 X Metal involved with a polar ionic bond present. No metals involved, bond is polar but is not H w/ N, O or F. Also asymmetrical charge distribution
23. SO2 X 38. XeF4 X No metals involved, polar bond exists between S─O. Also has a bent shape No metals involved, bond is polar but charge distribution is symmetrical.
24. CO2 X 39. He X No metals involved, no polar bond. Symmetrical charge distribution with its linear shape. No metals involved, no polar bond. Also this is a noble gas, so no permanent IMF’s exist
25. C6H6 X 40. Na X No metals involved, no polar bond. Metals involved, and just metals.
26. NO X 41. CO X
δ+ δ─ δ+ δ─ 0 0 0 0
+ ─ + ─ δ+ δ─ δ+ δ─
No metals involved, polar bond exists between N─O. Also an asymmetrical charge distribution No metals involved, polar bond exists between C─O. Also an asymmetrical charge distribution
27. H2SO4 X 42. Ar X No metals involved, bond is polar but is not H w/ N, O or F No metals involved, no polar bond. Also this is a noble gas, so no permanent IMF’s exist
28. WC X 43. Ba(OH)2 X Metal involved with a polar ionic bond present. The hint that this isn’t just dipole-dipole is lack of subscripts. Metal involved with a polar ionic bond present.
29. Si X 44. O2 X One of three covalent networks that we are supposed to recognize. No metals involved, no polar bond. As this is an element bonded to itself, no permanent IMF’s exist
30. SiO2 X 45. Hg X AKA quartz, one of three covalent networks that we are supposed to recognize. Metals involved, and just metals.
31. C(graphite( X 46. XeF2 X One of three covalent networks that we are supposed to recognize. No metals involved, bond is polar but charge distribution is asymmetrical unlike #22.
