المحاضرة الثانية عشر في الكيمياء اللاعضوية للصف الثاني لقسم الكيمياء

Homework 7 – Lewis structures and resonance
1.) Draw 4 resonance structures for N2O4 (no O-O bonds, there IS a bond between N-N): use formal charge calculations to determine the “best” resonance structure(s)
Valence electrons: N: 2 x 5 = 10 O: 4 x 6 = 24 10+ 24 = 34 valence electrons
I II III IV
N1N2O1O3O2O4 N1N2O1O4O2O3 N1N2O3O4O1O2 N1N2O2O3O1O4
2.)
3.)
4.)
5.)
6.)
7.)
8.)
9.)
10.)
11.)
12.)
13.)
All structures contribute equally – all have two Nitrogens at +1, two oxygens at -1, and 2 oxygens at 0. No structure is better than another: molecular shape: trigonal planar VSEPR: trigonal planar, hybridization: sp2, angle: 120o
2.) Draw 3 resonance structures for HN3 (NOT NH3!!!!) – (skeletal H-N-N-N) use formal charge calculations to determine the “best” resonance structure(s) – model your set-up as question 1 is displayed
N: 3 x 5 = 15 H: 1 x 1 = 1 15 + 1 = 16 valence electrons
I II III
HN1N2N3 HN1N2N3 HN1N2N3
H: 1 -1 = 0 H: 1 – 1 =0 H: 1 – 1 = 0
N1: 5 – 5 = 0 N1: 5 – 4 = +1 N1: 5 – 6 = -1 N2: 5 – 4 = +1 N2: 5 – 4 = +1 N2: 5 – 4 = +1
N3: 5 – 6 = -1 N3: 5 – 7 = -2 N3: 5 – 5 = 0
Structures I and III are equivalent and are better than structure II: Molecular shape: linear, VSEPR: linear, hybridization: sp , angle: 180o
Formal Charges
N1: 5 - 4 = +1
N2: 5 – 4 = +1
O1: 6 – 6 = 0
O2: 6 – 6 = 0
O3: 6 – 7 = -1
O4: 6 – 7 = -1
Formal Charges
N1: 5 – 4 = +1
N2: 5 – 4 = +1
O1: 6 – 6 = 0
O2: 6 – 7 = -1
O3: 6 – 7 = -1
O4: 6 – 6 = 0
Formal Charges
N1: 5 – 4 = +1
N2: 5 – 4 = +1
O1: 6 – 7 = -1
O2: 6 – 7 = -1
O3: 6 – 6 = 0
O4: 6 – 6 = 0
Formal Charges
N1: 5 – 4 = +1
N2: 5 – 4 = +1
O1: 6 – 7 = -1
O2: 6 – 6 = 0
O3: 6 – 6 = 0
O4: 6 – 7 = -1
3.) Draw 3 resonance structures for SO2 (no O-O bonds) – use formal charges to determine the “best” structure(s) – model your set-up as question 1 is displayed
S: 1 x 6 = 6 O: 2 x 6 = 12 6 + 12 = 18 valence electrons
I II III
O1SO2 O1SO2 O1SO2
S: 6 – 5 = +1 S: 6 – 5 = +1 S: 6 – 6 = 0
O1: 6 – 6 = 0 O1: 6 – 7 = -1 O1: 6 – 6 = 0
O2: 6 – 7 = -1 O2: 6 – 6 = 0 O2: 6 – 6 = 0
Structure 3 is the best structure – all formal charge values are zero! Molecular shape: bent, VSEPR: trigonal planar, hybridization: sp2 angle: 120o
4.) For SO4-2 draw at least 3 resonance structures – one of which MUST be the “best” structure for the ion (which means minimize the formal charge value on the central atom)! (hint: make your structure symmetrical)- use formal charge calculations to determine which ion is the “best” resonance structure – model your set-up as question 1 is displayed.
Valence: S: 6 x 1 = 6 O: 4 x 6 = 24 6 +24 = 30 + 2 = 32 valence electrons
I II III
SO1O2O3O4-2 SO2O1O3O4-2 SO2O4O1O3-2
S: 6 – 4 = +2 S: 6 – 5 = +1 S: 6 – 6 = 0
O1: 6 – 7 = -1 O1: 6 – 7 = -1 O1: 6 – 7 = -1
O2: 6 – 7 = -1 O2: 6 – 6 = 0 O2: 6 – 6 = 0
O3: 6 – 7 = -1 O3: 6 – 7 = -1 O3: 6 – 7 = -1
O4: 6 – 7 = -1 O4: 6 – 7 = -1 O4: 6 – 6 = 0
The best structure is structure III – central atom is minimized (formal charge = 0) and two oxygens have 0 formal charge values and only 2 have -1 formal charge values.
There are 3 additional resonance structures based on structure II, where the double bond rotates around the structure
There are 4 additional resonance structures for structure III where there are two sets of double bonds that rotate around the structure
5.) Draw 2 resonance structures for O3 ( skeletal structure is O - O – O it is not a ring!) use formal charge calculations to determine which ion is the “best” resonance structure – model your set-up as question 1 is displayed
O: 6 valence x 3 oxygens = 18 total valence electrons
OOO OOO
1 2 3 1 2 3
O1 : 6-7 = -1 O1 : 6-6 =0
O2 : 6-5 = +1 O2 : 6-5 = +1
O3 : 6-6 = 0 O3 : 6 -7 = -1
Neither structure is better than the other. Both have a zero formal charge Oxygen, a -1 formal charge oxygen, and a +1 formal charge oxygen. Since neither structure is better , they contribute equally to the overall resonance hybrid: molecular: bent, VSEPR: trigonal planar, hybridization is sp2, angle is 120o
6.) Draw 3 resonance structures for HNO3 (H bonded to O, no O-O bonds) use formal charge calculations to determine which ion is the “best” resonance structure – model your set-up as question 1 is displayed
Valence e-1 = 5 valence nitrogen + 1 valence hydrogen + (6 valence oxygen x 3) = 24 valence electrons
NOOOHNOOOHNOOOH
STRUCTURE 1 STRUCTURE 2 STRUCTURE 3
N : 5 – 4 = +1
O1 : 6 – 7 = -1
O2 : 6 – 7 = -1
O3 : 6 – 5 = + 1
H : 1 – 1 = 0
1
1
1
2
3
2
3
2
3
N : 5 – 4 = +1
O1 : 6 – 6 = 0
O2 : 6 – 7 = -1
O3 : 6 – 6 = 0
H : 1 – 1 = 0
N : 5 – 4 = +1
O1 : 6 – 7 = -1
O2 : 6 – 6 = 0
O3 : 6 – 6 = 0
H : 1 – 1 = 0
The first structure has 2 species at a +1 formal charge and 2 species at a -1 formal charge. Structures 2 and 3 only have 1 species at a +1 formal charge and 1 species at a -1 formal charge. The rest of the atoms are at a zero formal charge. This means that structures 2 and 3 are better than structure 1. BUT, there is again, no difference between structures 2 and 3. They are equal to one another (the nitrogen is the least electronegative atom and it carries the +1 formal charge value while the oxygen, which is more electronegative carries the -1 formal charge value). Therefore, structures 2 and 3 contribute equally to the overall resonance hybrid: molecular shape is trigonal planar, VSEPR is trigonal planar, angle is 120o, hybridization is sp2
7.) Draw 2 resonance structures for [NO2]-1 use formal charge calculations to determine which ion is the “best” resonance structure – model your set-up as question 1 is displayed
N = 5 valence + (6 valence x 2 oxygens) + 1e-1 = 18 total valence electrons
ONO ONO
1 2 1 2
O1 : 6-6 = 0 O1 : 6-7 = -1
N : 5-5 = 0 N : 5-5 = 0
O2 : 6-7 = -1 O2 : 6 -6 = 0
Neither structure is better than the other. Both have a zero formal charge Oxygen, a -1 formal charge oxygen, and a +1 formal charge nitrogen. Since neither structure is better , they contribute equally to the overall resonance hybrid: molecular shape is bent, VSEPR is trigonal planar, hybridization is sp2, angle is 120o