Chapter 13 : Bonding General Concepts
most substances: compounds and mixtures of compounds
rare examples: He, Ar – exist as atoms
example: graphite (soft) & diamond (hard)
CO2 (gas) & SiO2 (solid)
C : 2p orbital (small) Si : 3p orbital (large)
To form C=O double bond To form Si-O single bond
C=O bond energy : 192 kcal/mole Si-O bond energy : 110 kcal/mole
bonding – affects chemical & physical properties
Melting point, hardness, solubility …
molecular bonding &structure – important in determining
chemical reactions
The factors that control the structure of compounds
Chapter 13 : Bonding General Concepts
Bonding nature : the interaction of different wavefunctions (orbital)
To explain the bonding and structures :
Valence bond theory (VBT) & Molecular Orbital Theory (MO)
Bonding energy : different type of bonding
Covalent bond (strong) Ionic bond (relative weak)
Electron are shared
H+H
Repulsive (positive) term:
proton-proton and
electron –electron interaction
Attractive (negative) term:
electron-proton interaction
Electron are separated
Na+ + ClCation
Anion
The energy of interaction
between a pair of ions
calculated by the Coulomb’s
Law:
Chapter 13-1 : Type of Chemical Bonding
Covalent bond (strong)
Ionic bond (relative weak)
Bond Strength : Single bond < double bond < triple bond
Chapter 13 : Bonding General Concepts
當兩原子過度靠近時,
會增加核與核間以及電
子與電子間的排斥力
兩原子距離無窮遠時,兩者
間無作用力
H1 上的電子會與H2 核
有作用力,使其穩定
Chapter 13-1 : Type of Chemical Bonding
(partial)
e- transfer
Sharing electron
Na+ + Cl-
H-H
ionic bonding
covalent bonding
?
*electron transfer from one to another
*between two very different atoms
*equal sharing of electrons
*between two identical atoms
6
δδ+
δ+
1:9
determine
by electronegativity
δ-
2 : 8 or
3 : 7 or
4 : 6 …..
5:5
Chapter 13-2 : Electronegativity scale
electronegativity:
in a molecule, the ability of an atom to attract shared electrons
determining electronegativity – Linus Pauling (Figure 13.2)
relative electronegativity:
for H-X
實際的鍵能
∆= (B.E.H-X)act – (B.E.H-X)exp 預期的鍵能(
預期的鍵能(考慮電子共享:
考慮電子共享:共價鍵的狀態下)
共價鍵的狀態下)
expected H-X bond energy: (B.E.H-X)exp = (B.E.H-H + B.E.X-X)/2
*if ∆ = 0
H & X: same electronegativity
determine electronegativities of elements
其差值(
其差值(Δ) 代表鍵結被極化的程度
由此建立電負度表
Chapter 13-2 : Electronegativity scale
Chapter 13-2 : Electronegativity scale
H2C=CH2
<
<
Question Remark 1 :
Acidity : HC≡CH
H3C-CH3
Why ?
HC≡CH H2C=CH2 H3C-CH3
sp2
sp
sp3
50% (1/2) > 33%(1/3) > 25%(1/4)
Csp > Csp2 > Csp3
hybridization
S character ; the s electrons are closer to nuclear
Electronegativity
H-Csp > H-Csp2 > H-Csp3
Iconicity
H-Csp < H-Csp2 < H-Csp3
Bond energy
H2C=CH2
<
<
HC≡CH
H3C-CH3
Acidity
Chapter 13-3 : bond polarity and dipole moment
Chapter 13-3 : bond polarity and dipole moment
Chapter 13-3 : bond polarity and dipole moment
Molecular polarity : bond pair dipole & lone pair dipole
Lone pair are higher electron negativity
Question Remark 2 :
The molecular polarity : NH3 > NF3
Why ?
Electronegativity : N = 3.0 H= 2.1 F= 4.0
:
:
δ-
δ+
N
δ+ H
Cancellation
(lone pair dipole and
bond pair dipole)
H
H δ+
Dipole moment : 1.47 Debye
δ-
N
δ+
δ-
F
F
F δ-
Dipole moment : 0.24 Debye
Chapter 13-3 : bond polarity and dipole moment
Chapter 13-4 : Ions : Electron configuration and Sizes
Atoms in stable compounds usually have a noble gas arrangement of electrons
1. nonmetallic elements - other nonmetals(by sharing electrons to form covalent bonds)
2. nonmetallic elements - metals(by taking electrons from metals to form anions and cations)
Complete the valence configuration of both atoms.
The valence orbitals of the main group metal are empty.
Chapter 13-4 : Ions : Electron configuration and Sizes
in periodic table
elements in the same group: same valence electron configuration
similar chemical behavior (like 1A)
K (potassium) 1s22s22p63s23p64s1
= [Ar]4s1
Ca (calcium)
[Ar]4s2
Sc (scandium) [Ar]4s23d1
Ti (titanium)
[Ar]4s23d2
transition metals (elements)
39
Chapter 13-4 : Ions : Electron configuration and Sizes
Full of 4s2 and 3d10
Table list the outermost orbital
Full of 4s2
Half-filled
Sc : [Ar] 4s23d1
Half-filled
Ga : [Ar] 4s23d104p1
When d-block elements form ions, the 4s
electrons are lost first. r4s > r3d
Sc : [Ar] 4s23d1
Sc+ : [Ar] 4s13d1
Sc2+ : [Ar] 3d1
Cr : [Ar] 4s13d5
Cr+ : [Ar] 3d5
Cr2+ : [Ar] 3d4
Mn : [Ar] 4s23d5
Mn+ : [Ar] 4s13d5
Mn2+ : [Ar] 3d5
Chapter 13-4 : Ions : Electron configuration and Sizes
Chapter 13-4 : Ions : Electron configuration and Sizes
Chapter 13-4 : Ions : Electron configuration and Sizes
Chapter 13-5 : Formation of Binary : Ionic Compounds
Chapter 13-5 : Formation of Binary : Ionic Compounds
= ? ∆Hf
How to calculate the enthalpy of formation ?
∆Hs
Li(s)
Li(g)
½ F2(g)
Li(g)
Li+(g)
F(g)
F(g) ∆HD
F-(g)
EA
IE
U0
A cycle : (Born – Haber Cycle)
∆Hf = ∆Hs + ∆HD + IE + EA + U0
= 昇華能
Hess ‘ Law 黑斯定律
+ 斷鍵能 + 游離能 + 電子親和力 + 晶格能
Chapter 13-5 : Formation of Binary : Ionic Compounds
斷鍵能(吸熱
斷鍵能 吸熱):
吸熱
Dissociation energy
第一游離能(吸熱
第一游離能 吸熱)
吸熱 :
1st ionization energy
昇華能(吸熱
昇華能 吸熱):
吸熱
Sublimation energy
電子親合能
放熱):
電子親合能(放熱
放熱
Electron affinity
晶格能(放熱
晶格能 放熱):
放熱
Lattice energy
從自然界最穩定的狀態出發 : Li(s) , F2(g)
∆Hf = ∆Hs + ∆HD + IE + EA + U0 = 161 + 520 + 77 – 328 -1047 = -617 kJ(放熱
放熱)
放熱
Chapter 13-5 : Formation of Binary : Ionic Compounds
Example : NaF
Formation of enthalpy :
109 + 495 + 77 -328 – 923 = -570 kJ/mol (放熱
放熱)
放熱
Chapter 13-5 : Formation of Binary : Ionic Compounds
Example : MgO
O
O2-
need energy (吸熱
吸熱)
吸熱
Mg
Mg+
Mg
Mg2+
Mg+
Mg2+
1st ionization energy : 735 kJ/mol
2nd ionization energy : 1445 kJ/mol
1st + 2nd ionization energy (吸熱
吸熱)
吸熱
735+1445 = 2180 kJ/mol
Formation of enthalpy :
150 + 2180 + 247 +737 – 3916 = -602 kJ/mol (放熱
放熱)
放熱
Chapter 13-6 : Partial Ionic Character
If the charge are the same ?
use the Fajan’s Rule Polarizibility
Cation : radius
Polarizing Power
Stability
Anion : radius
Polarizing Power
Stability
How to determine the trend of the melting point for the following :
BeCO3 , MgCO3, CaCO3 , SrCO3
Radius trend : Be2+ < Mg2+ < Ca2+ < Sr2+
2A Cation
Polarizing Power : Be2+ > Mg2+ > Ca2+ > Sr2+
Iconicity trend : BeCO3 > MgCO3 > CaCO3 > SrCO3
Decomposed trend : BeCO3 < MgCO3 < CaCO3 < SrCO3
Chapter 13-6 : Partial Ionic Character
Chapter 13-1-13-4 : Exercise
1. Predict the order of increasing electronegativity in each of the following groups.
a. C, N, O b. S, Se, Cl c. Si, Ge, Sn d. Tl, S, Ge
2. Predict the type of bond(ionic, covalent, or polar covalent) one would expect to form
between the following pairs of elements.
a. Rb and Cl b. S and S c. C and F d. Ba and S e. N and P
3. For each of the following groups, place the atoms in order of decreasing size.
a. Cu, Cu+, Cu2+ b.Ni2+, Pd2+, Pt2+ c. Te2-, I-, Cs+, Ba2+, La3+
Chapter 13-7 : Covalent Bond
Covalent Bond : be viewed as forces that cause a group of atoms to
behave as a unit (Share the Electrons)
For example:
H atom + H atom →H2 molecule
The tendency of a system to seek its lowest possible energy
H+H
H2
∆ H = -458 KJ/mol
(Bonding Energy of H-H)
Lowest energy
Share the Electrons
A bonds represents a quantity of energy obtained from the
overall molecular energy of stabilization
Chapter 13-8 : Covalent Bond and Chemical Reaction
Then… we can calculate covalent bond energy (Diatomic Molecules)
O-O : 146 kJ/mol , Cl-Cl : 239 kJ/mol , F-F : 154 kJ/mol ….
How we evaluate the chemical bond in a Molecule ?
C+ 4H
CH4
bond energy for C-H?
CH3(g) + H(g)
CH4(g)
CH3(g)
CH2(g) + H(g)
CH2(g)
CH(g) + H(g)
CH(g)
C(g) + H(g)
1652 kJ/mol
1652 kJ/mol
∆ E: 435 kJ/mol
∆ E: 453 kJ/mol
∆ E: 425 kJ/mol
∆ E: 339 kJ/mol
average: 1652/4 = 413
Think of CH4 as containing four C-H bonds, each worth 413 kJ/mol
of bonds
Chapter 13-8 : Covalent Bond and Chemical Reaction
How we evaluate the enthalpy (∆H) for a chemical Reaction ?
(by considering all of the bonding energies involved in this Reaction)
Energy
2AB
Product : 產物
A2 + B 2
Adsorb energy
Reactant : 反應物
∆H > 0 (吸熱:endothermic)
C2 + D2
Reactant : 反應物
Release energy
2CD
Product : 產物
∆H < 0 (放熱:exothermic)
Bond Forming : (release energy) 生成鍵結 : 釋出能量
Bond Breaking : (adsorb energy) 破壞鍵結 : 吸收能量
Chapter 13-8 : Covalent Bond and Chemical Reaction
A2 + B2
2AB
∆H = ?
∆H = break bonds(斷鍵總能量) - formed bonds(生成鍵總能量)
= EAA + EBB - 2 EAB
B
A
A
+
B
B
A
B
A
假設鍵結全部破壞,重新生成 :
break the A-A and B-B bonds and formed the two A-B bond
Chapter 13-8 : Covalent Bond and Chemical Reaction
N ≡N + 3H-H
2 NH3
∆H = ?
∆H = break bonds(斷鍵總能量) - formed bonds(生成鍵總能量)
= EN ≡ N – 3 x EHH - 2 x (3 x ENH)
= 941 + 3 x 432 - 2 x 3 x 391
= -109 kJ/mol (放熱反應)
Chapter 13-8 : Covalent Bond and Chemical Reaction
Consider the following reaction :
H2C=CH2 + F2 CFH2-CFH2 ∆H = -549kJ/mol
Estimate the C-F bond energy given that the
C-C bond energy is 347 kJ/mol
C=C bond energy is 614 kJ/mol
F-F bond energy is 154 kJ/mol
∆H = Σbreaking bonds – Σforming bonds
( 4 x C-H + C=C + F-F ) – ( 2 x C-F + C-C + 4 x C-H ) = -549
C=C + F-F – C-C – 2 x C-F = -549
614 + 154 – 347 – 2 x C-F = -549
the C-F bond energy = 485 kJ/mol
Chapter 13
Exercise : Calculate the ∆H for each of the following reactions
HCN + 2H2
N2H4 + 2F2
CH3NH2
N2 + 4HF
Chapter 13-10 : Lewis Structures
Consider valence electrons :
Valence Bond Theory (VBT): use to determine molecular structure
Localized Electron Model : (Covalent bond)
Two type of electron pair :
1) Lone pairs (LP) : pairs of electrons localized on an atom
2) Bond pairs (BP) : pairs of electrons found in space between atoms
Lewis Structure : show how the valence electrons are arranged among
the atoms in the molecule
[
]
-
Br
Chapter 13-10 : Lewis Structures
Lewis Structure : show how the valence electrons are arranged among
the atoms in the molecule
Rule : to achieve noble gas electron configurations
Duet Rule : only H , He atom
(n=1 : only 1s orbital) : 2 electrons
full
Octet Rule (八隅體) : C, N, O, F, Ne …
(n=2 : only 2s 2p orbital) : 8 electrons
full
If n >= 3 (can break up the octet rule because 3d orbital)
H
He
C
v
X
v
C
S
X
v
Chapter 13-10 : Lewis Structures
Lewis Structure : show how the valence electrons are arranged among
the atoms in the molecule
For each atom :
need to obey the rules
n=1 : duet rule
n=2 : octet rule
Chapter 13-10 : Lewis Structures
CO2 molecule:
C have 4 valence electrons
O have 6 valence electrons
Total valence = 4 x 1 + 6 x 2 = 16 e-
1 ) Sum the valence electrons :
(including the charge)
2 ) How many pairs?
16 / 2 = 8 pairs
3 ) Obey the Rules
C,O
4 ) How many lone pairs? / bond pairs ?
4 bond pairs , 4 lone pairs
n=2 obey the octet rule (max. 8 e-)
Chapter 13-10 : Lewis Structures
Exercise : Write Lewis structure that obey the octet rule for each of following molecules
Example : NO2- , NO3- , OCN1 ) Sum the valence electrons : (including the charge)
2 ) How many pairs?
3 ) Obey the Rules
4 ) How many lone pairs? / bond pairs ?
O
NO2-
O N
O
NO3-
N
O
5 + 2 x 6 + 1 = 18 e-
5 + 3 x 6 + 1 = 24
18 / 2
24 / 2
OCNO C N
9 pairs (? BP , ? LP)
e-
12 pairs (? BP , ? LP)
O
Chapter 13-10 : Lewis Structures
Exception the octet rule : SF6
If n >= 3 (can break up the octet rule because 3d orbital)
Chapter 13-10 : Lewis Structures
http://www.kentchemistry.com/links/bonding/lewisdotstruct.htm
Chapter 13-11 : Resonance(共振)
More than one Lewis Structure can be written for a particular molecule
(有些分子的路易士結構可能大於一種)
1. 共振式的箭頭必為雙向
⟷
2. 若為帶電荷分子,必須加中括弧
3. 所有的共振結構須全部畫出
Experiments show that it exhibits only one type N-O bond with a length
and strength between those for a single and double bond.
Experimental :
( 2 N-O bond length + 1 N=O bond length ) / 3
the average result
[]
Chapter 13-11 : Resonance
原子排列方式一樣,
僅有電子分布不同
真實結構是所有共
振結構的平均
若其分子具有共振
形式,代表電子在其
中可delocalized(可
自由移動:非局域化)
3. Electron are delocalized
1. Arrangement the same
e-
2. The actual CC bond length of benzene
will between C-C and C=C
Chapter 13
Exercise : Draw the Lewis structure for CO32C-O : 1.43Å C=O : 1.23Å C≡O : 1.09Å
In the CO32- ion, all three C-O bonds have identical bond
lengths of 1.36Å , Why ?
Chapter 13-12 : Formal Charge
One method is to estimate the charge on each atom in various possible Lewis
structures and use these charges to select the most appropriate structure(s).
{用來評估最有可能的Lewis structure的一種方法}
1)
2)
3)
4)
5)
先畫出此分子可能的路易士結構 : draw the Lewis structures
假設電子均分至原子 : electron are totally shared for each atom
計算原子周圍所分配到的電子數 : calculate the electrons around atom
將原子本身的價電子數減去原子分配到的電子數 : calculate formal charge
加總各原子的形式電荷 : sum of | the formal charge |
Formal charge 總和最小者
即為最穩定的路易士結構
Charge 分離的程度代表鍵結的特性(separate 越大
離子鍵;separate越小
共價鍵)
Chapter 13-12 : Formal Charge
-1
-1
+2
-1
-1
Sum of the formal charge : 2+1+1+1+1 = 6
0
-1
0
0
-1
Sum of the formal charge : 1+1 = 2
Chapter 13-12 : Formal Charge
Sum of the formal charge
0 (minimize)
電負度高的原子其形式電荷允許
電負度高的原子其形式電荷允許 < 0
電負度低的原子其形式電荷允許 >0
For example : O = -1 , F , Cl … = -1 is acceptable
電負度高的原子其形式電荷應避免>0
電負度高的原子其形式電荷應避免
X
v
X
Formal Charge : Cl = 0
Formal Charge : Cl = +1
Formal Charge : O= +1
Chapter 13-12 : Formal Charge
For Example :
a. POCl3 b. SO42- c. ClO4- d. PO43Draw the possible Lewis structure and calculate the formal charge on each atom
O
Cl
P
Cl
2-
O
Cl
O
S
O
O
-
O
O
Cl
O
O
3-
O
O
P
O
O
Minimize the Formal charge and draw the most probable Lewis structure
Chapter 13
Exercise : Oxidation of the cyanide ion produces the stable cyanate
ion, OCN-. The fulminate ion, CNO-, is very unstable. Write the Lewis
structures and assign formal charges for the cyanate and fulminate
ions. Why is the fulminate ion so unstable?
Formal
charge
Formal
charge
O
C
N
O
C
N
O
C
N
0
0
-1
-1
0
0
+1
0
-2
C
N
O
C
N
O
C
N
O
-2
+1
0
-1
+1
-1
-3
+1
+1
All of the resonance structures for fulminate (CNO−) involve greater formal charges
than in cyanate (OCN−), making fulminate more reactive (less stable).
Chapter 13
Draw Lewis structures for Benzoic acid including resonance
forms. Label the formal charge on each atoms
Benzene 苯環
Acid 酸
Benzoic acid 苯酸(苯甲酸)
1. Draw the Lewis structure
2. Check all the possible resonances
3. Calculate the formal charge
Chapter 13-13 : VSEPR model
Lewis Structure :
show how the valence electrons are arranged among the atoms in the molecule
(路易士結構:僅能顯示出價電子在原子周圍的狀態:如lone pair, bond pair等)
Cant show actual structure in space (無法表示分子在空間中真實的結構)
For example :
1
O
OSO = 109.5o
Why ? How to explain ?
2
4
S
3
O1SO3 = 180o ?
O3SO4 = 90o ?
Lewis Structure
O
O
In Space
O
OSO = 109.5o
Chapter 13-13 : VSEPR model
three dimensional (3-D) molecular structure
VSEPR to determine approximate molecular structures
valence shell electron pair repulsion model:
e pairs are placed as far apart as possible
a simple way : 藉由中心原子的電子對數目來判斷結構
1. lone pair , bond pair 各算一對
2. 若遇到雙鍵,参鍵,則算一對
For example :
4對(3BP,1LP)
O
Cl
P
Cl
2對(4BP)
4對(4BP)
Cl
H
P
H
Lewis
H
H
C
N
4對(6BP)
Chapter 13-13 : VSEPR model
直線形(linear) : sp hybrid
平面三角形(Trigonal planar) : sp2
正四面體(Tetrahedral) : sp3
雙三角錐(Trigonal bipyramid) : dsp3
d : dz2
正八面體(Octahedral) : d2sp3
d : dx2-y2 , dz2
https://people.ok.ubc.ca/wsmcneil/vsepr.htm
Chapter 13-13 : VSEPR model
畫出最可能的路易士結構,判斷中心原子的電子對數目
優先考慮孤立電子的位置
擺入其他原子
決定分子結構名稱
畫出Lewis
Structure
中
心
原
子
有
四
對
角錐型
(trigonal pyramid)
正四面體
放入所有電子對
寫出最後分子結構時
不考慮lone pair
Chapter 13-13 : VSEPR model
畫出最可能的路易士結構,判斷中心原子的電子對數目
優先考慮孤立電子的位置
擺入其他原子
決定分子結構名稱
畫出Lewis
Structure
中
心
原
子
有
四
對
角型
(V(bend)-shape)
正四面體
放入所有電子對
寫出最後分子結構時
不考慮lone pair
Chapter 13-13 : VSEPR model
考慮孤立電子對(lone pairs)
repulsion
:
Chapter 13-13 : VSEPR model
考慮孤立電子對(lone
pairs)的擺放位置
的擺放位置
考慮孤立電子對
一對lone
pair
一對
兩對lone
pair
兩對
Lone pair擺任何地方均相同
擺任何地方均相同
Lone pair優先填入
優先填入x-y
plane上
上:
優先填入
電子對 赤力最小
Lone pair優先填入
優先填入z軸上
優先填入 軸上 : 電子對 赤力最小
三對lone
pair
三對
Chapter 13-13 : VSEPR model
考慮孤立電子對(lone
pairs)的擺放位置
的擺放位置
考慮孤立電子對
三對lone pairs
兩對lone
pairs
兩對
Chapter 13-13 : VSEPR model
無lone pair
Lone pair
一對lone
pair
一對
See-saw
翹翹板形
Lone pair
Lone pair
角錐形
兩對lone
pair
兩對
T形
形
Lone pair
Lone pair
Lone pair Lone pair
Lone pair
V形
形
Lone pair
直線形
Chapter 13-13 : VSEPR model
For Example :
a. XeCl2 b. ICl3 c. TeF4 , d. PCl5 e. XeC4, f. SeCl6
Draw the VSEPR structure , and determine the name of molecular structure
90o
F
Cl
Xe
Cl
≈ 120
F
o
Te
Cl
F
Cl
F
o
180
Cl
Xe
Cl
o
90
≈ 90o
Linear
Square planer
See-Saw
o
Cl ≈ 90
90 o
Cl
Cl
I
Cl
Cl
≈ 90
120o
Cl
Cl
P
Cl
Cl
T-shape
o
Cl
Se
Cl
Cl
Trigonal bipyramid
Cl
Cl
Octahedral
90o
Chapter 13
Draw Lewis structures for the following species. Assign the
formal charge to each central atom
SO2Cl2
PO43-
XeO4
SO3
Chapter 13
Draw Lewis structures for the following and predict whether
each is polar or nonpolar?
HOCN , COS , XeF2
Chapter 13
Exercise : Draw the Lewis structure : NO2
Nitrogen dioxide dimerizes to produce dinitrogen tetroxide
NO2 + NO2
N2O4
Chapter 13
Exercise : Draw the Lewis structure : BF3
Boron trifluoride accepts a pair of electrons from
ammonia, forming BF3NH3
BF3 + NH3
BF3NH3
Chapter 13
Place the species below in order of shortest to longest
nitrogen-nitrogen bond
N2 , N2F4
, N2F2