Subject : CHEMISTRY Topic : Chemical Bonding

28 Pages·2011·0.62 MB·English

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Subject : CHEMISTRY
Topic : Chemical Bonding
R
Index
1. Key Concepts
2. Exercise I
3. Exercise II
4. Exercise III
5. Exercise IV
6. Answer Key
7. 34 Yrs. Que. from IIT-JEE
8. 10 Yrs. Que. from AIEEE
Student’s Name :______________________
Class :______________________
Roll No. :______________________
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g
n
di
n
o
B
al
c
mi
e
h
C
8
2
of
2
e
g
a
P
L
A
P
O
H
B
om 1 ,
c 8
es. 88
s 5
as 0
cl 93
o 8
k 9
e 0
ww.t 0,
0
w 0
0
0
e: 2
t 3
bsi )-
5
e 5
w 7
m (0
o H:
r P
e f r)
g Si
ka K.
ac R.
y P S.
ud A (
Y
St RI
d A
a K
lo R.
n
w G
o A
D H
U
E S
RE or :
F ct
e
r
Di
S,
E
S
S
A
L
C
O
K
E
T
KEY CONCEPT g
n
di
n
o
B
Reasons for Bond formation: al
c
mi
(cid:2) e
Lowerisation of energy due to attractions. h
C
(cid:2) Attainment of Octet [ns2 np6], assumed to be most stable. 8
2
of
3
Types of bonds : Ionic, covalent, co-ordinate e
g
a
P
IONIC BOND [ELECTROVALENT]
(cid:2) Complete loss of e– to form ions.
L
(cid:2) A
Electrostatic attraction between ions. P
(cid:2) O
Elements of ‘p’ & ‘d’ block may show variable electrovalency due to H
B
om (a) Inert Pair effect (for p block): The reluctance of ‘s’ electron pair to take part in bond formation on1 ,
c moving down a group in ‘P’ block elements. 8
es. 88
s Finds application in 5
as 0
cl 93
o (cid:2) 8
k Stability of oxidation state of a particular metal atom. 9
e (cid:2) 0
ww.t Oxidizing & reducing power of compounds. 0,
(b) Unstability of core: For ‘d’ block elements the core may either have pseudo inert configuration (having0
w 0
18 electrons in outermost shell )or any other. 0
0
te: Properties of Ionic compounds 32
ebsi (cid:2) Ionic bonds are Non directional in nature 55)-
w (cid:2) 7
m High Melting points / Boiling points. (0
o (cid:2) In solid state they are conductor ( due to absence of charge carrier) while in aqueous & molten state theyH:
r P
e f are good conductor of electricity. r)
kag (cid:2)(cid:2) Soluble in Polar solvents K. Si
ac Show isomorphism. R.
y P No sp. theories to understand bond formation. S.
ud A (
Y
St characteristics like various crystal lattices to be done in solid state. RI
d A
a K
nlo COVALENT BOND R.
w (cid:2) G
o Sharing of electrons A
D (cid:2) H
Overlapping of orbitals U
E S
RE (cid:2)(cid:2) Types : single, double, triple, polar, non−polar bonds. or :
F Variable covalency : Shown by elements having vacant ‘d’ orbitals (caused due to excitation of thect
e
electron.) r
Di
Properties: S,
(cid:2) E
Covalent bonds are directional in nature S
S
(cid:2) Low melting point & boiling point. (except Diamond / Graphite, due to their peculiear structure) A
L
(cid:2) C
Electrical conductivity due to auto-protolysis or self ionisation.
O
(cid:2) Show isomerism. K
E
T
CO-ORDINATE BOND g
n
Bonding between lewis acid & lewis base or electron deficient & electron rich species. di
n
o
B
Lewis base: Species with lone pair on ‘central atom’ available for donation. eg. NH3, H2O al
c
charge mi
Lewis acid: Electron deficient due to incomplete octal, vacant p or d orbital & high + ve ratio. he
size C
8
2
Lewis Dot structures: of
(cid:2) 4
Arrangement of various atoms & types of bonding present but no idea of geometry. e
g
(cid:2) Selection of central atom [least E.N. of all elements excluding hydrogen] Pa
(cid:2)
In hydrogen containing proton donor oxy acids all ‘H’ atoms are attached to oxygen as –OH groups
except in H PO (dibasic), H PO (monobasic) & H P O (dibasic).
3 3 3 2 4 2 5 L
A
Applications: P
O
(cid:2) H
To know various linkages present B
om (cid:2) To calculate O.S. of various elements. 1 ,
c 8
es. 88
s Various Theories For Explaining Bonding 5
as 0
cl 93
o (cid:2) Electronic theory of valency (Kossel, Lewis) ;Singlat linkages 8
k 9
ww.te (cid:2)(cid:2) VMa.lOen.Tce. (bHounndd t ,h Meourlyl i k(Hene)i.tler, London, Pauling, Slater). (will be discussed in class XII th) 0, 0
0
w 0
0
e: DIPOLE MOMENT 2 0
bsit Dunipito) l1e mDo =m 1e0n–t1 i8s eas vue ccmto r= q 3u.a3n3ti t×y 1=0 µ–3 =0 cqo ×l. d m. Units = col m (S.I.) or esu cm(cgs) or Debey(common5)- 3
e 5
w 7
m observedD.M. (0
o % ionic character = × 100 % H:
r calculatedD.M.for100%ionic P
e f r)
g % ionic character = [ 16 (X − X ) + 3.5 (X − X )2 ] % [Hanny & Smyth equation] Si
ka A B A B K.
ac Dipole moment depends on R.
y P (cid:2) Electronegativity difference between atoms (cid:2) Direction of bond dipole moment S.
ud (cid:2) Angle between various bonds (cid:2) Influence of unshared e– pairs A (
St (cid:2) Magnetic of polarity of the molecule (cid:2) Symmetrical / Unsymmetrical shape. RIY
d A
a Bond Moments: K
lo R.
n
w H–F (1.9 D) H–O (1.5 D) C – C (0 D) C–F (1.4 D) G
Do H–Cl (1.1 D) H–N (1.3 D) C = O (2.3 D) C–Cl (1.5 D) HA
U
E H – Br (0.8 D) H– C (0.4 D) C – Br (1.4 D) S
RE H – I (0.4 D) C – I (1.2 D) or :
F ct
e
r
Di
S,
E
S
S
A
L
C
O
K
E
T
SHAPES OF MOLECULES BASED ON VSEPR THEORY g
n
di
Total no. No. of b.p. No. of General Type of Stereo Shape Exam. on
B
of hybrid (bond pairs) unshared formula hybridisations chemical al
c
orbitals pair i.e. l formula mi
p e
h
C
2 2 0 AB sp B–A–B linear BeCl 8
2 2 2
of
5
e
g
3 3 0 AB sp2 Trigonal BCl , a
3 3 P
planar GaF
3
3 2 1 AB sp2 Bent or GeF , AL
2 2 P
angular O O
3 H
B
om 4 4 0 AB4 sp3 Tetrahedral CH4 1 ,
c 8
es. 88
s 5
clas 4 3 1 AB3 sp3 Trigonal NH3 930
o pyramid 8
k 9
e 0
ww.t 0,
0
w 4 2 2 AB sp3 Bent or H O 0
2 2 0
angular 0
e: 2
t 3
bsi )-
5
we 4 1 3 AB sp3 linear HF 75
m (0
o H:
r P
e f 5 5 0 AB sp3d Trigonal PF , r)
g 5 5 Si
ka bipyramidal NbBr5 K.
ac R.
y P S.
ud 5 4 1 AB4 sp3d Seesaw SF4 YA (
St RI
d A
a K
lo R.
n
w 5 3 2 AB sp3d T-shaped ClF G
o 3 3, A
D BrF H
3 U
E S
FRE 5 2 3 AB2 sp3d Linear XICeFI2– ctor :
2 e
r
Di
S,
E
6 6 0 AB sp3d2 Octahedral SF S
6 6 S
A
L
C
O
K
E
6 5 1 AB sp3d2 Square IF T
5 5
pyramidal
g
n
6 4 2 AB sp3d2 Square IF di
4 4 on
planar XeF B
4 al
c
mi
e
h
7 7 0 AB sp3d3 Pentagonal IF C
7 7 8
bipyramidal of 2
6
e
g
a
P
HOW TO DECIDE THE TYPE OF HYBRIDISATION :
Type of hybridisation = (number of σ bonds + number of lone pairs)
L
A
RESONANCE P
O
(cid:2) Delocalisations of π electron cloud in between orbitals of various atoms in a molecule (provided all theH
B
om (cid:2) atoms are in the same plane) 1 ,
c Exists where more than one Lewis dot structure are possible for a molecule. 8
es. (cid:2) Resonance causes stablisation of the molecule & difference in the energies of hybrid & other structure is88
s 5
as termed as Resonance energy. 0
cl (cid:2) 93
o R.E. → Experimental heat of formation-Theoretical heat of formation. 8
k 9
e (cid:2) The properties of the actual structure (Resonance hybrid) are decided by the weighed average (depending0
ww.t on stability) of the contributing molecule. 0,
(cid:2) 0
w More the resonating structure more stable the molecule becomes. 0
0
e: FORCES OF ATTRACTION (WEAKER BONDS) 2 0
bsit (cid:2) Hydrogen bonding: When a hydrogen atom is linked to a highly electronegative atom (like F, O or N))- 3
e comes under the influence of another strongly electronegative atom, then a weak bond is developed55
w 7
m between them, which is called as hydrogen bond. (0
o Types of H-bonding: H:
e fr (cid:2) Intermolecular r) P
kag (cid:2) Intramolecular K. Si
ac Applications in: R.
y P (a) Abnormal behaviour of water. S.
ud (b) Association of a molecule as in carboxylic acid. A (
Y
St (c) Dissociation of a polar species. RI
d (d) Abnormal melting point & boiling point. A
a K
lo (e) Enhanced solubility in water. R.
n
w (cid:2) G
o Ion dipole attraction A
D (cid:2) Dipole-dipole attraction UH
E S
RE (cid:2)(cid:2) Ion-induced dipole attraction or :
F Dipole-Induced Dipole attraction ct
(cid:2) e
Induced -dipole Induced Dipole attraction r
Di
(cid:2) Metallic bonds: Electron gas model or sea model, with metal atom existing as kernels along with lessS,
firmly held valence e–s & bonds between various kernels ( at the lattice site) & valence e–s is known asE
S
S
metallic bonds. A
L
SOME TYPICAL BOND C
O
1. Back bonding:
K
(cid:2) E
If among the bonded atoms, one atom has a vacant orbital & another has excess ofT
e–s then a sort of π bonding takes place between he two. If this is between ‘P’ orbitals of the two, this is
known as pπ-pπ back bonding.
(cid:2) Most efficient when the atoms are very small & the orbitals involved of the two are of same energy level.g
n
di
2. Banana bond: on
B
(cid:2)(cid:2) TThhiiss tsytpreu cotfu broen sdhinogw iss ptrheaste ntht einr eB 2aHre6 .two types of mical
e
hydrogen atom-Terminals and bridging. h
C
8
2
MISCELLANEOUS CONCEPT of
7
e
g
a
1. Comparison of bond angles. P
(a) In case central atoms are having different hybridisation then it can be compared.
(b) If same hybridisation but different central atom then bond angle would be more of the molecule in which
L
C.A. is more E.N. eg. H S & H O. A
2 2 P
O
H
B
om 1 ,
c 8
es. 88
s (c) If C.A. is same & bonded atoms different then bond angle increases as the attached atom size increases. 5
as 0
cl 2. Bond strength , Bond length & Paramagnetic nature 93
o 8
k (a) Using concepts of resonance. 9
e 0
ww.t (b) Use of Lewis dot structure for the rest. 0,
0
w 0
0
0
e: 2
t 3
bsi )-
5
e 5
w 7
m (0
o H:
r P
e f r)
g Si
ka K.
ac R.
y P S.
ud A (
Y
St RI
d A
a K
lo R.
n
w G
o A
D H
U
E S
RE or :
F ct
e
r
Di
S,
E
S
S
A
L
C
O
K
E
T
EXERCISE - I g
n
di
IONIC BOND n
o
Q.1 The combination of atoms take place so that B
al
(A) They can gain two electrons in the outermost shell c
mi
(B) They get eight electrons in the outermost shell e
h
(C) They acquire stability by lowering of energy C
8
(D) They get eighteen electrons in the outermost shell. 2
of
8
e
Q.2 An ionic bond A+B− is most likely to be formed when : ag
P
(A) the ionization energy of A is high and the electron affinity of B is low
(B) the ionization energy of A is low and the electron affinity of B is high
(C) the ionization energy of A and the electron affinity of B is high
L
(D) the ionization energy of A and the electron affinity of B is low PA
O
H
Q.3 Which of the following compounds of elements in group IV is expected to be most ionic ? B
om (A) PbCl2 (B) PbCl4 (C) CCl4 (D) SiCl4 1 ,
c 8
es. Q.4 The compound which contains ionic as well as covalent bonds is 88
s 5
clas (A) C2H4Cl2 (B) CH3I (C) KCN (D) H2O2 930
o 8
k Q.5 The hydration of ionic compounds involves : 9
e 0
ww.t ((AC)) DEvisoslouctiioatni oonf hinetaot ions ((BD)) WAlel aokf etnhiensge of attractive forces 0,
0
w 0
Q.6 In which of the following species the bonds are Non-directional ? 0
0
e: (A) NCl (B) RbCl (C) BeCl (D) BCl 2
t 3 2 3 3
bsi Q.7 Which has the lowest anion to cation size ratio : 5)-
e 5
w (A) LiF (B) NaF (C) CsI (D) CsF 7
m (0
o Q.8 Which of the following statement(s) is/are correct regarding ionic compounds? H:
r P
e f (A) They are good conductors at room temperature in aqueous solution. r)
kag ((BC)) TThheeyy acroen gsiesnt eorfa ilolyn sso.luble in polar solvents. K. Si
y Pac (D) They generally have high melting and boiling points. S. R.
ud Q.9 Which of the following compounds contain/s both ionic and covalent bonds? A (
Y
St (A) NH Cl (B) KCN (C) CuSO ·5H O (D) NaOH RI
d 4 4 2 A
a K
lo Q.10 Among the following isostructural compounds, identify the compound, which has the highest LatticeR.
wn energy G
o (A) LiF (B) LiCl (C) NaCl (D) MgO A
D H
U
E Q.11 A bond formed between two like atoms cannot be S
RE (A) ionic (B) covalent (C) coordinate (D) metallic or :
F ct
e
Q.12 Which of the following, when dissolved in water forms a solution, which is Non-conductivity? r
Di
(A) Green Vitriol (B) Indian salt Petre S,
(C) Alcohol (D) Potash alum E
S
S
A
Q.13 Most ionic compounds have : L
C
(A) high melting points and low boiling points O
K
(B) high melting points and nondirectional bonds E
T
(C) high solubilities in polar solvents and low solubilities in nonpolar solvents
(D) three-dimensional network structures, and are good conductors of electricity in the molten state
Q.14 An electrovalent compound does not exhibit space isomerism because of g
n
(A) Presence of oppositively charged ions di
n
(B) High melting points Bo
(C) Non-directional nature of the bond al
c
(D) Crystalline nature mi
e
h
C
Q.15 Which of the following have an (18 + 2) electron configuration ? 8
2
(A) Pb2+ (B) Cd2+ (C) Bi3+ (D) SO42− e 9 of
g
a
Q.16 Which of the following contains (electrovalent) and non-polar (covalent) bonds ? P
(A) CH (B) H O (C) NH Cl (D) HCN
4 2 2 4
L
A
COVALENT BOND P
O
Q.17 A sigma bond may be formed by the overlap of 2 atomic orbitals of atoms A and B. If the bond is formedH
along as the x-axis, which of the following overlaps is acceptable ? B
om (A) s orbital of A and p orbital of B (B) p orbital of A and p orbital of B 1 ,
c z x y 8
es. (C) p orbital of A and p orbital of B (D) p orbital of A and s orbital of B 88
s z x x 5
as 0
cl Q.18 The maximum covalency is equal to 93
o 8
ek (A) the number of unpaired p-electrons 0 9
ww.t ((BC)) tthhee nnuummbbeerr ooff puanipreadir edd-e sl eacntdro pn-selectrons 0,
0
w 0
(D) the actual number of s and p-electrons in the outermost shell. 0
0
e: 2
bsit Q.19 How many bonded electron pairs are present in IF7 molecule : )- 3
e (A) 6 (B) 7 (C) 5 (D) 8 55
w 7
m (0
o Q.20 PCl5 exists but NCl5 does not because : H:
r P
e f (A) Nitrogen has no vacant 2d-orbitals (B) NCl5 is unstable r)
g (C) Nitrogen atom is much smaller than P (D) Nitrogen is highly inert Si
ka K.
ac Q.21 Which of the following has/have a strong covalent bond? R.
y P (A) Cl-F (B) F-F (C) C-Cl (D) C-F S.
ud A (
Y
St Q.22 Which of the following statements is/are true? RI
d (A) Covalent bonds are directional A
a K
lo (B) Ionic bonds are nondirectional R.
n
w (C) A polar bond is formed between two atoms which have the same electronegativity value. G
Do (D) The presence of polar bonds in a polyatomic molecule suggests that it has zero dipole moment HA
U
E S
RE Q.23 Rotation around the bond (between the underlined atoms) is restricted in : or :
F (A) C2H4 (B) H2O2 (C) Al2Cl6 (D) C2H6 ct
e
r
Di
Q.24 The octet rule is not obeyed in : S,
(A) CO (B) BCl (C) PCl (D) SiF E
2 3 5 4 S
S
A
Q.25 Which of the following two substances are expected to be more covalent : L
C
(A) BeCl (B) SnCl (C) ZnS (D) ZnCl O
2 4 2 K
E
T
Q.26 To which of the following species octet rule is not applicable :
(A) BrF (B) SF (C) IF (D) CO
5 6 7
Q.27 Which of the following species are hypervalent? g
n
1. ClO –, 2. BF , 3. SO 2– , 4. CO 2– di
4 3 4 3 n
(A) 1, 2, 3 (B) 1, 3 (C) 3, 4 (D) 1, 2 Bo
al
c
Q.28 The types of bond present in N O are mi
2 5 e
(A) only covalent (B) only ionic h
C
(C) ionic and covalent (D) covalent & coordinate 8
2
of
CO-ORDINATE BOND 10
e
Q.29 NH and BF combine readily because of the formation of : ag
3 3 P
(A) a covalent bond (B) a hydrogen bond (C) a coordinate bond (D) an ionic bond
L
Q.30 Which of the following species contain covalent coordinate bond : A
P
(A) AlCl3 (B) CO (C) [Fe(CN)6]4− (D) N3− HO
B
om Q.31 Which of the following molecules does not have coordinate bonds? 1 ,
c 8
es. (A) CH –NC (B) CO (C) O (D) CO2− 88
s 3 3 3 5
as 0
cl 93
o LEWIS STRUCTURE 8
k 9
e Q.32 Which of the following Lewis diagrams is(are) incorrect ? 0
www.t •• •• Cl  H| + H| H| 000,
te: (A) Na−O••−C••l•• (B) C l CC l Cl (C) H −HN| −H [ S]2- (D) H −N••−•N•−H 32 00
bsi 2 )-
5
e 5
w Q.33 The possible structure(s) of monothiocarbonate ion is : 7
m (0
o C C S S H:
r P
e f r)
kag (A) S (B) S (C) C (D) C K. Si
ac O O O O O O O O R.
y P S.
ud Q.34 The valency of sulphur in sulphuric acid is : A (
St (A) 2 (B) 8 (C) 4 (D) 6 RIY
d A
a K
nlo Q.35 The total number of valence electrons in 4.2g of N3− ion are : R.
w (A) 2.2 N (B) 4.2 N (C) 1.6 N (D) 3.2 N G
o A
D H
U
E Q.36 No X −X bond exists in which of the following compounds having general form of X H ? S
RE (A) B H (B) C H (C) Al H (D) Si H 2 6 or :
F 2 6 2 6 2 6 2 6 ct
e
r
Q.37 Pick out among the following species isoelectronic with CO : Di
2 S,
(A) N− (B) (CNO)− (C) (NCN)2− (D) NO− E
3 2 S
S
A
Q.38 Which of the following have a three dimensional network structure ? L
C
(A) SiO (B) (BN) (C) P (white) (D) CCl O
2 x 4 4 K
E
T
Q.39 Which of the following oxyacids of sulphur contain S−S bonds ?
(A) H S O (B) H S O (C) H S O (D) H S O
2 2 8 2 2 6 2 2 4 2 2 5