01 EP
02
=CVE + no. of MV
03 VALENCE BOND
VSEPR 2 HYBRIDISATION C - C = 1 bond
MV Remaining Monoatomic C = C = 1 + 1π
2,0- Linear - 180o surrounding 2- sp - Linear C -
= C = 1 + 2π
3,0- Trigonal planar - 120 o BP LP atoms 3- sp2 - Trigonal planar MOLECULAR ORB
2,1- Bent shape 4- sp 3 - Tetrahedral
EP =CVE + 0 Up to N2 the energy o
4,0- Tetrahedral - 109o28| 2 5- sp3d - Trigonal pyramidal
3,1- Pyramidal - 107o DV Remaining Diatomic 1s< *1s< 2s< *2
surrounding 6- sp3 d2 - Octahedral < 2pz<(π* 2px=π*2py)<
2,2- Bent - 104.5o
LP atoms 7- sp3 d3 - Pentagonal bipyramidal
BP
5,0- Trigonal Bipyramidal After N2 the energy orde
4,1- See saw CH3-CH3 - sp 3 1s< *1s< 2s< *2s<
EP =CVE + negative charge - positive charge
3,2- T-shape <(π* 2px=π*2py)< *2pz
2 CH2=CH2 - sp2
2,3- Linear - 180 o
MV/DV Remaining
B2-10 Paramagnetic
Charged Compound
CH=CH - sp
6,0- Octahedral - 90 o
O2-16 Paramagnetic
BP LP =C= - sp
5,1- Square pyramidal
4,2- Square planar
LP-LP > LP-BP > BP-BP
(Repulsion)
7,0- Pentagonal bipyramidal
6,1- Distorted octahedral
In XeF6, oxidation state and state of
Total no. of Lone pairs in I 3- is The no. of unp
Q3 Q4
hybridisation of Xe and shape of the
Q2 molecule are, respectively molecule is:
1) 3 2) 6 1) +6, sp3d3 , distorted octahedral 1) 0 2) 1
In which of the following pairs are 2) +4, sp3d2 , square planar
Q1 3) 9 4) 12 3) 2 4) 3
the two species trigonal pyramidal ? 3) +6, sp3 , pyramidal
4) +6, sp3d2 square pyramidal
1) BrO3 and XeO3
-
2) SF4 and XeF4
2) SO 3 and NO3-
2-
4) BF3 and NF3
CHEMICAL BONDING
1) Max B.A-18
1) Polar molecule, = 0 Resonance Changes B.O.
BOND 1) Hybridisation
DIPOLE Irregular geometry.
1) Benzene
BOND 2) Hybridisatio
MOMENT a) Different bonds RESONANCE ANGLE sp > sp2 > sp
ENERGY sp> sp2 >sp3 ( ) b) Lone pairs
3 NO2+ > NO2
2) B.E. B.O. (No. of Bonds)
c) Different surrounding atoms = 2 = 1.5 sp
eg: SF4, CH3Cl 3) if LP B.A
N2> O2 >F2 2) Non polar molecule, =0 CH4 > NH3 >
B.L.: C2H6 > C6H6 > C2H4 > C2H2
Regular geometry
3) B.E. E.N. difference 0 LP 1 LP
a) Same bonds B.O.: 1 1.5 2 3
HF> HCl > HBr > HI b) Zero lone pairs
4) If electrone
2) O3 3 C.A B.A
4) Halogens (B.E.) c) Same surrounding atoms = 2 = 1.5 NH3 > PH3 > A
eg: CH4, CO2
Cl2> Br2 > F2 > I2
CH3OH > CH3Cl > H2O> NH3> NF3> BF3 B.L.: H O > O > O 5) If electrone
02
=CVE + no. of MV
03 VALENCE BOND
VSEPR 2 HYBRIDISATION C - C = 1 bond
MV Remaining Monoatomic C = C = 1 + 1π
2,0- Linear - 180o surrounding 2- sp - Linear C -
= C = 1 + 2π
3,0- Trigonal planar - 120 o BP LP atoms 3- sp2 - Trigonal planar MOLECULAR ORB
2,1- Bent shape 4- sp 3 - Tetrahedral
EP =CVE + 0 Up to N2 the energy o
4,0- Tetrahedral - 109o28| 2 5- sp3d - Trigonal pyramidal
3,1- Pyramidal - 107o DV Remaining Diatomic 1s< *1s< 2s< *2
surrounding 6- sp3 d2 - Octahedral < 2pz<(π* 2px=π*2py)<
2,2- Bent - 104.5o
LP atoms 7- sp3 d3 - Pentagonal bipyramidal
BP
5,0- Trigonal Bipyramidal After N2 the energy orde
4,1- See saw CH3-CH3 - sp 3 1s< *1s< 2s< *2s<
EP =CVE + negative charge - positive charge
3,2- T-shape <(π* 2px=π*2py)< *2pz
2 CH2=CH2 - sp2
2,3- Linear - 180 o
MV/DV Remaining
B2-10 Paramagnetic
Charged Compound
CH=CH - sp
6,0- Octahedral - 90 o
O2-16 Paramagnetic
BP LP =C= - sp
5,1- Square pyramidal
4,2- Square planar
LP-LP > LP-BP > BP-BP
(Repulsion)
7,0- Pentagonal bipyramidal
6,1- Distorted octahedral
In XeF6, oxidation state and state of
Total no. of Lone pairs in I 3- is The no. of unp
Q3 Q4
hybridisation of Xe and shape of the
Q2 molecule are, respectively molecule is:
1) 3 2) 6 1) +6, sp3d3 , distorted octahedral 1) 0 2) 1
In which of the following pairs are 2) +4, sp3d2 , square planar
Q1 3) 9 4) 12 3) 2 4) 3
the two species trigonal pyramidal ? 3) +6, sp3 , pyramidal
4) +6, sp3d2 square pyramidal
1) BrO3 and XeO3
-
2) SF4 and XeF4
2) SO 3 and NO3-
2-
4) BF3 and NF3
CHEMICAL BONDING
1) Max B.A-18
1) Polar molecule, = 0 Resonance Changes B.O.
BOND 1) Hybridisation
DIPOLE Irregular geometry.
1) Benzene
BOND 2) Hybridisatio
MOMENT a) Different bonds RESONANCE ANGLE sp > sp2 > sp
ENERGY sp> sp2 >sp3 ( ) b) Lone pairs
3 NO2+ > NO2
2) B.E. B.O. (No. of Bonds)
c) Different surrounding atoms = 2 = 1.5 sp
eg: SF4, CH3Cl 3) if LP B.A
N2> O2 >F2 2) Non polar molecule, =0 CH4 > NH3 >
B.L.: C2H6 > C6H6 > C2H4 > C2H2
Regular geometry
3) B.E. E.N. difference 0 LP 1 LP
a) Same bonds B.O.: 1 1.5 2 3
HF> HCl > HBr > HI b) Zero lone pairs
4) If electrone
2) O3 3 C.A B.A
4) Halogens (B.E.) c) Same surrounding atoms = 2 = 1.5 NH3 > PH3 > A
eg: CH4, CO2
Cl2> Br2 > F2 > I2
CH3OH > CH3Cl > H2O> NH3> NF3> BF3 B.L.: H O > O > O 5) If electrone
