2 Flashcards
The shape of a molecule is only determine by
Sigma bond + lp…and not by pi bond
What is the shape of CO2 , SO2
A. Linear
B. Trigonalplaner
How many lobes dz2 have
8
D orbital pure or mixed me divide hote hai dono me 2 difference bto
Mixed has 2 conical nodes
Dz2 nucleus is more shielded
Delta bond koin banana hai
Dxy,dx2-y2
Carbon covalency
Phosphorus covalency
Carbon has 4 due to free excitation
Phosphorus covalency 3,5 dure to forced excitation
S
P
D
Orbitals
Free,or forced excitation
S to p free excitation
S to d , p to d forced excitation
Condition for excitation
Primary = vacent orbital present
Secondary = energy Gao b/w the orbital should be low
Contraction theory
Periodic contraction = jab hum left to right jaate hai to valence she’ll ke orbital ke beech distance increases Hota jata hai [2s…2p]
Charge contraction = jab nucleus ka change increase Kake hai to valence she’ll me ns aur ND pass pass aane lagte hai
Nitrogen and oxygen ki covalency
Nitrogen = normal 3 …but max 4 by forming coordination bond
Oxygen = normarmally 2 but maximum 3 donating lp by coordination bond
SCI6 NOT EXISTED BECAUSE OF
Steric crowding
Xef6 exist with covalency 6 while XeO4 covalency 8
Xef6 because of steric crowding
Mixing of orbital by two types
Hybridization
MOT
Hybridization types
Ideal = percentage of s,p equal in hybrid orbital
Non ideal = not equal
Hybrid orbital not form pi bond
Analysis of hybridization
Orbital
Steric
Energetic
Orbital analysis
i= –1/COS0
S%=COS0/COS0–1 X 100
BF3(120)=SP² IDEAL
NH3(107.5)=SP³.³³ NON IDEAL
BiH3(90)=INFINITI NOT HYBRIDIZED
CH4(109.5)= SP3
H2O(194)= SP3.9
PX3(104)= SP4.13
CO2(180)=SP¹
NO2(135)=SP1.414
Steric analysis
Ch4 = easily hybridization ( ideal hybridization)
Nh3= lp not easily undergoes hybridization ( non ideal)
Ph3 = not undergoes hybridization because p orbital need not s character to increase bond angle
Molecule not showing sp mixing
Central atom of large side ( 3rd period)
Side atom should small and less EN
Presence of lp
PH4+ ,NH4+
IDEAL OR NON IDEAL
Ideal
