Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
n =[ number of electron in valence shell + number of monovalent atoms attached(like H,Cl etc) - charge on molecule]/2 for n = 2 - hybridisation is SP , n =3 hybridisation is SP2 , for n = 4 - SP3 & so on ......... see in CH2CO2 we can use above method , left carbon bearing two monovalent atoms (2 Hydrogen) so for this carbon n = (4+2)/2 = 3, hybridisation is SP3 ... right carbon has no monovalent atoms so n = 4/2 = 2 , its hybridisation is SP ... for hydrocarbons , we have to use another method ... if carbon is connected to 1 triple bonds then its hybridisation is Sp ... if ,, ,, ,, 2double bonds then ,, ,, again SP .. if ,, ,, ,, 1 double bond then ,, ,, is SP2 if ,, ,, ,, no double or triple bond ,, ,, is SP3 organic compounds do not show hybridisation other than these three ... example - CH2 = C=CH-CH-CH3 from left , 1st carbon - SP2 (only one double bond) 2nd carbon - SP (two double bonds) 3rd - SP2 4rth & 5th -SP3 (no double or triple bond )
n =[ number of electron in valence shell + number of monovalent atoms attached(like H,Cl etc) - charge on molecule]/2
for n = 2 - hybridisation is SP ,
n =3 hybridisation is SP2 ,
for n = 4 - SP3 & so on .........
see in CH2CO2 we can use above method ,
left carbon bearing two monovalent atoms (2 Hydrogen) so for this carbon n = (4+2)/2 = 3,
hybridisation is SP3 ...
right carbon has no monovalent atoms so n = 4/2 = 2 , its hybridisation is SP ...
for hydrocarbons , we have to use another method ...
if carbon is connected to 1 triple bonds then its hybridisation is Sp ...
if ,, ,, ,, 2double bonds then ,, ,, again SP ..
if ,, ,, ,, 1 double bond then ,, ,, is SP2
if ,, ,, ,, no double or triple bond ,, ,, is SP3
organic compounds do not show hybridisation other than these three ...
example - CH2 = C=CH-CH-CH3
from left , 1st carbon - SP2 (only one double bond)
2nd carbon - SP (two double bonds)
3rd - SP2
4rth & 5th -SP3 (no double or triple bond )
N=(NO.OF VALENCE ELECTRON + NO. OF SURROUNDING ATOM EXCEPT<O2>+_ CHARGE) HENCE THE HYBRIDISATION GIVEN IS CORRECT FOR N=3 SP2 , FOR N=2 SP
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !