Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        how to find hybridisation?i want shortcut or any trick`
7 years ago

419 Points
```										Dear Pradeep
Following is the way to find hybridisation
1st step    :  count the no. of valence electrons[ V ]2nd step   :   now divide V by 8 ,if V >8                    or                    divide V by 2 ,if  V <=8                                   -[1]                    if the remainder is still  not zero then                   again divide it by 2,                                         - [2]                   therefore the final value = quotient in 1+ quotient in 2                   as    shown in the example3rd step    :  compare the above obtained value with the table given below :-                   sp          -   2                   sp2        -   3                   sp3        -   4                   sp3d      -   5                    sp3d2    -   6                   sp3d3    -   7ex:- for  c2h6step  1   -   2[4 ] + 6[1] = 4+6=10step  2    -   as 14>8                     14/8, quotient=1  as remainder  = 6 which is not equal to zero                    therefore 6 / 2   quotient=3                   therefore final value=1+3=4                   comparing with the table given above                    the hybridisation is sp3.note -  this is not applicable  to all compounds
All the bestAKASH GOYALAskiitiansExpert-IITDPlease feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation. Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
```
7 years ago
510 Points
```										you can use HON rule
according to this 2X=number of electrons in valence shell-charge+number of monovalent atoms attached(Cl,H etc)
if x comes 2 then hybridisation is sp
if 3 then sp2 ,if 4 then sp3 and so on
example ..hybridisation of NH4+
2x=5+4-1
x=4 so hybridisation is sp3
approve my ans if u like
```
7 years ago
Devasish Bindani
45 Points
```										to find the hybridisation of an element in a given compound  first count the number of σ bonds and the no. of lone pairs present then it becomes the steric no of the element then just get the hybridisation. hybridisation forsteric no.
2--sp
3--sp2
4--sp3
5--sp3d
6--sp3d2
7--sp3d3
thats it the short cut count the no of σ bonds  and ya got the hybridisation
note:- its applicable to all compounds provided ya know the structure or no of atoms connected to the given atom whose hybridisation is to be found hers an example check out

here no of σ bettween Cr and O are 5 there fore the hybridisation is sp3d
note sigma bond are the primary bonds so whereever you see double bond then there are one sigma and one pi bonds
for lone pair take NH3 nitrogen has 1 lone pair and 3 σ bonds thus the hybridisation is sp3
```
7 years ago
aishwarya aishu
18 Points
```										can u please tell me how to find out the hybridisation for ph3 please and aso give me the explanation thank you
```
4 years ago
Rahul John Joseph
31 Points
```										Just count the no.of sigma bonds and lone pairs (if any) Thats all!
```
3 years ago
siddharth singhal
19 Points
```										What is the hybridization of I 3 minus?
```
3 years ago
saksham agrawal
14 Points
```										for any compound identify the central atom then count the no. of sigma bonds and no. of lone pairs on it and then add them. eg: in a CO2 molecule carbon is the central atom, no. of sigma bonds are 2 and no. of lone pairs are 0 then sum is 2 which means sp hybridization is there.
```
3 years ago
Aazim Bill SE Yaswant
20 Points
```										for a shortcut. find the no of atoms attached to it. then write its corresponding hybridisation. if central atom is attached to 2.. sp.. 3..sp2.. so on
```
3 years ago
Parveen Khurana
31 Points
```										Countcount the no. Of sigma bond and no of lone of atom whose hybridisation is to be predicted . All single bonds ate sigma , in amultiple bond one will sigma others be pie
```
3 years ago
ashu
34 Points
```										Accr to VSEPR theory H={V+M-C+A} the simplest method V = valence electrons in central atomM=number of univalent atom attached to central atomC=charge on cationA=charge on anion
```
3 years ago
harit
14 Points
```										dont,t get confused just no. of sigma bond + lone pairs ,if sum is 2--sp 3--sp2 4--sp3 5--sp3d 6--sp3d2 7--sp3d3
```
3 years ago
Yashwardhan Bahure
24 Points
```										but what if remainder is not zero. it wont be divisible by 2 ex. CH3
```
2 years ago
anurag
15 Points
```										hybridization of h3po3
```
2 years ago
soliha noor
53 Points
```										calculate no. of bond pairs[sigma bonds only] and lone pairs and add them..the no. u get is the hybridisation of that atom...if h=2=sp,,if h=3=sp2,,if h=4=sp3,,if h=5=sp3d and so on
```
2 years ago
Nike Puma
24 Points
```										I hav a formula for finding hybridisationI call it "Magic Number"So......Magic Number = Number of connections   +1/2 ( Group no. of central atom   -         Valency of central atom   ±          Charge on molecule)**# Note that -ve charge is added      and +ve charge is subtractedAlso that this doesn`t apply for complexesSo group 13    group 4Now compare the magic no. with2--sp3--sp24--sp35--sp3d6--sp3d27--sp3d3 Eg.)   For    H3O+Magic Number = 3 + 1/2( 6 - 3 - 1 )                           =  3 + 1/2( 2 )                           =  3 + 1                           =  4Now 4 corresponds to sp3 hybridHence H3O+ is sp3 hybridised  group 3     group  14
```
7 months ago
Ashish Nayakidi
36 Points
```										You can calculate accordingly asH={V+M-(C+A)}V=valency of the central metal atomM=no. of monovalent atoms attached to the central metal atomC=charge on cationA=charge on anion
```
7 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Physical Chemistry

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
Get extra Rs. 3,180 off
USE CODE: CHEM20
Get extra Rs. 551 off
USE CODE: CHEM20