Hiii Puneet ya there is a very good method configured by our friends group during our preparation...
This method is actually applicable in few compunds only..and generally these compunds are asked in JEE ....
Read this ...you will get cleared about the method
No. of electrons in outermost orbit B.O(Bond order)
20 0.0
19 0.5
18 1
17 1.5
16 2
15 2.5
14 3
13 2.5
12 2.0
11 1.5
10 1.0
9 0.5
8 0
Magnetic Property :- Remember "DEPO" ...
D - Diamagnetic if Even no of electron
P-Paramagnetic if odd no of electron
i.e Dia if Even Para if Odd = = DEPO
Example
One negative charge will add one electron and one positive will subtract one electron.
CN- , N0+ N0- CO CN CO+ O2+ O2-
CN- = 6 + 7 + 1 = 14 ...B.O = 3 and no of electron are even so Diamagnetic in nature
NO+ = 7 + 8 - 1 = 14 ...B.O = 3 and no of electrons are even(14) so diamagnetic in nature
NO- = 7 + 8 + 1 = 16 B,O = 2 and again Dia in nature
CO = 6 + 8 = 14 B.O= 3 again diamagnetic in nature ( 14 e and is even)
O2+ = 2(8) - 1 = 15 B.O = 2.5 and is paramagnetic in nature ..coz odd no of electron ( 15 is odd)
O2- = 2(8) + 1 = 17 B.O = 1.5 and is paramagnetic in nature ..
CO+ = 6 + 8 - 1 = 13 B.O = 2.5 ....and is paramagnetic in nature due to odd no of electrons ( 13 electron)
EXCEPTION O2 and B2
O2 = 16 electron and B2 = 10 electron ....both have even no of electron but then also they are PARAMAGNETIC in nature...
Just remember these two exception and apply it for others ..
Solve for : N2+ N2- CN NO+ NO- ....and go through previous year question on B.O .....
All the mentioned compounds have been asked in JEE many times...
So i have told you a method not only how to calculate magnetic behaviour but also to calculate B.O
Now if let me know is this helpful to you or not ? do u get the method ?
With regards
Yagya
askiitians_expert