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A new trick for Bond Angle
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Hello guys ...this is my second article ...(Academic) .... Actually I am submitting this...because there is one question always in JEE, AIEEE or BITSAT...or any State Engineering Exam...on this topic....so make yourself comfortable ....and understand this... This method is applicable only for element N O P (mean Nitrogen Oxygen and Phosphorus ) ........... Trick is :- If central Atom is same ....then Higher the Electronegativity of Surrounding atom...lower the bond Angle ...and vice-versa If Surrounding Atom is same...then Lower the Electronegativity of Central atom Lower the bond angle and vice-versa Example ; NF3 NCl3 NBr3 NI3 Central atom is same in all the four case....i.e N...so higher the E.N of Surronding atom..lower the bond angle... So EN : F > Cl > Br > I Hence Bond Angle trend is NF3 < NCl3 < NBr3 < NI3
PH3 PF3 PCl3 PBr3 PI3
Central Atom is same.....again trend of EN of surrounding atom... F > Cl > Br > I > H ... So bond angle will be PH3 > PI3 > PBr3 > PCl3 > PF3
NH3 PH3
Here surrounding atom is same....but central is different ... So EN of N and P ... N > P So Bond angle of NH3 > PH3 .....( because higher the EN of central atom higher the Bond Angle )
H2O H2F
Again EN of F > O so H2F > H2O ( Bond angle ) ...this question is asked in JEE ( I don't remember the yr)
Solve for this using the above methods .... PF3 NF3 OF2 OCl2 OBr2 OH2 NCl3 PCl3 .....................and see some question from book..if they include some of this elements ...P N O ....and trend of Bond Angle is asked....then apply this..you will get right answer...
I will come up with more tricks...but first waiting for your comments...Did u get this method or not...Is it easier than what you mug up....daily ....?
With regards Yagya Read Other Helpful Articles: |
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