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Hello Guys, In this lecture we are going to discuss a very important test and complex analysis for jee mains, advanced, NEET and AIIMS.
The identification test of nitrate ( NO3-) using conc. H2SO4 and freshly prepared FeSO4 , is known as 'Brown Ring Test'( it is the confirmation test of NO3- too ). In a normal test tube take the supplied NO3- containing sample ( say KNO3 ) . Now add conc.H2SO4 dropwise with gentle shaking and then heat it shortly using Bunsen burner & add freshly prepared greenish FeSO4. Then, you can certainly see a brown ring is formed in that test tube bcz of formation of brown colred [Fe(H2O)5 (NO)]SO4 complex. The rxn equations are belows : I just avoid balance here but it's forbidden… balance is must required in all cases…
KNO3 + H2S04 = KHSO4 + HNO3
HNO3 + FeSO4 = NO + Fe2(SO4)3 + H2O. In this step chemists assumed that nassent oxygen is produced and the oxidation is done by [O]. Nextly, FeSO4 in aqueous solution of course exist as [Fe(H2O)6]SO4, It reacts with NO.
Central metal is Iron, Oxidation state is +1, hybridisation is sp3d2, outer octahedral complex, High spin complex, Magnetic moment is 3.87 BM because of 3 unpaired electron. Many more concept has been discussed in details inside the lecture.
This lecture is designed in accordance with the Higher secondary level and NCERT. But, in higher studies in University, things might change..
You are going to learn In 'brown ring' complex, the oxidation state of iron is +3 i.e. Fe(|||) and nitrosyl exists as NO(-), simply, anti-ferromagnetic coupling occur in the complex in b/w these two ions (please do calculate, total spin, we will get 5/2, & it had been proved or justified whatever you can say by Mössbeur Spectroscopy).
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We hope, this would be great help for you Guys.
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