Question

PARTA The ion [Ni(Pyridine), (H,0,12+ has d - d absorption bands at 27000, 16500 and 10150 cm-7. The 10 Dq value of Ni2+ ion is: (A) 10150 cm? (B) 10500 cm- (C) 16500 cm-1 D) 6350 cm- PART B Oxyhemoglobin is diamagnetic due to electron spin coupling between (A) Low spin Fe2+ and oxygen molecule (B) High spin Fe3+ and superoxide radical (C) High spin Fe2+ and oxygen molecule (D) Low spin Fe3+ and superoxide radical PARTC The most stable conformation of optically active butane-2,3-diol is CH, OH Hoh HOT он CH

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Answer 1

[willyridine),(4,0),12* No. of d electron = 8 & value = (2+2)+(141) +0-1-2 472-3 = 3 '&= 3 it means. f spin multiplicity - (25+1) TONI = 2x141=3 Hence Ground state term symbol = 36 first excited state term symbol = 3p stige p 3p - 3 Tig(F) (tzge) TO BAO 70.800 3F K- -0.2 Do 5 (-0.4x5+0.6*34. -0.2 Av 3Azy (tza.egy. E = (-0.446 0.6*2ja. - 1.240 Due to inteseledanie sepulsion (or mixing of two tig statel) two Trg states do not cross.

Electronic spectra of octahedial (kilo)]:- 12а — за 3 Tigt < 3 A2g = h его сам" 16500 can't 3the P да я V 27ооо см The lowest energy transition ie, 3Tage 3 Ay equal to toDq. a Hene 10 Dq = 10150 cm". - 0 2 А, -(-) 2 д. ) = Ao - (o P4 - 101 7 с т.

А seggen molemte Part B: oxy haemoglobin is diamagnetic нсами в Lous spin fe2" g бхчен тес coupling. fe 21 in low spin -- eg de 4 22 24 there is no unpaired electron in case of tow spin feel hence día magnetic while in case of fe 3t high spin there are s AIL unpaired electron - tag hence it can't be diamagnetie. low spin Pellas) - – eg. & umpaired 1 1 1 2011 2 unpaired & diamagnetic. High spin feat Cdo) 11 cog +29 e can't be

Part 1: conformation in B. The most stable CH3 Lord because their ip intramolecul ar Ho Bending which campansate the me-me. gauche intraetia. where as in other option A&D there is no intramolecular Ho Banding option Cip

diasteriones hence it can't be optically ack conformation of butane 2,3-diol.