The Crystal Field Stabilization Energy (CFSE) of [CoF3(H2O)3] (0 < P) is : (A) – 0.4 0 (B) – 0.8 0 ( — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

[Co(H2O)3F3] Co3+ = 3d64s0  1,1,2 g 2t , eg1,1 CFSE = [–0.4nt2g + 0.6neg]0 + n(P) = [–0.4 × 4 + 0.6 × 2]0 + 0 = –0.40 SECTION – 2 : (Maximum Marks : 20)  This section contains FIVE (05) questions. The answer to each question is NUMERICAL VALUE with two digit integer and decimal upto one digit.  If the numerical value has more than two decimal places truncate/round-off the value upto TWO decimal places.  Full Marks : +4 If ONLY the correct option is chosen.  Zero Marks : 0 In all other cases [kaM 2 ¼vf/kdre vad% 20)  bl [kaM esa ik¡p (05) iz'u gSA izR;sd iz'u dk mÙkj la[;kRed eku (NUMERICAL VALUE) gSa] tks f}&vadh; iw.kkZad rFkk n'keyo ,dy&vadu eas gSA  ;fn la[;kRed eku esa nks ls vf/kd n’'keyo LFkku gS ] rks la[;kRed eku dks n'keyo ds nks LFkkuksa rd VªadsV@jkmaM vkWQ (truncate/r