Question

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Answer 1

11. Strong John - Telecom chestonton cannot be observed in the following Octohedral complexes. A) CGI Bi COCO) C) CUCI) Coll) Answer: (O) CACI) Giv) 450 3d²0 HD unsym Slight distortion But cor II dh , Co (I) d7 ) Strong distortion Cu (21 d9 2. electron role Complex that obeys the is would be Ai (65-2546) Fe (Cove], B) [CMS-C5H5-) MOCC012] c) (In Cro? z Brz] © OS.CCO E Cph) (pph'), cl,

Answer bere o Os(60) C = (ph) cpph3), d. OS(I) 48306 = T electrons *Co=2 ECph = 3 (pph3) = 2X2=4 Cl=2@ +2+3+4+2=18eo. (3.) work function calculate the for the sodium metal 85 1.824 threshold frequency vo for sodiom. A) 2.75 X 1023 Hz B) 2.92 x 10 19 HZ @ 4.40 104 Hz ✓ b) 4.407 1052 Hz Answer (c) 440 X 1019 HZ w=hfo to = h = 1.82 ev ohh. = 1.82 4,135667 Xiots et Hel = 0.4400 x 105 Hz = 4040 x1014 HZ

a 4. show which compound con acts as - Niculeophile HÖV 5) HB"" c) H-NEW H - N A D A-d-H Amrrem :- (A) A- Nucleophile having Lone pain of electron ve charge (on) trot Acts ar Neutral Nucleophile BH Ī Mlectropusle It N H electrophile Neutral molecule. formed in the 5. What is major product following reaction, CH3 te ! " - - C Hy the Roor het - ch-ehty HCl

che CH3 E-CH CH-CH3 ✓ N-CH 13) HC-C-CH ₂ -CH3 сҢ C) HE - CH - CH₂ - CH2 -c D no-reaction Anrooy- (A) CASE s-c -CH2 Hel koos t+= ctty CH3-c = CH CH₂ a Chi -Ć -ct-chi • ' C с -2 -СН — . Stable radical н“ , сk - te - ch-ch (6) write the loPAC name of the following compound Goo (A) Dicyclohexyl Ethern Cydo heny Cyclohe nane- (C) Cyclohexyl ether ol dicadoheony ether Ausboy (53Cycloheory cyclo hesene Al : &s a commen nome