See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Sol. (A) The magnitude of splitting energy increases down the group. Thus Pt2+ – 5d8, Pd2+ – 4d8 and Ni2+ – 3d8. (B) NH3 is a borderline ligand which forces pairing of electrons in Co3+, whereas H2O is a weak- field ligand and it can not force the pairing in Ni2+. Co3+ in [Co(NH3)6]3+ – 3d6, 4s0 Ni2+ in [Ni(H2O)6]2+ – 3d8, 4s0 AITS-FT-VIII (Paper-1)-PCM(Sol.)-JEE(Advanced)/20 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 7 So, there is no unpaired electron in [Co(NH3)6]3+ whereas there are unpaired electron in [Ni(H2O)6]2+. Thus [Co(NH3)6]3+ is colourless and [Ni(H2O)6]2+ is coloured due to d–d transition. (C) CH2 NH2 CH2 NH2 M(n – 1) + O C O CH2 NH2 As en is symmetrical ligand thus there is no geometrical isomerism. (D) In K3[Fe(CN)6], the oxidation state of Fe is +3. [Fe(CN)6]3– – 3d5, 4s0. Number of unpaired electron = 1  = n(n 2)  B.M. = 1(1 2)  B.M. = 3 B.M.