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JEE Main 2016 · Chemistry · Coordination Compounds · hard

20 mol of a violet coordination compound of formula CoCl3·4NH3 is dissolved in water and treated with excess AgNO3, giving 28.7 g of AgCl (Ag=108…

Single correct option

0.20 mol0.20\ \mathrm{mol} of a violet coordination compound of formula CoCl3⋅4NH3\mathrm{CoCl_3\cdot4NH_3} is dissolved in water and treated with excess AgNO3\mathrm{AgNO_3}, giving 28.7 g28.7\ \mathrm{g} of AgCl\mathrm{AgCl} (Ag=108\mathrm{Ag}=108, Cl=35.5\mathrm{Cl}=35.5). The number of geometrical isomers of the complex cation and its spin-only magnetic moment are respectively
  1. A. 2 and 00 BM
  2. B. 2 and 4.904.90 BM
  3. C. 1 and 00 BM
  4. D. 3 and 4.904.90 BM
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Answer: (A) 2 and 00 BM

28.7/143.5=0.20028.7/143.5=0.200 mol AgCl from 0.200.20 mol compound, so only one Cl−\mathrm{Cl^-} per formula unit is ionisable. The complex is [Co(NH3)4Cl2]Cl[\mathrm{Co(NH_3)_4Cl_2}]\mathrm{Cl}, with a cation of type [MA4B2]\mathrm{[MA_4B_2]}, which exists as cis and trans: 2 geometrical isomers. Co3+^{3+} is d6d^6 and with strong-field NH3\mathrm{NH_3} forms a low-spin t2g6t_{2g}^6 complex with no unpaired electrons, so μ=0\mu=0 BM. A high-spin d6d^6 would have μ=24=4.90\mu=\sqrt{24}=4.90 BM, the trap in other options.
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