JEE Main 2018 · Chemistry · Coordination Compounds · hard

The ratio of the crystal field stabilisation energy (in the absence of pairing energy) of high-spin tetrahedral [CoCl4]^2- to that of high-spin…

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The ratio of the crystal field stabilisation energy (in the absence of pairing energy) of high-spin tetrahedral

\mathrm{[CoCl_4]^{2-}}

to that of high-spin octahedral

\mathrm{[Co(H_2O)_6]^{2+}}

is, taking

\Delta_t=\tfrac49\Delta_o

(same metal ion, same

d

-electron count),

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Answer: (C) $2/3$

\mathrm{Co^{2+}}

d^7

. Octahedral high-spin:

t_{2g}^5e_g^2

, CFSE

=5(-0.4)+2(+0.6)=-0.8\Delta_o

. Tetrahedral:

e^4t_2^3

, with

e

-0.6\Delta_t

and

t_2

+0.4\Delta_t

, so CFSE

=4(-0.6)+3(0.4)=-1.2\Delta_t=-1.2\times\tfrac49\Delta_o=-0.533\Delta_o

. Ratio

=0.533/0.8=2/3

. Using

\Delta_t=\Delta_o

would give

3/2

Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake