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B
0.4Δ0
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C
−0.4Δt
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D
0.6Δt
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Solution
The correct option is A−0.4Δ0
Option (B) is correct.
In [CoF6]3−, cobalt has 3d7,4s2 system in ground state but in excited state it loses three electrons in the formation of ions and two elctrons from 4s and one from 3d orbital so thus Cobalt gets 3d6 configuration. Now it is of high spin so 4 electrons go to t2g orbital and 2 electrons go to eg orbital. By applying formula,
Δ = no. of electrons in t2g⋅(−0.4)+ no. of electrons in eg(0.6)