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For Mn^(+3) pairing energy is 28000 cm^(...

For `Mn^(+3)` pairing energy is `28000 cm^(-1)` , `Delta_0` for `[Mn(CN)_6]^(3-)` is `38500 cm^(-1)` then which of the following is/are correct.

A

Complex will be coloured

B

Complex will be low spin complex

C

Net CFSE= -33600 `cm^(-1)`

D

Complex will be colourless

Text Solution

Verified by Experts

The correct Answer is:
B,C,D
`lambda_(ab)=1/38500=259xx10^(-7)` cm =259 nm (U.V. region)
and `t_(2g)^(2:1)eg^(0,0)`
C.F.S.E = `-1.6Delta_0+P=1.6xx38500+28000= -33600 cm^(-1)`
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