MCQ
Calculate crystal field stabilization energy in $[Co(CN)_6]^{3-}$
- ✓$ - 2.4\,{\Delta _0} + 2p$
- B$ + 2.4\,{\Delta _0} + 2p$
- C$ - 3.6{\Delta _0} + 2p$
- D$ - 1.8\,{\Delta _0} + 2p$
✓
Answer
Correct option: A.
$ - 2.4\,{\Delta _0} + 2p$
a
In $\left[ Fe ( CN )_6\right]^{4-}$, iron has $3 d ^6, 4 s ^2$ system in ground state but in excited state it loses two electrons in the formation of ions and two electrons from $4 s$, so thus Cobalt gets $3 d ^6$ configuration. Now it is of low spin complex due to $CN$ ligands so all 6 electrons will go to $t _{2 g }$ orbitals. and o electrons will be in $e _{ g }$ orbital. By applying formula,
In $\left[ Fe ( CN )_6\right]^{4-}$, iron has $3 d ^6, 4 s ^2$ system in ground state but in excited state it loses two electrons in the formation of ions and two electrons from $4 s$, so thus Cobalt gets $3 d ^6$ configuration. Now it is of low spin complex due to $CN$ ligands so all 6 electrons will go to $t _{2 g }$ orbitals. and o electrons will be in $e _{ g }$ orbital. By applying formula,
$\Delta=$ no. of electrons in $t _{2 g } \cdot(-0.4)+$ no. of electrons in $e _{ g }(0.6)$
$=6(-0.4)+0(0.6)$
$=-2.4 \Delta_0$
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