MCQ
Crystal field stablization energy for complex $[Co(CN)_6]^{-3}$ will be
- ✓$- 2.4{\Delta _0}\, + \,3P$
- B$+ 2.4{\Delta _0}\, + \,3P$
- C$- 3.6{\Delta _0}\, + \,3P$
- D$- 1.8{\Delta _0}\, + \,3P$
✓
Answer
Correct option: A.
$- 2.4{\Delta _0}\, + \,3P$
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 eg 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 eg orbital. By applying formula,
$\Delta= \text { no. of electrons in } t _{2 g } \cdot(-0.4)+\text { no. of electrons in } e _{ g }(0.6)$
$=6(-0.4)+0(0.6)$
$=-2.4 \Delta_0$
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