Question

The crystal field stabilisation energy [CFSE] and the spin only magnetic moment in Bohr magneton [BM] for the complex $K_3\left[Fe\right(CN{)}_6]$ respectively are:

• A. $0.0{\Delta }_o$ and $\sqrt{35}$ BM
• B. $2.0{\Delta }_o$ and $\sqrt{3}$ BM
• C. $0.4{\Delta }_o$ and $\sqrt{24}$ BM
• D. $-2.4{\Delta }_o$ and $0$ BM

Answer 1

Answer: (B)

$2.0{\Delta }_o$ and $\sqrt{3}$ BM

Electronic Configuration of ${}_{26}Fe$: ${}_{26}Fe\to 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}4s^2$

Electronic Configuration of ${}_{26}Fe{}^{3+}$ : ${}_{26}Fe{}^{3+}\to 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^0$

In $\left[fe\right(CN{)}_6]{}^{3-}$ ,hexacyanoferrate(iii) ion.We have 5-d electrons.

Here strong cyanide ligand causes greater d-orbital splitting.The octahedral splitting energy,becomes high.

During filling up process,the first 3 electrons go into $t_{2g}$ orbitals.Now however, the splitting energy is much greater so it is lesser energitically costly for electrons to pair up in the $t_{2g}$ orbitals than to go into the $e_g$ orbitals and low spin complex is formed.

Here the crystal field splitting energy:

= -no. of electrons in $e_g$ X0.6${\Delta }_0$+no. of electrons in $t_{2g}$ X0.4${\Delta }_0$

= -0 X0.6${\Delta }_0$+5X0.4${\Delta }_0$=2${\Delta }_0$

The spin only moment ${\mu }_s=\sqrt{n(n+2)}$

So..,

${\mu }_s=\sqrt{n(n+2)}$=${\mu }_s=\sqrt{1X(1+2)}$=$\sqrt{3}$=1.7BM