On the basis of crystal field theory explain why Co(III) forms paramagnetic octahedral complex with weak field ligands whereas it forms diamagnetic octahedral complex with strong field ligands.
With weak field ligands; $\Delta_0< P$, (pairing energy) so, the electronic configuration of Co (III) will be $t_{2 g}^4 e_g^2$ i.e., it has 4 unpaired electrons and is paramagnetic.
With strong field ligands, $\Delta_0>P$ (pairing energy), so pairing occurs thus, the electronic configuration will be $t_{2 g}^6 e_g^0$. It has no unpaired electrons and is diamagnetic.
Why are low spin tetrahedral complexes not formed?
In tetrahedral complex, the $d$-orbital is splitting to small as compared to octahedral. For same metal and same ligand $\Delta_t=\frac{4}{9} \Delta_0$.
Hence, the orbital splitting energies are not enough to force pairing. As a result, low spin configurations are rarely observed in tetrahedral complexes.
Give the electronic configuration of the following complexes on the basis of crystal field splitting theory. $\left[\mathrm{CoF}_6\right]^{3-},\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ and $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_6\right]^{2+}$.
According to spectrochemical series, ligands can be arranged in a series in the order of increasing field strength i.e., $\mathrm{F}^{-}<\mathrm{NH}_3<\mathrm{CN}^{-}$.
Hence, $\mathrm{CN}^{-}$and $\mathrm{NH}_3$ being strong field ligand pair up the $t_{2 g}$ electrons before filling $e_g$ set. $\left[\mathrm{CoF}_6\right]^{3-} ; \mathrm{Co}^{3+}=\left(d^6\right) t_{2 g}^4 e_g^2$
$\left[\mathrm{Fe}(\mathrm{CN})_6^{4-}, \mathrm{Fe}^{2+}=\left(\mathrm{d}^6\right) t_{2 g}^6 e_g^0\right.$
$\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_6\right]^{2+}, \mathrm{Cu}^{2+}=\left(d^9\right) t_{3 g}^6 e_g^3$
Explain why $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ has magnetic moment value of 5.92 BM whereas $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ has a value of only 1.74 BM ?
$$\begin{aligned} \text{As we know,}\quad\mu_m & =\sqrt{n(n+2} \mathrm{BM} \\ \text{where,}\quad\mu_m & =\text { magnetic moment } \\ \mu_n & =\text { number of unpaired electrons } \\ \text{It}\quad\mu_m & =1.74 \text { i.e., } n=1 \\ \text{and}\quad\mu_m & =5.92 \text { i.e., } n=5 \end{aligned}$$
$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ involves $d^2 s p^3$ hybridisation with one unpaired electron (as shown by its magnetic moment 1.74 BM ) and $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ involves $s p^3 d^2$ hybridisation with five unpaired electrons (because magnetic moment equal to 5.92 BM ). $\mathrm{CN}^{-}$is stronger ligand than $\mathrm{H}_2 \mathrm{O}$ according to spectrochemical series. $\Delta_0>P$ for $\mathrm{CN}^{-}$hence, fourth electron will pair itself. Whereas for water pairing will not happen for $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ the electronic configuration of $\mathrm{Fe}^{3+}$ is
One unpaired electron
For $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ the electronic configuration of $\mathrm{Fe}^{3+}$ is
Five unpaired electron Hence, $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ and $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ are inner orbital and outer orbital complex respectively.
Arrange following complex ions in increasing order of crystal field splitting energy $\left(\Delta_0\right)$.
$$\left[\mathrm{Cr}(\mathrm{Cl})_6\right]^{3-},\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+} .$$
CFSE is higher when the complex contains strong field ligand. Thus, crystal field splitting energy increases in the order
$$\left[\mathrm{Cr}(\mathrm{Cl})_6\right]^{3-}<\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}<\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-} .$$
Because according to spectrochemical series the order of field strength is
$$\mathrm{Cl}^{-}<\mathrm{NH}_3<\mathrm{CN}^{-}$$