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29
Subjective

Why are low spin tetrahedral complexes not formed?

Explanation

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.

30
Subjective

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 $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_6\right]^{2+}$.

Explanation

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. $\left[\mathrm{CoF}_6\right]^{3-} ; \mathrm{Co}^{3+}=\left(d^6\right) t_{2 g}^4 e_g^2$

$\left[\mathrm{Fe}(\mathrm{CN})_6^{4-}, \mathrm{Fe}^{2+}=\left(\mathrm{d}^6\right) t_{2 g}^6 e_g^0\right.$

$\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_6\right]^{2+}, \mathrm{Cu}^{2+}=\left(d^9\right) t_{3 g}^6 e_g^3$

31
Subjective

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 ?

Explanation

$$\begin{aligned} \text{As we know,}\quad\mu_m & =\sqrt{n(n+2} \mathrm{BM} \\ \text{where,}\quad\mu_m & =\text { magnetic moment } \\ \mu_n & =\text { number of unpaired electrons } \\ \text{It}\quad\mu_m & =1.74 \text { i.e., } n=1 \\ \text{and}\quad\mu_m & =5.92 \text { i.e., } n=5 \end{aligned}$$

$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ 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.

32
Subjective

Arrange following complex ions in increasing order of crystal field splitting energy $\left(\Delta_0\right)$.

$$\left[\mathrm{Cr}(\mathrm{Cl})_6\right]^{3-},\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+} .$$

Explanation

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

$$\mathrm{Cl}^{-}<\mathrm{NH}_3<\mathrm{CN}^{-}$$

33
Subjective

Why do compounds having similar geometry have different magnetic moment?

Explanation

It is due to the presence of weak and strong field ligands in complexes. If CFSE is high, the complex will show low value of magnetic moment and vice-versa, e.g. $\left[\mathrm{CoF}_6\right]^{3-}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$, the former is paramagnetic, and the latter is diamagnetic because $\mathrm{F}^{-}$is a weak field ligand and $\mathrm{NH}_3$ is a strong field ligand while both have similar geometry.