Assertion (A) Complexes of $M X_6$ and $M X_5 L$ type ( $X$ and $L$ are unidentate) do not show geometrical isomerism.
Reason (R) Geometrical isomerism is not shown by complexes of coordination number 6.
Assertion $(\mathrm{A})\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ ion shows magnetic moment corresponding to two unpaired electrons.
Reason (R) Because it has $d^2 s p^3$ type hybridisation.
Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following
(a) $\left[\mathrm{CoF}_6\right]^{3-},\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$
(b) $\mathrm{FeF}_6{ }^{3-},\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
(a) $\left[\mathrm{CoF}_6\right]^{3-}$.
$\mathrm{F}^{-}$is a weak field ligand.
Configuration of $\mathrm{Co}^{3+}=3 d^6$ (or $t_{2 g}^4 e_g^2$)
Number of unpaired electrons $(n)=4$
Magnetic moment $(\mu)=\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24}=4.9 \mathrm{BM}$
$\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$, $\mathrm{H}_2 \mathrm{O}$ is a weak field ligand.
Configuration of $\mathrm{Co}^{2+}=3 d^7\left(\right.$ or $\left.t_{2 g}^5 e_g^2\right)$
Number of unpaired electrons $(n)=3$
$$\mu=\sqrt{3(3+2)}=\sqrt{15}=3.87 \mathrm{BM}$$
$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$ i.e., $\mathrm{Co}^{3+}$
$\because \quad \mathrm{CN}$ is strong field ligand.
$$\mathrm{Co}^{3+}=3 \mathrm{~d}^6\left(\mathrm{or} t_{2 g}^6 \mathrm{e}_g^0\right)$$
There is no unpaired electron, so it is diamagnetic.
$$\mu=0$$
(b) $\left[\mathrm{FeF}_6\right]^{3-}$,
$\mathrm{Fe}^{3+}=3 d^5\left(\right.$ or $\left.t_{2 g}^3 e_g^2\right)$
Number of unpaired electrons, $n=5$
$$\begin{aligned} \mu & =\sqrt{5(5+2)} \\ & =\sqrt{35}=5.92 \mathrm{BM} \\ {\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} } & \end{aligned}$$
$$\mathrm{Fe}^{2+}=3 d^6\left(\operatorname{or} t_{2 g}^4 e_g^2\right)$$
Number of unpaired electrons, $n=4$
$$\begin{aligned} \mu & =\sqrt{4(4+2)} \\ Z & =\sqrt{24} \\ & =4.98 \mathrm{BM} \\ {\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} } & \end{aligned}$$
Since, $\mathrm{CN}^{-}$is a strong field ligand, all the electrons get paired.
$$\mathrm{Fe}^{2+}=3 d^6\left(\operatorname{or} t_{2 g}^6 e_g^0\right)$$
Because there is no unpaired electron, so it is diamagnetic in nature.
Using valence bond theory, explain the following in relation to the complexes given below
$$\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+},\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{FeCl}_6\right]^{4-}$$
(a) Type of hybridisation
(b) Inner or outer orbital complex
(c) Magnetic behaviour
(d) Spin only magnetic moment value.
(a) $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$
(i) $d^2 s p^3$ hybridisation
(ii) Inner orbital complex because $(n-1) d$-orbitals are used.
(iii) Paramagnetic, as two unpaired electrons are present.
(iv) Spin only magnetic moment $(\mu)=\sqrt{2(2+2)}=\sqrt{8}=2.82 \mathrm{BM}$
$$\begin{aligned} & \text{(b) }\quad{\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}} \\ & \mathrm{Co}^{3+}=3 d^6 4 s^0 \end{aligned}$$
( $\mathrm{NH}_3$ pair up the unpaired $3 d$ electrons.)
(i) $d^2 s p^3$ hybridisation
(ii) Inner orbital complex because of the involvement of $(n-1) d$-orbital in bonding.
(iii) Diamagnetic, as no unpaired electron is present.
(iv) $\mu=\sqrt{n(n+2)}=\sqrt{0(0+2)}=0$ (Zero)
(c) $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$
(i) $d^2 s p^3$ hybridisation
(ii) Inner orbital complex (as $(n-1) d$-orbital take part.)
(iii) Paramagnetic (as three unpaired electrons are present.)
(iv) $\mu=\sqrt{n(n+2)}=\sqrt{3(3+2)}=\sqrt{15}=3.87 \mathrm{BM}$
$\begin{aligned} & \text { (d) }\left[\mathrm{Fe}(\mathrm{Cl})_6\right]^{4-} \\ & \mathrm{Fe}^{2+}=3 d^6\end{aligned}$
(i) $s p^3 d^2$ hybridisation
(ii) Outer orbital complex because $n d$-orbitals are involved in hybridisation.
(iii) Paramagnetic (because of the presence of four unpaired electrons).
(iv) $\mu=\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24}=4.9 \mathrm{BM}$
$\mathrm{CoSO}_4 \mathrm{Cl} \cdot 5 \mathrm{NH}_3$ exists in two isomeric forms ' A ' and ' B '. Isomer ' A ' reacts with $\mathrm{AgNO}_3$ to give white precipitate, but does not react with $\mathrm{BaCl}_2$. Isomer ' B ' gives white precipitate with $\mathrm{BaCl}_2$ but does not react with $\mathrm{AgNO}_3$. Answer the following questions.
(a) Identify ' $A$ ' and ' $B$ ' and write their structural formulae.
(b) Name the type of isomerism involved.
(c) Give the IUPAC name of ' $A$ ' and ' $B$ '.
' $A$ ' gives precipitate with $\mathrm{AgNO}_3$, so in it Cl is present outside the coordination sphere. ' $B$ ' gives precipitate with $\mathrm{BaCl}_2$, so in it $\mathrm{SO}_4^{2-}$ is present outside the coordination sphere.
(a) $\mathrm{So}, \mathrm{A}-\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{SO}_4\right] \mathrm{Cl}$
$B-\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{SO}_4$
(b) Ionisation isomerism (as give different ions when subjected to ionisation.)
(c) $[A]$, Pentaamminesulphatocobalt (III) chloride.
[B], Pentaamminechloridocobalt (III) sulphate.