Coordination Compounds

[MnCl? ]³?
Mn (25) → [Ar] 4s²3d?
Cl? → weak ligand
Mn³? → 3d? i.e. t? g³, eg¹
Hybridization → sp³d² and paramagnetic

(A) [PtCl? (NH? )? ], Number of GI = 2

(B) [Ni (CO)? ], Number of G.I = 0 as it is tetrahedral.
(C) [Ru (H? O)? Cl? ] Number of G.I = 2 (Facial and Meridional)

(D) [CoCl? (NH? )? ]? Number of G.I = 2 (Cis and trans)

Kindly consider the following figure

Kindly consider the following figure

In strong field of octahedral complex of Fe²⁺, the electronic configuration is,

$F{e}_{26}^{2+}\to \left[Ar\right]3d^{6}4s^0$

Number of unpaired electrons are zero in the presence of strong field ligand.

Hence, spin only magnetic moment = $\sqrt[]{n\left(n+2\right)}=0BM\left(n=0\right)$

${\left[Fe{\left(CO\right)}_4\left(C_2{O}_4\right)\right]}^{+}$

x + 4 * 0 – 2 * 1 = +1

x = +3

Fe⁺³ = 4s⁰ 3d⁵

In presence of strong ligand, i.e, CO.

Spin only magnetic moment

$=\sqrt[]{n\left(n+2\right)}B.M=\sqrt[]{1\left(1+2\right)}B.M$

$\sqrt[]{3}B.M.=1.73B.M$

$A{g}^{+}\left(aq\right)+2N{H}_3\left(aq\right)?Ag{\left(N{H}_3\right)}_2^{+}\left(aq\right)$

t = 00.8 M $\left(\frac{a}{2}\right)M$                  

0 $=\infty$   5 * 10^-8 M   $\left(\frac{a}{2}-1.6\right)M$   0.8M

$k_f=\frac{\left[Ag{\left(N{H}_3\right)}_2^{+}\right]}{\left[A{g}^{+}\right]{\left[N{H}_3\right]}^2}$

$10^8=\frac{0.8}{\left(5*10^{-8}\right){\left(\frac{a}{2}-1.6\right)}^2}$

$\frac{a}{2}=2$

 Concentration of NH₃ added =   $\frac{a}{2}=2M$

 Volume of solution = 2L

 Moles of NH₃ added = 2 * 2 mol = 4 mol.

MCl₃.2L => octahedral complex

$MC{l}_3.2L\stackrel{AgN{O}_3\left(excess\right)}{\to }1moleofAgCl$

It means one Cl^- ion present in ionization sphere.

$\therefore \left[MC{l}_2{L}_2\right]Cl$

For octahedral complex co-ordination numbers is 6

$\therefore L$ is bidentate ligand i.e. its denticity is 2.

${\left[Fe{\left(CN\right)}_{5}NOS\right]}^{4-}$  ${{}_{}}^{}$ Violet colour

${{}_{}}^{}$  ${\left[Fe{\left(SCN\right)}_6\right]}^{4-}$ Blood red colour

$$ $F{e}_4{\left[Fe{\left(CN\right)}_6\right]}_3{H}_{2}O\to$  Prussian blue colour

$$ ${\left[Fe{\left(CN\right)}_6\right]}^{4-}\to$  Yellow colour

FeCl₃. 3H₂O i.e. [Fe(H₂O)₃], K₃[Fe(CN]₆ and [Co(NH₃)₆]Cl₃; the last two complex are inner-orbital complex due to presence of strong ligand.

$\therefore {\Delta }_{0}of{K}_3\left[Fe{\left(CN\right)}_6\right]>{\Delta }_{0}of\left[Co{\left(N{H}_3\right)}_6\right]C{l}_3$       

Because CN^- is strong ligand

$\therefore {\Delta }_0\propto \frac{1}{\lambda }$         

More ${\Delta }_0$  smaller value of absorbed $\lambda$ .

$K_3\left[Fe{\left(CN\right)}_6\right]\to F{e}^{+3}=4s^{0}3d^{5}4p^0$       

$\therefore u.e=1$      

$\therefore \mu =\sqrt[]{1*\left(1+2\right)}B.M=\sqrt[]{3}B.M$        

= 1.732 B.M

$\approx$ 2