Chemistry Coordination Compounds

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

x + 4 * 0 – 2 * 1 = +1

x = +3

Fe⁺³ = 4s⁰ 3d⁵

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

Spin only magnetic moment

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

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

$ln{\left[Co{\left(CN\right)}_6\right]}^{4-},$ Co has +2 oxidation state

Hence, electronic configuration of CO2+ is $\left[Ar\right]3d^{7}4s^0$ ${}^{}{}^{}$ . In complex   ${\left[Co{\left(CN\right)}_6\right]}^{4-},$ ${{}_{}}^{}$given ligand CN is strong hence, after pairing in d-subshell, total number of unpaired electron =

$\therefore$ spin magnetic moment =   $\sqrt[]{1\left(1+2\right)}=\sqrt[]{3}=1.73BM$ $\text{exception 25:}\text{exception 25:}$

$=1.73BM\approx 2.0BM$

Calamine is the ore of zinc i.e. ZnCO₃.

Malachite is the ore of copper i.e. CuCO₃.Cu (OH)₂.

Biuret is

Denticity is 2 of terminal -NH₂ only, because middle –NH will undergo in cross conjugation.

Two optical isomers of ${\left[Cr{\left(C_2{O}_4\right)}_3\right]}^{3-}$ are shown :

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

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.

Stability constant of complex = $\frac{1}{dissocationconstant}$  

$\therefore$ Dissociation constant = $\frac{1}{2.1*10^{13}}=4.76*10^{-14}$  

the nearest value of dissociation constant = 5 * 10^-14

Ans. = 5

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