Question

On treatment of $100mL$ of $0.1M$ solution of $CoC{l_3}.6{H_2}O$ with excess $AgN{O_3}$​; $1.2 \times {10^{22}}$ ions are precipitated. The complex is:
A) $[Co{({H_2}O)_3}C{l_3}].2{H_2}O$
B) $[Co{({H_2}O)_6}]C{l_3}$​
C) $[Co{({H_2}O)_5}Cl]C{l_2}.{H_2}O$
D) $[Co{({H_2}O)_4}Cl]C{l_2}.{H_2}O$

Answer 1

Millimoles of $AgN{O_3} = \dfrac{{1.2 \times {{10}^{22}}}}{{6 \times {{10}^{23}}}} \times 1000 = 20$
Millimoles of $CoC{l_3}\cdot6{H_2}O = 0.1{\text{ }} \times {\text{ }}100{\text{ }} = {\text{ }}10$
$\therefore$Each mole of $CoC{l_3}\cdot6{H_2}O$gives two chloride particles.
\therefore $$$$\left[ {Co{{\left( {{H_2}O} \right)}_5}Cl} \right]C{l_2}\cdot{H_2}O

Complete step by step answer:
Complex comprises a focal atom or ion, which is normally metallic and is known as the coordination centre, and an encompassing exhibit of bound atoms or particles, that are thus known as ligands or complexing specialists. Many metal-containing mixes, particularly those of progress metals, are coordination complexes. A coordination complex whose middle is a metal particle is known as a metal complex of the $d$ block element.
No of moles of $CoC{l_3}.6{H_2}O$ in arrangement are -
$n{\text{ }} = {\text{ }}M{\text{ }} \times {\text{ }}V$
$n{\text{ }} = {\text{ }}0.1{\text{ }} \times {\text{ }}0.1$
$n = {10^{ - 2}}$
Absolute no of particles of $C{l^ - }$ in assumed complex –
$n' = 3 \times {10^{ - 2}}$
No of $C{l^ - }$ particles accelerated -
${N_p} = \dfrac{{(1.2 \times {{10}^{22}})}}{{(6.022 \times {{10}^{23}})}}$
${N_p} = 2 \times {10^{ - 2}}$
No of $C{l^ - }$ particles unpredicted -
${N_u} = 3 \times {10^{ - 2}} - 2 \times {10^{ - 3}}$
${N_u} = {10^{ - 2}}$
Avogadro’s number is $6.023 \times {10^{23}}$
$1.2 \times {10^{22}}$ particles compare to $\dfrac{{1.2 \times {{10}^{22}}}}{{6.023 \times {{10}^{23}}}} = 0.02$ moles.
$100{\text{ }}mL$ of $0.1M$ arrangement of $CoC{l_3}.6{H_2}O$ relates to $\dfrac{{100mL}}{{1000mL/L}} \times 0.1M = 0.01moles$.
$0.01$ moles of $CoC{l_3}.6{H_2}O$ responds with excess $AgN{O_3}$ to frame $0.02$ moles of $AgCl$ accelerate, which implies $0.01$moles of $A{g^ + }$and $0.01$ moles of $C{l^ - }$ particles.
Consequently, from one particle of $CoC{l_3}.6{H_2}O$, one chlorine molecule is dislodged. Thus, the complex is
$\left[ {Co{{\left( {{H_2}O} \right)}_5}Cl} \right]C{l_2}.{H_2}O$ as number of chloride particle outside co-appointment circle ought to be one.
Werner's Theory and it's impediment -
Hypothesizes of Werner's hypothesis
1. Metals show two sorts of linkages, Primary and Secondary.
2. Secondary valences are non-ionisable.
3. Primary valences are ordinarily ionisable.
4. The gatherings bound by the secondary linkages have spatial plans.
Impediments of Warner's hypothesis
1.Did not explain why just certain components have capacity to frame complexes
2.Gave no clarification for directional properties of bonds in coordination mixes.
3.Could not clarify the attractive and optical properties
Hence, the correct option is (D).

Note:
That is accelerated particles are twofold that of unprecipitated particles. This implies that out of $3$,$\;2$ $C{l^ - }$ particles are out of the co-appointment circle.
Along these lines, the equation for a given complex is $\left[ {Co{{\left( {{H_2}O} \right)}_4}Cl} \right]C{l_2}.{H_2}O$.