Question

Example of neutral complex compound in the following is
A. $CoC{l_3}.6N{H_3}$
B. $CoC{l_3}.5N{H_3}$
C. $CoC{l_3}.4N{H_3}$
D. $CoC{l_3}.3N{H_3}$

Answer 1

A complex compound is referred to as a coordination compound in chemistry. It is represented in the form of coordination complexes where the central atom is a transition metal which is surrounded by different types of ligands.

Complete step by step answer:
Coordination compounds are classes of compounds in which the ligands or anions are bound to a central metal atom through a coordinate covalent bond. In general the transition metal forms several coordination complexes with various ligands.
The ligands are atoms, molecules or ions which are bound to central metal atom/ion. Ligands are anions like $C{l^ - }$ , $N{O_2}^ -$ , etc. or molecules like ${H_2}O$ , $N{H_3}$ etc.
The given sets of compounds are coordination compounds of cobalt. Cobalt is an element in the periodic table with atomic number $27$ . Its electronic configuration is $\left[ {Ar} \right]3{d^7}4{s^2}$. The common oxidation state of $Co$ found in coordination compounds is $+ 2$ and $+ 3$ .
A neutral complex is a coordination complex which does not ionize to produce charged species. In order to determine whether the given complexes are neutral or charged we have to check the ionizability of the complexes formed.
Let the complexes are octahedral complexes as the coordination number of $Co$ is six. Ammonia is a neutral ligand which satisfies the primary valency of the cobalt metal. The complexes are written as
$CoC{l_3}.6N{H_3}{\text{ = }}\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}$
$CoC{l_3}.5N{H_3}{\text{ = }}\left[ {Co{{\left( {N{H_3}} \right)}_5}Cl} \right]C{l_2}$
$CoC{l_3}.4N{H_3}{\text{ = }}\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]Cl$
$CoC{l_3}.3N{H_3} = \left[ {Co{{\left( {N{H_3}} \right)}_3}C{l_3}} \right]$
Except the complex $\left[ {Co{{\left( {N{H_3}} \right)}_3}C{l_3}} \right]$ , all the other three complexes are ionizable and produce cations and anions on dissociation. Hence they are not neutral complexes.
$\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3} \to {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^ + } + 3C{l^ - }$. Here three chloride ions are liberated or ionized so it is not a neutral complex.
$\left[ {Co{{\left( {N{H_3}} \right)}_5}Cl} \right]C{l_2} \to {\left[ {Co{{\left( {N{H_3}} \right)}_5}Cl} \right]^ + } + 2C{l^ - }$. Here two chloride ions are liberated or ionized so it is not a neutral complex.
$\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]Cl \to {\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]^ + } + C{l^ - }$. Here one chloride ion is liberated or ionized so it is not a neutral complex.
$\left[ {Co{{\left( {N{H_3}} \right)}_3}C{l_3}} \right] \to no{\text{ }}ionization$. Here no ionization takes place as the three ammonia and three chloride ions satisfy the primary valency of the central atom.
Thus option D is the correct answer, i.e. $\left[ {Co{{\left( {N{H_3}} \right)}_3}C{l_3}} \right]$ is a neutral complex.

Note:
Transition metal exhibits two types of valencies one is primary and other is secondary. In cationic complexes the coordination sphere is a cation e.g. $\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}$ . In anionic complex the coordination sphere is an anion ${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$ .