Basic Terms Questions in English

(C) The ligand structure contains one tertiary amine nitrogen atom,two phenolate oxygen atoms,and one diethylamino nitrogen atom.
Counting the potential donor sites:
$1$. One central tertiary amine nitrogen $(N)$.
$2$. Two phenolate oxygen atoms $(O^-)$.
$3$. One terminal diethylamino nitrogen atom $(NEt_2)$.
Thus,there are a total of $4$ donor atoms available for coordination.
Therefore,the ligand is tetradentate.

(A) The chemical formula of copper sulfate pentahydrate is written as $[Cu(H_2O)_4]SO_4 \cdot H_2O$.
In this structure,$4$ water molecules are directly coordinated to the central $Cu^{2+}$ ion as ligands.
The $5^{th}$ water molecule is held by hydrogen bonding in the crystal lattice and is not directly coordinated to the copper ion.

(A) The ligand shown is diethylenetriaminepentaacetate $(DTPA^{5-})$.
It contains $3$ nitrogen atoms and $5$ carboxylate oxygen atoms capable of coordination.
For common transition metal ions,the maximum coordination number is typically $6$,so the ligand acts as a hexadentate ligand (denticity = $6$).
For inner-transition metal ions (like lanthanides),the coordination number can be higher (up to $8$ or $9$) due to their larger size and availability of more vacant orbitals,allowing the ligand to act as an octadentate ligand (denticity = $8$).

(D) The coordination number is defined as the number of ligand donor atoms to which the metal is directly bonded.
$en$ (ethane$-1,2-$diamine) is a bidentate ligand,meaning it occupies $2$ coordination sites.
In $[Co(Cl)(en)_2]Cl$,the coordination number of $Co$ is $1(Cl) + 2(2 \times en) = 5$.
$C_2O_4^{2-}$ (oxalate) is also a bidentate ligand.
In $K_3[Al(C_2O_4)_3]$,the coordination number of $Al$ is $3(C_2O_4) \times 2 = 6$.
Therefore,the coordination numbers are $5$ and $6$.

(D) The brown ring test for nitrate ions involves the reaction of $NO_3^-$ with $Fe^{2+}$ in the presence of concentrated $H_2SO_4$ to form the complex $[Fe(H_2O)_5NO]SO_4$.
In this complex,the iron is in the $+1$ oxidation state $(Fe^+)$ and the nitric oxide ligand is present as $NO^+$.
Therefore,the statement that 'Iron has $+2$ oxidation state and $NO$ is neutral' is incorrect.
The brown colour arises due to charge transfer spectra.

(A) complex is non-ionisable if it does not contain any counter-ions outside the coordination sphere.
In the given options,$[Pt(NH_3)_2Cl_4]$ has no ions outside the square brackets,meaning it does not dissociate into ions in an aqueous solution.
Therefore,it is the non-ionisable complex.

(D) The coordination number of $Ni$ in $[Ni(C_2O_4)_3]^{4-}$ is $6$ because $C_2O_4^{2-}$ (oxalate) is a bidentate ligand and there are $3$ such ligands,so $2 \times 3 = 6$.
Let the oxidation state of $Ni$ be $x$.
$x + 3(-2) = -4$
$x - 6 = -4$
$x = +2$
Sum of coordination number and oxidation number = $6 + 2 = 8$.

(A) In $[Co(NH_3)_5(CO_3)]ClO_4$,the central metal atom is $Co$. The ligands are five $NH_3$ (monodentate) and one $CO_3^{2-}$ (bidentate). However,in this specific complex,$CO_3^{2-}$ acts as a monodentate ligand to satisfy the coordination sphere. Thus,the coordination number is $5 + 1 = 6$.
Let $X$ be the oxidation number of $Co$. The charge on $NH_3$ is $0$,$CO_3$ is $-2$,and $ClO_4$ is $-1$. Therefore,$X + 5(0) + (-2) + (-1) = 0$,which gives $X = +3$.
The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$. For $Co^{3+}$,the configuration is $[Ar] 3d^6 4s^0$. Thus,there are $6$ $d$-electrons.
$NH_3$ is a strong field ligand. In the presence of strong ligands,the $6$ electrons in the $3d$ orbitals of $Co^{3+}$ pair up,resulting in $0$ unpaired $d$-electrons.

(B) The molar mass of $AgCl$ is $143.5 \ g/mol$.
Given mass of $AgCl$ precipitate = $28.7 \ g$.
Number of moles of $AgCl$ = $\frac{28.7 \ g}{143.5 \ g/mol} = 0.2 \ mol$.
Since $0.1 \ mol$ of the complex produces $0.2 \ mol$ of $AgCl$,each mole of the complex releases $2 \ mol$ of $Cl^-$ ions.
This indicates that $2 \ Cl^-$ ions are ionizable (outside the coordination sphere).
Therefore,the complex is $[CrCl(H_2O)_5]Cl_2 \cdot H_2O$.

(D) $1$. The formula is $CoBr_2Cl \cdot 4NH_3$.
$2$. Conductance measurements show $2$ ions per formula unit,which implies the complex dissociates as $[Complex]^+ + [Counterion]^-$.
$3$. Silver nitrate $(AgNO_3)$ gives an immediate precipitate of $AgCl$,meaning $Cl^-$ is outside the coordination sphere (as a counterion).
$4$. It gives no $AgBr$ precipitate,meaning $Br^-$ ions are inside the coordination sphere.
$5$. Therefore,the structure must be $[CoBr_2(NH_3)_4]Cl$.
$6$. This complex dissociates as $[CoBr_2(NH_3)_4]^+ + Cl^-$,providing $2$ ions in solution.

(A) To determine the number of ions produced,we look at the dissociation of the complex in water.
$A$: $[Pt(NH_3)_5Cl]Cl_3 \rightarrow [Pt(NH_3)_5Cl]^{3+} + 3Cl^{-}$. This yields $1 + 3 = 4$ ions.
$B$: $[Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^{-}$. This yields $1 + 4 = 5$ ions.
$C$: $[Pt(NH_3)_2Cl_4]$ is a neutral complex and does not ionize.
$D$: $[Pt(NH_3)_4Cl_2]Cl_2 \rightarrow [Pt(NH_3)_4Cl_2]^{2+} + 2Cl^{-}$. This yields $1 + 2 = 3$ ions.
Therefore,the complex that produces four ions per molecule is $[Pt(NH_3)_5Cl]Cl_3$.

(D) The coordination number of a central metal atom in a coordination complex is defined as the total number of $\sigma-$ bonds formed by the ligands with the central metal atom.
Since each ligand donates at least one lone pair to form a coordinate covalent bond (which is a $\sigma-$ bond),the coordination number is the sum of the number of $\sigma-$ bonds formed by all ligands attached to the metal ion.
Therefore,it is the number of monodentate ligands (or the equivalent number of donor atoms) bonded to the metal ion by $\sigma-$ bonds.

(D) The coordination number of a metal ion in a complex is defined as the number of ligand donor atoms to which the metal is directly bonded.
For most cations,the coordination number is not fixed; it depends on several factors,including the size of the central metal ion,the charge of the cation,and the steric/electronic properties of the ligands.
Option $D$ is the most accurate statement as it highlights the dependence on the physical and chemical properties of the cation.

(A) The complex is $[Co(NH_3)_4 CO_3]ClO_4$.
$NH_3$ is a monodentate ligand $(4 \times 1 = 4)$ and $CO_3^{2-}$ is a bidentate ligand $(1 \times 2 = 2)$.
Coordination number $= 4 + 2 = 6$.
Let the oxidation state of $Co$ be $x$.
$x + 4(0) + (-2) = +1$ (since $ClO_4^-$ has a charge of $-1$).
$x - 2 = +1 \implies x = +3$.
The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
Thus,the number of $d$-electrons is $6$.

(A) The reaction between sodium thiosulphate and $AgBr$ is given by:
$AgBr(s) + 2Na_2S_2O_3(aq) \rightarrow Na_3[Ag(S_2O_3)_2](aq) + NaBr(aq)$
In the complex ion $[Ag(S_2O_3)_2]^{3-}$,the silver ion $(Ag^+)$ is bonded to two thiosulphate ligands.
Since each thiosulphate ligand acts as a monodentate ligand (binding through the sulphur atom),the coordination number of silver is $2$.

(C) Double salts are molecular compounds that exist only in the solid state (crystal lattice) and dissociate into their constituent ions when dissolved in water.
Among the given options,$(NH_4)_2SO_4 \cdot ZnSO_4 \cdot 6H_2O$ is not a known stable double salt.
Other options like Mohr's salt $(NH_4)_2SO_4 \cdot FeSO_4 \cdot 6H_2O$ are well-known examples of double salts.

(C) The $NO$ molecule is a unique ligand. While most ligands donate two electrons to the central metal atom,$NO$ can act as a three-electron donor in certain complexes (e.g.,in $[Fe(H_2O)_5(NO)]SO_4$),where it donates three electrons to the metal center.

(D) In the complex $Fe(CO)_5$,$CO$ is a neutral ligand with an oxidation state of $0$.
Let the oxidation state of $Fe$ be $x$.
$x + 5(0) = 0$
$x = 0$
Therefore,the oxidation state of $Fe$ in $Fe(CO)_5$ is $0$.

(D) The chemical formula $K_4[Fe(CN)_6]$ contains $6$ atoms of carbon per molecule.
Therefore,$1 \, mol$ of $K_4[Fe(CN)_6]$ contains $6 \, mol$ of carbon atoms.
The molar mass of carbon is $12 \, g/mol$.
So,$1 \, mol$ of $K_4[Fe(CN)_6]$ contains $6 \times 12 = 72 \, g$ of carbon.
For $0.5 \, mol$ of $K_4[Fe(CN)_6]$,the mass of carbon is $0.5 \times 72 = 36 \, g$.

(B) To determine the oxidation state of the transition metal,we consider the charge of the ligands:
$1$. In $[Fe(H_2O)_3(OH)_3]$,$Fe + 3(0) + 3(-1) = 0 \implies Fe = +3$.
$2$. In $[Ni(CO)_4]$,$Ni + 4(0) = 0 \implies Ni = 0$. Here,$CO$ (carbonyl) is a neutral ligand.
$3$. In $[Fe(H_2O)_6]SO_4$,$Fe + 6(0) = +2 \implies Fe = +2$.
$4$. In $[Co(NH_3)_6]Cl_3$,$Co + 6(0) = +3 \implies Co = +3$.
Thus,the transition metal with an oxidation state of zero is in $[Ni(CO)_4]$.

(B) Ligands are species that can donate a pair of electrons to a central metal atom or ion to form a coordinate bond.
For a species to act as a ligand,it must possess at least one lone pair of electrons.
In $NH_4^+$,the nitrogen atom has used all its valence electrons to form four covalent bonds with hydrogen atoms.
Since there is no lone pair of electrons available on the nitrogen atom in $NH_4^+$,it cannot donate electrons to a metal ion.
Therefore,$NH_4^+$ is not expected to act as a ligand.

(D) An ambidentate ligand is a ligand that can coordinate to a central metal atom through two different donor atoms.
$OCN^{-}$ (cyanate) can coordinate through $N$ or $O$.
$SCN^{-}$ (thiocyanate) can coordinate through $S$ or $N$.
$NO_2^{-}$ (nitrite) can coordinate through $N$ or $O$.
$NH_2CONH_2$ (urea) is a monodentate ligand that coordinates only through the oxygen atom. Therefore,it is not an ambidentate ligand.

(A) The given coordination compound is $[Co(NH_3)_5Cl]Cl_2$.
When this compound dissolves in water,it dissociates into ions as follows:
$[Co(NH_3)_5Cl]Cl_2 \rightarrow [Co(NH_3)_5Cl]^{2+} + 2Cl^-$.
Here,the complex ion $[Co(NH_3)_5Cl]^{2+}$ acts as one unit and there are two chloride ions $(Cl^-)$.
Thus,the ratio of the complex ion to the chloride ions is $1:2$.
Therefore,it is a $1:2$ type electrolyte.

(C) In the complex $K_4[Fe(CN)_6]$,the potassium ions $(K^+)$ and the complex anion $[Fe(CN)_6]^{4-}$ are held together by ionic bonds.
Inside the complex anion $[Fe(CN)_6]^{4-}$,the $C$ and $N$ atoms in the cyanide ligand $(CN^-)$ are connected by a covalent bond (triple bond).
The $Fe^{2+}$ ion and the $CN^-$ ligands are linked by coordinate covalent bonds (dative bonds),where the $CN^-$ ligand donates a lone pair of electrons to the $Fe^{2+}$ center.
Therefore,the compound contains ionic,covalent,and coordinate covalent bonds.

(C) To determine the number of ions produced,we look at the dissociation of the coordination compounds in water:
$1$. $[Pt(NH_3)_4]Cl_4 \rightarrow [Pt(NH_3)_4]^{4+} + 4Cl^-$ ($5$ ions total)
$2$. $[Pt(NH_3)_2Cl_4]$ does not dissociate into ions as it is a neutral complex.
$3$. $[Pt(NH_3)_5Cl]Cl_3 \rightarrow [Pt(NH_3)_5Cl]^{3+} + 3Cl^-$ ($4$ ions total)
$4$. $[Pt(NH_3)_4Cl_2]Cl_2 \rightarrow [Pt(NH_3)_4Cl_2]^{2+} + 2Cl^-$ ($3$ ions total)
Since the question states that the complex produces $4$ ions per molecule,the correct structure is $[Pt(NH_3)_5Cl]Cl_3$.

(A) According to Werner's theory:
- Primary valencies are ionisable and non-directional.
- Secondary valencies are non-ionisable and possess directional properties,which determine the geometry of the coordination complex.
- Ligands are connected to the central metal ion by coordinate bonds.
- Therefore,all three statements ($1$,$2$,and $3$) are correct.

(B) The reaction between $Zn$ and $NaOH$ is given by: $Zn + 2NaOH \to Na_2[Zn(OH)_4]$ or $Na_2ZnO_2 + H_2$.
In the complex $Na_2[Zn(OH)_4]$,the complex ion is $[Zn(OH)_4]^{2-}$.
Since the complex ion carries a negative charge,the $Zn$ atom is present in the anionic part of the complex.

(D) The complex $PtCl_4 \cdot 4NH_3$ can be written as $[Pt(NH_3)_4]Cl_4$ according to Werner's theory.
$1$. The primary valency is satisfied by the $4$ chloride ions $(Cl^-)$ outside the coordination sphere,which are ionizable.
$2$. The secondary valency is satisfied by the $4$ $NH_3$ molecules inside the coordination sphere,which are non-ionizable.
$3$. Therefore,statement $III$ is correct because all $4$ $NH_3$ molecules are bonded by secondary valency.
$4$. Statement $I$ is correct because all $4$ chloride ions are bonded by primary valency.
$5$. Statements $II$ and $IV$ are incorrect because the chloride ions are not bonded by secondary valency in this specific complex.
Thus,the correct statements are $I$ and $III$. However,based on the provided options,the most accurate choice reflecting the secondary valency behavior is $III$.

(C) ligand is defined as a species that can donate a lone pair of electrons to a central metal atom or ion to form a coordinate bond.
In $NH_3$,the nitrogen atom has one lone pair of electrons available for donation.
In $NH_4^+$,the nitrogen atom has used all its valence electrons to form four $N-H$ covalent bonds,leaving no lone pair of electrons available for donation.
Therefore,$NH_4^+$ cannot act as a ligand.

(B) In coordination compounds,the species inside the square brackets $[...]$ do not dissociate into individual ions in an aqueous solution.
$A$. $[Fe(CN)_6]^{3-}$ is a complex ion and does not release $Fe^{3+}$ ions.
$B$. $Fe_2(SO_4)_3$ is an ionic salt that dissociates completely in water as $Fe_2(SO_4)_3 \rightarrow 2Fe^{3+} + 3SO_4^{2-}$,thus providing $Fe^{3+}$ ions.
$C$. $[Fe(CN)_6]^{4-}$ is a complex ion and does not release $Fe^{3+}$ ions.
$D$. $(NH_4)_2SO_4 \cdot FeSO_4 \cdot 6H_2O$ (Mohr's salt) dissociates to give $Fe^{2+}$ ions,not $Fe^{3+}$ ions.

(A) The charge on a coordination complex is the sum of the oxidation states of the central metal atom and the charges on the ligands.
For $[Co(NH_3)_3Cl_3]$:
$Co$ is in the $+3$ oxidation state.
$NH_3$ is a neutral ligand (charge $= 0$).
$Cl^-$ is an anionic ligand (charge $= -1$).
Total charge $= (+3) + 3(0) + 3(-1) = +3 + 0 - 3 = 0$.
Since the total charge is $0$,the complex is neutral.

(D) double salt is an addition compound that dissociates into its constituent ions in aqueous solution.
$1.$ Carnallite $(KCl \cdot MgCl_2 \cdot 6H_2O)$ is a double salt.
$2.$ Mohr's salt $(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O)$ is a double salt.
$3.$ Alum $(K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O)$ is a double salt.
Since all the given options are double salts,the correct answer is $D$.

(A) $Werner's$ theory states that in coordination compounds,the central metal atom exhibits two types of valencies: $1$. Primary valency and $2$. Secondary valency.
Primary valency corresponds to the oxidation state of the metal and is ionizable,meaning it is satisfied by negative ions.
Secondary valency corresponds to the coordination number and is non-ionizable,meaning it is satisfied by ligands (neutral molecules or negative ions) attached to the metal atom.

(B) In the coordination compound $[Co(NH_3)_6]Cl_3$,the central metal ion is $Co^{3+}$.
Each $NH_3$ ligand is a monodentate ligand,meaning it donates one lone pair of electrons to the central metal ion to form one coordinate bond.
Since there are $6$ $NH_3$ ligands attached to the $Co^{3+}$ ion,the total number of coordinate bonds formed is $6 \times 1 = 6$.

(A) According to Werner's coordination theory:
$1$. Primary valency corresponds to the oxidation state of the metal ion and is satisfied by anions.
$2$. Secondary valency corresponds to the coordination number and is satisfied by ligands.
In $K_3[Fe(CN)_6]$,the $Fe^{3+}$ ion has an oxidation state of $+3$ (primary valency) and a coordination number of $6$ (secondary valency). The $CN^-$ ions act as ligands,satisfying both the primary valency (by balancing the charge) and the secondary valency (by coordinating to the metal).
Statement $A$ claims the ligand satisfies *only* the secondary valency,which is incorrect because $CN^-$ ions also satisfy the primary valency of the $Fe^{3+}$ ion.

(D) Diethylenetriamine,represented as $NH_2CH_2CH_2NHCH_2CH_2NH_2$,contains three nitrogen donor atoms.
It can form three coordinate bonds with a central metal ion,making it a $tridentate$ ligand.
Since it can form a ring structure with the metal ion,it acts as a $chelating$ agent.
Because it has more than one donor atom,it is also classified as a $polydentate$ ligand.
Therefore,all the given options are correct.

(A) The ligand $SCN^-$ is an ambidentate ligand,meaning it can coordinate through either the sulfur atom or the nitrogen atom.
When it coordinates through the sulfur atom,it is named as $Thiocyanato-S$.
When it coordinates through the nitrogen atom,it is named as $Thiocyanato-N$ (also known as $Isothiocyanato$).

(C) Lithium tetrahydroaluminate is represented by the formula $LiAlH_4$.
In this complex,the central metal atom is $Al^{3+}$ and the ligands attached to it are hydride ions $(H^{-})$.
Therefore,the ligand present is $H^{-}$.

(D) The glycinato ligand is derived from glycine $(NH_2-CH_2-COOH)$ by the loss of a proton from the carboxylic acid group.
Its formula is $NH_2-CH_2-COO^-$.
It acts as a bidentate ligand because it coordinates through both the nitrogen atom of the amine group and one of the oxygen atoms of the carboxylate group.
Therefore,it has two donor sites ($N$ and $O$).

(C) ligand is an ion or molecule that binds to a central metal atom or ion to form a coordination complex. This requires the ligand to have at least one lone pair of electrons to donate to the metal center.
$PH_3$ has a lone pair on $P$.
$NO^{+}$ is a nitrosonium ion,which acts as a ligand in many complexes.
$Cl^{-}$ is a classic anionic ligand.
$BF_3$ is an electron-deficient molecule (Lewis acid) with an incomplete octet on the boron atom. It does not have a lone pair to donate to a metal center,so it is not considered a ligand.

(D) $NH_2-NH_2$ (hydrazine) has two nitrogen atoms,each with a lone pair of electrons.
It can act as a monodentate ligand by donating one lone pair to a metal center.
It can also act as a bridging ligand by using both nitrogen atoms to connect two metal centers.
Therefore,it can function as both a monodentate and a bridging ligand.

(C) The coordination number is defined as the number of ligand donor atoms to which the metal is directly bonded.
In the complex $[Co(en)_2Br_2]Cl_2$,the ligands are $en$ (ethylenediamine) and $Br^-$.
$en$ is a bidentate ligand,meaning it donates two donor atoms per molecule.
There are $2$ $en$ ligands,so they provide $2 \times 2 = 4$ donor atoms.
There are $2$ $Br^-$ ligands,which are monodentate,providing $2 \times 1 = 2$ donor atoms.
Total coordination number = $4 + 2 = 6$.

(C) $1$. The coordination number is the total number of ligand donor atoms bonded to the central metal atom. Here,$SO_4^{2-}$ acts as a bidentate ligand (or monodentate depending on bonding,but typically in this complex,it occupies one coordination site) and $5$ $NH_3$ molecules are monodentate. However,in $[X(SO_4)(NH_3)_5]Cl$,$SO_4$ is a bidentate ligand in some contexts,but standard coordination chemistry for this formula assumes $SO_4$ acts as a monodentate ligand. Thus,$1 (SO_4) + 5 (NH_3) = 6$ coordination sites.
$2$. To find the oxidation state of $X$,let it be $x$. The charge on $SO_4$ is $-2$,$NH_3$ is $0$,and $Cl$ is $-1$. The total charge of the complex is $0$.
$3$. $x + (-2) + 5(0) + (-1) = 0$
$4$. $x - 3 = 0$,so $x = +3$.
$5$. Therefore,the coordination number is $6$ and the oxidation state is $+3$.

(A) In oxyhemoglobin,the $Fe(II)$ ion is coordinated to four nitrogen atoms of the porphyrin ring,one nitrogen atom from the proximal histidine residue,and one oxygen molecule $(O_2)$.
Thus,the total coordination number of $Fe(II)$ is $4 + 1 + 1 = 6$.

(B) The primary valency of a central metal atom in a coordination compound corresponds to its oxidation state.
In $K_4[Fe(CN)_6]$,let the oxidation state of $Fe$ be $x$.
$4(+1) + x + 6(-1) = 0$
$4 + x - 6 = 0$
$x - 2 = 0$
$x = +2$
Therefore,the primary valency of iron is $2$.

(A) The given coordination compound is $[Co(NH_3)_4Cl_2]Cl$.
In an aqueous solution,the coordination sphere remains intact,and the ionizable counter-ion dissociates.
The dissociation reaction is: $[Co(NH_3)_4Cl_2]Cl(aq) \rightarrow [Co(NH_3)_4Cl_2]^+(aq) + Cl^-(aq)$.
This results in two ions: one complex cation $[Co(NH_3)_4Cl_2]^+$ and one chloride anion $Cl^-$.
Therefore,the total number of ions produced is $2$.

(B) $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$ is a double salt known as Mohr's salt.
When dissolved in water,it dissociates completely into its constituent ions.
The dissociation reaction is:
$FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O (aq) \rightarrow Fe^{2+} (aq) + 2NH_4^+ (aq) + 2SO_4^{2-} (aq) + 6H_2O (l)$
The total number of ions produced is $1 (Fe^{2+}) + 2 (NH_4^+) + 2 (SO_4^{2-}) = 5$ ions.

(D) The reaction of coordination compounds with silver nitrate $(AgNO_3)$ depends on the presence of ionizable chloride ions $(Cl^-)$ outside the coordination sphere.
If $Cl^-$ ions are present outside the coordination sphere,they react with $Ag^+$ to form a white precipitate of silver chloride $(AgCl)$.
$1$. $[Co(NH_3)_6]Cl_3$ gives $3$ moles of $Cl^-$ ions.
$2$. $[Co(NH_3)_5Cl]Cl_2$ gives $2$ moles of $Cl^-$ ions.
$3$. $[Co(NH_3)_4Cl_2]Cl$ gives $1$ mole of $Cl^-$ ions.
$4$. $[Co(NH_3)_3Cl_3]$ has no $Cl^-$ ions outside the coordination sphere,so it will not give a white precipitate with $AgNO_3$.

(C) The reaction of $AgNO_3$ with the aqueous solution of a coordination compound precipitates only the chloride ions $(Cl^-)$ that are present outside the coordination sphere (ionizable chloride).
Given that $1/3$ of the total chlorine is precipitated,it means that out of $3$ chlorine atoms,only $1$ is present outside the coordination sphere.
Therefore,the formula must be of the type $[CrCl_2(H_2O)_4]Cl \cdot 2H_2O$.
In this structure,there is $1$ ionizable $Cl^-$ ion,which reacts with $AgNO_3$ to form $AgCl$ precipitate.