Basic Terms Questions in English

(C) $DMG$ (Dimethylglyoxime) is a bidentate ligand that forms a stable five-membered chelate ring with metal ions like $Ni^{2+}$. Therefore,it acts as a chelating ligand.

(A) 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 or ion.
It represents the number of ligand donor atoms to which the metal is directly attached.

(D) $K_4[Fe(CN)_6]$ is a coordination compound that dissociates in water as follows:
$K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$.
Thus,it produces $4$ potassium ions $(K^+)$ and $1$ complex ion $([Fe(CN)_6]^{4-})$.
Total number of ions = $4 + 1 = 5$.

(B) The metal ion $Ag^{+}$ forms a stable linear complex $[Ag(CN)_2]^-$ with cyanide ions $(CN^-)$.
In this complex,the coordination number of the silver ion is $2$.

(D) The compound $[Pt(NH_3)_4]Cl_2$ ionizes in water to give $2Cl^-$ ions as follows:
$[Pt(NH_3)_4]Cl_2 \to [Pt(NH_3)_4]^{2+} + 2Cl^-$.
These free chloride ions react with $AgNO_3$ to form a white precipitate of silver chloride $(AgCl)$:
$Ag^+ + Cl^- \to AgCl \downarrow$ (white precipitate).
Other options do not contain ionizable chloride ions outside the coordination sphere.

(A) The amount of $AgCl$ precipitated depends on the number of ionizable $Cl^-$ ions outside the coordination sphere.
$1$. $[Co(NH_3)_6]Cl_3$ gives $3 \ mol$ of $Cl^-$ ions.
$2$. $[Co(NH_3)_5Cl]Cl_2$ gives $2 \ mol$ of $Cl^-$ ions.
$3$. $[Co(NH_3)_4Cl_2]Cl$ gives $1 \ mol$ of $Cl^-$ ions.
$4$. $Na_2[PtCl_4]$ gives $0 \ mol$ of ionizable $Cl^-$ ions.
Since $[Co(NH_3)_6]Cl_3$ provides the highest number of $Cl^-$ ions $(3 \ mol)$,it will yield the maximum amount of $AgCl$.

(A) The reaction with $AgNO_3$ depends on the presence of ionizable $Cl^-$ ions outside the coordination sphere.
In the complex $[Co(NH_3)_3Cl_3]$,all three $Cl^-$ ions are directly bonded to the central metal atom inside the coordination sphere.
Therefore,it does not provide any $Cl^-$ ions in the solution to react with $Ag^+$ to form $AgCl$ precipitate.
Thus,the correct answer is $[Co(NH_3)_3Cl_3]$.

(D) The compound ${K_4[Fe(CN)_6]}$ consists of $4K^+$ ions and the complex ion $[Fe(CN)_6]^{4-}$.
$1$. The bond between $K^+$ and $[Fe(CN)_6]^{4-}$ is ionic.
$2$. Within the cyanide ligand $(CN^-)$,there is a triple bond between $C$ and $N$,which is covalent.
$3$. The bond between the $Fe^{2+}$ central metal ion and the $C$ atom of the $CN^-$ ligand is a coordinate covalent bond (dative bond),where the $CN^-$ ligand donates a lone pair of electrons to the $Fe^{2+}$ ion.
Therefore,the compound contains ionic,covalent,and coordinate covalent bonds.

(D) double salt is an addition compound that dissociates into its constituent ions in aqueous solution.
$(D)$ Mohr’s salt,$FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$,is a classic example of a double salt.

(A) The complex $[Pt(NH_3)_6]Cl_4$ dissociates in an aqueous solution as follows:
$[Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^-$
Here,one complex cation $[Pt(NH_3)_6]^{4+}$ and four chloride anions $Cl^-$ are produced.
Total number of ions = $1 + 4 = 5$ ions.

(C) The charge $x$ on the complex ion $[Co^{III}(NH_3)_5Cl]^x$ is calculated as follows:
$x = (+3) + (5 \times 0) + (-1) = +2$.
To form a neutral complex salt,the complex cation must combine with anions having a total charge of $-2$.
Therefore,it combines with $2Cl^-$ to form the neutral complex salt $[Co(NH_3)_5Cl]Cl_2$.

(B) The given coordination compound is $[Cr(NH_3)_4Cl_2]Cl$.
In this complex,there is only $1$ chloride ion outside the coordination sphere,which acts as an ionizable group.
When excess $AgNO_3$ is added,the ionizable $Cl^-$ reacts with $Ag^+$ to form $AgCl$ precipitate according to the reaction: $[Cr(NH_3)_4Cl_2]Cl + AgNO_3 \rightarrow [Cr(NH_3)_4Cl_2]NO_3 + AgCl \downarrow$.
Since there is $1$ mole of ionizable $Cl^-$ per mole of the complex,$1$ mole of $AgCl$ precipitate is formed.

(B) The reaction of coordination complexes with excess $AgNO_3$ precipitates the chloride ions present outside the coordination sphere (ionizable chlorides).
$1$. $[Co(NH_3)_6]Cl_3$ contains $3$ ionizable $Cl^-$ ions,yielding $3 \ mol \ AgCl$.
$2$. $[Co(NH_3)_5Cl]Cl_2$ contains $2$ ionizable $Cl^-$ ions,yielding $2 \ mol \ AgCl$.
$3$. $[Co(NH_3)_4Cl_2]Cl$ contains $1$ ionizable $Cl^-$ ion,yielding $1 \ mol \ AgCl$.
Thus,the correct order is $3 \ AgCl, 2 \ AgCl, 1 \ AgCl$.

(D) Let the oxidation state of metal $M$ be $x$.
The complex is $[M(en)_2(C_2O_4)]Cl$. The charge on $en$ is $0$,the charge on $C_2O_4^{2-}$ is $-2$,and the charge on the complex ion is $+1$ (since $Cl$ is $-1$).
$x + 2(0) + 1(-2) = +1$
$x - 2 = 1$
$x = +3$
Thus,the oxidation number of $M$ is $3$.
Now,for the coordination number:
$en$ (ethylenediamine) is a bidentate ligand,and $C_2O_4^{2-}$ (oxalate) is also a bidentate ligand.
Coordination Number $(C.N.) = (2 \times 2) + (1 \times 2) = 4 + 2 = 6$.
The sum of the coordination number and the oxidation number is $6 + 3 = 9$.

(C) The formula of dichlorotetraaqua-chromium$(III)$ chloride is $[Cr(H_2O)_4Cl_2]Cl$.
In this coordination compound,only the chloride ion outside the coordination sphere is ionizable.
$[Cr(H_2O)_4Cl_2]Cl \rightarrow [Cr(H_2O)_4Cl_2]^+ + Cl^-$
One mole of the complex yields $1 \ mole$ of $Cl^-$ ions,which reacts with $AgNO_3$ to form $1 \ mole$ of $AgCl$ precipitate.
Number of moles of complex $= M \times V_{(L)} = 0.01 \times 0.1 = 0.001 \ mol$.
Therefore,the number of moles of $AgCl$ precipitated is $0.001 \ mol$.

(C) Acetylacetone $(acac)$ is $CH_3COCH_2COCH_3.$ Upon deprotonation,it forms the acetylacetonate anion,$CH_3COCHCOCH_3^-.$
When this anion coordinates with $Co^{3+},$ the metal ion binds to the two oxygen atoms.
This creates a ring consisting of the metal atom,two oxygen atoms,and three carbon atoms from the ligand backbone.
Counting these atoms $(Co, O, C, C, C, O)$,we find that the chelate ring is a six-membered ring.

(A) $EDTA$ is a hexadentate ligand,meaning it has $6$ donor atoms ($4$ oxygen atoms and $2$ nitrogen atoms) that can coordinate to a central metal ion.
Since an octahedral complex requires a coordination number of $6$,a single molecule of $EDTA$ is sufficient to satisfy all $6$ coordination sites of the $Ca^{2+}$ ion.
Therefore,$1$ molecule of $EDTA$ is required.

(C) Trioxalato aluminate $(III)$: The central metal is $Al$ with an oxidation state of $+3$. The oxalate ion $(C_2O_4^{2-})$ is a bidentate ligand with a charge of $-2$. The coordination entity is $[Al(C_2O_4)_3]^{x}$.
Calculating the charge: $x = +3 + 3(-2) = +3 - 6 = -3$. Thus,the ion is $[Al(C_2O_4)_3]^{3-}$.
Tetrafluoro borate $(III)$: The central atom is $B$ with an oxidation state of $+3$. It is bonded to $4$ fluoride ions $(F^-)$,each with a charge of $-1$. The coordination entity is $[BF_4]^{y}$.
Calculating the charge: $y = +3 + 4(-1) = +3 - 4 = -1$. Thus,the ion is $[BF_4]^{-}$.

(B) The complex $[Co(NH_3)_5Br]SO_4$ ionizes in an aqueous solution to give $[Co(NH_3)_5Br]^{2+}$ and $SO_4^{2-}$ ions.
When $BaCl_2$ is added to the solution,the $SO_4^{2-}$ ions react with $Ba^{2+}$ ions to form a white precipitate of $BaSO_4$.
The chemical reaction is: $[Co(NH_3)_5Br]SO_4 + BaCl_2 \rightarrow [Co(NH_3)_5Br]Cl_2 + BaSO_4 \downarrow$ (white precipitate).
Therefore,the correct reagent is $BaCl_2$.

(D) Diethylene triamine,represented as $NH_2CH_2CH_2NHCH_2CH_2NH_2$,contains three nitrogen donor atoms.
Since it has three donor atoms capable of binding to a central metal ion,it is classified as a $Tridentate$ ligand.
Because it is a polydentate ligand,it can form ring structures with the metal ion,making it a $Chelating$ agent.
Since it is a polydentate ligand,a tridentate ligand,and a chelating agent,all the given options are correct.

(B) The coordination complex is $[Co(NH_3)_5Cl]Cl_2$. In this complex,there are $2$ ionizable $Cl^-$ ions outside the coordination sphere.
The reaction with excess $AgNO_3$ is:
$[Co(NH_3)_5Cl]Cl_2 + 2AgNO_3 \rightarrow [Co(NH_3)_5Cl](NO_3)_2 + 2AgCl \downarrow$
Number of moles of complex $= M \times V(L) = 0.1 \ M \times 0.1 \ L = 0.01 \ mol$.
Since $1 \ mole$ of the complex yields $2 \ moles$ of $AgCl$,then $0.01 \ mole$ of the complex will yield $0.01 \times 2 = 0.02 \ moles$ of $AgCl$.

(A) Chelating agents are ligands that can form ring structures with a central metal atom by binding through two or more donor atoms.
$Oxalato$ $(C_2O_4^{2-})$,$glycinato$ $(NH_2CH_2COO^-)$,and $ethylene diamine$ $(en)$ are all polydentate (specifically bidentate) ligands capable of forming chelate rings.
$Thiosulphato$ $(S_2O_3^{2-})$ acts as a monodentate ligand in most coordination complexes,binding through a single sulfur atom.
Since it cannot form a ring,it is not a chelating agent.

(C) donor site is an atom that can donate a lone pair of electrons to the central metal ion to form a coordinate bond.
$1$. Triethylene tetramine $(trien)$ has $4$ nitrogen donor atoms (tetradentate).
$2$. Ethylenediamine tetracetate ion $(EDTA^{4-})$ has $6$ donor atoms ($4$ oxygen and $2$ nitrogen) (hexadentate).
$3$. Ethylenediamine triacetate ion $(EDTA^{3-})$ has $5$ donor atoms ($3$ oxygen and $2$ nitrogen) (pentadentate).
$4$. Diethylene triamine $(dien)$ has $3$ nitrogen donor atoms (tridentate).
Therefore,the correct answer is Ethylenediamine triacetate ion.

(D) Number of moles of the given complex $= 10 \ mL \times 1 \ M = 10 \ mmol = 0.01 \ mol$.
Number of moles of $AgCl$ formed $= \frac{4.305 \ g}{143.5 \ g/mol} = 0.03 \ mol$.
Since $AgNO_3$ is in excess,all ionizable $Cl^-$ ions react to form $AgCl$.
Ratio of moles of $AgCl$ to moles of complex $= \frac{0.03 \ mol}{0.01 \ mol} = 3$.
This indicates that each mole of the complex releases $3 \ mol$ of $Cl^-$ ions.
Therefore,the complex must be $[Cr(H_2O)_6]Cl_3$.

(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 the ammonium ion $(NH_4^{+})$,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.

(B) The number of donor sites (denticity) for the given ligands are as follows:
$1$. Dimethyl glyoxime $(dmg^-)$: It is a bidentate ligand,so it has $2$ donor sites.
$2$. Glycinato $(gly^-)$: It is a bidentate ligand,so it has $2$ donor sites.
$3$. Diethylene triamine $(dien)$: It is a tridentate ligand,so it has $3$ donor sites.
$4$. Ethylenediaminetetraacetate $(EDTA^{4-})$: It is a hexadentate ligand,so it has $6$ donor sites.
Therefore,the number of donor sites are $2, 2, 3$ and $6$ respectively.

(C) The complex hydridotrimethoxyborate$(III)$ ion consists of one hydride ion $(H^-)$ and three methoxide ions $(OCH_3^-)$ coordinated to a central boron atom $(B)$.
The oxidation state of boron is given as $+3$.
Let the oxidation state of boron be $x$.
The sum of oxidation states in the complex is equal to the charge on the ion $(n)$:
$x + (1 \times -1) + (3 \times -1) = n$
$x - 1 - 3 = n$
$x - 4 = n$
Given that the oxidation state of boron $(x)$ is $+3$:
$3 - 4 = n$
$n = -1$
Therefore,the charge on the complex ion is $-1$,and the formula is $[BH(OCH_3)_3]^-$.
Thus,option $C$ is correct.

(A) bidentate ligand is a ligand that has two donor atoms that coordinate directly to the central metal atom in a complex.
$dien$ (diethylenetriamine,$NH_2CH_2CH_2NHCH_2CH_2NH_2$) is a tridentate ligand because it has three nitrogen donor atoms.
$tn$ (trimethylenediamine or propane$-1,3-$diamine) is a bidentate ligand.
$bn$ (butylenediamine or butane$-1,2-$diamine) is a bidentate ligand.
$gly^{-}$ (glycinate ion,$NH_2CH_2COO^{-}$) is a bidentate ligand.

(C) bidentate ligand is a ligand that has two donor atoms that coordinate directly to the central metal atom in a complex.
$8$-hydroxyquinolinato ion,$N, N$-diethyldithiocarbamato ion,and salicylate ion are all bidentate ligands.
$Hydrazine$ $(NH_2-NH_2)$ acts as a monodentate ligand because it typically coordinates through only one nitrogen atom due to steric hindrance and the nature of the lone pairs,although it has the potential to be a bridging ligand,it is not classified as a standard bidentate chelating ligand.

(B) An ambidentate ligand is a ligand that can coordinate to a central metal atom through two different donor atoms.
$SCN^-$ is a classic example of an ambidentate ligand because it can coordinate through the sulfur atom (thiocyanato-$S$) or the nitrogen atom (isothiocyanato-$N$).
The other options ($NO_3^-$,$SO_4^{2-}$,and $CO_3^{2-}$) are polyatomic ions that typically coordinate through specific oxygen atoms and do not exhibit ambidentate behavior in the same manner.

(C) To determine the number of $5$-membered rings formed,we analyze the denticity and structure of each ligand:
$1$. $en$ (Ethylenediamine) is a bidentate ligand forming $1$ five-membered ring.
$2$. $dmg^-$ (Dimethylglyoximate) is a bidentate ligand forming $2$ five-membered rings.
$3$. $edta^{4-}$ (Ethylenediaminetetraacetate) is a hexadentate ligand. It forms $5$ five-membered chelate rings with the central metal ion.
$4$. $trien$ (Triethylenetetramine) is a tetradentate ligand forming $3$ five-membered rings.
Comparing these,$edta^{4-}$ forms the maximum number of $5$-membered rings ($5$ rings).

(C) The given complex is $[Co(NH_3)_4Cl_2]Cl$.
In aqueous solution,it ionizes as: $[Co(NH_3)_4Cl_2]Cl \rightarrow [Co(NH_3)_4Cl_2]^+ + Cl^-$.
This produces two ions in solution,so option $A$ is incorrect.
The primary valency of a metal in a coordination compound corresponds to its oxidation state. Here,$x + 4(0) + 3(-1) = 0$,so $x = +3$. Thus,the primary valency is $3$,not $6$,making option $B$ incorrect.
Since there is one $Cl^-$ ion outside the coordination sphere,it reacts with $AgNO_3$ to give one mole of $AgCl$ precipitate: $Cl^- + AgNO_3 \rightarrow AgCl(s) + NO_3^-$. Thus,option $C$ is correct.
The primary valency is the oxidation state,which is $3$,not $1$,making option $D$ incorrect.

(A) Complex salts contain two different metallic elements but dissociate in such a way that they only provide the test for the ions present in the ionization sphere,not the central metal ion.
Double salts,on the other hand,completely dissociate into their constituent simple ions in an aqueous solution,thereby giving a test for all the ions present.

(A) According to Werner's theory of coordination compounds:
$1$. Metal atoms exhibit two types of valencies: primary and secondary.
$2$. Primary valencies are ionisable and are satisfied by negative ions.
$3$. Secondary valencies are non-ionisable,fixed in number,and are satisfied by ligands (neutral or negative) connected to the metal ion by coordinate bonds.
$4$. Secondary valencies have specific directional properties,which determine the geometry of the coordination complex.
Therefore,all three statements $(a), (b),$ and $(c)$ are correct.

(C) The reaction with $AgNO_3$ precipitates only the ionizable $Cl^-$ ions present outside the coordination sphere.
For $25\%$ mole of $AgCl$ to be formed from $1$ mole of the complex,there must be $1$ ionizable $Cl^-$ ion out of $4$ total $Cl^-$ atoms.
This corresponds to the formula $[Pt(NH_3)_4Cl_3]Cl$,which is represented by $PtCl_4 \cdot 4NH_3$.
Thus,$PtCl_4 \cdot 4NH_3$ dissociates as $[Pt(NH_3)_4Cl_3]^+ + Cl^-$,yielding $1$ mole of $AgCl$ per mole of complex.

(A) To determine the number of chelate rings,we identify the denticity of the ligands:
$(i) \, [Co(en)_2(Ox)]^+$: $en$ is bidentate ($2$ rings each),$Ox$ is bidentate ($1$ ring). Total rings = $(2 \times 2) + 1 = 5$.
$(ii) \, [Fe(Trien)(Ox)]^+$: $Trien$ (triethylenetetramine) is tetradentate ($3$ rings),$Ox$ is bidentate ($1$ ring). Total rings = $3 + 1 = 4$.
$(iii) \, [Ca(EDTA)]^{2-}$: $EDTA$ is hexadentate,forming $5$ chelate rings. Total rings = $5$.
$(iv) \, [Pt(Ox)_2(H_2O)_2]$: $Ox$ is bidentate ($1$ ring each),$H_2O$ is monodentate ($0$ rings). Total rings = $(2 \times 1) + 0 = 2$.
Comparing the values: $(i) = 5$,$(ii) = 4$,$(iii) = 5$,$(iv) = 2$.
Note: While $(i)$ and $(iii)$ both have $5$ rings,in standard textbook problems,$EDTA$ complexes are often cited as having the maximum number of chelate rings. Re-evaluating the options provided,the sequence $iii > i > ii > iv$ is the most logical fit.

(C) The total number of chlorine atoms in $CrCl_3 \cdot 6H_2O$ is $3$.
If $\frac{2}{3}$ of the total chlorine is precipitated by $AgNO_3$,it means $2$ chlorine atoms are present as free $Cl^-$ ions outside the coordination sphere.
Therefore,the coordination formula must contain $2$ chloride ions outside the brackets.
The formula is $[CrCl(H_2O)_5]Cl_2 \cdot H_2O$.

(D) Ferrous ammonium sulphate,also known as Mohr's salt,has the chemical formula $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$.
In an aqueous solution,it dissociates completely into its constituent ions: $Fe^{+2}$,$SO_4^{-2}$,and $NH_4^+$.
Therefore,the solution contains all of these ions.

(B) The reaction of the coordination compound with $AgNO_3$ precipitates only the chloride ions $(Cl^-)$ present outside the coordination sphere.
Given that $0.1 \ mol$ of the complex yields $0.2 \ mol$ of $AgCl$,the ratio of $AgCl$ to the complex is $2:1$.
This indicates that there are $2$ ionizable $Cl^-$ ions outside the coordination sphere.
Therefore,the formula is $[Co(NH_3)_5Cl]Cl_2$.

(B) To produce four ions per molecule,the complex must dissociate as $[Complex]^{3+} + 3Cl^-$.
This corresponds to the formula $[Pt(NH_3)_5Cl]Cl_3$.
Therefore,the complex is $PtCl_4 \cdot 5NH_3$.

(B) In the complex $[Ni(dmg)_2]$,the dimethylglyoximate $(dmg^-)$ ligand acts as a bidentate ligand.
Each $dmg^-$ ligand forms one five-membered chelate ring with the $Ni^{2+}$ ion.
Since there are two $dmg^-$ ligands,they form $2$ chelate rings.
The hydrogen bonds present in the structure stabilize the complex but are not considered chelate rings.
Therefore,the total number of chelate rings is $2$.

(D) According to Werner's coordination theory:
$1$. Primary valency corresponds to the oxidation state of the central metal atom $(CMA)$.
$2$. It is satisfied by negative ions (anions).
$3$. It is ionisable and nondirectional in nature.
Therefore,all the given statements are correct.

(C) Mohr's salt is represented by the formula $(NH_4)_2SO_4 \cdot FeSO_4 \cdot 6H_2O$.
It is a double salt that dissociates completely in water into its constituent ions: $2NH_4^+$,$Fe^{2+}$,and $2SO_4^{2-}$.
$1$. It has $2 + 1 + 2 = 5$ ions per formula unit,so option $A$ is correct.
$2$. It is a double salt,so option $B$ is correct.
$3$. Since it dissociates into $Fe^{2+}$ ions in aqueous solution,it gives a positive test for $Fe^{2+}$,making option $C$ incorrect.
$4$. It contains two types of cations,$NH_4^+$ and $Fe^{2+}$,so option $D$ is correct.
Therefore,the wrong statement is $C$.

(C) The number of chelate rings in a coordination complex is equal to the number of chelate rings formed by polydentate ligands.
$A$. $[Co(ox)_2Cl_2]^{3-}$ has two oxalate ligands,each forming one ring. Total rings = $2$.
$B$. $[Ni(dmg)_2]$ has two dimethylglyoximate ligands,each forming one ring. Total rings = $2$.
$C$. $[Ca(EDTA)]^{2-}$ has one $EDTA^{4-}$ ligand,which is hexadentate and forms $5$ chelate rings.
$D$. The brown ring complex $[Fe(H_2O)_5(NO)]SO_4$ contains no chelate rings.
Thus,the correct match is $C$.

(B) The coordination compound $[CrCl(NH_3)_5]Cl_2$ dissociates in water to release the ions present outside the coordination sphere.
The formula can be written as $[CrCl(NH_3)_5]^{2+} + 2Cl^-$.
When $AgNO_3$ is added,it reacts with the ionizable chloride ions: $2Cl^- + 2AgNO_3 \rightarrow 2AgCl(s) + 2NO_3^-$.
Total chlorine atoms in the compound = $3$.
Ionizable chlorine atoms = $2$.
Therefore,the fraction of chlorine precipitated as $AgCl$ is $2/3$.

(D) The consumption of $AgNO_3$ depends on the number of ionizable $Cl^-$ ions present outside the coordination sphere.
$Na_2[CrCl_5(H_2O)]$ provides $0$ ionizable $Cl^-$ ions.
$Na_3[CrCl_6]$ provides $0$ ionizable $Cl^-$ ions.
$[Cr(H_2O)_5Cl]Cl_2$ provides $2$ ionizable $Cl^-$ ions.
$[Cr(H_2O)_6]Cl_3$ provides $3$ ionizable $Cl^-$ ions.
Since $[Cr(H_2O)_6]Cl_3$ furnishes the highest number of $Cl^-$ ions ($3$ moles per mole of complex),it will consume the most equivalents of $AgNO_3$.

(A) homoleptic complex is a coordination compound in which the metal atom or ion is bonded to only one type of donor atom or ligand.
In the complex $[Co(NH_3)_6]Cl_3$,the central metal ion $Co^{3+}$ is bonded to six $NH_3$ ligands,which are all of the same type.
Therefore,$[Co(NH_3)_6]Cl_3$ is a homoleptic complex.

(A) ligand is an atom,molecule,or ion that donates 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 $CH_4$ (methane),the carbon atom is $sp^3$ hybridized and forms four covalent bonds with hydrogen atoms,completing its octet.
Since $CH_4$ has no lone pair of electrons available for donation,it cannot act as a ligand.

(C) In the formation of the coordination complex $[Co(NH_3)_6]^{3+}$,the central metal ion $Co^{3+}$ acts as the Lewis acid because it accepts lone pairs of electrons from the ligands.
$NH_3$ acts as the Lewis base because it donates lone pairs of electrons to the central metal ion.
Therefore,the reaction is: $\mathop {Co^{3+}}\limits_{\text{Lewis acid}} + \mathop {6NH_3}\limits_{\text{Lewis base}} \to \mathop {[Co(NH_3)_6]^{3+}}\limits_{\text{adduct}}$

(D) In the complex $K_4[Th(C_2O_4)_4(OH_2)_2]$,the central metal atom is $Th$.
The ligands present are:
$1$. Four oxalate ions $(C_2O_4^{2-})$,which are bidentate ligands. Contribution = $4 \times 2 = 8$.
$2$. Two water molecules $(OH_2)$,which are monodentate ligands. Contribution = $2 \times 1 = 2$.
The coordination number is the sum of the contributions of all ligands attached to the central metal atom.
Coordination number = $8 + 2 = 10$.