Basic Terms Questions in English

(N/A) $(i)$ Ammonium molybdate is $(NH_4)_2MoO_4$.
$(ii)$ Ammonium phosphomolybdate is $(NH_4)_3[PMo_{12}O_{40}]$.
$(iii)$ Ferric thiocyanate ion is $[Fe(SCN)]^{2+}$.
$(iv)$ Sodium sulphide is $Na_2S$.

(C) $EDTA^{4-}$ (ethylenediaminetetraacetate) is a hexadentate ligand.
It contains $2$ nitrogen atoms and $4$ oxygen atoms as donor sites.
Therefore,the total number of coordination sites (denticity) is $6$.

(C) The complex $trans-CoCl_{3} \cdot 4NH_{3}$ can be written as $trans-[Co(NH_{3})_{4}Cl_{2}]Cl$.
In this complex,there are $4$ neutral $NH_{3}$ ligands in the coordination sphere.
Ethylene diamine $(en)$ is a bidentate ligand,meaning one molecule of $en$ occupies two coordination sites.
To replace $4$ monodentate $NH_{3}$ ligands,we require $2$ molecules of the bidentate ligand $en$ ($2 \times 2 = 4$ coordination sites).
Therefore,the number of equivalents of ethylene diamine required is $2$.

(D) The reaction of $FeCl_3$ with oxalic acid in the presence of $KOH$ leads to the formation of potassium trioxalatoferrate$(III)$:
$FeCl_3 + 3H_2C_2O_4 + 6KOH \rightarrow K_3[Fe(C_2O_4)_3] + 3KCl + 6H_2O$
The product $A$ is $K_3[Fe(C_2O_4)_3]$.
In this coordination complex,the oxalate ion $(C_2O_4^{2-})$ is a bidentate ligand.
Since there are $3$ oxalate ligands,the coordination number (secondary valency) of $Fe$ is $3 \times 2 = 6$.

(B) In the structure of $CuSO_{4} \cdot 5H_{2}O,$ the copper ion $(Cu^{2+})$ is coordinated to $4$ water molecules directly,which defines its secondary valency as $4$.
The fifth water molecule is not directly coordinated to the $Cu^{2+}$ ion but is held in the crystal lattice by hydrogen bonding between the water molecule and the sulfate $(SO_{4}^{2-})$ group.
Therefore,the secondary valency is $4$ and the number of hydrogen-bonded water molecules is $1$.

(C) Statement $I$ is true because dimethyl glyoxime $(DMG)$ is used as a specific reagent for the detection of $Ni^{2+}$ ions in an ammoniacal medium $(NH_{4}OH)$.
Statement $II$ is false because dimethyl glyoxime $(C_{4}H_{8}N_{2}O_{2})$ acts as a bidentate monoanionic ligand $(dmg^{-})$ after losing a proton,not as a neutral ligand.
The reaction is: $Ni^{2+}_{(aq)} + 2dmg^{-} \rightarrow [Ni(dmg)_{2}] \text{ (rosy red precipitate)}$.

(C) The complex $MCl_{3} \cdot 2 \ L$ reacts with excess $AgNO_{3}$ to yield $1 \ mol$ of $AgCl$,which indicates that only $1 \ mol$ of $Cl^{-}$ ion is present in the ionization sphere.
Thus,the coordination formula can be written as $[MCl_{2} \ L_{2}]Cl$.
For an octahedral complex,the coordination number is $6$.
Let the denticity of ligand $L$ be $x$.
The coordination number is calculated as: $(2 \times \text{denticity of } Cl^{-}) + (2 \times x) = 6$.
Since $Cl^{-}$ is a monodentate ligand (denticity $= 1$),we have: $(2 \times 1) + 2x = 6$.
$2 + 2x = 6 \implies 2x = 4 \implies x = 2$.
Therefore,the denticity of ligand $L$ is $2$.

(A) Biuret $(NH_2CONHCONH_2)$ acts as a bidentate ligand in coordination complexes.
It coordinates to the central metal ion through the two oxygen atoms of the carbonyl groups.
Therefore,its denticity is $2$.

(A) The structure of the ethylene diaminetetraacetate $(EDTA^{4-})$ ion consists of two nitrogen atoms and four oxygen atoms (from the four carboxylate groups) that can act as donor sites.
Since it has six donor atoms capable of binding to a central metal ion,it is classified as a hexadentate ligand.
Therefore,the correct option is $A$.

(B) The number of moles of $AgCl$ precipitated corresponds to the number of $Cl^{-}$ ions present in the ionization sphere.
Given the empirical formula $CrCl_{3} \cdot 3NH_{3} \cdot 3H_{2}O$,and the fact that $3$ moles of $AgCl$ are precipitated,all $3$ chloride ions must be outside the coordination sphere.
Thus,the complex is $[Cr(H_{2}O)_{3}(NH_{3})_{3}]Cl_{3}$.
In this complex,the coordination number of the central metal ion $Cr^{3+}$ is $6$,which is satisfied by $3$ $H_{2}O$ molecules and $3$ $NH_{3}$ molecules.
Since all $3$ $Cl^{-}$ ions are in the ionization sphere,the number of chloride ions satisfying the secondary valency (coordination number) is $0$.

(A) The complex $Co(en)_2 Cl_3$ reacts with $AgNO_3$ to precipitate $AgCl$. Since $3$ moles of $AgCl$ are produced from $3$ moles of the complex,each mole of the complex releases $1$ mole of $Cl^-$ ions.
This indicates the formula is $[Co(en)_2 Cl_2]Cl$.
The secondary valency corresponds to the coordination number $(C.N.)$.
In $[Co(en)_2 Cl_2]Cl$,the central metal $Co$ is bonded to two bidentate $en$ ligands ($2 \times 2 = 4$ coordination sites) and two monodentate $Cl^-$ ligands ($2 \times 1 = 2$ coordination sites).
Therefore,the coordination number $= 4 + 2 = 6$.
The secondary valency of $Co$ is $6$.

(N/A) When a polydentate ligand coordinates to a metal ion,forming a five- or six-membered closed ring structure that includes the metal ion,it is called a chelating ligand,and the phenomenon is known as chelation. The resulting compounds are known as chelates.
Chelates are highly stable. The stabilization of coordination compounds due to chelation is called the chelate effect.
For example: $[Pt(en)_2]^{2+}$
In the above example,the bidentate ligand ethylene diamine $(en)$ forms a five-membered closed ring structure with the central metal ion $Pt^{2+}$.

(N/A) The following rules are applied while writing the formulas:
$1$. The central atom is listed first.
$2$. The ligands are then listed in alphabetical order. The placement of a ligand in the list does not depend on its charge.
$3$. Polydentate ligands are sorted alphabetically. In case of an abbreviated ligand,the first letter of the abbreviation is used to determine the position of the ligand in the alphabetical order.
$4$. The formula for the entire coordination entity,whether charged or not,is enclosed in square brackets. When ligands are polyatomic,their formulas are enclosed in parentheses. Ligand abbreviations are also enclosed in parentheses.
$5$. There should be no space between the ligands and the metal within a coordination sphere.
$6$. When the formula of a charged coordination entity is to be written without that of the counter ion,the charge is indicated outside the square brackets as a right superscript with the number before the sign. For example,$[Co(CN)_{6}]^{3-}$,$[Cr(H_{2}O)_{6}]^{3+}$,etc.
$7$. The charge of the cation$(s)$ is balanced by the charge of the anion$(s)$.

(N/A) Werner's theory could not explain the following:
$1$. The directional properties of bonds in coordination compounds.
$2$. The ability of certain elements to form coordination compounds.
$3$. The magnetic and optical properties of coordination compounds.
The nature of bonding in coordination compounds is better explained by Valence Bond Theory $(VBT)$,Crystal Field Theory $(CFT)$,Ligand Field Theory $(LFT)$,and Molecular Orbital Theory $(MOT)$.

(A) The complex $CoCl_3(NH_3)_4$ acts as a $1:1$ electrolyte,which means it dissociates into two ions in solution.
This indicates the formula is $[Co(NH_3)_4Cl_2]Cl$.
The primary valency of the central metal ion $(Co)$ corresponds to its oxidation state.
In $[Co(NH_3)_4Cl_2]Cl$,let the oxidation state of $Co$ be $x$.
$x + 4(0) + 2(-1) + 1(-1) = 0$
$x - 3 = 0$
$x = +3$.
Therefore,the primary valency is $3$.

(C) The complex is $Na[Co(bpy)Cl_4]$.
$bpy$ $(2,2'-bipyridine)$ is a bidentate ligand,and $Cl^-$ is a monodentate ligand.
Coordination number of $Co = (1 \times 2) + (4 \times 1) = 6$.
Let the oxidation state of $Co$ be $x$.
The charge on $Na$ is $+1$,$bpy$ is $0$,and $Cl$ is $-1$.
$1 + x + 0 + 4(-1) = 0$
$1 + x - 4 = 0$
$x - 3 = 0$
$x = +3$.
Magnitude of oxidation state $= 3$.
Sum $= 6 + 3 = 9$.

(A) The reaction is: $CoH_{12}N_{4}Cl_{3} + AgNO_{3} \rightarrow AgCl$ (white ppt).
Molar mass of $CoH_{12}N_{4}Cl_{3} = 59 + 12(1) + 4(14) + 3(35.5) = 59 + 12 + 56 + 106.5 = 233.5 \ g/mol$.
Moles of $X = \frac{2.33 \ g}{233.5 \ g/mol} = 0.01 \ mol$.
Moles of $AgCl = \frac{1.435 \ g}{143.5 \ g/mol} = 0.01 \ mol$.
Since $0.01 \ mol$ of $X$ produces $0.01 \ mol$ of $AgCl$,there is $1 \ Cl^-$ ion outside the coordination sphere.
The complex is $[Co(NH_{3})_{4}Cl_{2}]Cl$.
Primary valency (oxidation state of $Co$): $x + 4(0) + 2(-1) = +1 \Rightarrow x = +3$.
Secondary valency (coordination number of $Co$): $4(NH_{3}) + 2(Cl) = 6$.

(B) The structure of the complex with the empirical formula $CoCl_3 \cdot 4NH_3$ is represented as $[Co(NH_3)_4Cl_2]Cl$.
Upon dissolution in water,it dissociates as follows:
$[Co(NH_3)_4Cl_2]Cl \rightarrow [Co(NH_3)_4Cl_2]^+ + Cl^-$.
This complex produces $2$ ions in total ($1$ complex cation and $1$ chloride anion).

(D) In the complex $[Co(NH_3)_5Cl]Cl_2$,the oxidation state of $Co$ is calculated as: $x + 5(0) + 1(-1) = +2$,so $x = +3$.
Primary valency corresponds to the oxidation state of the central metal atom,which is $3$.
Secondary valency corresponds to the coordination number,which is the total number of ligands directly bonded to the central metal atom. Here,$5$ $NH_3$ molecules and $1$ $Cl^-$ ion are bonded,so the coordination number is $5 + 1 = 6$.
Thus,the primary and secondary valencies are $3$ and $6$ respectively.

(D) The number of $AgCl$ moles precipitated depends on the number of ionizable chloride ions $(Cl^-)$ outside the coordination sphere.
$1$. $[Co(NH_3)_4Cl_2]Cl$ dissociates to give $1$ $Cl^-$ ion,so it produces $1$ mole of $AgCl$.
$2$. $[Ni(H_2O)_6]Cl_2$ dissociates to give $2$ $Cl^-$ ions,so it produces $2$ moles of $AgCl$.
$3$. $[Pt(NH_3)_2Cl_2]$ has no ionizable $Cl^-$ ions outside the coordination sphere,so it produces $0$ moles of $AgCl$.
$4$. $[Pd(NH_3)_4]Cl_2$ dissociates to give $2$ $Cl^-$ ions,so it produces $2$ moles of $AgCl$.
Total moles of $AgCl = 1 + 2 + 0 + 2 = 5$ moles.

(A) Fehling's reagent contains a copper$(II)$ complex with tartrate ions $(C_4H_4O_6^{2-})$.
In this complex,the tartrate ion acts as a bidentate ligand,meaning it coordinates to the central metal ion through two donor atoms (typically the carboxylate oxygen and the hydroxyl oxygen).
Therefore,the denticity of the tartrate ligand is $2$.

(A) Double salts are addition compounds that dissociate into their constituent ions completely when dissolved in water.
$FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$ (Mohr's salt) and $K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$ (Potash alum) are examples of double salts.
$CuSO_4 \cdot 4NH_3 \cdot H_2O$ and $Fe(CN)_2 \cdot 4KCN$ (which is $K_4[Fe(CN)_6]$) are coordination compounds (complex salts) that do not dissociate completely into their constituent ions in solution.
Therefore,$A$ and $C$ are double salts.

(C) An ambidentate ligand is a ligand that can coordinate to the central metal atom through two different donor atoms.
In the complex $[M(en)(SCN)_4]$,the ligand $en$ (ethylenediamine) is a bidentate chelating ligand,which coordinates through two nitrogen atoms.
The ligand $SCN^-$ (thiocyanate) is an ambidentate ligand because it can coordinate through either the sulfur atom $(S-CN^-)$ or the nitrogen atom $(N-CS^-)$.
There are $4$ $SCN^-$ ligands present in the complex.
Therefore,the number of ambidentate ligands in the complex is $4$.

(A) An ambidentate ligand is a ligand that can coordinate to a metal atom through more than one donor atom (e.g.,$NO_2^-$ can coordinate through $N$ or $O$,and $NCS^-$ can coordinate through $N$ or $S$).
Analyzing the options:
$A$: $C_2O_4^{2-}$ (oxalate) is a didentate ligand,ethylene diamine $(en)$ is a didentate ligand,and $H_2O$ is a monodentate ligand. None of these are ambidentate.
$B$: $NCS^-$ is an ambidentate ligand.
$C$: $NO_2^-$ is an ambidentate ligand.
$D$: $NO_2^-$ and $NCS^-$ are ambidentate ligands.
Therefore,the set that does not contain any ambidentate ligand is $A$.

(A) The complex $[Cr(H_2O)_5Cl]Cl_2$ dissociates in water as: $[Cr(H_2O)_5Cl]Cl_2 \rightarrow [Cr(H_2O)_5Cl]^{2+} + 2Cl^-$.
Each mole of the complex provides $2$ moles of ionizable chloride ions $(Cl^-)$.
Number of millimoles of the complex $= Molarity \times Volume (mL) = 0.01 \, M \times 20 \, mL = 0.2 \, mmol$.
Since $1$ mole of complex gives $2$ moles of $Cl^-$,$0.2 \, mmol$ of complex gives $0.4 \, mmol$ of $Cl^-$.
The precipitation reaction is: $Ag^+ + Cl^- \rightarrow AgCl(s)$.
Thus,$0.4 \, mmol$ of $AgNO_3$ is required to precipitate $0.4 \, mmol$ of $Cl^-$.
Using $Molarity = \frac{millimoles}{Volume (mL)}$,we get $0.1 = \frac{0.4}{V}$.
$V = \frac{0.4}{0.1} = 4 \, mL$.

(B) homoleptic complex is a coordination compound in which the central metal atom or ion is bound to only one type of donor atom or ligand.
- Triamminetriaquachromium $(III)$ chloride: $[Cr(NH_3)_3(H_2O)_3]Cl_3$ (Heteroleptic)
- Potassium trioxalatoaluminate $(III)$: $K_3[Al(C_2O_4)_3]$ (Homoleptic,as all ligands are oxalate ions)
- Diamminechloridonitrito-$N$-platinum $(II)$: $[Pt(NH_3)_2Cl(NO_2)]$ (Heteroleptic)
- Pentaamminecarbonatocobalt $(III)$ chloride: $[Co(NH_3)_5(CO_3)]Cl$ (Heteroleptic)
Therefore,the correct option is $B$.

(A) The brown ring complex formed is $[Fe(H_2O)_5(NO)]^{2+}$.
In this complex,$H_2O$ is a neutral ligand (charge $0$) and $NO$ exists as $NO^+$ (nitrosonium ion) with a charge of $+1$.
Let the oxidation state of $Fe$ be $x$.
$x + 5(0) + 1 = +2$
$x + 1 = +2$
$x = +1$.
Therefore,the oxidation number of iron is $+1$.

(A) homoleptic complex is a coordination compound in which all the ligands attached to the central metal atom or ion are of the same type.
In the complex $[Ni(CN)_4]^{2-}$,all four ligands are cyanide $(CN^-)$ ions.
Therefore,$[Ni(CN)_4]^{2-}$ is a homoleptic complex.

(C) The reaction of the complex with excess $AgNO_3$ yields $2$ moles of $AgCl$,which indicates that there are $2$ chloride ions outside the coordination sphere.
The formula of the complex is $[Co(NH_3)_nCl]Cl_2$.
Since the complex is octahedral,the coordination number of $Co$ is $6$. Thus,$n + 1 = 6$,which gives $n = 5$.
The complex is $[Co(NH_3)_5Cl]Cl_2$.
Let the oxidation state of $Co$ be $x$. The sum of oxidation states in the complex is $x + 5(0) + 1(-1) + 2(-1) = 0$,which gives $x - 3 = 0$,so $x = +3$.
Therefore,$x + n = 3 + 5 = 8$.

(C) Ambidentate ligands are those which possess two different donor atoms but can coordinate to the central metal atom through only one of them at a time.
In the given list:
$1$. $NO_2^{-}$ (can coordinate through $N$ or $O$)
$2$. $SCN^{-}$ (can coordinate through $S$ or $N$)
$3$. $CN^{-}$ (can coordinate through $C$ or $N$)
Other ligands like $C_2O_4^{2-}$,$NH_3$,$SO_4^{2-}$,and $H_2O$ are not ambidentate.
Therefore,the total number of ambidentate ligands is $3$.

(D) homoleptic complex is one in which a metal is bound to only one kind of donor groups.
$[Co(NH_3)_6]^{3+}$ contains only $NH_3$ ligands,so it is homoleptic.
Statement $I$ is true.
$A$ heteroleptic complex is one in which a metal is bound to more than one kind of donor groups.
$[Co(NH_3)_4 Cl_2]^{+}$ contains both $NH_3$ and $Cl^{-}$ ligands,so it is heteroleptic.
Statement $II$ correctly explains the definitions of homoleptic and heteroleptic complexes.
Therefore,both Statement $I$ and Statement $II$ are true.

(C) In $CuSO_4 \cdot 5 H_2 O$,the copper ion $(Cu^{2+})$ is six-coordinated.
Four water molecules are directly coordinated to the $Cu^{2+}$ ion.
The remaining fifth water molecule is held by hydrogen bonding between the coordinated water molecules and the sulphate $(SO_4^{2-})$ ions.
Therefore,the number of water molecules directly bonded to the metal centre is $4$.

(C) Ethylenediaminetetraacetic acid $(EDTA)$ is a hexadentate ligand.
Its structure consists of an ethylenediamine backbone $(H_2N-CH_2-CH_2-NH_2)$ where the four hydrogen atoms on the nitrogen atoms are replaced by four acetic acid groups $(CH_2COOH)$.
Therefore,the structure is $(HOOCCH_2)_2N-CH_2-CH_2-N(CH_2COOH)_2$.
This corresponds to option $C$.

(C) The complex is acetylbromidodicarbonylbis(triethylphosphine)iron$(II)$.
From the structure,the ligands attached to the central $Fe$ atom are:
$1$. Two $CO$ (carbonyl) ligands,which are $Fe-C$ bonded.
$2$. One acetyl group $(-COCH_3)$,which is $Fe-C$ bonded.
$3$. Two $PEt_3$ (triethylphosphine) ligands,which are $Fe-P$ bonded.
$4$. One $Br^-$ (bromide) ligand,which is $Fe-Br$ bonded.
Therefore,the total number of $Fe-C$ bonds is $2$ (from $CO$) $+ 1$ (from acetyl) $= 3$.

(C) The chemical formula of the complex formed between $Ni^{2+}$ and dimethylglyoxime $(dmg)$ is $[Ni(dmg)_2]$.
Each dimethylglyoxime ligand has the formula $C_4H_7N_2O_2^-$.
In the complex $[Ni(dmg)_2]$,there are two such ligands.
Total number of $H$ atoms = $2 \times 7 = 14$.
These $14$ hydrogen atoms include $12$ hydrogen atoms from the four methyl groups $(-CH_3)$ and $2$ hydrogen atoms involved in intramolecular hydrogen bonding.

(A) The complex $Co(NH_3)_5Cl_3$ dissociates in water to give $3$ moles of ions,which implies the formula is $[Co(NH_3)_5Cl]Cl_2$.
The dissociation reaction is: $[Co(NH_3)_5Cl]Cl_2 \rightarrow [Co(NH_3)_5Cl]^{2+} + 2Cl^{-}$.
This results in $1$ complex ion and $2$ chloride ions,totaling $3$ ions.
The reaction with excess $AgNO_3$ confirms the presence of $2$ ionizable chloride ions: $[Co(NH_3)_5Cl]Cl_2 + 2AgNO_3 \rightarrow [Co(NH_3)_5Cl](NO_3)_2 + 2AgCl_{(s)}$.

(A) Primary valency is equal to the oxidation state of the central metal atom.
Secondary valency is equal to the coordination number of the central metal atom.
$1$. $[Co(en)_2 Cl_2] Cl$: $Co$ is in $+3$ state,$CN = 2(2) + 2 = 6$. So,$(A-I)$.
$2$. $[Pt(NH_3)_2 Cl(NO_2)]$: $Pt$ is in $+2$ state,$CN = 2 + 1 + 1 = 4$. So,$(B-II)$.
$3$. $Hg[Co(SCN)_4]$: $Co$ is in $+3$ state (considering $Hg^{2+}$),$CN = 4$. So,$(C-III)$.
$4$. $[Mg(EDTA)]^{2-}$: $Mg$ is in $+2$ state,$EDTA$ is hexadentate,$CN = 6$. So,$(D-IV)$.
Thus,the correct match is $A-I, B-II, C-III, D-IV$.

(D) In both complexes,the oxidation state of $Ni$ is calculated as follows:
For $[NiCl_4]^{2-}$: $x + 4(-1) = -2 \Rightarrow x = +2$.
For $[Ni(CN)_4]^{2-}$: $x + 4(-1) = -2 \Rightarrow x = +2$.
The oxidation state of the central metal ion is referred to as the primary valency.
Therefore,both complexes show similarity in the primary valency of $Ni$.

(B) Mohr salt is a double salt with the chemical formula $(NH_4)_2Fe(SO_4)_2 \cdot 6 H_2O$.
It is prepared by mixing equimolar aqueous solutions of ferrous sulfate $(FeSO_4 \cdot 7 H_2O)$ and ammonium sulfate $((NH_4)_2SO_4)$.
The reaction is: $FeSO_4 + (NH_4)_2SO_4 + 6 H_2O \rightarrow (NH_4)_2Fe(SO_4)_2 \cdot 6 H_2O$.

(C) The ligand $N(CH_2-CH_2-NH_2)_3$ is known as $tren$ (tris($2$-aminoethyl)amine).
It contains one tertiary amine nitrogen atom and three primary amine nitrogen atoms,all of which have lone pairs available for coordination.
Since it has four donor nitrogen atoms that can bind to a central metal ion simultaneously,it acts as a tetradentate ligand.

(C) chelate complex is formed when a polydentate ligand binds to a central metal atom through two or more donor atoms,forming a ring structure.
$A$: bis(dimethylglyoximato)nickel$(II)$ contains the bidentate ligand dimethylglyoximate,which forms a chelate.
$B$: Potassium ethylenediaminetetrathiocyanatochromate$(III)$ contains the bidentate ligand ethylenediamine,which forms a chelate.
$C$: Tetraamminediazidocobalt$(III)$ nitrate contains monodentate ligands ($NH_3$ and $N_3^-$). Monodentate ligands cannot form chelate rings.
$D$: trans-diglycinatoplatinum$(II)$ contains the bidentate ligand glycinate,which forms a chelate.

(D) According to Werner's theory,the chloride ions present outside the coordination sphere (ionizable chlorides) react with $AgNO_3$ to form $AgCl$ precipitate.
$1$. $PtCl_4 \cdot 4 NH_3$ can be written as $[Pt(NH_3)_4Cl_2]Cl_2$,which gives $2$ moles of $AgCl$.
$2$. $PtCl_4 \cdot 5 NH_3$ can be written as $[Pt(NH_3)_5Cl]Cl_3$,which gives $3$ moles of $AgCl$.
$3$. $PtCl_4 \cdot 2 NH_3$ can be written as $[Pt(NH_3)_2Cl_4]$,which gives $0$ moles of $AgCl$.
$4$. $PtCl_4 \cdot 3 NH_3$ can be written as $[Pt(NH_3)_3Cl_3]Cl$,which gives $1$ mole of $AgCl$.
Therefore,the complex that gives only one mole of $AgCl$ is $PtCl_4 \cdot 3 NH_3$.

(B) homoleptic complex is a coordination compound in which all the ligands attached to the central metal atom are of the same type.
In $Fe_4[Fe(CN)_6]_3$ (Prussian blue),the coordination entity is $[Fe(CN)_6]^{4-}$,where only one type of ligand,$CN^-$,is present.
Therefore,it is a homoleptic complex.

(B) The name of the complex is Pentaaquachloridochromium $(III)$ chloride monohydrate.
$1$. The central metal atom is Chromium $(Cr)$ with an oxidation state of $+3$.
$2$. The ligands are five water molecules $(pentaaqua = 5 \ H_2O)$ and one chloride ion $(chlorido = 1 \ Cl^-)$ inside the coordination sphere.
$3$. The counter ion is chloride $(Cl^-)$ outside the coordination sphere.
$4$. There is one water molecule of hydration $(monohydrate = 1 \ H_2O)$.
$5$. Balancing the charges: $[Cr(H_2O)_5Cl]^{x} + Cl^- = 0$. Since $Cr$ is $+3$ and $Cl$ is $-1$,the complex ion charge is $+2$. Thus,two chloride ions are needed to balance the charge: $[Cr(H_2O)_5Cl]Cl_2$.
$6$. Adding the monohydrate: $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$.

(C) The reaction for the detection of $NO_2^{-}$ (or $NO_3^{-}$) is the brown ring test.
In this test,$Fe^{2+}$ ions react with $NO$ in the presence of water to form the brown-coloured complex $[Fe(H_2O)_5NO]^{2+}$.
In this complex,the oxidation state of $Fe$ is calculated as: $x + 0 + 0 = +2$,so $x = +2$.
Therefore,the correct statement is that the complex formed is $[Fe(H_2O)_5NO]^{2+}$.

(B) The ligand $en$ stands for ethylenediamine,which is a bidentate ligand.
In the given options,$Bis(ethylenediamine)dithiocyanatoplatinum(IV)$ contains the ligand $en$ as indicated by the name 'ethylenediamine'.
The formula for this complex is $[Pt(en)_2(SCN)_2]^{2+}$.
Therefore,the correct option is $B$.

(B) The chemical formula for the Tetracyanonickelate$(II)$ ion is $[Ni(CN)_4]^{2-}$.
In this complex,the ligand is the cyanide ion $(CN^-)$.
Each cyanide ion acts as a monodentate ligand,meaning it donates one electron pair through the carbon atom.
Since there are $4$ cyanide ligands,the total number of donor atoms is $4 \times 1 = 4$.

(B) The chemical formula of cisplatin is $[Pt(NH_{3})_{2}Cl_{2}]$.
In this coordination complex,the central metal ion is platinum $(Pt^{2+})$.
The ligands attached to the central metal ion are two ammonia $(NH_{3})$ molecules and two chloride $(Cl^{-})$ ions.
Therefore,the correct option is $B$.

(A) Cisplatin is a coordination complex with the chemical formula $[Pt(NH_3)_2Cl_2]$.
In this complex,the central metal atom is $Pt$ (Platinum).
It is bonded to two $NH_3$ ligands and two $Cl^-$ ligands.
Since each ligand is monodentate,the total number of coordinate bonds formed by the central metal atom is $2 + 2 = 4$.
Therefore,the coordination number of $Pt$ in cisplatin is $4$.

(D) The chemical formula for pentaamminecarbonatocobalt$(III)$ chloride is $[Co(NH_3)_5(CO_3)]Cl$.
In this coordination compound,the chloride ion $(Cl^-)$ is present outside the coordination sphere as a counter ion.
When this compound reacts with excess silver nitrate $(AgNO_3)$,the chloride ion outside the coordination sphere reacts to form silver chloride $(AgCl)$ precipitate:
$[Co(NH_3)_5(CO_3)]Cl + AgNO_3 \rightarrow [Co(NH_3)_5(CO_3)]NO_3 + AgCl(s)$.
Since there is only $1$ mole of ionizable $Cl^-$ per mole of the complex,$1$ mole of $AgCl$ will be precipitated.