Nomenclature and oxidation State Questions in English

(B) The name diammine silver $(I)$ chloride indicates:
$1$. The central metal atom is silver $(Ag)$ with an oxidation state of $+1$.
$2$. The ligand is ammonia $(NH_3)$,and the prefix 'di-' indicates there are $2$ ammonia molecules.
$3$. The counter ion is chloride $(Cl^-)$.
Combining these,the coordination entity is $[Ag(NH_3)_2]^+$.
Balancing the charge with the chloride ion,the final chemical formula is $[Ag(NH_3)_2]Cl$.

(A) In the complex $[Co(NH_3)_5NO_2]Cl_2$,the oxidation state of cobalt $(Co)$ is calculated as: $x + 5(0) + 1(-1) + 2(-1) = 0$,which gives $x - 3 = 0$,so $x = +3$.
There are $5$ ammine $(NH_3)$ ligands and $1$ nitro $(NO_2^-)$ ligand attached to the central metal ion.
Following $IUPAC$ nomenclature rules for coordination compounds,the name is written as: Pentaamminenitrocobalt $(III)$ chloride.

(A) In the complex ion $[Cr(NH_3)_6]^{3+}$,let the oxidation state of $Cr$ be $x$.
Since $NH_3$ is a neutral ligand,its charge is $0$.
Therefore,$x + 6(0) = +3$,which gives $x = +3$.
The ligand $NH_3$ is named as 'ammine' and there are $6$ such ligands,so the prefix 'hexa' is used.
The central metal is chromium,followed by its oxidation state in Roman numerals in parentheses.
Thus,the $IUPAC$ name is $\text{Hexamminechromium}(III) \text{ ion}$.

(A) $1$. Identify the ligands: $NH_3$ (ammine),$CN^-$ (cyano),$O_2^{2-}$ (peroxo),and $O^{2-}$ (oxo).
$2$. Determine the oxidation state of $Cr$: Let it be $x$. $2(+1) + x + 2(-1) + 1(-2) + 2(-2) + 1(0) = 0$.
$3$. $2 + x - 2 - 2 - 4 = 0$,so $x = +6$.
$4$. Alphabetical order of ligands: ammine,cyano,dioxo,peroxo.
$5$. The complex is anionic,so the metal is named chromate.
$6$. The correct $IUPAC$ name is Potassium ammine dicyano dioxo peroxo chromate $(VI)$.

(B) $1$. Identify the cation: $K^+$ is potassium.
$2$. Identify the anion: $[Fe(CN)_6]^{3-}$ is the hexacyanoferrate $(III)$ ion.
$3$. Calculate the oxidation state of $Fe$: Let $x$ be the oxidation state of $Fe$. Then $x + 6(-1) = -3$,which gives $x = +3$.
$4$. Combine the names: The $IUPAC$ name is Potassium hexacyanoferrate $(III)$.

(B) The complex is $[Ni(NH_3)_4]SO_4$.
Let the oxidation state of $Ni$ be $x$.
The oxidation state of $NH_3$ is $0$ and $SO_4$ is $-2$.
$x + 4(0) + (-2) = 0$
$x - 2 = 0 \Rightarrow x = +2$.
The coordination number is determined by the number of ligands directly bonded to the central metal ion. Here,$4$ $NH_3$ molecules are bonded to $Ni$,so the coordination number is $4$.
Thus,the oxidation state is $2$ and the coordination number is $4$.

(B) The given complex is ${K_3}[Fe(CN)_6]$.
Let the oxidation state of $Fe$ be $x$.
The oxidation state of $K$ is $+1$ and the oxidation state of $CN^-$ is $-1$.
Sum of oxidation states in a neutral complex is $0$.
$3(+1) + x + 6(-1) = 0$
$3 + x - 6 = 0$
$x - 3 = 0$
$x = +3$
Therefore,the oxidation state of $Fe$ is $+3$.

(A) In metal carbonyls,the metal is always in the zero oxidation state.
Since $CO$ is a neutral ligand,the oxidation state of $Fe$ in $Fe(CO)_5$ is $0$.

(A) In metal carbonyls,the oxidation state of the metal is $0$.
Since $CO$ is a neutral ligand,the oxidation state of $Cr$ in $Cr(CO)_6$ is calculated as:
$x + 6(0) = 0$
$x = 0$.

(A) In the complex $[Fe(H_2O)_5(NO)]SO_4$,the $NO$ ligand acts as $NO^+$.
Let the oxidation state of $Fe$ be $x$.
The sum of oxidation states is: $x + 5(0) + 1(+1) + 1(-2) = 0$.
$x + 1 - 2 = 0$.
$x - 1 = 0$.
$x = +1$.
Therefore,the oxidation state of iron is $+1$.

(B) In the complex $[Cr(CO)_6]$,the ligand $CO$ (carbonyl) is a neutral ligand with an oxidation state of $0$.
Let the oxidation state of $Cr$ be $x$.
Then,$x + 6(0) = 0$,which gives $x = 0$.
Therefore,the oxidation state of the metal in $[Cr(CO)_6]$ is $0$.

(C) $1$. The name $dibromidobis(ethylenediamine)chromium(III)$ bromide indicates the following components:
$2$. $Chromium(III)$ is the central metal ion: $Cr^{3+}$.
$3$. $Bis(ethylenediamine)$ means two $en$ ligands: $(en)_2$.
$4$. $Dibromido$ means two $Br^-$ ligands: $Br_2$.
$5$. The coordination sphere is $[Cr(en)_2Br_2]$.
$6$. The oxidation state of $Cr$ is $x + 2(0) + 2(-1) = +1$ (for the whole complex),so $x - 2 = +1$,$x = +3$.
$7$. The counter ion is $Bromide$ $(Br^-)$,which balances the $+1$ charge of the complex.
$8$. Thus,the formula is $[Cr(en)_2Br_2]Br$.

(C) The coordination entity is hexa-aquamanganese $(II)$,which is represented as $[Mn(H_2O)_6]^{2+}$.
The phosphate ion is $PO_4^{3-}$.
To balance the charges,we use the criss-cross method: the charge of the cation $(+2)$ becomes the subscript of the anion,and the charge of the anion $(-3)$ becomes the subscript of the cation.
Thus,the formula is $[Mn(H_2O)_6]_3(PO_4)_2$.

(B) $1$. The name 'dibromidobis(ethylenediamine)chromium$(III)$ bromide' indicates the following components:
$2$. 'bis(ethylenediamine)' means two ethylenediamine $(en)$ ligands are present: $[Cr(en)_2]$.
$3$. 'dibromido' means two bromide ions are present as ligands inside the coordination sphere: $[Cr(en)_2Br_2]$.
$4$. 'chromium$(III)$' indicates the oxidation state of $Cr$ is $+3$. Let $x$ be the oxidation state of $Cr$. Then $x + 2(0) + 2(-1) = +1$ (since the total charge of the complex must balance the outer bromide ion). Thus,$x - 2 = +1$,so $x = +3$.
$5$. 'bromide' outside the coordination sphere indicates one $Br^-$ ion is present to balance the charge of the complex ion $[Cr(en)_2Br_2]^+$.
$6$. Therefore,the formula is $[Cr(en)_2Br_2]Br$.

(C) To name the coordination complex $[CO(NH_3)_5Cl]^{2+}$:
$1$. The ligand $NH_3$ is named as 'ammine' and $Cl^-$ is named as 'chloro'.
$2$. Since there are $5$ ammine ligands,we use the prefix 'penta'.
$3$. The metal is cobalt.
$4$. To find the oxidation state of $Co$: $x + 5(0) + 1(-1) = +2$,which gives $x = +3$.
$5$. Thus,the name is 'pentaamminechlorocobalt $(III)$ ion'.

(B) $1$. The complex is $K_3[Cr(C_2O_4)_3]$.
$2$. Let the oxidation state of $Cr$ be $x$.
$3$. The oxalate ion $(C_2O_4)^{2-}$ has a charge of $-2$.
$4$. Potassium $(K)$ has a charge of $+1$.
$5$. The sum of oxidation states in a neutral complex is zero: $3(+1) + x + 3(-2) = 0$.
$6$. $3 + x - 6 = 0$,which gives $x = +3$.
$7$. The oxalate ion $(C_2O_4)^{2-}$ is a bidentate ligand,meaning it forms $2$ coordinate bonds per ligand.
$8$. With $3$ oxalate ligands,the coordination number is $3 \times 2 = 6$.
$9$. Thus,the oxidation state is $+3$ and the coordination number is $6$.

(C) $1$. Identify the ligands: $NH_3$ is ammine,$H_2O$ is aqua,and $Br^-$ is bromo.
$2$. Arrange ligands in alphabetical order: ammine,aqua,bromo.
$3$. The coordination sphere is $[Co(NH_3)_4(H_2O)Br]^{2+}$.
$4$. Oxidation state of $Co$: $x + 4(0) + 1(0) + 1(-1) = +2$,so $x = +3$.
$5$. The name is tetraammineaquabromocobalt $(III)$ nitrate.

(A) $1$. The central metal atom is Nickel $(Ni)$ with an oxidation state of $+2$.
$2$. The ligands are two cyano groups $(CN^-)$ and two oxalato groups $(C_2O_4^{2-})$.
$3$. The charge on the coordination sphere is calculated as: $x + 2(-1) + 2(-2) = +2 - 2 - 4 = -4$. Wait,let's re-calculate: $Ni$ is $+2$,$2(CN^-)$ is $-2$,$2(C_2O_4^{2-})$ is $-4$. Total charge = $+2 - 2 - 4 = -4$.
$4$. To balance a charge of $-4$,we need $4$ potassium ions $(K^+)$.
$5$. Thus,the formula is $K_4[Ni(CN)_2(C_2O_4)_2]$.

(A) $1$. Identify the ligands: $NH_3$ is ammine,$NO_2^-$ is nitro,$Cl^-$ is chloro,and $CN^-$ is cyano.
$2$. Arrange ligands in alphabetical order: Ammine,Chloro,Cyano,Nitro.
$3$. Since there are three $NH_3$ ligands,use the prefix 'tri'.
$4$. The central metal is Cobalt $(Co)$. The oxidation state is calculated as: $x + 3(0) + (-1) + (-1) + (-1) = 0$,so $x = +3$.
$5$. Combining these,the name is Triamminochlorocyanonitrocobalt $(III)$.

(C) $1$. The name 'diamminesilver $(I)$ chloride' indicates a coordination compound.
$2$. 'Diammine' means two $NH_3$ ligands are attached to the central metal ion.
$3$. 'Silver $(I)$' indicates the central metal is $Ag^+$.
$4$. The coordination sphere is $[Ag(NH_3)_2]^+$.
$5$. 'Chloride' is the counter ion,$Cl^-$.
$6$. Combining these,the formula is $[Ag(NH_3)_2]Cl$.

(A) The $IUPAC$ name of the coordination compound $K_3[Fe(ONO)_6]$ is determined as follows:
$1$. The cation is $K^+$,named as Potassium.
$2$. The coordination entity is $[Fe(ONO)_6]^{3-}$.
$3$. The ligand is $ONO^-$,which is nitrito-$O$. Since there are $6$ such ligands,it is named as hexanitrito-$O$.
$4$. The central metal atom is $Fe$ in an anionic complex,so it is named as ferrate.
$5$. The oxidation state of $Fe$ is calculated as: $x + 6(-1) = -3$,which gives $x = +3$. Thus,it is ferrate$(III)$.
$6$. Combining these,the name is Potassium hexanitrito-$O$-ferrate$(III)$.

(C) $1$. Identify the ligands: $NH_3$ (ammine),$Br^-$ (bromo),$NO_2^-$ (nitro),$Cl^-$ (chloro).
$2$. Arrange ligands in alphabetical order: ammine,bromo,chloro,nitro.
$3$. The coordination sphere is $[Pt(NH_3)_3(Br)(NO_2)Cl]$. The central metal is $Pt$.
$4$. Calculate the oxidation state of $Pt$: $x + 3(0) + (-1) + (-1) + (-1) = +1$ (since $Cl^-$ is outside the sphere),so $x - 3 = +1$,$x = +4$.
$5$. Combine the parts: Triammine (for $3$ $NH_3$) + bromo + chloro + nitro + platinum $(IV)$ + chloride.
$6$. The correct name is Triamminebromochloronitroplatinum $(IV)$ chloride.

(B) $1$. In $CrO_2Cl_2$: Let the oxidation state of $Cr$ be $x$. $x + 2(-2) + 2(-1) = 0 \implies x - 4 - 2 = 0 \implies x = +6$. This is the maximum oxidation state for $Cr$ $(3d^5 4s^1)$.
$2$. In $MnO_4^-$: Let the oxidation state of $Mn$ be $y$. $y + 4(-2) = -1 \implies y - 8 = -1 \implies y = +7$. This is the maximum oxidation state for $Mn$ $(3d^5 4s^2)$.
$3$. Both $Cr$ in $CrO_2Cl_2$ and $Mn$ in $MnO_4^-$ are in their highest possible oxidation states.

(A) To name the coordination compound $[Co(NH_3)_3(H_2O)_2Cl_2]$:
$1$. Identify the ligands: $NH_3$ is ammine,$H_2O$ is aqua,and $Cl^-$ is chlorido.
$2$. Arrange ligands alphabetically: Ammine,Aqua,Chlorido.
$3$. Count the ligands: $3$ ammine,$2$ aqua,$1$ chlorido.
$4$. The name is Triamminediaquachloridocobalt $(III)$.
$5$. Since the complex is neutral,the oxidation state of $Co$ is $x + 3(0) + 2(0) + 1(-1) = 0$,so $x = +3$.

(B) In the complex $K_3[Fe(CN)_6]$,let the oxidation state of $Fe$ be $x$.
$3(+1) + x + 6(-1) = 0$
$3 + x - 6 = 0$
$x = +3$.
Thus,the $IUPAC$ name is Potassium hexacyanoferrate $(III)$.
Commonly,this compound is known as Potassium ferricyanide.
Therefore,statements $(2)$ and $(3)$ are correct.

(B) $1$. The formula is $[VO(acac)_2]$.
$2$. Here,$V$ is the central metal atom,$O$ is the oxo ligand (charge $-2$),and $acac$ is the acetylacetonato ligand (charge $-1$ each).
$3$. Let the oxidation state of $V$ be $x$.
$4$. $x + (-2) + 2(-1) = 0 \implies x - 4 = 0 \implies x = +4$.
$5$. The name is derived as: $Bis(acetylacetonato)oxovanadium(IV)$.

(A) The complex is hexaaquamanganese $(II)$ phosphate.
The coordination sphere is $[Mn(H_2O)_6]^{2+}$ because the oxidation state of $Mn$ is $+2$ and $H_2O$ is a neutral ligand.
The phosphate ion is $PO_4^{3-}$.
To balance the charges,we need three $[Mn(H_2O)_6]^{2+}$ ions and two $PO_4^{3-}$ ions.
Therefore,the formula is $[Mn(H_2O)_6]_3(PO_4)_2$.

(A) $1$. The central metal ion is Cobalt $(Co)$ with an oxidation state of $+3$.
$2$. The ligands are 'triammine' $(3 \ NH_3)$ and 'diaqua' $(2 \ H_2O)$.
$3$. The coordination sphere is $[Co(NH_3)_3(H_2O)_2]$.
$4$. The charge on the coordination sphere is calculated as: $x + 3(0) + 2(0) = +3$,so the complex ion is $[Co(NH_3)_3(H_2O)_2]^{3+}$.
$5$. To balance the $+3$ charge,$3$ chloride ions $(Cl^-)$ are required.
$6$. Thus,the formula is $[Co(NH_3)_3(H_2O)_2]Cl_3$.

(C) $1$. Identify the central metal atom: Cobalt $(Co)$.
$2$. Identify the ligands: 'chloro' $(Cl^-)$,'bis(ethylenediamine)' $(2 \times en)$,and 'nitro' $(NO_2^-)$.
$3$. Determine the oxidation state: The complex is a cobalt $(III)$ ion,so the charge on $Co$ is $+3$.
$4$. Calculate the total charge: $Co(+3) + Cl(-1) + 2 \times en(0) + NO_2(-1) = +3 - 1 - 1 = +1$.
$5$. Combine the components: The formula is $[Co(en)_2(NO_2)Cl]^{+}$.
$6$. Therefore,the correct option is $C$.

(D) $1$. The formula is $FeCl_3 \cdot 4NH_3$. Since it gives a white precipitate with $AgNO_3$,at least one $Cl^-$ ion must be outside the coordination sphere.
$2$. The coordination number is $6$. With $4$ $NH_3$ ligands,we need $2$ more ligands to satisfy the coordination number of $6$. These must be $Cl^-$ ions.
$3$. The complex can be written as $[Fe(NH_3)_4Cl_2]Cl$.
$4$. The $IUPAC$ name for $[Fe(NH_3)_4Cl_2]Cl$ is tetraamminedichloridoiron $(III)$ chloride.

(A) $1$. The complex is a coordination entity consisting of a cationic part $[Co(NH_3)_6]^{3+}$ and an anionic part $[Cr(CN)_6]^{3-}$.
$2$. For the cationic part,the ligand is $NH_3$ (ammine) and the metal is $Co$ (cobalt). The oxidation state of $Co$ is $x + 6(0) = +3$,so $x = +3$. The name is hexaamminecobalt $(III)$.
$3$. For the anionic part,the ligand is $CN^-$ (cyano) and the metal is $Cr$ (chromium). Since it is an anion,the suffix '-ate' is added to the metal name,becoming chromate. The oxidation state of $Cr$ is $x + 6(-1) = -3$,so $x = +3$. The name is hexacyanochromate $(III)$.
$4$. Combining these,the $IUPAC$ name is Hexaamminecobalt $(III)$ hexacyanochromate $(III)$.

(B) Let the oxidation state of $Fe$ be $x$.
For $K_3[Fe(CN)_6]$,the sum of oxidation states of all atoms is equal to $0$.
$3 \times (+1) + x + 6 \times (-1) = 0$
$3 + x - 6 = 0$
$x - 3 = 0$
$x = +3$

(A) Let the oxidation state of $Cr$ be $x$.
The oxidation state of $NH_3$ is $0$ and $Cl$ is $-1$.
The total charge on the complex is $+1$.
$x + 4(0) + 2(-1) = +1$
$x - 2 = +1$
$x = +3$
Therefore,the oxidation state of $Cr$ is $+3$.

(A) In the complex $[Co(NH_3)_5NO_2]Cl_2$,the oxidation state of $Co$ is calculated as: $x + 5(0) + 1(-1) = 2(1) \implies x - 1 = 2 \implies x = +3$.
The ligand $NH_3$ is named as 'ammine' and $NO_2^-$ is named as 'nitrito-$N$'.
Since there are $5$ ammine ligands,it is 'pentaammine'.
Combining these,the $IUPAC$ name is 'Pentaamminenitrito-$N$-cobalt$(III)$ chloride'.

(D) The charge on the complex ion is equal to the sum of the oxidation state of the metal and the charge on the ligands.
Let the oxidation state of $Pt$ be $x$.
The charge on the $Cl^-$ ligand is $-1$.
Therefore,$x + 6 \times (-1) = -2$.
$x - 6 = -2$.
$x = +4$.
Thus,the oxidation state of $Pt$ in $[PtCl_6]^{2-}$ is $+4$.

(D) The $IUPAC$ name for the complex $Li[AlH_4]$ is determined by naming the cation first,followed by the anion.
$1$. The cation is $Li^+$,which is named as Lithium.
$2$. The anion is $[AlH_4]^-$,which is a complex anion.
$3$. The ligand is $H^-$,named as hydrido. Since there are $4$ ligands,it is named as tetrahydrido.
$4$. The central metal atom is Aluminum,which in an anionic complex is named as Aluminate.
$5$. The oxidation state of $Al$ is calculated as: $x + 4(-1) = -1$,so $x = +3$. Thus,it is Aluminate $(III)$.
Combining these,the name is Lithium tetrahydridoaluminate $(III)$.

(A) $1$. Identify the bridging ligand: The $OH^-$ group acts as a bridge between two $Cr$ centers,denoted as $\mu$-hydroxo.
$2$. Identify the coordination spheres: Each $Cr$ is bonded to $5$ $NH_3$ molecules,so we use the prefix 'bis' for two units of 'pentaamminechromium'.
$3$. Calculate the oxidation state of $Cr$: Let the oxidation state of $Cr$ be $x$. The total charge is $+5$. The charge of $NH_3$ is $0$ and $OH^-$ is $-1$. Thus,$2x + (-1) = +5$,which gives $2x = +6$,so $x = +3$.
$4$. Combine the parts: The name is $\mu$-hydroxo bis(pentaamminechromium$(III)$) ion.

(A) $1$. The given complex is $[Ni(NH_3)_4][NiCl_4]$.
$2$. The cation is $[Ni(NH_3)_4]^{2+}$ and the anion is $[NiCl_4]^{2-}$.
$3$. In the cation $[Ni(NH_3)_4]^{2+}$,the oxidation state of $Ni$ is $x + 4(0) = +2$,so $x = +2$. The name is tetraamminenickel $(II)$.
$4$. In the anion $[NiCl_4]^{2-}$,the oxidation state of $Ni$ is $x + 4(-1) = -2$,so $x = +2$. Since it is an anionic complex,the metal name ends in '-ate',so it is tetrachloridonickelate $(II)$.
$5$. Combining these,the $IUPAC$ name is tetraamminenickel $(II)$ tetrachloridonickelate $(II)$.

(C) In the complex $Fe(CO)_5$,$CO$ is a neutral ligand (carbonyl).
Let the oxidation state of $Fe$ be $x$.
Since $CO$ is neutral,its charge is $0$.
$x + 5(0) = 0$
$x = 0$
According to $IUPAC$ nomenclature for neutral complexes,the metal name remains as it is (Iron) followed by its oxidation state in parentheses.
Therefore,the name is Pentacarbonyliron $(0)$.

(C) The complex $Fe(CO)_5$ is a neutral coordination compound.
In this complex,$CO$ is a neutral ligand (carbonyl).
Since the overall charge of the complex is $0$,the oxidation state of the central metal atom $Fe$ is $0$.
According to $IUPAC$ nomenclature rules for neutral complexes,the metal name is written as it is (iron) followed by its oxidation state in parentheses.
Therefore,the $IUPAC$ name is $Pentacarbonyliron(0)$.

(C) $1$. The central metal ion is Cobalt with an oxidation state of $+III$,denoted as $Co^{3+}$.
$2$. The ligand is ethylenediamine (en),which is a neutral bidentate ligand. The prefix 'tris' indicates there are $3$ such ligands,so the coordination sphere is $[Co(en)_3]^{3+}$.
$3$. The counter ion is sulfate,which has a charge of $-2$,denoted as $SO_4^{2-}$.
$4$. To balance the charges,we need $2$ units of the complex cation $[Co(en)_3]^{3+}$ and $3$ units of the sulfate anion $SO_4^{2-}$.
$5$. Thus,the formula is $[Co(en)_3]_2(SO_4)_3$.

(A) The $IUPAC$ name is dichloridobis(urea)copper $(II)$.
$1$. The central metal ion is Copper $(II)$,which means $Cu^{2+}$.
$2$. 'Dichlorido' indicates two chloride ligands $(Cl^-)$.
$3$. 'Bis(urea)' indicates two urea ligands $(NH_2CONH_2)$.
$4$. Combining these,the coordination entity is $[Cu(NH_2CONH_2)_2Cl_2]$.
Thus,the correct formula is $[Cu(NH_2CONH_2)_2Cl_2]$.

(B) $1$. The central metal atom is platinum $(Pt)$ with an oxidation state of $+II$.
$2$. The ligands are two chloride ions $(Cl^-)$ and two ammonia molecules $(NH_3)$.
$3$. According to $IUPAC$ nomenclature rules for coordination compounds,the ligands are listed in alphabetical order: ammine $(NH_3)$ comes before chloride $(Cl^-)$.
$4$. The prefix 'di' is used for both ligands,resulting in 'diamminedichlorido'.
$5$. Combining these,the coordination entity is $[Pt(NH_3)_2Cl_2]$.
$6$. Since the complex is neutral,the formula is $[Pt(NH_3)_2Cl_2]$.

(D) The formula of sodium nitroprusside is $Na_2[Fe(CN)_5NO]$.
In this complex,the ligand $NO$ is present as $NO^+$ (nitrosonium ion).
Let the oxidation state of $Fe$ be $x$.
$2(+1) + x + 5(-1) + 1 = 0$
$2 + x - 5 + 1 = 0$
$x - 2 = 0$
$x = +2$.
Therefore,the $IUPAC$ name is sodium pentacyanonitrosylferrate $(II)$.
Note: Option $(D)$ is often cited in older textbooks as $(III)$ due to the assumption of $NO$ as a neutral ligand,but the correct oxidation state of $Fe$ is $+2$ when $NO$ is treated as $NO^+$.

(C) To name the coordination compound $K_2[Zn(OH)_4]$:
$1$. The cation is potassium $(K^+)$.
$2$. The coordination entity is $[Zn(OH)_4]^{2-}$.
$3$. The ligand is hydroxide $(OH^-)$,which is named as 'hydroxo'. Since there are four,it is 'tetrahydroxo'.
$4$. The central metal is zinc,and since the complex is anionic,it is named as 'zincate'.
$5$. The oxidation state of $Zn$ is calculated as: $x + 4(-1) = -2$,so $x = +2$.
$6$. Combining these,the $IUPAC$ name is Potassium tetrahydroxozincate $(II)$.

(D) Let the oxidation state of $Cr$ be $x$.
For the complex $[Cr(NH_3)_4Cl_2]^+$,the sum of oxidation states of all atoms equals the charge on the complex.
$x + 4(0) + 2(-1) = +1$
$x + 0 - 2 = +1$
$x = +3$
Therefore,the oxidation state of $Cr$ is $+3$.

(C) $1$. Identify the coordination entity: $[Ni(CN)_4]^{4-}$.
$2$. The oxidation state of $Ni$ is calculated as: $x + 4(-1) = -4$,which gives $x = 0$.
$3$. The complex is an anionic complex,so the metal name ends in '-ate'.
$4$. The correct name is Potassium tetracyanonickelate $(0)$.

(C) Let the oxidation state of $Pt$ be $x$.
In the complex $[Pt(NH_3)_5Cl]Cl_3$,the ligands are $5 \ NH_3$ (neutral,charge $= 0$) and $1 \ Cl^-$ (charge $= -1$).
The total charge of the coordination sphere is balanced by $3 \ Cl^-$ ions outside the bracket (each with charge $-1$).
Thus,$x + 5(0) + 1(-1) + 3(-1) = 0$.
$x - 1 - 3 = 0$.
$x - 4 = 0$.
$x = +4$.

(C) $1$. Identify the coordination entity: $[OsCl_5N]^{2-}$.
$2$. The ligand $N^{3-}$ is named as 'nitrido'.
$3$. The ligand $Cl^-$ is named as 'chloro'.
$4$. The oxidation state of $Os$ is calculated as: $x + 5(-1) + (-3) = -2$,which gives $x - 8 = -2$,so $x = +6$.
$5$. Since the complex ion is anionic,the metal name ends in '-ate',so $Os$ becomes 'osmate'.
$6$. Combining these,the name is Potassium pentachloronitridoosmate $(VI)$.

(A) In the complex $Ni(CO)_4$,$CO$ is a neutral ligand with an oxidation state of $0$.
Let the oxidation state of $Ni$ be $x$.
Then,$x + 4(0) = 0$,which gives $x = 0$.
Therefore,the oxidation state of $Ni$ is $0$.
Statement $A$ claims the oxidation state is $+4$,which is incorrect.