Nomenclature and oxidation State Questions in English

(A) In the complex $Na_2[Fe(CN)_5NO]$,the oxidation state of $Fe$ is calculated as follows:
Let the oxidation state of $Fe$ be $x$.
The oxidation state of $Na$ is $+1$,$CN$ is $-1$,and $NO$ (nitrosonium ligand) is $+1$.
$2(+1) + x + 5(-1) + 1 = 0$
$2 + x - 5 + 1 = 0$
$x - 2 = 0$
$x = +2$
Therefore,the oxidation state of $Fe$ is $+2$.

(A) The complex is $Hg[Co(CNS)_4]$.
In this complex,the cation is mercury $(Hg^{2+})$ and the anion is the coordination entity $[Co(CNS)_4]^{2-}$.
The ligand $CNS^-$ is named as thiocyanato-$S$ or simply thiocyanato.
Since there are $4$ such ligands,it is named as tetrathiocyanato.
The central metal atom is cobalt,and since it is in an anionic complex,it is named as cobaltate.
The oxidation state of $Co$ is calculated as: $x + 4(-1) = -2$,which gives $x = +2$.
Thus,the $IUPAC$ name is mercury tetrathiocyanatocobaltate $(II)$.

(A) The name of the complex is triamminediaquachlorocobalt $(III)$ chloride.
$1$. The central metal ion is $Co^{3+}$.
$2$. The ligands are three ammine $(NH_3)$,two aqua $(H_2O)$,and one chloro $(Cl^-)$ groups.
$3$. The coordination sphere is $[Co(NH_3)_3(H_2O)_2Cl]$.
$4$. To balance the charge of $Co^{3+}$ $(+3)$ and $Cl^-$ $(-1)$,the total charge of the coordination sphere is $+2$. Thus,two chloride ions are required outside the sphere.
$5$. The formula is $[CoCl(NH_3)_3(H_2O)_2]Cl_2$.

(D) $1$. Identify the ligands: $NH_3$ is ammine,$Cl^-$ is chlorido,and $NH_2CH_3$ is methanamine.
$2$. Arrange ligands in alphabetical order: ammine,chlorido,methanamine.
$3$. Determine the oxidation state of $Pt$: Let $x$ be the oxidation state of $Pt$. The complex is $[Pt(NH_3)_2 Cl(NH_2CH_3)] Cl$. The total charge is $0$. So,$x + 2(0) + (-1) + 0 + (-1) = 0$,which gives $x = +2$.
$4$. Name the complex: The ligands are listed alphabetically as diammine,chlorido,and methanamine. The metal is platinum followed by its oxidation state in Roman numerals $(II)$,and the counter ion is chloride.
$5$. The correct $IUPAC$ name is Diamminechlorido(methanamine)platinum $(II)$ chloride.

(N/A) $[Co(NH_{3})_{4}(H_{2}O)Cl]Cl_{2}$
$(b)$ $K_{2}[Zn(OH)_{4}]$
$(c)$ $K_{3}[Al(C_{2}O_{4})_{3}]$
$(d)$ $[CoCl_{2}(en)_{2}]^{+}$
$(e)$ $[Ni(CO)_{4}]$

(N/A) Diamminechloridonitrito-$N$-platinum$(II)$
$(b)$ Potassium trioxalatochromate$(III)$
$(c)$ Dichloridobis(ethane-$1,2$-diamine)cobalt$(III)$ chloride
$(d)$ Pentaamminecarbonatocobalt$(III)$ chloride
$(e)$ Mercury$(II)$ tetrathiocyanato-$S$-cobaltate$(III)$

(N/A) $(i)$ $[Co(NH_3)_4(H_2O)_2]Cl_3$
$(ii)$ $K_2[Ni(CN)_4]$
$(iii)$ $[Cr(en)_3]Cl_3$
$(iv)$ $[Pt(NH_3)BrCl(NO_2)]$
$(v)$ $[PtCl_2(en)_2](NO_3)_2$
$(vi)$ $Fe_4[Fe(CN)_6]_3$

(N/A) $(i)$ Hexaamminecobalt $(III)$ chloride
$(ii)$ Pentaamminechloridocobalt $(III)$ chloride
$(iii)$ Potassium hexacyanoferrate $(III)$
$(iv)$ Potassium trioxalatoferrate $(III)$
$(v)$ Potassium tetrachloridopalladate $(II)$
$(vi)$ Diamminechlorido(methylamine)platinum $(II)$ chloride

(N/A) $(i) [Co(H_{2}O)(CN)(en)_{2}]^{2+}$
Let the oxidation number of $Co$ be $x$.
The charge on the complex is $+2$.
$x + 0 + (-1) + 2(0) = +2$
$x - 1 = +2$
$x = +3$
$(ii) [CoBr_{2}(en)_{2}]^{+}$
Let the oxidation number of $Co$ be $x$.
The charge on the complex is $+1$.
$x + 2(-1) + 2(0) = +1$
$x - 2 = +1$
$x = +3$
$(iii) [PtCl_{4}]^{2-}$
Let the oxidation number of $Pt$ be $x$.
The charge on the complex is $-2$.
$x + 4(-1) = -2$
$x - 4 = -2$
$x = +2$
$(iv) K_{3}[Fe(CN)_{6}]$
Let the oxidation number of $Fe$ be $x$.
The charge on the complex $[Fe(CN)_{6}]$ is $-3$.
$x + 6(-1) = -3$
$x - 6 = -3$
$x = +3$
$(v) [Cr(NH_{3})_{3}Cl_{3}]$
Let the oxidation number of $Cr$ be $x$.
The complex is neutral,so its total charge is $0$.
$x + 3(0) + 3(-1) = 0$
$x - 3 = 0$
$x = +3$

(N/A) $(i) \quad [Zn(OH)_4]^{2-}$
$(ii) \quad K_2[PdCl_4]$
$(iii) \quad [Pt(NH_3)_2Cl_2]$
$(iv) \quad K_2[Ni(CN)_4]$
$(v) \quad [Co(ONO)(NH_3)_5]^{2+}$
$(vi) \quad [Co(NH_3)_6]_2(SO_4)_3$
$(vii) \quad K_3[Cr(C_2O_4)_3]$
$(viii) \quad [Pt(NH_3)_6]^{4+}$
$(ix) \quad [CuBr_4]^{2-}$
$(x) \quad [Co(NO_2)(NH_3)_5]^{2+}$

(A) $(i)$ Hexaamminecobalt $(III)$ chloride
$(ii)$ Diamminechlorido(methylamine)platinum $(II)$ chloride
$(iii)$ Hexaquatitanium $(III)$ ion
$(iv)$ Tetraamminichloridonitrito-$N$-cobalt $(III)$ chloride
$(v)$ Hexaquamanganese $(II)$ ion
$(vi)$ Tetrachloridonickelate $(II)$ ion
$(vii)$ Hexaamminenickel $(II)$ chloride
$(viii)$ Tris(ethane$-1, 2-$diamine)cobalt $(III)$ ion
$(ix)$ Tetracarbonylnickel $(0)$

(C) We know that $CO$ is a neutral ligand and $K$ carries a charge of $+1$.
Therefore,the complex can be written as $K^{+}[Co(CO)_4]^{-}$.
Let the oxidation number of $Co$ be $x$.
For the complex ion $[Co(CO)_4]^{-}$,the sum of oxidation numbers is equal to the charge on the ion:
$x + 4(0) = -1$
$x = -1$
Therefore,the oxidation number of $Co$ in the given complex is $-1$. Hence,option $(iii)$ is correct.

(B) In $K_2Cr_2O_7$,let the oxidation state of $Cr$ be $x$.
The sum of oxidation states is $2(+1) + 2(x) + 7(-2) = 0$.
$2 + 2x - 14 = 0$.
$2x = 12$.
$x = +6$.
Therefore,the name according to the stock notation is Potassium dichromate $(VI)$.

(C) Let the oxidation number of $Pt$ be $x$.
The complex is $[Pt(C_2H_4)Cl_4]^-$.
$C_2H_4$ (ethylene) is a neutral ligand with an oxidation number of $0$.
$Cl$ is a chloro ligand with an oxidation number of $-1$.
The overall charge on the complex is $-1$.
Therefore,the equation is: $x + 0 + 4(-1) = -1$.
$x - 4 = -1$.
$x = +3$.
Thus,the oxidation number of $Pt$ is $+3$.

(A) Let the oxidation state of $Cu$ be $x$.
The oxidation state of the cyanide ion $(CN^-)$ is $-1$.
The overall charge on the complex $[Cu(CN)_4]^{3-}$ is $-3$.
Setting up the equation: $x + 4(-1) = -3$.
$x - 4 = -3$.
$x = -3 + 4 = +1$.
Therefore,the oxidation state of $Cu$ is $+1$.

(N/A) $(i)$ In both cationic and anionic coordination entities,the cation is named first.
$(ii)$ Ligands are named in alphabetical order before the name of the central atom/ion.
$(iii)$ Names of anionic ligands end in $"o"$,while neutral and cationic ligands retain their names,with exceptions like $H_{2}O$ (aqua),$NH_{3}$ (ammine),$CO$ (carbonyl),and $NO$ (nitrosyl).
$(iv)$ Prefixes like $mono, di, tri$ are used to indicate the number of individual ligands. If the ligand name already contains a numerical prefix,terms like $bis, tris, tetrakis$ are used,and the ligand name is placed in parentheses.
Example: $[NiCl_{2}(PPh_{3})_{2}]$ is named as $dichlorobis(triphenylphosphine)nickel(II)$.
$(v)$ The oxidation state of the metal in cationic,anionic,or neutral coordination entities is indicated by a Roman numeral in parentheses.
$(vi)$ If the complex ion is a cation,the metal is named as the element itself. For example,$Co$ is named as $cobalt$ and $Pt$ as $platinum$. If the complex ion is an anion,the suffix $-ate$ is added to the metal name. For example,in $[Co(SCN)_{4}]^{2-}$,it is named as $cobaltate$. For some metals,Latin names are used in anionic complexes,e.g.,$Fe$ as $ferrate$.
$(vii)$ Neutral complex molecules are named similarly to cationic complexes.

(N/A) $(i)$ $\text{Triamminetriaquachromium(III) chloride}$.
$(ii)$ $\text{Tris(ethane-1,2-diamine)cobalt(III) sulfate}$.
$(iii)$ $\text{Diamminesilver(I) dicyanoargentate(I)}$.
$(iv)$ $\text{Sodium trioxalatochromate(III)}$.

(N/A) $(i)$ ${K_2}[Zn(Cl)_4]$: Potassium tetrachlorozincate$(II)$
$(ii)$ ${K_4}[Fe(CN)_6]$: Potassium hexacyanoferrate$(II)$
$(iii)$ $[Co(NH_3)_4Cl_2]Cl$: Tetraamminedichlorocobalt$(III)$ chloride
$(iv)$ $[Fe(NH_3)_6]^{3+}$: Hexaammineiron$(III)$ ion

(N/A) Oxidation number of central atom: The oxidation number of the central atom is defined as the charge it would carry if all the ligands are removed along with the electron pairs that are shared with the central atom. The oxidation number is represented by a Roman numeral in parenthesis following the name of the coordination entity.
Example: In $[Cu(CN)_4]^{3-}$,the $Cu$ has an oxidation number of $(+1)$ and it is written as $Cu(I)$.

(D) For compound $(A): Na_4[Fe(CN)_5(NOS)]$
Let the oxidation state of $Fe$ be $x.$
$(+1) \times 4 + x + (-1) \times 5 + (-1) = 0$
$4 + x - 5 - 1 = 0 \implies x = +2$
For compound $(B): Na_4[FeO_4]$
Let the oxidation state of $Fe$ be $y.$
$(+1) \times 4 + y + (-2) \times 4 = 0$
$4 + y - 8 = 0 \implies y = +4$
For compound $(C): [Fe_2(CO)_9]$
Let the oxidation state of $Fe$ be $z.$
$2z + 0 \times 9 = 0 \implies z = 0$
Sum of oxidation states $(x + y + z) = 2 + 4 + 0 = 6.$

(C) Let the oxidation state of $Pt$ be $x$ in $[Pt(Py)_{4}][PtCl_{4}]$.
For the cationic part $[Pt(Py)_{4}]^{2+}$,$x + 4(0) = +2$,so $x = +2$.
For the anionic part $[PtCl_{4}]^{2-}$,$x + 4(-1) = -2$,so $x = +2$.
Since the oxidation state of $Pt$ is $+2$ in both parts,option $C$ is correct.

(B) The oxidation state of the central metal atom $Pt$ is calculated as follows:
Let the oxidation state of $Pt$ be $x$.
The sum of oxidation states of all ligands and the central metal equals the charge on the complex ion.
$(x) + 3(0) + (-1) + (-1) + (-1) = +1$
$x - 3 = +1$
$x = +4$
According to $IUPAC$ rules,ligands are named in alphabetical order: $ammine$ $(NH_3)$,$bromido$ $(Br^-)$,$chlorido$ $(Cl^-)$,and $nitro$ $(NO_2^-)$.
Thus,the name is triamminebromidochloridonitroplatinum $(IV)$ chloride.

(D) $1$. Identify the ligands: There are $5$ ammine $(NH_3)$ ligands and $1$ nitrito$-N$ $(NO_2^-)$ ligand.
$2$. Determine the oxidation state of the central metal atom $(Co)$: Let the oxidation state be $x$. The charge on $NH_3$ is $0$,$NO_2^-$ is $-1$,and $Cl^-$ is $-1$. The total charge of the complex is $0$. Thus,$x + 5(0) + 1(-1) + 2(-1) = 0$,which gives $x - 3 = 0$,so $x = +3$.
$3$. Name the complex: The ligands are named in alphabetical order (ammine before nitrito$-N$). The metal is cobalt followed by its oxidation state in Roman numerals in parentheses. The counter ion is chloride.
$4$. The correct name is Pentaammine nitrito$-N$-cobalt$(III)$ chloride.

(C) The given complex is $[Ag(NH_{3})_{2}][Ag(CN)_{2}]$.
This is a coordination compound consisting of a cationic part $[Ag(NH_{3})_{2}]^+$ and an anionic part $[Ag(CN)_{2}]^-$.
In the cationic part $[Ag(NH_{3})_{2}]^+$,let the oxidation state of $Ag$ be $x$. Since $NH_{3}$ is a neutral ligand,$x + 2(0) = +1$,so $x = +1$.
In the anionic part $[Ag(CN)_{2}]^-$,let the oxidation state of $Ag$ be $y$. Since $CN^-$ has a charge of $-1$,$y + 2(-1) = -1$,so $y - 2 = -1$,which gives $y = +1$.
The sum of the oxidation states of the two silver ions is $x + y = +1 + 1 = 2$.

(N/A) . Rules to name ions: If the coordination entity is anionic,the name of the cation is written first,followed by the name of the anionic complex.
For example: In $K_{2}[HgI_{4}]$,the name of $K^{+}$ is written first,followed by the name of $[HgI_{4}]^{2-}$. If the coordination entity is cationic,the name of the positive part is written first,followed by the name of the anion.
For example: In $[Ag(NH_{3})_{2}]OH$,the name of the positive ion $[Ag(NH_{3})_{2}]^{+}$ is written first,followed by the name of the anion,i.e.,$OH^{-}$ ion.
$B$. Rule to name ligands: If the ligands are anionic,the name must end with the suffix $-o$. If the ligands are neutral,they are given the same name as the uncoordinated molecule.

$Cl^{-}$ : Chlorido	${}^{-}NH_{2}$ : Amido
$Br^{-}$ : Bromido	$-NC$ : Isocyano
$I^{-}$ : Iodido	$-SCN$ : Thiocyanato
$-CN$ : Cyanido	$-NCS$ : Isothiocyanato
$-OH$ : Hydroxo	$ONO^{-}$ : Nitrito
$H^{-}$ : Hydrido	$SO_{4}^{2-}$ : Sulphato
$O^{2-}$ : Oxo	$SO_{3}^{2-}$ : Sulphito
$O_{2}^{-}$ : Superoxo	$PO_{4}^{3-}$ : Phosphato
$O_{2}^{2-}$ : Peroxo	$CO_{3}^{2-}$ : Carbonato
$S^{2-}$ : Sulphido	$CH_{3}COO^{-}$ : Acetato
$N_{3}^{-}$ : Azido	$C_{2}O_{4}^{2-}$ : Oxalato

(N/A) The chemical formula for Sodium nitroprusside is $Na_{2}[Fe(CN)_{5}(NO)]$.
In this complex,the oxidation state of iron $(Fe)$ is calculated as follows:
$2(+1) + x + 5(-1) + 0 = 0$
$2 + x - 5 = 0$
$x = +3$
However,in the context of the nitroprusside ion $[Fe(CN)_{5}(NO)]^{2-}$,the $NO$ ligand is considered as $NO^{+}$ (nitrosonium ion),which makes the iron center $Fe(II)$.
Thus,the $IUPAC$ name is Sodium pentacyanidonitrosylferrate$(II)$.

(N/A) Nessler's reagent is the chemical compound $K_{2}[HgI_{4}]$.
In this complex,the central metal ion is mercury $(Hg)$ with an oxidation state of $+2$.
The ligand is iodide $(I^-)$,which is named as 'iodido' in the $IUPAC$ system.
Since there are four iodide ligands,the prefix 'tetra' is used.
The complex anion is $[HgI_{4}]^{2-}$,which is named as 'tetraiodidomercurate$(II)$' because the metal is part of an anionic complex.
Combining these,the full $IUPAC$ name is Potassium tetraiodidomercurate$(II)$.

(N/A) The chemical formula for Sodium cobaltinitrite is $Na_{3}[Co(NO_{2})_{6}]$.
In this complex,the central metal ion is $Co^{3+}$,and the ligand is $NO_{2}^{-}$ (nitrito-$N$).
There are $6$ nitrito-$N$ ligands,so the prefix is 'hexanitrito-$N$'.
The oxidation state of $Co$ is calculated as: $x + 6(-1) = -3$,which gives $x = +3$.
Thus,the $IUPAC$ name is Sodium hexanitrito-$N$-cobaltate$(III)$.

(B) $Cis-platin$ is a coordination complex with the formula $[Pt(NH_3)_2Cl_2]$.
According to $IUPAC$ nomenclature rules for coordination compounds:
$1$. The ligands are named in alphabetical order: $ammine$ $(NH_3)$ comes before $chlorido$ $(Cl^-)$.
$2$. The metal is named as $platinum$ followed by its oxidation state in Roman numerals in parentheses.
$3$. The oxidation state of $Pt$ is calculated as: $x + 2(0) + 2(-1) = 0$,so $x = +2$.
$4$. The $Cis-$ isomerism is specified as a prefix.
Therefore,the $IUPAC$ name is $cis-diamminedichloridoplatinum(II)$.

(B) $1$. Identify the ligands: $Cl^-$ is named 'chlorido' and $CH_3NH_2$ is named 'methanamine'.
$2$. Determine the order: Ligands are named alphabetically. 'Chlorido' comes before 'methanamine'.
$3$. Apply prefixes: Since there are two of each,use 'di' for chlorido and 'bis' for methanamine.
$4$. Determine oxidation state: Let $x$ be the oxidation state of $Cu$. $x + 2(-1) + 2(0) = 0$,so $x = +2$.
$5$. Combine: The name is Dichloridobis(methanamine)copper$(II)$.

(B) $1$. The complex is $[Fe(NH_3)_6][Fe(CN)_6]$.
$2$. The cation is $[Fe(NH_3)_6]^{3+}$ and the anion is $[Fe(CN)_6]^{3-}$.
$3$. In the cation,$Fe$ is in $+3$ oxidation state: $x + 6(0) = +3 \implies x = +3$.
$4$. In the anion,$Fe$ is in $+3$ oxidation state: $x + 6(-1) = -3 \implies x = +3$.
$5$. The $IUPAC$ name is Hexammineiron$(III)$ hexacyanoferrate$(III)$.

(A) $1$. The complex is $K_{3}[Fe(CN)_{5}(NO)] \cdot 2 H_{2}O$.
$2$. The oxidation state of $Fe$ is calculated as: $3(+1) + x + 5(-1) + 0 = 0$,which gives $x - 2 = 0$,so $x = +2$.
$3$. The ligands are $5$ cyano groups (cyanido) and $1$ nitrosyl group $(NO)$.
$4$. The $IUPAC$ name is Potassium pentacyanidonitrosylferrate$(II)$ dihydrate.

(A) $1$. The cation is $[Ph_4As]^+$,which is named as Tetraphenylarsonium.
$2$. The anion is $[PtCl_2(H)(CH_3)]^-$.
$3$. In the anion,the ligands are named in alphabetical order: $Cl^-$ (chlorido),$H^-$ (hydrido),and $CH_3^-$ (methyl).
$4$. The central metal is Platinum $(Pt)$,which is in the anionic complex,so it is named as Platinate.
$5$. The oxidation state of $Pt$ is calculated as: $x + 2(-1) + (-1) + (-1) = -1$,so $x - 4 = -1$,which gives $x = +II$.
$6$. Combining these,the name is Tetraphenylarsonium dichloridohydridomethylplatinate$(II)$.

(A) $1$. The complex is $[Co(NH_3)_3(NO_2)_3]$.
$2$. The ligands are $3$ ammine $(NH_3)$ and $3$ nitrito-$N$ $(NO_2^-)$ groups.
$3$. The oxidation state of $Co$ is calculated as: $x + 3(0) + 3(-1) = 0$,so $x = +3$.
$4$. According to $IUPAC$ nomenclature,ligands are named alphabetically: ammine before nitrito-$N$.
$5$. The name is Triamminetrinitrito-$N$-cobalt$(III)$.

(B) The complex is $Li[AlH_4]$.
In this complex,the cation is $Li^+$ and the anion is $[AlH_4]^-$.
The ligand is hydride $(H^-)$,which is named as 'hydrido' in $IUPAC$ nomenclature.
Since there are $4$ hydride ligands,it is named as 'tetrahydrido'.
The central metal atom is aluminium,and since it is in an anionic complex,it is named as 'aluminate'.
The oxidation state of $Al$ is calculated as: $x + 4(-1) = -1$,so $x = +3$.
Thus,the $IUPAC$ name is Lithium tetrahydridoaluminate$(III)$.

(B) $1$. The complex ion is $[Cr(C_2O_4)_3]^{3-}$.
$2$. The ligand is oxalate $(C_2O_4^{2-})$,which is an anionic ligand,so the suffix '-ate' is added to the metal name.
$3$. There are three oxalate ligands,so the prefix 'tri' is used.
$4$. The oxidation state of $Cr$ is calculated as: $x + 3(-2) = -3$,which gives $x = +3$.
$5$. Combining these,the name is trioxalatochromate$(III)$ ion.

(A) The complex is $[Co(en)_{2}(ONO)Cl]Cl$.
$1$. The ligand $en$ (ethane-$1,2$-diamine) is a neutral bidentate ligand.
$2$. The ligand $Cl^-$ is named as chlorido.
$3$. The ligand $ONO^-$ is an ambidentate ligand bonded through oxygen,named as nitrito-$O$.
$4$. The oxidation state of $Co$ is calculated as: $x + 2(0) + (-1) + (-1) = 0$,so $x = +3$.
$5$. Following alphabetical order for ligands: chlorido,bis(ethane-$1,2$-diamine),nitrito-$O$.
$6$. The full name is Chloridobis(ethane-$1,2$-diamine)nitrito-$O$-cobalt$(III)$ chloride.

(A) $1$. The ligand $H_{2}NCH_{2}CH_{2}NH_{2}$ is ethane-$1,2-$diamine (en),which is a neutral bidentate ligand.
$2$. Let the oxidation state of $Co$ be $x$. The complex is $[Co(en)_{3}]_{2}(SO_{4})_{3}$.
$3$. The total charge of the complex is $2x + 3(-2) = 0$,so $2x = 6$,which means $x = +3$.
$4$. The name is written as: Tris(ligand name)metal(oxidation state) anion.
$5$. Thus,the $IUPAC$ name is Tris(ethane-$1,2-$diamine)cobalt$(III)$ sulphate.

(A) $1$. Identify the ligands: $NH_{3}$ is ammine and $H_{2}O$ is aqua. Alphabetically,ammine comes before aqua.
$2$. Count the ligands: There are $3$ ammine and $3$ aqua ligands,so the prefix is tri-.
$3$. Determine the oxidation state of the metal: Let the oxidation state of $Cr$ be $x$. The charge on $NH_{3}$ is $0$,$H_{2}O$ is $0$,and $Cl$ is $-1$. Thus,$x + 3(0) + 3(0) + 3(-1) = 0$,which gives $x = +3$.
$4$. Name the complex: The name is triamminetriaquachromium$(III)$ chloride.

(A) The complex $[Ag(NH_{3})_{2}][Ag(CN)_{2}]$ is a coordination compound consisting of a cationic part $[Ag(NH_{3})_{2}]^+$ and an anionic part $[Ag(CN)_{2}]^-$.
$1$. The cationic part $[Ag(NH_{3})_{2}]^+$ is named as diamminesilver$(I)$.
$2$. The anionic part $[Ag(CN)_{2}]^-$ is named as dicyanidoargentate$(I)$.
$3$. Combining these,the $IUPAC$ name is diamminesilver$(I)$ dicyanidoargentate$(I)$.

(A) $1$. The ligand is $NCS^-$,which is an ambidentate ligand. In the complex $[Zn(NCS)_4]^{2+}$,the ligand is bonded through the nitrogen atom,so it is named as thiocyanato-$N$.
$2$. There are four such ligands,so the prefix is 'tetra'.
$3$. The central metal ion is zinc $(Zn)$. Since the complex ion has a positive charge of $+2$ and the four $NCS^-$ ligands contribute a charge of $-4$,the oxidation state of $Zn$ is calculated as: $x + 4(-1) = +2$,which gives $x = +6$. However,for $Zn^{2+}$,the oxidation state is $II$. Note: The formula provided $[Zn(NCS)_4]^{2+}$ implies $Zn$ is in the $+6$ state,which is chemically impossible for Zinc. Assuming the complex is $[Zn(NCS)_4]^{2-}$,the oxidation state is $x - 4 = -2$,so $x = +2$. The correct name is Tetrathiocyanato-$N$-zinc$(II)$ ion.

(A) The given complex is $\left[ Ag(H_{2}O)_{2} \right] \left[ Ag(CN)_{2} \right]$.
In this complex,the cation is $\left[ Ag(H_{2}O)_{2} \right]^+$ and the anion is $\left[ Ag(CN)_{2} \right]^-$.
For the cation $\left[ Ag(H_{2}O)_{2} \right]^+$,the oxidation state of $Ag$ is $x + 2(0) = +1$,so $x = +1$.
The name is diaquasilver$(I)$.
For the anion $\left[ Ag(CN)_{2} \right]^-$,the oxidation state of $Ag$ is $x + 2(-1) = -1$,so $x = +1$.
The name is dicyanidoargentate$(I)$.
Combining these,the $IUPAC$ name is diaquasilver$(I)$ dicyanidoargentate$(I)$.

(D) The chemical formula of ammonium phosphomolybdate is $(NH_4)_3[P(Mo_{12}O_{40})]$.
In this complex,the phosphate ion is $PO_4^{3-}$ and the molybdate unit is $MoO_3$.
Let the oxidation state of $Mo$ be $x$.
In $MoO_3$,the oxidation state of oxygen is $-2$.
$x + 3(-2) = 0$
$x - 6 = 0$
$x = +6$.
Thus,the oxidation state of $Mo$ is $+6$.

(C) $1$. Identify the cation: $K^+$ is Potassium.
$2$. Identify the ligand: $C_2O_4^{2-}$ is oxalate,and there are $3$ of them,so it is tris(oxalato).
$3$. Identify the central metal atom: Since the complex ion $[Co(C_2O_4)_3]^{3-}$ is anionic,the metal $Co$ is named as cobaltate.
$4$. Determine the oxidation state of $Co$: $x + 3(-2) = -3$,so $x = +3$.
$5$. Combine the parts: Potassium tris(oxalato)cobaltate$(III)$.

(A) $1$. In $Fe(CO)_5$,the oxidation state of $Fe$ is $0$ because $CO$ is a neutral ligand.
$2$. In $VO^{2+}$,let the oxidation state of $V$ be $x$. Then $x + (-2) = +2$,so $x = +4$.
$3$. In $WO_3$,let the oxidation state of $W$ be $y$. Then $y + 3(-2) = 0$,so $y = +6$.
$4$. The sum of the oxidation states $= 0 + 4 + 6 = 10$.

(B) In $K_2MnO_4$,let the oxidation state of $Mn$ be $x$.
$2(+1) + x + 4(-2) = 0$
$2 + x - 8 = 0$
$x = +6$
The oxidation state of $Mn$ is $+6$.
According to $IUPAC$ nomenclature for coordination compounds and oxoanions,the name is Potassium tetraoxidomanganate $(VI)$.

(C) In the complex $[PtBr_2(PMe_3)_2]$:
$1$. The ligand $Br^-$ is named as bromo.
$2$. The ligand $PMe_3$ is named as trimethylphosphine.
$3$. Since there are two $Br^-$ ligands,we use the prefix 'di' to get 'dibromo'.
$4$. Since there are two $PMe_3$ ligands,we use the prefix 'bis' to get 'bis(trimethylphosphine)'.
$5$. The metal is platinum,and its oxidation state is calculated as $x + 2(-1) + 2(0) = 0$,so $x = +2$.
$6$. Combining these,the correct $IUPAC$ name is dibromobis(trimethylphosphine)platinum$(II)$.

(C) The complex $[Ni(NH_3)_4][NiCl_4]$ is an ionic coordination compound consisting of a cationic part $[Ni(NH_3)_4]^{2+}$ and an anionic part $[NiCl_4]^{2-}$.
In the cationic part,the oxidation state of $Ni$ is $x + 4(0) = +2$,so $x = +2$.
In the anionic part,the oxidation state of $Ni$ is $x + 4(-1) = -2$,so $x = +2$.
The cationic part is named as tetraamminenickel $(II)$.
The anionic part is named as tetrachloronickelate $(II)$ because the metal is in an anionic complex.
Therefore,the correct $IUPAC$ name is tetraamminenickel $(II)$ tetrachloronickelate $(II)$.

(D) $1$. Identify the ligands: $H_2O$ is named 'aqua' and $NH_3$ is named 'ammine'.
$2$. Alphabetical order: 'Ammine' comes before 'aqua'.
$3$. Number of ligands: There are $2$ ammine ligands (di) and $4$ aqua ligands (tetra).
$4$. Central metal: The complex is cationic,so the metal is 'cobalt'.
$5$. Oxidation state: Let $x$ be the oxidation state of $Co$. $x + 4(0) + 2(0) = +3$ (since $Cl_3$ gives $-3$ charge),so $x = +3$.
$6$. Combining these: The name is Diamminetetraaquacobalt $(III)$ chloride.