Nomenclature and oxidation State Questions in English

(C) In $F_2 + O_2 \to O_2F_2$,fluorine is more electronegative than oxygen,so fluorine is reduced and oxygen is oxidized. Thus,oxygen acts as a reducing agent,not an oxidizing agent. Statement $(a)$ is $False$.
$(b)$ The acidic strength of oxoacids of chlorine increases with the number of oxygen atoms: $HClO < HClO_2 < HClO_3 < HClO_4$. Thus,$HClO_4$ is the strongest acid. Statement $(b)$ is $False$.
$(c)$ In $[Ni(CO)_4]$,$CO$ is a neutral ligand (oxidation state $0$). Therefore,the oxidation state of $Ni$ is $0$. Statement $(c)$ is $True$.
$(d)$ In $BaO_2 + H_2SO_4 \to BaSO_4 + H_2O_2$,the oxidation states of all elements remain unchanged $(Ba: +2, O: -1, H: +1, S: +6)$. It is a double displacement reaction,not a redox reaction. Statement $(d)$ is $False$.
Therefore,the correct order is $F, F, T, F$.

(A) The complex is $[(H_2O)_4Cr(\mu-OH)(\mu-O_2)Cr(H_2O)_4]^{n+}$ (Note: The provided structure is a $\mu$-peroxo and $\mu$-hydroxo bridged complex).
Assuming the structure contains one $\mu$-peroxo $(O_2^{2-})$ bridge and one $\mu$-hydroxo $(OH^-)$ bridge:
Let the oxidation state of $Cr$ be $x$.
Total charge on the complex = $3+$.
Sum of oxidation states = $2x + 4(0) + 4(0) + (-1) + (-2) = +3$.
$2x - 3 = +3$.
$2x = 6$.
$x = +3$.
Thus,the oxidation state of $Cr$ is $+3$.

(C) For the complex $[Pt(NH_3)_2Cl(NO_2)]$:
$1$. Let the oxidation state of $Pt$ be $x$.
$2$. The charge on $NH_3$ is $0$,$Cl^-$ is $-1$,and $NO_2^-$ is $-1$.
$3$. $x + 0 + (-1) + (-1) = 0$,which gives $x = +2$.
$4$. The ligands are named in alphabetical order: $NH_3$ (ammine),$Cl^-$ (chlorido),and $NO_2^-$ (nitrito-$N$).
$5$. Combining these,the $IUPAC$ name is Diamminechloridonitrito-$N$-platinum$(II)$.

(B) $1$. Identify the cation: $Na^+$ is sodium.
$2$. Identify the coordination entity: $[Co(ONO)_6]^{3-}$.
$3$. The ligand is $ONO^-$,which is named 'nitrito-$O$'. Since there are $6$ such ligands,it is 'hexanitrito-$O$'.
$4$. The central metal is cobalt. Since the complex ion is anionic,it is named 'cobaltate'.
$5$. Calculate the oxidation state of $Co$: $x + 6(-1) = -3$,so $x = +3$.
$6$. Combining these,the name is Sodium hexanitrito-$O$-cobaltate$(III)$.

(B) The chemical formula for Prussian blue is $Fe_4[Fe(CN)_6]_3$.
In this complex,the coordination entity is $[Fe(CN)_6]^{4-}$,where the oxidation state of $Fe$ is $+2$ (ferrocyanide).
To balance the charge,the four $Fe$ atoms outside the bracket must have a total charge of $+12$,meaning each $Fe$ atom is in the $+3$ oxidation state (ferric).
Therefore,the $IUPAC$ name is Iron $(III)$ hexacyanoferrate $(II)$.

(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 $0$. So,$x + 5(0) + 1(-1) + 2(-1) = 0$,which gives $x - 3 = 0$,so $x = +3$.
$3$. Apply $IUPAC$ naming rules: List ligands in alphabetical order (ammine before nitrito-$N$),followed by the metal name with its oxidation state in Roman numerals,and finally the counter ion.
$4$. The name is: Pentaammine nitrito-$N$-cobalt$(III)$ chloride.

(D) The central metal is chromium in the $+3$ oxidation state.
Ligands present are: two ammine $(NH_3)$ which are neutral,two chloro $(Cl^-)$ which have a charge of $-1$ each,and two cyano $(CN^-)$ which have a charge of $-1$ each.
The total charge on the coordination entity is calculated as: $x + 2(0) + 2(-1) + 2(-1) = +3 + 0 - 2 - 2 = -1$.
Therefore,the correct formula is $[CrCl_2(CN)_2(NH_3)_2]^-$.

(C) The $IUPAC$ name for $[Ni(NH_3)_4][NiCl_4]$ is determined as follows:
$1$. The complex cation $[Ni(NH_3)_4]^{2+}$ is named as $tetraamminenickel(II)$.
$2$. The complex anion $[NiCl_4]^{2-}$ is named as $tetrachloronickelate(II)$.
$3$. Combining these,the full name is $tetraamminenickel(II) tetrachloronickelate(II)$.
$4$. In coordination compounds,the cation is named first,followed by the anion. Since the anion is a complex,the suffix $-ate$ is added to the metal name.

(A) The complex $[Pt(NH_3)_4][PtCl_4]$ consists of a cationic part $[Pt(NH_3)_4]^{2+}$ and an anionic part $[PtCl_4]^{2-}$.
In the cationic part,$NH_3$ is a neutral ligand,so the oxidation state of $Pt$ is $+2$. It is named as $tetraammineplatinum(II)$.
In the anionic part,$Cl^-$ is an anionic ligand,so the oxidation state of $Pt$ is $+2$. Since it is an anion,the metal name ends in $-ate$,making it $tetrachloridoplatinate(II)$.
Combining these,the correct $I.U.P.A.C.$ name is $tetraammineplatinum(II)$ $tetrachloridoplatinate(II)$.

(A) To determine the highest oxidation state,we calculate the oxidation number for each metal:
$1$. $MnO_4^-$: Let $x$ be the oxidation state of $Mn$. $x + 4(-2) = -1 \implies x = +7$. $CrO_2Cl_2$: Let $y$ be the oxidation state of $Cr$. $y + 2(-2) + 2(-1) = 0 \implies y = +6$. Both $Mn$ $(+7)$ and $Cr$ $(+6)$ are in their highest possible oxidation states.
$2$. $[NiCl_4]^{2-}$: $Ni$ is $+2$; $[CoCl_4]^{2-}$: $Co$ is $+2$.
$3$. $[Fe(CN)_6]^{3-}$: $Fe$ is $+3$; $[Cu(CN)_4]^{2-}$: $Cu$ is $+2$.
$4$. $[FeCl_4]^-$: $Fe$ is $+3$; $Co_2O_3$: $Co$ is $+3$.
Thus,the pair $MnO_4^-$ and $CrO_2Cl_2$ contains metals in their highest oxidation states.

(B) For $[Mn(CN)_5]^{2-}$,let the oxidation state of $Mn$ be $x$.
$x + 5(-1) = -2$
$x - 5 = -2$
$x = +3$.
Thus,the correct $IUPAC$ name is Pentacyanomanganate$(III)$ ion.
Therefore,the combination in option $B$ is incorrect.

(D) The oxidation number of $K$ is $+1$ and the ligand $CO$ is neutral (oxidation number $= 0$).
Let the oxidation number of $Co$ be $x$.
In the complex $K[Co(CO)_4]$,the sum of oxidation states is zero:
$1 + (x + 4 \times 0) = 0$
$x + 1 = 0$
$x = -1$
Thus,the oxidation number of $Co$ is $-1$.
The coordination number is determined by the number of donor atoms attached to the central metal atom. Since $CO$ is a monodentate ligand and there are $4$ of them,the coordination number is $4$.
Therefore,the oxidation number is $-1$ and the coordination number is $4$.

(A) $1$. Identify the ligands: There are $4$ ammine $(NH_3)$ ligands and $2$ chlorido $(Cl^-)$ ligands. According to alphabetical order,'ammine' comes before 'chlorido'.
$2$. Name the coordination sphere: The ligands are named as 'tetraammine' and 'dichlorido'. The metal is chromium. Since the complex ion is cationic,the metal name remains 'chromium'.
$3$. Determine the oxidation state of $Cr$: Let the oxidation state be $x$. $x + 4(0) + 2(-1) = +1$ (since $NO_3$ is $-1$). $x - 2 = +1$,so $x = +3$.
$4$. Combine the parts: The name is 'tetraamminedichloridochromium $(III)$ nitrate'.

(A) The complex $K_2Fe[Fe(CN)_6]$ is also known as Prussian blue or Turnbull's blue depending on the synthesis,but here we analyze the oxidation states.
Let the oxidation state of $Fe$ be $x$.
$2(+1) + x + [x + 6(-1)] = 0$
$2 + x + x - 6 = 0$
$2x = 4$
$x = +2$
Both iron atoms are in the $+2$ oxidation state.
The coordination number of the central $Fe$ atom in the $[Fe(CN)_6]^{4-}$ ion is $6$ because it is surrounded by $6$ $CN^-$ ligands.
$CN^-$ is a strong field ligand,so the complex is a low spin complex.

(C) In sodium nitroprusside,$Na_2[Fe(CN)_5NO]$,the complex ion is $[Fe(CN)_5NO]^{2-}$.
Let the oxidation state of $Fe$ be $x$.
The ligand $CN^-$ has a charge of $-1$ and $NO$ acts as a nitrosonium ion $(NO^+)$ with a charge of $+1$.
Applying the charge balance: $x + 5(-1) + 1 = -2$.
$x - 5 + 1 = -2$.
$x - 4 = -2$.
$x = +2$.
Thus,the oxidation state of $Fe$ is $+2$ and the ligand is $NO^+$. Both statements are correct.

(C) In the complex ion $[Fe(H_2O)_5(NO)]^{2+}$,$H_2O$ is a neutral ligand with an oxidation state of $0$.
$NO$ (nitrosyl) acts as a positive ligand with a charge of $+1$.
Let the oxidation state of $Fe$ be $x$.
The total charge on the complex is $+2$.
Setting up the equation: $x + 5(0) + 1 = +2$.
Solving for $x$: $x + 1 = +2$,which gives $x = +1$.

(B) The compound given is $Na_{4}[Fe(CN)_{5}(NOS)]$.
Let the oxidation state of $Fe$ be $x$.
The oxidation state of $Na$ is $+1$,$CN$ is $-1$,and $NOS$ is $-1$.
The sum of oxidation states in a neutral complex is $0$.
$4(+1) + x + 5(-1) + 1(-1) = 0$.
$4 + x - 5 - 1 = 0$.
$x - 2 = 0$.
$x = +2$.

(D) $1$. The complex ion is $[CoCl(ONO)(en)_2]^+$.
$2$. The ligands are $Cl^-$ (chloro),$ONO^-$ (nitrito-$O$),and $en$ (ethylenediamine).
$3$. Since there are two ethylenediamine ligands,we use the prefix 'bis'.
$4$. The oxidation state of $Co$ is calculated as: $x + (-1) + (-1) + 2(0) = +1$,so $x = +3$.
$5$. The complex is cationic,so the metal name remains 'cobalt'.
$6$. Arranging ligands alphabetically: chloro,bis(ethylenediamine),nitrito-$O$.
$7$. The correct $IUPAC$ name is chlorobis(ethylenediamine)nitrito-$O$-cobalt$(III)$ ion.

(C) The $IUPAC$ name of $[Fe(O_2)(CN)_4Cl]^{4-}$ is Chlorotetracyano superoxoferrate $(II)$ ion.
$1$. Ligands are named in alphabetical order: Chloro $(Cl^-)$,Cyano $(CN^-)$,and Superoxo $(O_2^-)$.
$2$. The prefix 'tetra' is used for four cyanide ligands.
$3$. The complex is anionic,so the metal iron is named as 'ferrate'.
$4$. Oxidation state calculation: $x + (-1) + 4(-1) + (-1) = -4$,which gives $x - 6 = -4$,so $x = +2$.
$5$. Thus,the name is Chlorotetracyano superoxoferrate $(II)$ ion.

(A) In the complex $[RhCl(PPh_3)_3]$,the ligand $Cl^-$ has a charge of $-1$ and $PPh_3$ is a neutral ligand.
Let the oxidation state of $Rh$ be $x$.
$x + (-1) + 3(0) = 0$
$x - 1 = 0$
$x = +1$
Thus,the oxidation state of rhodium is $(I)$.
The name is constructed as: Chlorido (or Chloro) + tris(triphenylphosphine) + rhodium$(I)$.
Therefore,the correct name is Chlorotris(triphenylphosphine)rhodium$(I)$.

(D) The complex is named as diammine dichlorodicyano chromate $(III)$.
$1$. The central metal is Chromium $(Cr)$ with an oxidation state of $+3$.
$2$. The ligands are: two ammine ($NH_3$,neutral),two chloro ($Cl^-$,charge $-1$),and two cyano ($CN^-$,charge $-1$).
$3$. The total charge on the coordination entity is calculated as: $Charge = (Oxidation \ state \ of \ Cr) + (Charge \ of \ ligands) = (+3) + 2(0) + 2(-1) + 2(-1) = 3 - 2 - 2 = -1$.
$4$. Therefore,the formula is $[CrCl_2(CN)_2(NH_3)_2]^{-}$.

(D) To determine the $IUPAC$ name of $K_2[Cr(CN)_2O_2(O_2)NH_3]$,we first calculate the oxidation state of $Cr$.
Let the oxidation state of $Cr$ be $x$.
The charge on $K$ is $+1$,$CN$ is $-1$,$O$ is $-2$,$O_2$ (peroxo) is $-2$,and $NH_3$ is $0$.
$2(+1) + x + 2(-1) + 1(-2) + 1(-2) + 0 = 0$
$2 + x - 2 - 2 - 2 = 0$
$x - 4 = 0$,so $x = +6$.
The complex anion is $[Cr(CN)_2O_2(O_2)NH_3]^{2-}$.
Naming the ligands in alphabetical order: ammine $(NH_3)$,cyano $(CN^-)$,dioxo $(O^{2-})$,and peroxo $(O_2^{2-})$.
Since there are two cyano,two oxo,and one peroxo ligand,the name is Potassium amminedicyanodioxoperoxo chromate $(VI)$.

(D) $H_2[PtCl_6]$ is an acid,not a salt.
According to $IUPAC$ nomenclature rules for coordination compounds,when the complex anion contains the metal,the name of the acid is derived by naming the anion followed by the word acid.
Here,the anion is $[PtCl_6]^{2-}$,which is named hexachloroplatinate $(IV)$.
Therefore,the name is Hexachloroplatinic $(IV)$ acid.
Thus,the correct option is $(D)$.

(A) The complex is $[PtCl(NH_2CH_3)(NH_3)_2]Cl$.
$1$. The ligands are $NH_3$ (ammine),$Cl^-$ (chloro),and $NH_2CH_3$ (methylamine).
$2$. Alphabetical order of ligands: ammine,chloro,methylamine.
$3$. Since there are two $NH_3$ groups,the prefix is 'di'.
$4$. The central metal is platinum $(Pt)$.
$5$. Let the oxidation state of $Pt$ be $x$.
$x + (-1) + 0 + 2(0) = +1$ (since the counter ion is $Cl^-$),so $x = +2$.
$6$. The name is diamminechloromethylamineplatinum $(II)$ chloride.

(C) The $IUPAC$ name of the complex $Na[PtBrCl(NO_2)(NH_3)]$ is determined as follows:
$1$. The cation is $Na^+$ (sodium ion) and the anion is the complex ion $[PtBrCl(NO_2)(NH_3)]^-$.
$2$. The ligands are named in alphabetical order: ammine $(NH_3)$,bromo $(Br^-)$,chloro $(Cl^-)$,and nitro-$N$ $(NO_2^-)$.
$3$. Since the complex ion is an anion,the metal $Pt$ is named as platinate.
$4$. The oxidation state of $Pt$ is calculated: $x + (-1) + (-1) + (-1) + 0 = -1$,so $x = +2$.
$5$. Combining these,the name is sodium amminebromochloronitro-$N$-platinate$(II)$.

(B) The compound $Xe[PtF_6]$ is an ionic complex consisting of the $Xe^+$ cation and the $[PtF_6]^-$ anion.
In the anion $[PtF_6]^-$,let the oxidation state of $Pt$ be $x$.
$x + 6(-1) = -1 \implies x = +5$.
Thus,the anion is named as hexafluoroplatinate$(V)$.
The cation $Xe^+$ is named as xenon.
Therefore,the $IUPAC$ name is Xenon hexafluoroplatinate$(V)$.

(A) $1$. In $Fe(\eta^5-C_5H_5)_2$,the oxidation state of $Fe$ is $+2$,not $0$. The correct name is $Bis(\eta^5-cyclopentadienyl)iron(II)$. Thus,option $A$ is incorrectly written.
$2$. In $Cr(\eta^6-C_6H_6)_2$,the oxidation state of $Cr$ is $0$. The name is correct.
$3$. In $[CoCl_2(H_2O)_4]Cl \cdot 2H_2O$,the oxidation state of $Co$ is $x + 2(-1) + 4(0) = +1$,so $x = +3$. The name is correct.
$4$. In $[Zn(NCS)_4]^{2-}$,the oxidation state of $Zn$ is $x + 4(-1) = -2$,so $x = +2$. The name is correct.

(C) The $IUPAC$ nomenclature rules for coordination compounds require naming the ligand alphabetically followed by the metal name and its oxidation state.
Option $(A)$ is incorrect because the name should be $Tris(acetylacetonato)iron(III) chloride$ (if it were a salt) or $Tris(acetylacetonato)iron(III)$ (if neutral).
Option $(B)$ is incorrect due to improper naming structure for a bridged or complex salt.
Option $(C)$ is correct: $cis$-dichloro(ethylenediamine)platinum $(II)$ follows the alphabetical order of ligands ($chloro$ before $ethylenediamine$) and correctly specifies the geometry and oxidation state.
Option $(D)$ is incorrect due to the lack of proper alphabetical ordering and oxidation state formatting.

(A) The complex compound is $[Cr(NH_3)_5(CN)][Ir(NO_2)_6]$.
$1$. The cation is $[Cr(NH_3)_5(CN)]^{2+}$,named as pentaamminecyanidochromium$(III)$.
$2$. The anion is $[Ir(NO_2)_6]^{3-}$,named as hexanitrito-$N$-iridate$(III)$.
$3$. Combining these,the correct $IUPAC$ name is Pentaamminecyanidochromium$(III)$ hexanitrito-$N$-iridate$(III)$.

(A) To determine the oxidation state of nickel $(Ni)$ in each compound:
$1$. In $Ni(CO)_4$,$CO$ is a neutral ligand,so the oxidation state of $Ni$ is $0$.
$2$. In $(CH_3COO)_2Ni$,the acetate ion $(CH_3COO^-)$ has a charge of $-1$,so $Ni + 2(-1) = 0$,which gives $Ni = +2$.
$3$. In $NiO$,oxygen $(O)$ has a charge of $-2$,so $Ni + (-2) = 0$,which gives $Ni = +2$.
$4$. In $NiCl_2(PPh_3)_2$,$Cl$ has a charge of $-1$ and $PPh_3$ is a neutral ligand,so $Ni + 2(-1) + 2(0) = 0$,which gives $Ni = +2$.
Comparing these values,$Ni(CO)_4$ has the lowest oxidation state of $0$.

(A) In the Brown ring complex $[Fe(H_2O)_5NO]SO_4$,the $NO$ ligand exists as the nitrosonium ion $(NO^+)$.
Let the oxidation state of $Fe$ be $x$.
The charge on $H_2O$ is $0$,and the charge on $SO_4$ is $-2$.
Thus,$x + 5(0) + 1(+1) + (-2) = 0$.
$x + 1 - 2 = 0$.
$x - 1 = 0$.
$x = +1$.
Therefore,the oxidation state of $Fe$ is $+1$.

(D) The given coordination compound is $[Co(NH_3)_4Cl(ONO)]Cl$.
$1$. The ligands are $4$ ammine $(NH_3)$,$1$ chlorido $(Cl^-)$,and $1$ nitrito-$O$ $(ONO^-)$.
$2$. Alphabetical order of ligands: ammine,chlorido,nitrito-$O$.
$3$. The oxidation state of $Co$ is calculated as: $x + 4(0) + 1(-1) + 1(-1) + 1(-1) = 0$,so $x = +3$.
$4$. Combining these,the name is Tetraamminechloridonitrito-$O$-cobalt $(III)$ chloride.

(C) Let the oxidation state of $Fe$ be $x$.
The oxidation state of $Na$ is $+1$,$NH_3$ is $0$,$CN^-$ is $-1$,and $H_2O$ is $0$.
The sum of oxidation states in a neutral complex is zero:
$3(+1) + x + 1(0) + 5(-1) + 2(0) = 0$
$3 + x - 5 = 0$
$x - 2 = 0$
$x = +2$
Therefore,the oxidation state of $Fe$ is $+2$.

(B) Let the oxidation state of $Pt$ be $x$.
$C_2H_4$ (ethylene) is a neutral ligand,so its charge is $0$.
$Cl$ is a chloro ligand with a charge of $-1$.
The overall charge on the complex $[Pt(C_2H_4)Cl_3]^-$ is $-1$.
Setting up the equation: $x + (0) + 3(-1) = -1$.
$x - 3 = -1$.
$x = +2$.
Therefore,the oxidation state of $Pt$ is $+2$.

(A) In the brown ring complex $[Fe(H_2O)_5NO]SO_4$,the complex ion is $[Fe(H_2O)_5NO]^{2+}$.
In this complex,$NO$ exists as $NO^+$. The oxidation state of $H_2O$ is $0$.
Let the oxidation state of $Fe$ be $x$.
$x + 5(0) + 1(+1) = +2$
$x + 1 = +2$
$x = +1$
Therefore,the oxidation state of $Fe$ in the brown ring complex is $+1$.

(C) $1$. Identify the ligands: $NH_3$ is 'ammine' and $Cl^-$ is 'chlorido'.
$2$. Determine the oxidation state of the central metal atom $Cr$: Let the oxidation state be $x$. The charge on $NH_3$ is $0$,$Cl$ is $-1$,and the counter ion $NO_3$ is $-1$. Thus,$x + 4(0) + 2(-1) + (-1) = 0$,which gives $x - 3 = 0$,so $x = +3$.
$3$. Apply $IUPAC$ nomenclature rules: The ligands are listed alphabetically. 'Ammine' comes before 'chlorido'. Since there are four $NH_3$ and two $Cl$,the prefix is 'tetraammine' and 'dichlorido'.
$4$. Combine the parts: The complex cation is named first,followed by the anion. The name is 'tetraamminedichloridochromium $(III)$ nitrate'.

(B) $1$. Identify the ligands: $NH_3$ is ammine,$Cl^-$ is chloro,and $NO_2^-$ is nitro.
$2$. Arrange ligands alphabetically: ammine,chloro,nitro.
$3$. The coordination entity is $[Pt(NH_3)_4 NO_2Cl]^{2+}$,which is a cation.
$4$. The oxidation state of $Pt$ is calculated as: $x + 4(0) + (-1) + (-1) = +2$,so $x = +4$.
$5$. Combining these,the name is Tetraamminechloronitroplatinum$(IV)$ sulfate.

(B) $1$. The name 'dichlorobis(urea)copper$(II)$' indicates a coordination complex.
$2$. 'Copper$(II)$' means the central metal ion is $Cu^{2+}$.
$3$. 'Dichlorido' means two chloride ligands $(Cl^-)$ are attached.
$4$. 'Bis(urea)' means two urea molecules $(O=C(NH_2)_2)$ are attached as ligands.
$5$. Combining these,the coordination sphere is $[Cu(urea)_2Cl_2]$.
$6$. The oxidation state of $Cu$ is $x + 2(0) + 2(-1) = 0$,so $x = +2$,which matches the name.

(A) The given complex is $Fe_2(CO)_9$,which can be represented as $(CO)_3Fe(\mu-CO)_3Fe(CO)_3$.
In this structure,there are $3$ bridging carbonyl groups $(\mu-CO)$ and $6$ terminal carbonyl groups $(CO)$.
Each iron atom is bonded to $3$ terminal carbonyls and $3$ bridging carbonyls.
The $IUPAC$ name is constructed by identifying the bridging ligands first,followed by the metal centers.
The correct name is $Tri-\mu-carbonyl-bis(tricarbonyliron(0))$.
Thus,the correct option is $A$.

(A) The given complex is $Hg[Co(CNS)_4]$.
In this complex,the cation is $Hg^{2+}$ and the anion is $[Co(CNS)_4]^{2-}$.
For the anion $[Co(CNS)_4]^{2-}$,the ligand is $CNS^-$,which is thiocyanate. Since it is bonded through the sulfur atom,it is named as thiocyanato-$S$.
There are $4$ such ligands,so the prefix is 'tetra'.
The central metal atom is Cobalt,and since the complex ion is anionic,it is named as 'cobaltate'.
The oxidation state of $Co$ is calculated as: $x + 4(-1) = -2$,which gives $x = +2$.
Thus,the name is Mercury tetrathiocyanato-$S$-cobaltate$(II)$.

(B) The compound $Fe(C_5H_5)_2$ is an organometallic sandwich complex.
According to $IUPAC$ nomenclature for coordination compounds,the ligand $C_5H_5^-$ is named as 'cyclopentadienyl'.
Since there are two such ligands,the prefix 'bis' is used.
The central metal atom is iron,and its oxidation state is calculated as $x + 2(-1) = 0$,so $x = +2$.
Thus,the correct $IUPAC$ name is 'bis(cyclopentadienyl)iron $(II)$'.

(D) $1$. The complex is $K_3[Ir(C_2O_4)_3]$.
$2$. The cation is $K^+$,named as Potassium.
$3$. The coordination sphere is $[Ir(C_2O_4)_3]^{3-}$.
$4$. The ligand is oxalate $(C_2O_4^{2-})$,which is a bidentate ligand. Since there are $3$ oxalate ligands,we use the prefix 'tris' instead of 'tri'.
$5$. Since the complex ion is anionic,the central metal $Ir$ is named as Iridate.
$6$. The oxidation state of $Ir$ is calculated as: $x + 3(-2) = -3$,so $x - 6 = -3$,$x = +3$.
$7$. Combining these,the $IUPAC$ name is Potassium tris(oxalato)iridate$(III)$.

(A) $1$. Identify the central metal ion: Cobalt $(Co)$ with oxidation state $+III$.
$2$. Identify the ligands: 'triammine' means $3 \ NH_3$,'diaqua' means $2 \ H_2O$,and 'chloro' means $1 \ Cl^-$.
$3$. The coordination sphere is $[Co(NH_3)_3(H_2O)_2Cl]$.
$4$. Calculate the charge on the coordination sphere: $x + 3(0) + 2(0) + 1(-1) = +3$,so $x = +4$. Wait,let's re-evaluate: The oxidation state of $Co$ is $+3$. The sum of charges is $3 + 0 + 0 - 1 = +2$. Thus,the complex ion is $[Co(NH_3)_3(H_2O)_2Cl]^{2+}$.
$5$. To balance the $+2$ charge,we need $2 \ Cl^-$ ions outside the coordination sphere.
$6$. The final formula is $[Co(NH_3)_3(H_2O)_2Cl]Cl_2$.

(D) The chemical formula of sodium nitroprusside is $Na_2[Fe(CN)_5(NO)]$.
In this complex,the ligand $NO$ is treated as $NO^+$ (nitrosonium ion) because the iron is in the $+2$ oxidation state.
Let the oxidation state of $Fe$ be $x$.
$2(+1) + x + 5(-1) + 1(+1) = 0$
$2 + x - 5 + 1 = 0$
$x - 2 = 0$
$x = +2$.
Thus,the name is sodium pentacyanonitrosylferrate $(II)$.

(B) $1$. Identify the cation: $K^+$ is potassium.
$2$. Identify the complex anion: $[Zn(OH)_4]^{2-}$.
$3$. Name the ligands: Four $OH^-$ groups are named as 'tetrahydroxo'.
$4$. Name the central metal: Since the complex is an anion,$Zn$ is named as 'zincate'.
$5$. Determine the oxidation state of $Zn$: $x + 4(-1) = -2$,so $x = +2$.
$6$. Combine the parts: Potassium tetrahydroxozincate $(II)$.

(C) $1$. Identify the ligands: $NH_3$ (ammine),$CN^-$ (cyano),$O^{2-}$ (oxo),$O_2^{2-}$ (peroxo).
$2$. Determine the oxidation state of $Cr$: Let it be $x$. The charge on $K$ is $+1$,$CN$ is $-1$,$O$ is $-2$,$O_2$ is $-2$,and $NH_3$ is $0$.
$2(+1) + x + 2(-1) + 1(-2) + 1(-2) + 0 = 0$
$2 + x - 2 - 2 - 2 = 0$
$x - 4 = 0 \implies x = +4$.
$3$. Order the ligands alphabetically: ammine,cyano,dioxo,peroxo.
$4$. Combine: Potassium amminedicyanodioxoperoxochromate $(IV)$.

(C) The given complex is a metal-metal bonded carbonyl complex $[(CO)_5Mn-Mn(CO)_5]$.
In this complex,the manganese atoms are in the $0$ oxidation state.
According to $IUPAC$ nomenclature for metal-metal bonded carbonyls,it is named as decacarbonyl dimanganese $(0)$.
However,looking at the options provided,the most appropriate name describing the structure is bis(pentacarbonyl manganese).

(B) In the reactant complex $Na_2[Fe(CN)_5NO]$,the oxidation state of $Fe$ is calculated as: $2(+1) + x + 5(-1) + 0 = 0$ (assuming $NO$ is neutral $NO^+$ or $NO$ depending on convention,but standard nitroprusside has $Fe$ in $+2$ state). $x = +2$.
In the product complex $Na_4[Fe(CN)_5NOS]$,the ligand $NOS$ is $NOS^{3-}$. The oxidation state of $Fe$ is calculated as: $4(+1) + x + 5(-1) + (-2) = 0$ $\Rightarrow x - 3 = 0$ $\Rightarrow x = +3$.
However,in the specific reaction of sodium nitroprusside with sulfide,the $Fe$ center remains in the $+2$ oxidation state as the reaction involves the conversion of $NO^+$ to $NOS^-$. Thus,the oxidation state of $Fe$ in both complexes is $+2$.

(A) In the complex $[RhCl(PPh_3)_3]$,the ligand $Cl^-$ has a charge of $-1$ and $PPh_3$ (triphenylphosphine) is a neutral ligand.
Let the oxidation state of $Rh$ be $x$.
$x + (-1) + 3(0) = 0$
$x - 1 = 0$
$x = +1$.
Therefore,the oxidation state of $Rh$ is $(I)$.
The name is constructed as: Chlorido (for $Cl^-$) + tris(triphenylphosphine) (for three $PPh_3$ ligands) + rhodium$(I)$ (metal name with oxidation state).
Thus,the correct $IUPAC$ name is Chloridotris(triphenylphosphine)rhodium$(I)$.