Nomenclature and oxidation State Questions in English

(A) The name "dichlorido-bis-(ethane-$1, 2$-diamine)-platinum$(IV)$ nitrate" indicates the following components:
$1$. Central metal: $Pt$ (Platinum) with oxidation state $+4$.
$2$. Ligands: Two chloride ions $(Cl^-)$ and two ethane-$1, 2$-diamine $(en)$ molecules.
$3$. Counter ion: Nitrate $(NO_3^-)$.
Calculation of oxidation state for $[PtCl_2(en)_2](NO_3)_2$:
Let the oxidation state of $Pt$ be $x$.
$x + 2(-1) + 2(0) + 2(-1) = 0$
$x - 2 - 2 = 0$
$x = +4$.
Thus,the complex is $[PtCl_2(en)_2](NO_3)_2$.

(A) $1$. Identify the ligands: $NH_3$ is ammine and $CO_3^{2-}$ is carbonato. Since there are $5$ ammine ligands,we use the prefix 'penta'.
$2$. Determine the oxidation state of the central metal atom $(Co)$: Let the oxidation state be $x$. The charge on $NH_3$ is $0$,$CO_3$ is $-2$,and $Cl$ is $-1$. Thus,$x + 5(0) + (-2) + (-1) = 0$,which gives $x = +3$.
$3$. Assemble the name: The ligands are listed alphabetically (ammine before carbonato). The name is Pentaamminocarbonatocobalt$(III)$ chloride.

(C) For $K_3[Cr(C_2O_4)_3]$: Oxidation state $= +3$,Coordination number $= 6$,Sum $= 3 + 6 = 9$
For $[Cr(CO)_6]$: Oxidation state $= 0$,Coordination number $= 6$,Sum $= 0 + 6 = 6$
For $K_2[PtCl_6]$: Oxidation state $= +4$,Coordination number $= 6$,Sum $= 4 + 6 = 10$
For $K_4[Fe(CN)_6]$: Oxidation state $= +2$,Coordination number $= 6$,Sum $= 2 + 6 = 8$
Thus,the sum is maximum for $K_2[PtCl_6]$.

(C) For $[Pt(NH_3)_2Cl(NO_2)]$:
Let the oxidation number of $Pt = x$.
$NH_3$ is a neutral ligand $(0)$,$Cl^-$ is $-1$,and $NO_2^-$ is $-1$.
$x + 2(0) + (-1) + (-1) = 0$
$x - 2 = 0 \implies x = +2$.
For $[CoCl_2(en)_2]^{\oplus}$:
Let the oxidation number of $Co = y$.
$en$ (ethylenediamine) is a neutral ligand $(0)$ and $Cl^-$ is $-1$.
$y + 2(-1) + 2(0) = +1$
$y - 2 = +1 \implies y = +3$.
Thus,the oxidation states are $+2$ and $+3$ respectively.

(C) The given compound is $[Ag(NH_3)_2][Ag(CN)_2]$.
In this coordination entity,the cation is $[Ag(NH_3)_2]^+$ and the anion is $[Ag(CN)_2]^-$.
For the cation $[Ag(NH_3)_2]^+$,the name is diammine silver $(I)$.
For the anion $[Ag(CN)_2]^-$,the name is dicyanoargentate $(I)$.
Combining these,the $IUPAC$ name is diammine silver $(I)$ dicyanoargentate $(I)$.

(D) The name tris(ethane-$1,2$-diamine)cobalt$(III)$ sulphate indicates a coordination complex.
$1$. The central metal is Cobalt $(Co)$ with an oxidation state of $+3$.
$2$. The ligand is ethane-$1,2$-diamine,abbreviated as $en$,which is a neutral bidentate ligand. Since there are three such ligands,the coordination sphere is $[Co(en)_3]^{3+}$.
$3$. The anion is sulphate,$SO_4^{2-}$.
$4$. To balance the charges,we need two $[Co(en)_3]^{3+}$ cations and three $SO_4^{2-}$ anions.
$5$. Thus,the chemical formula is $[Co(en)_3]_2(SO_4)_3$.
Hence,option $D$ is the correct answer.

(C) The rules for naming a coordination compound are:
$1.$ Name the counterion first: $Ammonium$.
$2.$ Name the ligands in alphabetical order: $diaqua$ $(H_2O)$ and $oxalato$ $(C_2O_4^{2-})$. Since $oxalato$ is a polydentate ligand,we use $bis$ for the prefix.
$3.$ Name the central metal atom: Since the complex is anionic,we add the suffix $-ate$ to $Nickel$,resulting in $nickelate$.
$4.$ Determine the oxidation state of $Ni$: Let $x$ be the oxidation state of $Ni$. $2(+1) + x + 2(-2) + 2(0) = 0 \implies 2 + x - 4 = 0 \implies x = +2$.
$5.$ Combine the parts: $Ammonium$ $diaquabis(oxalato)nickelate(II)$.

(C) When $Pt$ reacts with a $3:1$ mixture of concentrated hydrochloric acid $(HCl)$ and concentrated nitric acid $(HNO_3)$, it forms a solution known as aqua regia. Aqua regia is capable of dissolving platinum by forming chlorocomplexes. The reaction typically forms $H_2[PtCl_6]$.
In this complex, the $Pt$ is in the $+4$ oxidation state. The balanced chemical equation for the reaction is:
$Pt + 4HNO_3 + 6HCl \rightarrow H_2[PtCl_6] + 4NO_2 + 4H_2O$
So, in $[PtCl_6]^{2-}$ the oxidation state of $Pt$ is:
$x + 6(-1) = -2$
$x - 6 = -2$
$x = +4$

(D) $1$. Identify the ligands: There are four $NH_3$ (ammine) ligands and two $Cl^-$ (chloro) ligands attached to the central $Cr$ atom.
$2$. Determine the geometry: The two $Cl^-$ ligands are adjacent to each other (at $90^\circ$ angle), which indicates a $cis$ configuration.
$3$. Apply $IUPAC$ nomenclature rules:
- List ligands alphabetically: $ammine$ comes before $chloro$.
- Use prefixes for the number of ligands: $tetra$ for four $NH_3$ and $di$ for two $Cl$.
- The name is $cis-tetraamminodichlorochromium(III)$ chloride.
$4$. Oxidation state of $Cr$: $x + 4(0) + 2(-1) = +1$ (since $Cl^-$ is outside the coordination sphere), so $x = +3$.

(A) $1$. The given coordination compound is $\text{Hg[Co(SCN)}_4]$.
$2$. The compound dissociates into $\text{Hg}^{2+}$ and $\text{[Co(SCN)}_4]^{2-}$.
$3$. In the complex anion $\text{[Co(SCN)}_4]^{2-}$, the ligand is thiocyanate $(SCN^-)$, which is bonded through the sulfur atom, hence it is named 'thiocyanato-$S$'.
$4$. Let the oxidation state of cobalt $(Co)$ be $x$. Since the charge on the complex is $-2$ and the charge on $SCN^-$ is $-1$, we have: $x + 4(-1) = -2$, which gives $x = +2$.
$5$. The cation is mercury, which is in the $+2$ oxidation state, so it is named 'Mercury$(II)$'.
$6$. The complex anion is named 'tetrathiocyanato-$S$-cobaltate$(II)$' because the metal is in an anionic complex.
$7$. Combining these, the correct $IUPAC$ name is 'Mercury$(II)$ tetrathiocyanato-$S$-cobaltate$(II)$'.

(D) To determine the formula of the coordination compound tetraammineaquachloridocobalt$(III)$ chloride:
$1$. The central metal ion is Cobalt$(III)$, denoted as $Co^{3+}$.
$2$. The ligands are: 'tetraammine' $(4 \times NH_3)$,'aqua' $(1 \times H_2O)$, and 'chlorido' $(1 \times Cl^-)$.
$3$. The coordination sphere is $[Co(NH_3)_4(H_2O)Cl]$.
$4$. Calculate the charge on the coordination sphere: $Charge = (+3) + 4(0) + 1(0) + 1(-1) = +2$.
$5$. To balance the $+2$ charge of the complex ion, two chloride ions $(Cl^-)$ are required outside the coordination sphere.
$6$. Thus, the formula is $[Co(NH_3)_4(H_2O)Cl]Cl_2$.

(D) Step $1$: Calculate the oxidation state of the metal in each species.
Step $2$: For $MnO_4^{-}$, let the oxidation state of $Mn$ be $x$. $x + 4(-2) = -1 \implies x = +7$.
Step $3$: For $[Cr(CN)_6]^{3-}$, let the oxidation state of $Cr$ be $x$. $x + 6(-1) = -3 \implies x = +3$.
Step $4$: For $Cr_2O_3$, let the oxidation state of $Cr$ be $x$. $2x + 3(-2) = 0 \implies 2x = 6 \implies x = +3$.
Step $5$: For $CrO_2Cl_2$, let the oxidation state of $Cr$ be $x$. $x + 2(-2) + 2(-1) = 0 \implies x - 4 - 2 = 0 \implies x = +6$.
Step $6$: Thus, $CrO_2Cl_2$ contains the metal in the $+6$ oxidation state.

(C) $1$. The central metal ion is $\text{Platinum}$ in oxidation state $+IV$, denoted as $Pt^{4+}$.
$2$. $\text{Bis(ethylenediamine)}$ indicates two $en$ ligands, which are neutral ($0$ charge).
$3$. $\text{Dithiocyanato}$ indicates two $SCN^-$ ligands, each with a $-1$ charge.
$4$. The total charge on the complex ion is calculated as: $\text{Charge} = \text{Oxidation state of } Pt + (2 \times \text{charge of } en) + (2 \times \text{charge of } SCN^-)$.
$5$. $\text{Charge} = (+4) + (2 \times 0) + (2 \times -1) = +4 - 2 = +2$.
$6$. Thus, the formula is $[Pt(en)_2(SCN)_2]^{2+}$.

(A) $1$. The coordination number of $M$ is calculated by considering the denticity of the ligands. $en$ (ethylenediamine) is a bidentate ligand and $C_2O_4^{2-}$ (oxalate) is also a bidentate ligand.
$2$. Coordination number $= (2 \times 2) + (1 \times 2) = 4 + 2 = 6$.
$3$. Let the oxidation number of $M$ be $x$. The charge on $en$ is $0$, the charge on $C_2O_4^{2-}$ is $-2$, and the charge on $Cl^-$ is $-1$.
$4$. The total charge of the complex $[M(en)_2C_2O_4]Cl$ is $0$. Thus, $x + 2(0) + 1(-2) - 1 = 0$.
$5$. $x - 2 - 1 = 0 \implies x = +3$.
$6$. The sum of the coordination number and the oxidation number $= 6 + 3 = 9$.

(B) $1$. The complex $[Co(NH_3)_5Br]SO_4$ dissociates into $[Co(NH_3)_5Br]^{2+}$ and $SO_4^{2-}$.
$2$. Let the oxidation state of $Co$ be $x$.
$3$. The oxidation state of $NH_3$ is $0$ and $Br$ is $-1$.
$4$. The charge on the complex ion $[Co(NH_3)_5Br]^{2+}$ is $+2$.
$5$. Setting up the equation: $x + 5(0) + (-1) = +2$.
$6$. Solving for $x$: $x - 1 = +2$, which gives $x = +3$.

(C) Let the oxidation state of cobalt $(Co)$ be $x$.
The oxidation state of the neutral ligand ammonia $(NH_3)$ is $0$.
The overall charge on the complex ion $[Co(NH_3)_6]^{3+}$ is $+3$.
Setting up the equation: $x + 6(0) = +3$.
Solving for $x$: $x = +3$.
Therefore, the oxidation state of cobalt is $+3$.