Nomenclature and oxidation State Questions in English

(B) Let the oxidation number of $Cr$ be $x$.
In the complex $[Cr(H_2O)_4Cl_2]^+$,the oxidation state of $H_2O$ is $0$ and $Cl$ is $-1$.
The overall charge on the complex is $+1$.
Therefore,the equation is: $x + 4(0) + 2(-1) = +1$.
$x - 2 = +1$.
$x = +3$.
Thus,the oxidation number of $Cr$ is $+3$.

(C) The complex is $K[Co(CO)_4]$.
Let the oxidation state of $Co$ be $x$.
The oxidation state of $K$ is $+1$ and the ligand $CO$ (carbonyl) is neutral $(0)$.
Setting the sum of oxidation states to zero: $1 + x + 4(0) = 0$.
Solving for $x$: $x = -1$.

(A) In the coordination compound $Ni(CO)_4$,the ligand is carbonyl $(CO)$,which is a neutral ligand with an oxidation state of $0$.
Let the oxidation state of $Ni$ be $x$.
The sum of oxidation states of all atoms in a neutral complex is $0$.
$x + 4 \times (0) = 0$
$x = 0$.
Therefore,the oxidation number of nickel in $Ni(CO)_4$ is $0$.

(A) Let the oxidation state of $Cr$ be $x$.
The oxidation state of $NH_3$ is $0$ and $Cl$ is $-1$.
The overall charge on the complex $[Cr(NH_3)_4Cl_2]^+$ is $+1$.
Therefore,$x + 4(0) + 2(-1) = +1$.
$x - 2 = +1$.
$x = +3$.

(D) Let the oxidation number of iron $(Fe)$ be $x$.
In the complex $K_4[Fe(CN)_6]$,the oxidation number of potassium $(K)$ is $+1$ and the cyanide ligand $(CN^-)$ is $-1$.
The sum of oxidation numbers in a neutral compound is $0$.
$4(+1) + x + 6(-1) = 0$
$4 + x - 6 = 0$
$x - 2 = 0$
$x = +2$.
Therefore,the oxidation state of iron is $+2$.

(B) In the brown ring complex $[Fe(H_2O)_5NO]SO_4$,the oxidation state of iron is calculated by considering the ligands.
Water $(H_2O)$ is a neutral ligand with a charge of $0$.
The nitric oxide $(NO)$ ligand in this specific complex acts as $NO^+$.
The sulfate ion $(SO_4^{2-})$ has a charge of $-2$.
Let the oxidation state of $Fe$ be $x$.
$x + 5(0) + 1(+1) + (-2) = 0$
$x + 1 - 2 = 0$
$x - 1 = 0$
$x = +1$.
Therefore,the oxidation state of iron is $+1$.

(A) Let the oxidation number of $Co$ be $x$.
The oxidation number of $NH_3$ is $0$,$Cl^-$ is $-1$,and $NO_2^-$ is $-1$.
The sum of oxidation numbers in a neutral complex is $0$.
$x + 4(0) + (-1) + (-1) = 0$
$x - 2 = 0$
$x = +2$.

(D) The complex is $K_4[Ni(CN)_4]$.
Let the oxidation number of $Ni$ be $x$.
The oxidation number of $K$ is $+1$ and $CN$ is $-1$.
Applying the rule: $4 \times (+1) + x + 4 \times (-1) = 0$.
$4 + x - 4 = 0$.
$x = 0$.
Therefore,the oxidation number of nickel is $0$.

(B) To find the oxidation number of $Fe$ in ${K_3}[Fe{(CN)_6}]$,let the oxidation state of $Fe$ be $x$.
The oxidation state of $K$ is $+1$ and the oxidation state of $CN^-$ is $-1$.
The 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 number of $Fe$ is $+3$.

(A) For $H_2S_2O_7$:
$2(+1) + 2(x) + 7(-2) = 0$
$2 + 2x - 14 = 0$
$2x = 12$
$x = + 6$ for $S$.
For $K_4[Fe(CN)_6]$:
$4(+1) + x + 6(-1) = 0$
$4 + x - 6 = 0$
$x = + 2$ for $Fe$.
Thus,the oxidation numbers are $+ 6$ and $+ 2$ respectively.

(A) The chemical formula for potassium ferrocyanide is $K_4[Fe(CN)_6]$.
Let the oxidation number of iron $(Fe)$ be $x$.
The oxidation number of potassium $(K)$ is $+1$ and the cyanide ligand $(CN)$ is $-1$.
Applying the rule that the sum of oxidation states in a neutral complex is zero:
$4(+1) + x + 6(-1) = 0$
$4 + x - 6 = 0$
$x - 2 = 0$
$x = +2$.
Therefore,the oxidation number of iron is $2$.

(D) The $IUPAC$ name of $K_3[Fe(CN)_6]$ is Potassium hexacyanoferrate$(III)$.
It is also commonly known as Potassium ferricyanide.
It is historically referred to as Red prussiate of potash.
Therefore,all the given names are correct.

(A) The compound $[Fe(H_2O)_5(NO)]SO_4$ dissociates into $[Fe(H_2O)_5(NO)]^{2+}$ and $SO_4^{2-}$ ions.
In the complex ion $[Fe(H_2O)_5(NO)]^{2+}$,the ligand $H_2O$ is neutral (oxidation state $0$).
The ligand $NO$ in the brown ring complex is considered to be in the $NO^+$ form (nitrosonium ion),which has an oxidation state of $+1$.
Let the oxidation state of $Fe$ be $x$.
$x + 5(0) + 1 = +2$
$x = +2 - 1 = +1$.
Therefore,the oxidation state of $Fe$ is $+1$.

(C) To name the coordination compound $[Pt(NH_3)_3(Br)(NO_2)Cl]Cl$,we follow the $IUPAC$ rules:
$1$. Name the ligands in alphabetical order: Ammine $(NH_3)$,Bromo $(Br^-)$,Chloro $(Cl^-)$,Nitro $(NO_2^-)$.
$2$. The prefix for three $NH_3$ ligands is 'triammine'.
$3$. Combining these,we get 'triamminebromochloronitro'.
$4$. The metal is platinum $(Pt)$. Since the complex is cationic,we use the name 'platinum'.
$5$. Calculate the oxidation state of $Pt$: $x + 3(0) + 1(-1) + 1(-1) + 1(-1) = +1$ (for the outer $Cl^-$),so $x - 3 = 1$,which gives $x = +4$.
$6$. The final name is Triamminebromochloronitroplatinum $(IV)$ chloride.

(B) In this coordination complex,the central metal ion is $Cu^{2+}$.
There are $2$ chloro ligands $(Cl^-)$ and $2$ urea ligands $(O=C(NH_2)_2)$.
According to $IUPAC$ nomenclature rules for coordination compounds,ligands are named in alphabetical order.
'Chloro' comes before 'urea'.
Since there are $2$ urea ligands,the prefix 'bis' is used for the ligand name 'urea'.
Thus,the formula is written as $[CuCl_2\{O=C(NH_2)_2\}_2]$.
Therefore,the correct option is $B$.

(D) $1$. Identify the ligands: $NH_3$ is named as 'ammine' and $Cl^-$ is named as 'chlorido'.
$2$. Determine the oxidation state of the central metal atom $Pt$: Let the oxidation state be $x$. Then $x + 2(0) + 2(-1) = 0$,which gives $x = +2$.
$3$. Arrange the ligands alphabetically: 'ammine' comes before 'chlorido'.
$4$. Combine the parts: 'diammine' (for two $NH_3$) and 'dichlorido' (for two $Cl^-$).
$5$. The final $IUPAC$ name is 'diamminedichloridoplatinum$(II)$'.

(C) The name 'diammine silver $(I)$ chloride' indicates a coordination compound.
'Diammine' refers to two $NH_3$ ligands attached to the central metal ion.
'Silver $(I)$' indicates the central metal is $Ag^+$.
'Chloride' is the counter ion outside the coordination sphere.
Combining these,the coordination entity is $[Ag(NH_3)_2]^+$,and with the chloride ion,the formula becomes $[Ag(NH_3)_2]Cl$.

(A) $1$. The complex is a salt consisting of a cationic part $[Pt(NH_3)_4Cl_2]^{2+}$ and an anionic part $[PtCl_4]^{2-}$.
$2$. In the cation $[Pt(NH_3)_4Cl_2]^{2+}$,let the oxidation state of $Pt$ be $x$. Then $x + 4(0) + 2(-1) = +2$,which gives $x = +4$. The name is tetraamminedichloroplatinum $(IV)$.
$3$. In the anion $[PtCl_4]^{2-}$,let the oxidation state of $Pt$ be $y$. Then $y + 4(-1) = -2$,which gives $y = +2$. Since it is an anion,the metal is named platinate $(II)$.
$4$. Combining these,the correct name is tetraamminedichloroplatinum $(IV)$ tetrachloroplatinate $(II)$.

(A) Potassium ferrocyanide is a coordination compound containing the hexacyanidoferrate$(II)$ ion.
In this complex,the iron atom is in the $+2$ oxidation state.
According to the $IUPAC$ nomenclature rules,the formula is $K_4[Fe(CN)_6]$.

(A) $1$. Identify the ligands: $NH_3$ is ammine and $NO_2^-$ is nitro-$N$.
$2$. Determine the oxidation state of the central metal atom $(Co)$: Let $x$ be the oxidation state of $Co$. $x + 3(0) + 3(-1) = 0$,so $x = +3$.
$3$. Apply $IUPAC$ naming rules for coordination compounds: List ligands in alphabetical order (ammine before nitro-$N$).
$4$. The name is Triamminetrinitro-$N$-cobalt $(III)$.

(B) The chemical formula for sodium tetrafluoro oxochromate is $Na_2[CrF_4O]$.
Let the oxidation number of chromium be $x$.
The oxidation state of sodium $(Na)$ is $+1$,fluorine $(F)$ is $-1$,and oxygen $(O)$ is $-2$.
Setting the sum of oxidation states to zero: $2(+1) + x + 4(-1) + (-2) = 0$.
$2 + x - 4 - 2 = 0$.
$x - 4 = 0$.
$x = +4$.

(A) To name the coordination compound $K_4[Fe(CN)_6]$ according to $IUPAC$ rules:
$1$. Name the cation first: $Potassium$.
$2$. Name the ligands in alphabetical order: $hexacyano$ (for $6$ $CN^-$ ions).
$3$. Name the central metal atom: Since the complex ion $[Fe(CN)_6]^{4-}$ is anionic,the metal $Fe$ is named as $ferrate$.
$4$. Determine the oxidation state of $Fe$: $x + 6(-1) = -4$,so $x = +2$. Thus,it is $ferrate(II)$.
Combining these,the name is $Potassium hexacyanoferrate(II)$.

(B) In the complex $[Ni(CO)_4]$,$CO$ is a neutral ligand with an oxidation state of $0$.
Let the oxidation state of $Ni$ be $x$.
$x + 4(0) = 0$,which gives $x = 0$.
Since the complex is neutral,the metal is named as nickel,not nickelate.
Therefore,the $IUPAC$ name is Tetracarbonylnickel $(0)$.

(C) The compound $Fe_4[Fe(CN)_6]_3$ is a coordination compound.
In this complex,the anion is $[Fe(CN)_6]^{4-}$ and the cation is $Fe^{3+}$.
According to $IUPAC$ nomenclature rules for coordination compounds,the cation is named first,followed by the anion.
The oxidation state of $Fe$ in the anion $[Fe(CN)_6]^{4-}$ is calculated as: $x + 6(-1) = -4$,which gives $x = +2$.
Thus,the anion is hexacyanoferrate $(II)$.
The cation $Fe^{3+}$ is named as Iron $(III)$.
Therefore,the correct name is Iron $(III)$ hexacyanoferrate $(II)$.

(D) The compound is $Na_3[Co(ONO)_6]$.
$1$. The cation is $Na^+$,named as $Sodium$.
$2$. The coordination entity is $[Co(ONO)_6]^{3-}$.
$3$. The ligand $ONO^-$ is named as $nitrito-O$ or simply $nitrito$.
$4$. Since the coordination entity is anionic,the central metal $Co$ is named as $cobaltate$.
$5$. The oxidation state of $Co$ is calculated as: $x + 6(-1) = -3$,so $x = +3$.
$6$. Combining these,the $IUPAC$ name is $Sodium \ hexanitritocobaltate(III)$.

(D) Let the oxidation number of $Cr$ be $x$.
The oxidation number of $NH_3$ is $0$ and the oxidation number of $Cl$ is $-1$.
The sum of oxidation numbers in a neutral complex is $0$.
$x + 6 \times (0) + 3 \times (-1) = 0$
$x - 3 = 0$
$x = +3$
Therefore,the oxidation number of $Cr$ is $+3$.

(B) The complex is $[Co(ONO)(NH_3)_5]Cl_2$.
$1$. The ligand $NH_3$ is named as 'ammine' and there are $5$ of them,so it is 'pentaammine'.
$2$. The ligand $ONO^-$ is an ambidentate ligand attached through oxygen,named as 'nitrito-$O$' or simply 'nitrito'.
$3$. The central metal is Cobalt $(Co)$.
$4$. The oxidation state of $Co$ is calculated as: $x + (-1) + 5(0) = +2$,so $x = +3$.
$5$. The counter ion is chloride $(Cl^-)$.
Combining these,the $IUPAC$ name is 'Pentaammine nitrito cobalt $(III)$ chloride'.

(A) The complex compound $[Co(NH_3)_6]Cl_3$ dissociates as $[Co(NH_3)_6]Cl_3 \to [Co(NH_3)_6]^{3+} + 3Cl^-$.
Let the oxidation state of $Co$ be $x$.
The oxidation state of the neutral ligand $NH_3$ is $0$.
Therefore,$x + 6(0) = +3$,which gives $x = +3$.
Thus,the oxidation state of cobalt is $+3$.

(B) The chemical formula for potassium cuprochloride is $K_3[CuCl_4]$.
According to $IUPAC$ nomenclature rules for coordination compounds:
$1$. The cation (potassium) is named first.
$2$. The ligand (chloro) is named with the prefix 'tetra' for four ligands.
$3$. The central metal atom (copper) is named as 'cuprate' because the complex ion $[CuCl_4]^{3-}$ is anionic.
$4$. The oxidation state of copper is calculated as: $x + 4(-1) = -3$,so $x = +1$.
Thus,the correct name is Potassium tetrachlorocuprate $(I)$.

(D) $1$. Identify the cation: $K^+$ is potassium.
$2$. Identify the coordination sphere: $[Fe(CN)_6]^{3-}$.
$3$. Name the ligands: Six $CN^-$ ligands are named as 'hexacyano'.
$4$. Name the central metal: Since the complex ion is anionic,$Fe$ is named as 'ferrate'.
$5$. Determine the oxidation state of $Fe$: $x + 6(-1) = -3$,so $x = +3$.
$6$. Combine: The name is Potassium hexacyanoferrate $(III)$.

(B) Let the oxidation state of $Ni$ be $x$.
In $K_4[Ni(CN)_4]$,the oxidation state of $K$ is $+1$ and $CN$ is $-1$.
The sum of oxidation states of all atoms in a neutral complex is $0$.
$4(+1) + x + 4(-1) = 0$
$4 + x - 4 = 0$
$x = 0$
Therefore,the oxidation state of $Ni$ is $0$.

(A) The chemical formula for sodium nitroprusside is $Na_2[Fe(CN)_5(NO)] \cdot 2H_2O$.
In this complex,the ligand $NO$ is treated as $NO^+$ (nitrosonium ion).
The oxidation state of $Fe$ is calculated as: $2(+1) + x + 5(-1) + (+1) = 0$,which gives $x = +2$.
Thus,the central metal is $Fe(II)$.
The ligands are five cyanide ions (pentacyanido) and one nitrosyl group (nitrosyl).
According to $IUPAC$ rules,the name is sodium pentacyanidonitrosylferrate$(II)$.

(A) In the complex $Ni(CO)_4$,the ligand $CO$ (carbonyl) is a neutral ligand with an oxidation state of $0$.
Therefore,the oxidation number of $Ni$ is calculated as: $x + 4(0) = 0$,which gives $x = 0$.
Thus,statement $(A)$ is incorrect because the oxidation number of $Ni$ is $0$,not $+4$.

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

(B) The complex $K_2[PtCl_6]$ is named according to $IUPAC$ rules:
$1$. The cation $K^+$ is named first as $Potassium$.
$2$. The coordination entity $[PtCl_6]^{2-}$ is named next.
$3$. The ligand $Cl^-$ is named as $chloro$ and since there are $6$ of them,it is $hexachloro$.
$4$. The central metal $Pt$ is in the anion,so it is named as $platinate$.
$5$. The oxidation state of $Pt$ is calculated as $x + 6(-1) = -2$,so $x = +4$.
$6$. Combining these,the name is $Potassium \ hexachloroplatinate(IV)$.

(C) To name the coordination compound $K_3[Al(C_2O_4)_3]$ according to $IUPAC$ rules:
$1$. Name the cation first: $Potassium$.
$2$. Name the ligands in alphabetical order: $trioxalato$ (since there are $3$ oxalate ions).
$3$. Name the central metal atom: Since the complex ion $[Al(C_2O_4)_3]^{3-}$ is anionic,the metal $Al$ is named as $aluminate$.
$4$. Determine the oxidation state of $Al$: Let $x$ be the oxidation state of $Al$. $x + 3(-2) = -3$,which gives $x = +3$.
$5$. Combine the parts: $Potassium$ $trioxalato$ $aluminate(III)$.

(B) The coordination entity is an anion,so the central metal $Ir$ is named as iridate.
Since the ligand is a bidentate ligand (oxalate),we use the prefix 'tris' instead of 'tri'.
The oxidation state of $Ir$ is calculated as: $x + 3(-2) = -3$,which gives $x = +3$.
Therefore,the correct $I.U.P.A.C.$ name is Potassium tris(oxalato)iridate$(III)$.

(A) $1$. Identify the ligand: $NH_3$ is named as 'ammine'. Since there are $6$ ligands,it is 'hexaammine'.
$2$. Identify the metal: The central metal is Cobalt $(Co)$.
$3$. Calculate the oxidation state: Let the oxidation state of $Co$ be $x$. The charge on $NH_3$ is $0$ and on $Cl$ is $-1$. So,$x + 6(0) + 3(-1) = 0$,which gives $x = +3$.
$4$. Name the complex: The complex is a cation,so the metal name remains 'cobalt'. The oxidation state is written in Roman numerals in parentheses: 'cobalt$(III)$' .
$5$. Name the anion: The counter ion is $Cl^-$,which is 'chloride'.
$6$. Combine: The full name is 'Hexaamminecobalt$(III)$ chloride'.

(B) $1$. Identify the ligands in alphabetical order: $NH_3$ (ammine),$H_2O$ (aqua),and $Cl^-$ (chloro).
$2$. There are $3$ ammine ligands,$2$ aqua ligands,and $1$ chloro ligand.
$3$. The name of the coordination sphere is $Triamminediaquachlorocobalt(III)$.
$4$. The counter ion is chloride.
$5$. Combining these,the $IUPAC$ name is $Triamminediaquachlorocobalt(III)$ chloride.

(C) The $IUPAC$ name 'diamminedichloroplatinum $(II)$' indicates that the central metal ion is platinum $(Pt)$ with an oxidation state of $+2$.
The ligands are two ammine $(NH_3)$ groups and two chloro $(Cl^-)$ groups.
The coordination sphere is represented by square brackets $[...]$.
Thus,the formula is $[Pt(NH_3)_2Cl_2]$.

(A) To determine the formula,we analyze the components of the name:
$1$. The central metal is nickel $(Ni)$ with an oxidation state of $+II$.
$2$. The ligands are two cyano groups $(CN^-)$ and two oxalato groups $(C_2O_4^{2-})$.
$3$. The coordination sphere charge is calculated as: $x = (+2) + 2(-1) + 2(-2) = 2 - 2 - 4 = -4$.
$4$. To balance the charge of $-4$,we need four potassium ions $(K^+)$.
$5$. Thus,the formula is $K_4[Ni(CN)_2(C_2O_4)_2]$.

(B) The complex is $[Ni(CN)_4]^x$.
In this complex,the oxidation state of $Ni$ is typically $+2$.
The charge on the ligand $CN^-$ is $-1$.
The total charge $x$ on the complex is the sum of the oxidation states of the central metal atom and the ligands:
$x = (+2) + 4(-1)$
$x = 2 - 4$
$x = -2$
Therefore,the correct option is $(B)$.

(B) $1$. Identify the ligands: There are $5$ ammine $(NH_3)$ ligands and $1$ chloro $(Cl^-)$ ligand. According to $IUPAC$ rules,ligands are named alphabetically: ammine before chlorido.
$2$. Name the coordination sphere: $5$ ammine ligands are named as 'pentaammine' and $1$ chloro ligand is named as 'chlorido'.
$3$. Determine the oxidation state of the central metal atom $(Co)$: Let the oxidation state be $x$. The charge on $NH_3$ is $0$ and $Cl$ is $-1$. The total charge of the complex is $+2$ (since there are $2$ chloride ions outside the sphere). Thus,$x + 5(0) + 1(-1) = +2$,which gives $x = +3$.
$4$. Combine the parts: The name is 'pentaamminechloridocobalt $(III)$ chloride'.

(D) Solution:
$(1)$ In options $B$ and $C$,there are no transition metals,so they are eliminated.
$(2)$ In $CrO_5$ (chromium pentoxide),the oxidation state of $Cr$ is $+6$ due to the presence of peroxo linkages.
$(3)$ In $[Fe(CO)_5]$,carbon monoxide $(CO)$ is a neutral ligand. Since the overall charge of the complex is zero,the oxidation state of $Fe$ is calculated as $x + 5(0) = 0$,which gives $x = 0$.

(D) The name of the complex is chlorodiaquatriamminecobalt $(III)$ chloride.
In this complex,the central metal atom is $Co$ with an oxidation state of $+3$.
The ligands are $3$ ammine $(NH_3)$,$2$ aqua $(H_2O)$,and $1$ chloro $(Cl^-)$ ligand inside the coordination sphere.
To balance the $+3$ charge of $Co$ and the $-1$ charge of the inner chloro ligand,the total charge of the complex ion is $[+3 + (-1)] = +2$.
Therefore,$2$ chloride ions $(Cl^-)$ are required outside the coordination sphere to neutralize the complex.
The formula is $[CoCl(NH_3)_3(H_2O)_2]Cl_2$.

(A) According to $IUPAC$ nomenclature rules for coordination compounds:
$1$. Ligands are named in alphabetical order: ammine $(NH_3)$,chloro $(Cl^-)$,cyano $(CN^-)$,and nitro $(NO_2^-)$.
$2$. The prefix 'tri' is used for three ammine ligands.
$3$. The metal name is 'cobalt' followed by its oxidation state in parentheses.
$4$. Oxidation state of $Co$: $x + 3(0) + (-1) + (-1) + (-1) = 0$,so $x = +3$.
$5$. Combining these,the name is triamminechlorocyanonitrocobalt $(III)$.

(B) Let $x$ be the oxidation state of $Pt$ in $[Pt(C_2H_4)Cl_3]^{-}$.
Since the overall charge on the complex is $-1$,the sum of oxidation states of all elements in it should be equal to $-1$.
Ethylene $(C_2H_4)$ is a neutral ligand with an oxidation state of $0$.
Chlorine $(Cl)$ as a ligand has an oxidation state of $-1$.
Therefore,$x + 0 + 3(-1) = -1$.
$x - 3 = -1$.
$x = +2$.

(C) The $IUPAC$ name for $K[Ag(CN)_2]$ is Potassium dicyanoargentate $(I)$.
In this coordination compound,$K^+$ is the counter ion.
The complex anion is $[Ag(CN)_2]^-$.
Here,$Ag$ is the central metal atom.
Let the oxidation state of $Ag$ be $x$.
$x + 2(-1) = -1 \implies x = +1$.
Since the complex part is an anion,the suffix '-ate' is added to the metal name,so $Silver$ becomes $Argentate$.

(B) Let the oxidation state of $Co$ be $x$.
In the complex $[Co(H_2O)_5Cl]^{2+}$,the oxidation state of $H_2O$ is $0$ and $Cl$ is $-1$.
The sum of oxidation states equals the charge on the complex:
$x + 5(0) + (-1) = +2$
$x - 1 = +2$
$x = +3$
Therefore,the oxidation state of $Co$ is $+3$.