Oxidation number and Oxidation state Questions in English

(C) In $N_2H_4$,the oxidation state of $N$ is $-2$.
Total oxidation state of two $N$ atoms in $N_2H_4 = 2 \times (-2) = -4$.
When $N_2H_4$ loses $10 \ mol$ of electrons,the total oxidation state of the two $N$ atoms in $Y$ becomes $-4 + 10 = +6$.
Therefore,the oxidation state of one $N$ atom in $Y = \frac{+6}{2} = +3$.

(C) To calculate the oxidation state of iodine $(x)$ in these compounds,we use the rule that the sum of oxidation states in a neutral molecule is $0$. The oxidation state of $H$ is $+ 1$ and $O$ is $- 2$.
For $HIO_4$: $1 + x + 4(- 2) = 0 \implies 1 + x - 8 = 0 \implies x = + 7$.
For $H_3IO_5$: $3(1) + x + 5(- 2) = 0 \implies 3 + x - 10 = 0 \implies x = + 7$.
For $H_5IO_6$: $5(1) + x + 6(- 2) = 0 \implies 5 + x - 12 = 0 \implies x = + 7$.
Thus,the oxidation states of iodine in all three compounds are $+ 7, + 7, + 7$.

(A) $1$. For $S_8$: Since it is an elemental form of sulphur,the oxidation state is $0$.
$2$. For $S_2F_2$: Let the oxidation state of $S$ be $x$. The oxidation state of $F$ is $-1$. Thus,$2x + 2(-1) = 0$,which gives $2x = 2$,so $x = +1$.
$3$. For $H_2S$: Let the oxidation state of $S$ be $x$. The oxidation state of $H$ is $+1$. Thus,$2(+1) + x = 0$,which gives $2 + x = 0$,so $x = -2$.
Therefore,the oxidation states are $0, +1, -2$.

(A) To determine the number of electrons transferred,we calculate the change in the oxidation number $(O.N.)$ of the central metal atom.
For option $A$: In $MnO_4^-$,the $O.N.$ of $Mn$ is $+7$. In $Mn^{2+}$,the $O.N.$ is $+2$. The change is $|7 - 2| = 5$ electrons.
For option $B$: In $CrO_4^{2-}$,the $O.N.$ of $Cr$ is $+6$. In $Cr^{3+}$,the $O.N.$ is $+3$. The change is $3$ electrons.
For option $C$: In $MnO_4^{2-}$,the $O.N.$ of $Mn$ is $+6$. In $MnO_2$,the $O.N.$ of $Mn$ is $+4$. The change is $2$ electrons.
For option $D$: In $Cr_2O_7^{2-}$,the $O.N.$ of $Cr$ is $+6$. In $Cr^{3+}$,the $O.N.$ is $+3$. Since there are two $Cr$ atoms,the total change is $2 \times (6 - 3) = 6$ electrons.
Thus,the reaction involving the transfer of $5$ electrons is $MnO_4^- \to Mn^{2+}$.

(D) For $PO_4^{3-}$: Let the oxidation number of $P$ be $x$. Then $x + 4(-2) = -3$,which gives $x - 8 = -3$,so $x = +5$.
For $SO_4^{2-}$: Let the oxidation number of $S$ be $x$. Then $x + 4(-2) = -2$,which gives $x - 8 = -2$,so $x = +6$.
For $Cr_2O_7^{2-}$: Let the oxidation number of $Cr$ be $x$. Then $2x + 7(-2) = -2$,which gives $2x - 14 = -2$,so $2x = 12$,and $x = +6$.
Thus,the oxidation numbers are $+5, +6, +6$.

(C) The compound $CrO_6$ is a peroxide derivative of chromium.
In this structure,$Cr$ is bonded to four oxygen atoms via two peroxide linkages $(-O-O-)$ and one double-bonded oxygen atom,or more accurately,it is a butterfly structure where $Cr$ is in the $+6$ oxidation state.
Since the maximum oxidation state of $Cr$ is $+6$,it maintains this state in $CrO_6$.

(A) To find the decreasing order of oxidation states,we calculate the oxidation state of nitrogen $(N)$ in each compound:
$1$. In $HNO_{3}$: $1 + x + 3(-2) = 0 \implies x = +5$
$2$. In $NO$: $x + (-2) = 0 \implies x = +2$
$3$. In $N_{2}$: The oxidation state of an element in its free state is $0$.
$4$. In $NH_{4}Cl$: $x + 4(1) + (-1) = 0 \implies x + 3 = 0 \implies x = -3$
Comparing the values: $+5 > +2 > 0 > -3$.
Therefore,the correct order is $HNO_{3} > NO > N_{2} > NH_{4}Cl$.

(C) Potassium $(K)$ is an alkali metal belonging to Group $1$ of the periodic table.
It always exhibits an oxidation state of $+1$ in its combined compounds.
In $K_2O$ (potassium oxide),$K$ is $+1$.
In $K_2O_2$ (potassium peroxide),$K$ is $+1$.
In $KO_2$ (potassium superoxide),$K$ is $+1$.
Therefore,the oxidation number of potassium in all three compounds is $+1$.

(N/A) $NaH_2\underline{P}O_4$: Let oxidation number of $P$ be $x$. $1(+1) + 2(+1) + x + 4(-2) = 0$ $\Rightarrow 3 + x - 8 = 0$ $\Rightarrow x = +5$.
$(b)$ $NaH\underline{S}O_4$: Let oxidation number of $S$ be $x$. $1(+1) + 1(+1) + x + 4(-2) = 0$ $\Rightarrow 2 + x - 8 = 0$ $\Rightarrow x = +6$.
$(c)$ $H_4\underline{P_2}O_7$: Let oxidation number of $P$ be $x$. $4(+1) + 2x + 7(-2) = 0$ $\Rightarrow 4 + 2x - 14 = 0$ $\Rightarrow 2x = 10$ $\Rightarrow x = +5$.
$(d)$ $K_2\underline{Mn}O_4$: Let oxidation number of $Mn$ be $x$. $2(+1) + x + 4(-2) = 0$ $\Rightarrow 2 + x - 8 = 0$ $\Rightarrow x = +6$.
$(e)$ $Ca\underline{O_2}$: Let oxidation number of $O$ be $x$. $1(+2) + 2x = 0$ $\Rightarrow 2 + 2x = 0$ $\Rightarrow x = -1$.
$(f)$ $Na\underline{B}H_4$: Let oxidation number of $B$ be $x$. $1(+1) + x + 4(-1) = 0$ $\Rightarrow 1 + x - 4 = 0$ $\Rightarrow x = +3$.
$(g)$ $H_2\underline{S_2}O_7$: Let oxidation number of $S$ be $x$. $2(+1) + 2x + 7(-2) = 0$ $\Rightarrow 2 + 2x - 14 = 0$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$.
$(h)$ $KAl(\underline{S}O_4)_2 \cdot 12H_2O$: Let oxidation number of $S$ be $x$. Ignoring neutral $12H_2O$,$1(+1) + 1(+3) + 2(x + 4(-2)) = 0$ $\Rightarrow 4 + 2x - 16 = 0$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$.

(N/A) $(a)$ $K\underline {I_3}$: The average oxidation number of $I$ is $-1/3$. Since $O.N.$ cannot be fractional,we consider the structure: $K^+ [I-I \leftarrow I]^-$. The two terminal $I$ atoms have $O.N.$ of $0$,and the central $I$ atom has $O.N.$ of $-1$.
$(b)$ $H_2\underline {S_4}O_6$: The average $O.N.$ of $S$ is $+2.5$. In the structure $HO_3S-S-S-SO_3H$,the two terminal $S$ atoms have $O.N.$ of $+5$ and the two central $S$ atoms have $O.N.$ of $0$.
$(c)$ $\underline {Fe_3}O_4$: The average $O.N.$ of $Fe$ is $+8/3$. $Fe_3O_4$ is a mixed oxide $FeO \cdot Fe_2O_3$,where one $Fe$ is $+2$ and two $Fe$ atoms are $+3$.
$(d)$ $\underline {C}H_3\underline {C}H_2OH$: The average $O.N.$ of $C$ is $-2$. Specifically,the $CH_3$ carbon is $-3$ and the $CH_2OH$ carbon is $-1$.
$(e)$ $\underline {C}H_3\underline {C}OOH$: The average $O.N.$ of $C$ is $0$. Specifically,the $CH_3$ carbon is $-3$ and the $COOH$ carbon is $+3$.

(A) $(i)$ For $H_2SO_5$: Let the oxidation number of $S$ be $x$. $2(+1) + x + 5(-2) = 0$ $\Rightarrow 2 + x - 10 = 0$ $\Rightarrow x = +8$. However,the oxidation number of $S$ cannot exceed its valence electrons,which is $6$. This fallacy arises because $H_2SO_5$ (Caro's acid) contains a peroxide linkage. The correct structure is $HO-S(=O)_2-O-OH$. Assigning $-1$ to each peroxide oxygen: $2(+1) + x + 3(-2) + 2(-1) = 0 \Rightarrow x = +6$.
$(ii)$ For $Cr_2O_7^{2-}$: Let the oxidation number of $Cr$ be $x$. $2x + 7(-2) = -2$ $\Rightarrow 2x - 14 = -2$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$. The structure is $[O_3Cr-O-CrO_3]^{2-}$,where each $Cr$ is in the $+6$ oxidation state.
$(iii)$ For $NO_3^-$: Let the oxidation number of $N$ be $x$. $x + 3(-2) = -1$ $\Rightarrow x - 6 = -1$ $\Rightarrow x = +5$. The structure is a trigonal planar arrangement with $N$ at the center,exhibiting an oxidation state of $+5$.

(N/A) The substances where carbon can exhibit oxidation states from $-4$ to $+4$ are listed in the following table:

Substance	$O.N.$ of carbon
$CH_4$	$-4$
$CH_3CH_3$	$-3$
$CH_3Cl$	$-2$
$CH_2Cl_2$	$0$
$CHCl_3$	$+2$
$CCl_4$	$+4$

The substances where nitrogen can exhibit oxidation states from $-3$ to $+5$ are listed in the following table:

Substance	$O.N.$ of nitrogen
$NH_3$	$-3$
$N_2H_4$	$-2$
$N_2H_2$	$-1$
$N_2$	$0$
$N_2O$	$+1$
$NO$	$+2$
$N_2O_3$	$+3$
$NO_2$	$+4$
$N_2O_5$	$+5$

(N/A) The oxidation number is defined as the charge that an atom would have if all bonds to atoms of different elements were considered to be $100\%$ ionic.
It represents the number of electrons that an atom either gains or loses when it forms a chemical bond with other atoms.
In simple ionic compounds,the oxidation number of an element is equivalent to its actual electric charge.

(N/A) $1$. In elements,in the free or uncombined state,each atom bears an oxidation number of $0$. e.g.,$H_{2}, O_{2}, Cl_{2}, N_{2}, Na, Mg, Al, S_{8}, P_{4}$ have an oxidation number of $0$.
$2$. For ions composed of only one atom,the oxidation number is equal to the charge on the ion. e.g.,$Na^{+} = +1, Mg^{+2} = +2, Al^{+3} = +3, Cl^{-} = -1, O^{-2} = -2, F^{-} = -1$.
$3$. Compounds of all alkali metals have an oxidation number of $+1$ and all alkaline earth metals have an oxidation number of $+2$. e.g.,$Na^{+}, K^{+}, Cs^{+}, Rb^{+}, Li^{+}$ and $Mg^{+2}, Ca^{+2}, Be^{+2}, Sr^{+2}$.
$4$. The oxidation number of oxygen in most compounds is $-2$. e.g.,$H_{2}O$ $\rightarrow O^{-2}; CuO$ $\rightarrow O^{-2}$.
Exception: In peroxide compounds,each oxygen atom is assigned an oxidation number of $-1$. e.g.,$H_{2}O_{2}$ $\rightarrow O^{-1}; Na_{2}O_{2}$ $\rightarrow O^{-1}; BaO_{2}$ $\rightarrow O^{-1}$.
$5$. In all its compounds,fluorine has an oxidation number of $-1$. Other halogens $(Cl, Br, I)$ also have an oxidation number of $-1$,except when combined with oxygen in oxoacids and oxoanions,where they have positive oxidation numbers. e.g.,$HClO$ $\rightarrow Cl^{+1}; HClO_{4}$ $\rightarrow Cl^{+7}$.
$6$. The algebraic sum of the oxidation numbers of all the atoms in a neutral compound must be $0$. In a polyatomic ion,the algebraic sum of the oxidation numbers of all atoms must equal the charge on the ion. e.g.,for $(CO_{3})^{-2}$,the sum of oxidation numbers of $C$ and $3O$ must equal $-2$.

Let the oxidation state of $P$ be $x$.
$(i)$ $H_{3}PO_{3}$: $3(+1) + x + 3(-2) = 0 \implies 3 + x - 6 = 0 \implies x = +3$.
$(ii)$ $PCl_{3}$: $x + 3(-1) = 0 \implies x - 3 = 0 \implies x = +3$.
$(iii)$ $Ca_{3}P_{2}$: $3(+2) + 2(x) = 0 \implies 6 + 2x = 0 \implies 2x = -6 \implies x = -3$.
$(iv)$ $Na_{3}PO_{4}$: $3(+1) + x + 4(-2) = 0 \implies 3 + x - 8 = 0 \implies x - 5 = 0 \implies x = +5$.
$(v)$ $POF_{3}$: $x + (-2) + 3(-1) = 0 \implies x - 5 = 0 \implies x = +5$.

For $NaH_{2}PO_{4}$:
$1(+1) + 2(+1) + P + 4(-2) = 0$
$1 + 2 + P - 8 = 0$
$P = +5$
For $NaHSO_{4}$:
$1(+1) + 1(+1) + S + 4(-2) = 0$
$2 + S - 8 = 0$
$S = +6$
For $H_{4}P_{2}O_{7}$:
$4(+1) + 2(P) + 7(-2) = 0$
$4 + 2P - 14 = 0$
$2P = 10$
$P = +5$

(N/A) $1$. $K_2MnO_4$: $2(+1) + Mn + 4(-2) = 0 \implies Mn = +6$
$2$. $CaO_2$: $Ca$ is $+2$,so $2(O) = -2 \implies O = -1$ (Peroxide ion)
$3$. $NaBH_4$: $1(+1) + B + 4(-1) = 0 \implies B = +3$
$4$. $KAl(SO_4)_2 \cdot 12H_2O$: Since $H_2O$ is neutral,$1(+1) + 3 + 2(S) + 8(-2) = 0 \implies 4 + 2S - 16 = 0 \implies 2S = 12 \implies S = +6$

(N/A) $(1)$ $KI_3$: $K$ has an oxidation number of $+1$. The $I_3^-$ ion consists of an $I_2$ molecule coordinated to an $I^-$ ion. The oxidation numbers of the two iodine atoms in $I_2$ are $0$,and the oxidation number of the $I^-$ ion is $-1$. Thus,the oxidation numbers are $0, 0, -1$.
$(2)$ $H_2S_4O_6$ (Tetrathionic acid): Based on the structure $HO_3S-S-S-SO_3H$,the two central sulfur atoms have an oxidation number of $0$,while the two terminal sulfur atoms bonded to three oxygen atoms and one sulfur atom have an oxidation number of $+5$.
$(3)$ $Fe_3O_4$: This is a mixed oxide $FeO \cdot Fe_2O_3$. In $FeO$,$Fe$ is $+2$. In $Fe_2O_3$,$Fe$ is $+3$. The average oxidation number is $8/3$,but the actual oxidation states are $+2$ and $+3$.
$(4)$ $CH_3CH_2OH$: For $CH_3$,$C$ is bonded to $3H$ ($+1$ each) and $1C$ $(0)$. Thus,$C + 3(+1) = 0 \Rightarrow C = -3$. For $CH_2OH$,$C$ is bonded to $2H$ ($+1$ each),$1O$ $(-2)$,$1H$ $(+1)$,and $1C$ $(0)$. Thus,$C + 2(+1) + 1(-2) + 1(+1) = 0 \Rightarrow C = -1$.
$(5)$ $CH_3COOH$: For $CH_3$,$C$ is $-3$. For $COOH$,$C$ is bonded to $2O$ ($-2$ each) and $1OH$ $(-1)$. Thus,$C + 2(-2) + (-1) = 0 \Rightarrow C = +3$.

(N/A) $(i)$ $H_2SO_5$ (Caro's acid):
Applying the standard rule: $2(+1) + S + 5(-2) = 0 \implies S = +8$.
Since the maximum oxidation state of sulfur is $+6$,this result is a fallacy.
The structure contains a peroxide linkage $(-O-O-)$. Thus,two oxygen atoms have an oxidation state of $-1$ and three have $-2$.
Calculation: $2(+1) + S + 2(-1) + 3(-2) = 0 \implies 2 + S - 2 - 6 = 0 \implies S = +6$.
$(ii)$ $Cr_2O_7^{2-}$ (Dichromate ion):
$2(Cr) + 7(-2) = -2 \implies 2Cr - 14 = -2 \implies 2Cr = +12 \implies Cr = +6$.
$(iii)$ $NO_3^-$ (Nitrate ion):
$N + 3(-2) = -1 \implies N - 6 = -1 \implies N = +5$.

(N/A) The oxidation states of carbon and nitrogen vary across different compounds based on their bonding and electronegativity. The following tables provide examples for each oxidation state:

Substance	Oxidation number of $C$
$CH_{4}$	$-4$
$C_{2}H_{6}$	$-3$
$CH_{2}Cl_{2}$	$-2$
$C_{2}H_{2}$	$-1$
$C_{6}H_{12}O_{6}$	$0$
$C_{2}Cl_{2}$	$+1$
$CO$	$+2$
$C_{2}Cl_{6}$	$+3$
$CO_{2}$	$+4$

Substance	Oxidation number of $N$
$NH_{3}$	$-3$
$N_{2}H_{4}$	$-2$
$NH_{2}OH$	$-1$
$N_{2}$	$0$
$N_{2}O$	$+1$
$NO$	$+2$
$N_{2}O_{3}$	$+3$
$NO_{2}$	$+4$
$N_{2}O_{5}$	$+5$

The oxidation number or state of a metal in a compound is sometimes presented according to the notation given by the German chemist,Alfred Stock,known as Stock notation.
In this system,the oxidation number is expressed by placing a Roman numeral in parentheses immediately after the symbol of the metal in the molecular formula.
This theory is particularly useful for naming metal oxides and other coordination compounds.
Examples:
$Cu_{2}O \rightarrow \text{Copper}(I) \text{Oxide}$
$CuO \rightarrow \text{Copper}(II) \text{Oxide}$
$FeO \rightarrow \text{Iron}(II) \text{Oxide}$
$Fe_{2}O_{3} \rightarrow \text{Iron}(III) \text{Oxide}$
$Na_{2}CrO_{4} \rightarrow \text{Sodium chromate}(VI)$
$K_{2}Cr_{2}O_{7} \rightarrow \text{Potassium dichromate}(VI)$
$V_{2}O_{5} \rightarrow \text{Vanadium}(V) \text{Oxide}$
$Cr_{2}O_{3} \rightarrow \text{Chromium}(III) \text{Oxide}$
$FeSO_{4} \rightarrow \text{Iron}(II) \text{Sulphate}$
$KMnO_{4} \rightarrow \text{Potassium Permanganate}(VII)$
$Mn_{2}O_{7} \rightarrow \text{Manganese}(VII) \text{Oxide}$

(B) Let the oxidation number of $S$ be $x$.
For $H_2SO_4$,the sum of oxidation numbers is $0$.
$2(+1) + x + 4(-2) = 0$
$2 + x - 8 = 0$
$x - 6 = 0$
$x = +6$.
Therefore,the oxidation number of $S$ in $H_2SO_4$ is $+6$.

(A) In $HClO_4$: Let the oxidation number of $Cl$ be $x$.
$1 + x + 4(-2) = 0 \implies x - 7 = 0 \implies x = +7$
In $HClO_3$: Let the oxidation number of $Cl$ be $x$.
$1 + x + 3(-2) = 0 \implies x - 5 = 0 \implies x = +5$
Therefore,the oxidation numbers are $+7$ and $+5$ respectively.

(A) In $C_3O_2$ (carbon suboxide),the structure is $O=C=C=C=O$.
Let the oxidation number of $C$ be $x$.
Since the oxidation number of $O$ is $-2$,we have:
$3(x) + 2(-2) = 0$
$3x - 4 = 0$
$3x = 4$
$x = +\frac{4}{3}$.
Thus,the average oxidation number of $C$ in $C_3O_2$ is $+\frac{4}{3}$.

(A) $1$. For $SO_3^{2-}$: Let the oxidation number of $S$ be $x$. Then $x + 3(-2) = -2$,which gives $x - 6 = -2$,so $x = +4$.
$2$. For $S_2O_4^{2-}$: Let the oxidation number of $S$ be $y$. Then $2y + 4(-2) = -2$,which gives $2y - 8 = -2$,so $2y = +6$,$y = +3$.
$3$. For $S_2O_6^{2-}$: Let the oxidation number of $S$ be $z$. Then $2z + 6(-2) = -2$,which gives $2z - 12 = -2$,so $2z = +10$,$z = +5$.
$4$. Comparing the values: $+3 < +4 < +5$.
$5$. Therefore,the increasing order is $S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$.

(B) In $HNO_4$ (peroxynitric acid),the structure contains a peroxy linkage $(-O-O-)$.
The oxidation state of $H$ is $+1$.
There are two oxygen atoms in the peroxy linkage with an oxidation state of $-1$ each,and two other oxygen atoms with an oxidation state of $-2$ each.
Let the oxidation number of $N$ be $x$.
Sum of oxidation numbers $= 1 + x + 2(-1) + 2(-2) = 0$
$1 + x - 2 - 4 = 0$
$x - 5 = 0$
$x = +5$

(A) The molecular formula of bleaching powder is $CaOCl_{2}$.
It is an ionic compound consisting of $Ca^{2+}$,$OCl^{-}$,and $Cl^{-}$ ions.
In the $OCl^{-}$ ion,the oxidation number of $O$ is $-2$,so $x + (-2) = -1$,which gives $x = +1$ for $Cl$.
In the $Cl^{-}$ ion,the oxidation number of $Cl$ is $-1$.
Therefore,the oxidation numbers of the two chlorine atoms in bleaching powder are $+1$ and $-1$.

(N/A) The structure of acetic acid is $CH_3-COOH$.
In $CH_3COOH$,the carbon atom in the methyl group $(CH_3)$ has an oxidation number of $-3$.
Let the oxidation number of the carbon atom in the carboxyl group $(-COOH)$ be $x$.
The sum of oxidation numbers of all atoms in the molecule is zero.
$(-3) + x + 2(-2) + (+1) = 0$
$-3 + x - 4 + 1 = 0$
$x - 6 = 0$
$x = +3$
Therefore,the oxidation number of the carbon atom in the carboxyl group is $+3$.

(N/A) For $KO_{3}$ (Potassium ozonide):
$1(K) + 3(O) = 0$
$1(+1) + 3(O) = 0$
$3(O) = -1$
$O = -\frac{1}{3}$
For $Na_{2}O_{2}$ (Sodium peroxide):
$2(Na) + 2(O) = 0$
$2(+1) + 2(O) = 0$
$2 + 2(O) = 0$
$2(O) = -2$
$O = -1$

(A) The chemical formula of glucose is $C_{6}H_{12}O_{6}$.
Let the oxidation number of $C$ be $x$.
Using the oxidation states of $H$ $(+1)$ and $O$ $(-2)$,we set the sum of oxidation numbers to $0$:
$6(x) + 12(+1) + 6(-2) = 0$
$6x + 12 - 12 = 0$
$6x = 0$
$x = 0$
Therefore,the average oxidation number of $C$ in glucose is $0$.

(N=+3, CL=+7) The compound $NOClO_4$ is an ionic compound consisting of $NO^+$ and $ClO_4^-$ ions.
For the $NO^+$ ion:
Let the oxidation number of $N$ be $x$.
$x + 1(-2) = +1$
$x - 2 = +1$
$x = +3$
Thus,the oxidation number of $N$ is $+3$.
For the $ClO_4^-$ ion:
Let the oxidation number of $Cl$ be $y$.
$y + 4(-2) = -1$
$y - 8 = -1$
$y = +7$
Thus,the oxidation number of $Cl$ is $+7$.

(B) Let the oxidation number of $Fe$ be $x$.
In the compound $Fe_{0.94}O$,the total charge is $0$.
Therefore,$(0.94 \times x) + (-2) = 0$.
$0.94x = 2$.
$x = \frac{2}{0.94} \approx 2.127$.
Rounding to two decimal places,we get $x \approx +2.13$.

(B) The structure of $CrO_5$ is a butterfly-like structure.
In this structure,the central $Cr$ atom is bonded to one double-bonded oxygen atom and two peroxide groups $(-O-O-)$.
Each peroxide group contains two oxygen atoms linked by a single bond.
Therefore,there are $2$ peroxide groups present in $CrO_5$.

(N/A) The structure of $Br_3O_8$ shows three $Br$ atoms in a chain.
$1$. The terminal $Br$ atoms are each bonded to four oxygen atoms (three double bonds and one single bond,or equivalent resonance structures). Each terminal $Br$ atom is bonded to four $O$ atoms,resulting in an oxidation state of $+6$.
$2$. The central $Br$ atom is bonded to two oxygen atoms and two other $Br$ atoms. The oxidation state of the central $Br$ atom is $+4$.
Therefore,the consecutive oxidation numbers of $Br$ are $+6, +4, +6$.

(A) The chemical formula for potassium ozonide is $KO_3$.
Let the oxidation number of oxygen be $x$.
The oxidation number of potassium $(K)$ is $+1$.
The sum of oxidation numbers in a neutral molecule is $0$.
$1(+1) + 3(x) = 0$
$1 + 3x = 0$
$3x = -1$
$x = -\frac{1}{3}$

(C) The initial oxidation state of the metal ion is $+3$.
When an atom or ion loses electrons,its oxidation number increases.
Since the ion loses $3$ additional electrons,the change in oxidation number is $+3$.
Therefore,the final oxidation number $= (+3) + (+3) = +6$.

(A) The oxidation numbers are calculated as follows:
$1$. In $C_2H_6$,$2x + 6(+1) = 0 \implies x = -3$.
$2$. In $H_2O_2$,$2(+1) + 2x = 0 \implies x = -1$.
$3$. In $LiAlH_4$,$+1 + x + 4(-1) = 0 \implies x = +3$.
$4$. In $HO_2^-$,$(+1) + 2x = -1 \implies 2x = -2 \implies x = -1$. (Note: The provided option $e$ is $+1$,which is incorrect for $O$ in $HO_2^-$. However,based on standard matching,$4$ matches $d$ or $e$ depending on the intended element. Given the options,$4$ matches $d$ $(-1)$).
$5$. In $PbSO_4$,$Pb^{2+} + SO_4^{2-}$,$S$ in $SO_4^{2-}$ is $x + 4(-2) = -2 \implies x = +6$.
Correct matching: $1-c, 2-d, 3-a, 4-d, 5-b$.

(A) The oxidation states are calculated as follows:
$1$. $CrO_5$ (Butterfly structure): The $Cr$ atom is bonded to four peroxo oxygen atoms ($-1$ each) and one double-bonded oxygen $(-2)$. Thus,$x + 4(-1) + 1(-2) = 0 \implies x = +6$. So,$1-a$.
$2$. $H_2SO_4$: $2(+1) + x + 4(-2) = 0 \implies x = +6$. So,$2-a$.
$3$. $C_4OCl_2$: This appears to be a typo in the question. Assuming the intended compound is $COCl_2$ (Phosgene),$x + (-2) + 2(-1) = 0 \implies x = +4$. Given the options provided,there is no match. If we consider the oxidation state of $Cl$ in $C_4OCl_2$,it is $-1$. So,$3-c$.
$4$. $(CH_3)_2SO$: $S$ is bonded to two methyl groups and one oxygen. $x + 2(0) + (-2) = 0 \implies x = +2$. So,$4-d$.
Correct matching: $1-a, 2-a, 3-c, 4-d$.

(B) $1$. In $C_6H_{12}O_6$,the average oxidation state of $C$ is $6x + 12(+1) + 6(-2) = 0 \implies 6x = 0 \implies x = 0$.
$2$. In $CCl_4$,$x + 4(-1) = 0 \implies x = +4$.
$3$. In $NH_4Cl$,$x + 4(+1) + (-1) = 0 \implies x + 3 = 0 \implies x = -3$.
$4$. In $Ba(H_2PO_2)_2$,$Ba$ is $+2$. For $(H_2PO_2)^-$,$2(+1) + x + 2(-2) = -1 \implies x - 2 = -1 \implies x = +1$.
Thus,the correct match is $1-b, 2-c, 3-d, 4-a$.

(N/A) $(1)$ In $Na_3Cr_3O_{10}$,let the oxidation number of $Cr$ be $x$. $3(+1) + 3(x) + 10(-2) = 0 \implies 3 + 3x - 20 = 0 \implies 3x = 17 \implies x = +\frac{17}{3}$.
$(2)$ In $LiAlH_4$,$Li$ is $+1$,$Al$ is $+3$,and $H$ is $-1$.
$(3)$ In potassium ozonide $(KO_3)$,$K$ is $+1$. Thus,$3(O) = -1 \implies O = -\frac{1}{3}$.
$(4)$ In elemental forms like fullerene and graphite,the oxidation number of $C$ is $0$.

(N/A) $(1)$ $2 \ e^{-}$ (as the oxidation state of $C$ changes from $+2$ to $+4$ and $N$ remains $-3$,the change is $2$ units).
$(2)$ $+6$ (chalcogens belong to group $16$ and can exhibit a maximum oxidation state of $+6$ by utilizing their $ns$ and $np$ electrons).

(A) $(1)$ True: In $NH_4Cl$,the oxidation state of $N$ is $x + 4(+1) + (-1) = 0$,so $x + 3 = 0$,$x = -3$.
$(2)$ False: In $CH_3Cl$,let the oxidation state of $C$ be $x$. $x + 3(+1) + (-1) = 0$,so $x + 2 = 0$,$x = -2$. Wait,the statement says $-2$,which is correct. Let's re-evaluate: $x + 3 - 1 = 0 \implies x = -2$. The statement is True.
$(3)$ True: In $Br_2$,the oxidation state of $Br$ is $0$. In $BrO_3^-$,$x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5$. The statement is True.

(A) In $Li_2O$,oxygen is an oxide ion with an oxidation state of $-2$. Let the oxidation state of $Li$ be $x$. The sum of oxidation states in a neutral molecule is $0$. Therefore,$2x + (-2) = 0$,which gives $2x = 2$,so $x = +1$.

(A) In $Na_2O_2$ (sodium peroxide),oxygen exists as a peroxide ion,$O_2^{2-}$,where the oxidation state of each oxygen atom is $-1$.
Let the oxidation state of sodium be $x$.
Applying the rule that the sum of oxidation states in a neutral molecule is $0$:
$2(x) + 2(-1) = 0$
$2x - 2 = 0$
$2x = 2$
$x = +1$.
Therefore,the oxidation state of sodium in $Na_2O_2$ is $+1$.

(A) In $OF_2$,fluorine is more electronegative than oxygen. The oxidation state of fluorine is always $-1$. Let the oxidation state of oxygen be $x$. The sum of oxidation states in a neutral molecule is $0$. Therefore,$x + 2(-1) = 0$,which gives $x - 2 = 0$,so $x = +2$. Thus,the oxidation state of oxygen in $OF_2$ is $+2$.

(N/A) The Lewis structure of the $O_2^-$ ion is $[:\ddot{O}-\ddot{O}:]^-$.
In this structure,one oxygen atom is neutral (oxidation state $0$) and the other oxygen atom carries a formal charge of $-1$ (oxidation state $-1$).
The average oxidation state of oxygen in $O_2^-$ is calculated as $\frac{0 + (-1)}{2} = -\frac{1}{2}$.

(N/A) For $HPO_3^{2-}$:
Let the oxidation number of $P$ be $x$.
$1 + x + 3(-2) = -2$
$x + 1 - 6 = -2$
$x - 5 = -2$
$x = +3$
$(b)$ For $PO_4^{3-}$:
Let the oxidation number of $P$ be $x$.
$x + 4(-2) = -3$
$x - 8 = -3$
$x = +5$

(N/A) $Na_2S_2O_3$: In the structure,one $S$ atom is bonded to the central $S$ atom via a coordinate bond (acting as an acceptor),so its oxidation number is $-2$. For the central $S$ atom $(x)$:
$2(+1) + x + (-2) + 3(-2) = 0$
$2 + x - 2 - 6 = 0$
$\therefore x = +6$
Thus,the oxidation numbers are $-2$ and $+6$.
$(b)$ $Na_2S_4O_6$: In the structure,the two central $S$ atoms are bonded to each other,so their oxidation number is $0$. For the other two $S$ atoms $(x)$:
$2(+1) + 2(x) + 2(0) + 6(-2) = 0$
$2 + 2x - 12 = 0$
$2x = 10$
$\therefore x = +5$
Thus,the oxidation numbers are $0$ and $+5$.
$(c)$ $Na_2SO_3$: Using the average oxidation state method:
$2(+1) + x + 3(-2) = 0$
$2 + x - 6 = 0$
$\therefore x = +4$
$(d)$ $Na_2SO_4$: Using the average oxidation state method:
$2(+1) + x + 4(-2) = 0$
$2 + x - 8 = 0$
$\therefore x = +6$