Oxidation number and Oxidation state Questions in English

(D) In $HNO_3$,the oxidation state of $N$ is calculated as: $1 + x + 3(-2) = 0 \implies x = +5$.
In $N_2O$,the oxidation state of $N$ is calculated as: $2x + (-2) = 0 \implies 2x = +2 \implies x = +1$.
The change in oxidation state is $|(+5) - (+1)| = 4$.

(D) The thermal decomposition of ammonium nitrate $(NH_4NO_3)$ upon heating is given by the reaction:
$NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$
In the product $N_2O$ (nitrous oxide),let the oxidation state of $N$ be $x$.
$2(x) + (-2) = 0$
$2x = 2$
$x = +1$
Thus,the oxidation state of nitrogen in $N_2O$ is $+1$.

(C) The compound is $Ba(H_2PO_2)_2$. The barium ion is $Ba^{2+}$,so the hypophosphite ion is $H_2PO_2^-$.
Let the oxidation state of $P$ be $x$.
In the ion $H_2PO_2^-$,the sum of oxidation states equals the charge on the ion:
$(2 \times (+1)) + x + (2 \times (-2)) = -1$
$2 + x - 4 = -1$
$x - 2 = -1$
$x = +1$
Therefore,the oxidation state of phosphorus is $+1$.

(B) The structure of the thiosulfate ion $(S_2O_3^{2-})$ consists of a central sulfur atom bonded to three oxygen atoms and one terminal sulfur atom.
In this structure,the central sulfur atom is in the $+6$ oxidation state,while the terminal sulfur atom is in the $-2$ oxidation state.
Due to this disproportionation tendency and the presence of these distinct oxidation states,the acidified solution is unstable and tends to decompose into sulfur and sulfur dioxide.

(B) $1$. For $S_2O_4^{2-}$: Let the oxidation state of $S$ be $x$. $2x + 4(-2) = -2 \implies 2x - 8 = -2 \implies 2x = 6 \implies x = +3$.
$2$. For $SO_3^{2-}$: Let the oxidation state of $S$ be $x$. $x + 3(-2) = -2 \implies x - 6 = -2 \implies x = +4$.
$3$. For $S_2O_6^{2-}$: Let the oxidation state of $S$ be $x$. $2x + 6(-2) = -2 \implies 2x - 12 = -2 \implies 2x = 10 \implies x = +5$.
$4$. Comparing the oxidation states: $+3 < +4 < +5$.
$5$. Therefore,the correct order is $S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$.

(C) The structure of thiosulfuric acid $(H_2S_2O_3)$ is $HO-S(=O)_2-SH$.
In this structure,one sulfur atom is the central atom bonded to three oxygen atoms and one sulfur atom.
The central sulfur atom has an oxidation state of $+6$.
The terminal sulfur atom (bonded to the central sulfur) has an oxidation state of $-2$.
Therefore,the oxidation states are $+6$ and $-2$.

(B) In $FeS_2$ (iron pyrite),the sulfur exists as the disulfide ion $(S_2^{2-})$.
Each sulfur atom in the $S_2^{2-}$ unit has an oxidation state of $-1$.
In $FeS$,$S$ is $-2$.
In $Na_2SO_3$,$S$ is $+4$.
In $Na_2S_2O_3$,the average oxidation state of $S$ is $+2$,with individual atoms being $0$ and $+4$.

(B) To determine the oxidation state of iodine $(I)$ in each compound:
$1$. In $HIO_4$: Let the oxidation state of $I$ be $x$. $1 + x + 4(-2) = 0 \implies x = +7$.
$2$. In $ICl_5$: Let the oxidation state of $I$ be $x$. $x + 5(-1) = 0 \implies x = +5$.
$3$. In $I_2$: The oxidation state of iodine in its elemental form is $0$.
$4$. In $HI$: Let the oxidation state of $I$ be $x$. $1 + x = 0 \implies x = -1$.
Thus,the decreasing order of oxidation states is $+7 (HIO_4) > +5 (ICl_5) > 0 (I_2) > -1 (HI)$.

(D) The chemical formula for hypochlorous acid is $HOCl$.
Let the oxidation number of chlorine be $x$.
The oxidation number of hydrogen $(H)$ is $+1$ and oxygen $(O)$ is $-2$.
Since the molecule is neutral,the sum of oxidation numbers is zero:
$(+1) + x + (-2) = 0$
$x - 1 = 0$
$x = +1$.
Therefore,the oxidation number of chlorine in $HOCl$ is $+1$.

(D) The oxidation states of phosphorus $(P)$ in the given oxoacids are calculated as follows:
$1$. For $H_4P_2O_5$ (Pyrophosphorous acid): The structure contains $P-O-P$ linkage. Each $P$ atom is bonded to two $OH$ groups and one $O$ atom. The oxidation state of $P$ is $+3$.
$2$. For $H_4P_2O_6$ (Hypophosphoric acid): The structure contains $P-P$ linkage. Each $P$ atom is bonded to two $OH$ groups,one $O$ atom,and the other $P$ atom. The oxidation state of $P$ is $+4$.
$3$. For $H_4P_2O_7$ (Pyrophosphoric acid): The structure contains $P-O-P$ linkage. Each $P$ atom is bonded to two $OH$ groups and two $O$ atoms (one double-bonded). The oxidation state of $P$ is $+5$.
Thus,the oxidation states are $+3, +4, +5$ respectively.

(D) To find the oxidation state of phosphorus $(P)$ in each acid,we use the rule that the sum of oxidation states in a neutral molecule is $0$. Let the oxidation state of $P$ be $x$.
$A$. Hypophosphoric acid $(H_4P_2O_6)$: $4(1) + 2x + 6(-2) = 0$ $\Rightarrow 4 + 2x - 12 = 0$ $\Rightarrow 2x = 8$ $\Rightarrow x = +4$.
$B$. Orthophosphorous acid $(H_3PO_3)$: $3(1) + x + 3(-2) = 0$ $\Rightarrow 3 + x - 6 = 0$ $\Rightarrow x = +3$.
$C$. Pyrophosphorous acid $(H_4P_2O_5)$: $4(1) + 2x + 5(-2) = 0$ $\Rightarrow 4 + 2x - 10 = 0$ $\Rightarrow 2x = 6$ $\Rightarrow x = +3$.
$D$. Hypophosphorous acid $(H_3PO_2)$: $3(1) + x + 2(-2) = 0$ $\Rightarrow 3 + x - 4 = 0$ $\Rightarrow x = +1$.
Comparing the oxidation states: $+4, +3, +3, +1$. The lowest oxidation state is $+1$,which corresponds to Hypophosphorous acid.

(B) The correct answer is $B$. Hydrogen does not always show $+1$ oxidation state. In metal hydrides like $NaH$,$CaH_2$,and $LiH$,hydrogen exhibits a $-1$ oxidation state. Fluorine is the most electronegative element and always shows $-1$ in compounds. Sodium,being an alkali metal,always shows $+1$. Calcium,an alkaline earth metal,always shows $+2$.

(C) Ammonium nitrate is $NH_4NO_3$,which dissociates into $NH_4^+$ and $NO_3^-$.
For $NH_4^+$: Let the oxidation state of $N$ be $x$. Since the oxidation state of $H$ is $+1$,we have $x + 4(+1) = +1$,which gives $x = -3$.
For $NO_3^-$: Let the oxidation state of $N$ be $y$. Since the oxidation state of $O$ is $-2$,we have $y + 3(-2) = -1$,which gives $y - 6 = -1$,so $y = +5$.
Therefore,the oxidation states of $N$ in $NH_4NO_3$ are $-3$ and $+5$.

(A) The chemical formula $K_3CrO_8$ can be written as $K_3[Cr(O_2)_4]$.
In this complex,the central metal atom is $Cr$ and it is surrounded by $4$ peroxide ions $(O_2^{2-})$.
$A$ peroxy linkage is defined as an $O-O$ bond.
Since there are $4$ peroxide ions $(O_2^{2-})$,each containing one $O-O$ bond,there are $4$ peroxy linkages present in the molecule.

(C) To determine the oxidation number of oxygen in each compound:
$1$. In $BaO_2$ (barium peroxide),oxygen is in the peroxide form,so its oxidation number is $-1$.
$2$. In $H_2O_2$ (hydrogen peroxide),oxygen is in the peroxide form,so its oxidation number is $-1$.
$3$. In $CO_2$ (carbon dioxide),oxygen is in the oxide form,so its oxidation number is $-2$.
$4$. In $SO_3$ (sulfur trioxide),oxygen is in the oxide form,so its oxidation number is $-2$.
$5$. In $H_2O_2$,the oxidation number of oxygen is $-1$.
$6$. In $O_2F_2$ (dioxygen difluoride),fluorine is more electronegative than oxygen,so fluorine has an oxidation number of $-1$. Calculating for oxygen: $2x + 2(-1) = 0$,which gives $2x = +2$,so $x = +1$.
Since $H_2O_2$ has an oxidation number of $-1$ and $O_2F_2$ has an oxidation number of $+1$,this pair has different oxidation numbers for oxygen.

(A) The oxidation state of $Cr$ in $K_3CrO_8$ is given as $+5$.
$Cr$ belongs to group $6$,so its maximum possible oxidation state is $+6$.
The number of peroxy linkages can be calculated using the formula:
$\text{Number of peroxy linkages} = \frac{\text{Theoretical Oxidation State} - \text{Actual Oxidation State}}{2}$
However,a more direct method for $K_3CrO_8$ (which is $K_3[Cr(O_2)_4]$) is to observe the structure.
In $K_3[Cr(O_2)_4]$,the $Cr$ atom is bonded to $4$ peroxy groups $(O_2^{2-})$.
Therefore,there are $4$ peroxy linkages present in the molecule.

(B) The reaction is: $Cl_2O + H_2O \longrightarrow 2HClO$.
In $Cl_2O$,the oxidation state of $Cl$ is $+1$.
In the product $HClO$ (hypochlorous acid),the oxidation state of $Cl$ is $+1$.
Since the oxidation state remains the same,the product is an oxyacid of chlorine with the $-ous$ suffix (hypochlorous acid).
Therefore,the correct option is $B$.

(A) The reaction is $HNO_4 + H_2O \longrightarrow HNO_3 + H_2O_2$.
In the product $HNO_3$ (Nitric acid),the oxidation state of $N$ is calculated as: $1 + x + 3(-2) = 0 \implies x = +5$.
The suffix $-ic$ is used for $HNO_3$ (Nitric acid) where the central atom is in its highest oxidation state.
Therefore,the product $HNO_3$ is an oxyacid with an $-ic$ suffix.

(B) The reaction is: $\underline{Cl}F + H_2O \longrightarrow HOCl + HF$.
In the product $HOCl$ (hypochlorous acid),the oxidation state of chlorine is $+1$.
The suffix $-ous$ is used for acids with lower oxidation states,while $-ic$ is used for higher oxidation states.
Since $HOCl$ is hypochlorous acid,it corresponds to the $-ous$ suffix category.
Therefore,the correct option is $B$.

(A) The reaction for the complete hydrolysis of sulfuryl chloride $(SO_2Cl_2)$ is:
$SO_2Cl_2 + 2H_2O \longrightarrow H_2SO_4 + 2HCl$
In the product $H_2SO_4$ (sulfuric acid),the sulfur atom is in its $+6$ oxidation state.
The suffix $-ic$ is used for the oxyacid where the central atom is in its higher oxidation state (e.g.,sulfuric acid).
Since $H_2SO_4$ is an oxyacid with an $-ic$ suffix,the correct option is $A$.

(A) To find the oxidation state of nitrogen $(x)$ in each compound:
$1$. In $N_3H$: $3x + 1 = 0 \implies x = -1/3 \approx -0.33$
$2$. In $NH_2OH$: $x + 2(+1) + (-2) + 1 = 0 \implies x + 2 - 2 + 1 = 0 \implies x = -1$
$3$. In $N_2H_4$: $2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$
$4$. In $NH_3$: $x + 3(+1) = 0 \implies x = -3$
Comparing the values: $-0.33 > -1 > -2 > -3$. Therefore,nitrogen exhibits the highest oxidation state in $N_3H$.

(A) $1$. In Peroxodisulphuric acid $(H_2S_2O_8)$,there is a peroxide linkage $(-O-O-)$. Let the oxidation state of $S$ be $x$. The structure is $HO_3S-O-O-SO_3H$. The oxidation state of $S$ is calculated as: $2(1) + 2(x) + 6(-2) + 2(-1) = 0$,which gives $2 + 2x - 12 - 2 = 0$,so $2x = 12$,$x = +6$.
$2$. In Sodium tetrathionate $(Na_2S_4O_6)$,the structure is $Na^+O_3S-S-S-SO_3^-Na^+$. The two terminal $S$ atoms are bonded to three oxygen atoms and one sulfur atom,having an oxidation state of $+5$. The two central $S$ atoms are bonded to two other sulfur atoms,having an oxidation state of $0$. The average oxidation state is $(2 \times 5 + 2 \times 0) / 4 = +2.5$. However,the question asks for the oxidation states present. The values are $+5$ and $0$ for the atoms,or $+2.5$ as an average. Given the options,the most appropriate representation for the set is $+6$ and $+2.5$.

(A) Caro's acid is $H_2SO_5$ (Peroxomonosulphuric acid).
In $H_2SO_5$,there is one peroxy linkage $(-O-O-)$. Let the oxidation state of $S$ be $x$. The oxidation states of $H$ is $+1$ and $O$ is $-2$ (for three atoms) and $-1$ (for two peroxy atoms).
$2(+1) + x + 3(-2) + 2(-1) = 0 \implies 2 + x - 6 - 2 = 0 \implies x = +6$.
Marshall's acid is $H_2S_2O_8$ (Peroxodisulphuric acid).
In $H_2S_2O_8$,there is one peroxy linkage $(-O-O-)$. Let the oxidation state of $S$ be $x$. The oxidation states of $H$ is $+1$ and $O$ is $-2$ (for six atoms) and $-1$ (for two peroxy atoms).
$2(+1) + 2x + 6(-2) + 2(-1) = 0 \implies 2 + 2x - 12 - 2 = 0 \implies 2x = 12 \implies x = +6$.
Thus,the oxidation state of $S$ in both acids is $+6$.

(D) The chemical formula for tetrathionic acid is $H_2S_4O_6$.
Its structure is $HO-SO_2-S-S-SO_2-OH$.
In this structure,the two central sulfur atoms are bonded only to other sulfur atoms,so their oxidation state is $0$.
The two terminal sulfur atoms are each bonded to three oxygen atoms and one sulfur atom.
Calculating the oxidation state for terminal sulfur $(x)$: $x + 3(-2) + 1(-1) = -1$ (considering the $S-OH$ bond),or more simply,by formal charge assignment,the terminal sulfur atoms have an oxidation state of $+5$.
Thus,the oxidation states of sulfur in $H_2S_4O_6$ are $+5$ and $0$.

(D) The oxidation states of chlorine in the given compounds are:
$1$. In $HOCl$,$O.S. = +1$
$2$. In $HClO_2$,$O.S. = +3$
$3$. In $HClO_4$,$O.S. = +7$
For chlorine oxyacids,the acidic strength increases with the increase in oxidation state. $HClO_4$ $(+7)$ is a strong acid,$HClO_2$ $(+3)$ is an intermediate acid,and $HOCl$ $(+1)$ is the weakest acid. None of these act as amphoteric in the context of these specific oxyacids,as they all primarily exhibit acidic character.

(C) To find the oxidation number of phosphorus $(P)$ in each compound:
$1$. In $PH_3$: $x + 3(+1) = 0 \implies x = -3$.
$2$. In $H_4P_2O_6$: $4(+1) + 2x + 6(-2) = 0 \implies 4 + 2x - 12 = 0 \implies 2x = 8 \implies x = +4$.
$3$. In $H_3PO_4$: $3(+1) + x + 4(-2) = 0 \implies 3 + x - 8 = 0 \implies x = +5$.
$4$. In $H_3PO_3$: $3(+1) + x + 3(-2) = 0 \implies 3 + x - 6 = 0 \implies x = +3$.
Comparing the values $(-3, +4, +5, +3)$,the maximum oxidation number is $+5$ in $H_3PO_4$.

(D) To determine the oxidation state of the underlined atoms:
$1$. In $Fe_3O_4$,the average oxidation state of $Fe$ is $3x + 4(-2) = 0$,so $x = +8/3$. This is fractional.
$2$. In $N_3H$ (hydrazoic acid),the average oxidation state of $N$ is $3x + 1 = 0$,so $x = -1/3$. This is fractional.
$3$. In $KO_2$ (potassium superoxide),the oxidation state of $O$ is $+1 + 2x = 0$,so $x = -1/2$. This is fractional.
$4$. In $Na_2S_2O_3$ (sodium thiosulfate),the average oxidation state of $S$ is $2(1) + 2x + 3(-2) = 0$,which gives $2x = 4$,so $x = +2$. This is an integer,not a fractional oxidation state.
Therefore,$Na_2\underline{S}_2O_3$ does not have a fractional oxidation state.

(A) To find the oxidation state of the central atom:
$1$. In $CrO_2Cl_2$: $x + 2(-2) + 2(-1) = 0 \implies x = +6$.
$2$. In $MnO_4^-$: $x + 4(-2) = -1 \implies x = +7$.
$3$. In $MnO_2$: $x + 2(-2) = 0 \implies x = +4$.
$4$. In $[Fe(CN)_6]^{3-}$: $x + 6(-1) = -3 \implies x = +3$.
$5$. In $[Co(CN)_6]^{3-}$: $x + 6(-1) = -3 \implies x = +3$.
$6$. In $MnO$: $x + (-2) = 0 \implies x = +2$.
The maximum oxidation state is $+7$ in $MnO_4^-$,which is present in option $A$.

(B) $NH_4NO_3$ exists as $NH_4^+$ and $NO_3^-$.
In $NH_4^+$,let the oxidation state of $N$ be $x$. Then $x + 4(+1) = +1$,which gives $x = -3$.
In $NO_3^-$,let the oxidation state of $N$ be $y$. Then $y + 3(-2) = -1$,which gives $y = +5$.
Thus,the oxidation states are $-3$ and $+5$.

(C) In $N_2H_4$,the oxidation state of $N$ is $-2$.
Total moles of $N$ atoms in $2$ moles of $N_2H_4$ is $2 \times 2 = 4$ moles of $N$.
Total electrons lost = $16$ moles.
Electrons lost per mole of $N$ atom = $\frac{16 \text{ moles of } e^-}{4 \text{ moles of } N} = 4$ moles of $e^-$.
Since oxidation involves an increase in oxidation state,the new oxidation state of $N$ = (Initial oxidation state) + (Electrons lost per $N$ atom) = $-2 + 4 = +2$.

(A) In $PbO_2$,lead is in the $+4$ oxidation state. The oxygen atoms are in the $-2$ oxidation state,making it a dioxide,not a peroxide.
In $H_2O_2$,$SrO_2$,and $BaO_2$,the oxygen atoms are linked as $O_2^{2-}$,which is the peroxide ion.
Therefore,$PbO_2$ does not contain a peroxide ion.

(A) To find the oxidation state of phosphorus $(P)$ in each acid,let the oxidation state be $x$:
$1$. Hypophosphorous acid $(H_3PO_2)$: $3(+1) + x + 2(-2) = 0 \implies 3 + x - 4 = 0 \implies x = +1$.
$2$. Orthophosphoric acid $(H_3PO_4)$: $3(+1) + x + 4(-2) = 0 \implies 3 + x - 8 = 0 \implies x = +5$.
$3$. Pyrophosphoric acid $(H_4P_2O_7)$: $4(+1) + 2x + 7(-2) = 0 \implies 4 + 2x - 14 = 0 \implies 2x = 10 \implies x = +5$.
$4$. Metaphosphoric acid $(HPO_3)$: $1(+1) + x + 3(-2) = 0 \implies 1 + x - 6 = 0 \implies x = +5$.
Comparing the values,the minimum oxidation state is $+1$,which corresponds to Hypophosphorous acid.

(D) The thermal decomposition of ammonium nitrate $(NH_4NO_3)$ is given by the reaction:
$NH_4NO_3 \xrightarrow{\Delta} N_2O + 2H_2O$
In nitrous oxide $(N_2O)$,let the oxidation state of nitrogen be $x$.
$2x + (-2) = 0$
$2x = 2$
$x = +1$
Therefore,the oxidation state of nitrogen in the product is $+1$.

(A) The structure of the thiosulfate ion $(S_2O_3^{2-})$ consists of a central sulfur atom bonded to three oxygen atoms and one terminal sulfur atom.
In this structure,the central sulfur atom is in the $+6$ oxidation state,while the terminal sulfur atom is in the $-2$ oxidation state.
When acidified,the thiosulfate ion undergoes disproportionation to form elemental sulfur $(S)$ and sulfur dioxide $(SO_2)$,making the acidic solution unstable.

(A) The structure of thiosulfuric acid $(H_2S_2O_3)$ is $HO-SO_2-SH$.
In this structure,one sulfur atom is at the center,bonded to three oxygen atoms (two via double bonds and one via a single bond with hydrogen) and one sulfur atom via a double bond.
The central sulfur atom is bonded to three oxygen atoms and one sulfur atom.
Assigning oxidation states:
$1$. The central sulfur atom is bonded to three oxygen atoms (each $-2$) and one hydroxyl group $(-1)$.
$2$. The terminal sulfur atom is bonded to the central sulfur atom via a double bond.
Using the structure,the central sulfur atom has an oxidation state of $+6$ and the terminal sulfur atom has an oxidation state of $-2$.

(A) Let the oxidation state of $P$ be $x$.
The oxidation states of $K$,$H$,and $O$ are $+1$,$+1$,and $-2$ respectively.
The sum of oxidation states in a neutral molecule is $0$.
$(+1) + 2(+1) + x + 2(-2) = 0$
$1 + 2 + x - 4 = 0$
$x - 1 = 0$
$x = +1$
Therefore,the oxidation state of $P$ in $KH_2PO_2$ is $+1$.

(B) $HNO_4$ is peroxynitric acid,which contains one peroxy linkage $(-O-O-)$.
Let the oxidation state of $N$ be $x$.
The structure is $HO-O-NO_2$.
Assigning oxidation states: $H = +1$,$O$ (in $-OH$) $= -2$,$O$ (in peroxy linkage) $= -1$,$O$ (in $NO_2$) $= -2$.
Calculation: $(+1) + (-2) + (-1) + x + 2(-2) = 0$
$1 - 2 - 1 + x - 4 = 0$
$x - 6 = 0$
$x = +5$

(A) The compound is $Ba(H_2PO_2)_2$.
Let the oxidation number of $P$ be $x$.
The oxidation number of $Ba$ is $+2$,$H$ is $+1$,and $O$ is $-2$.
Setting the sum of oxidation numbers to $0$:
$+2 + 2 \times [2 \times (+1) + x + 2 \times (-2)] = 0$
$+2 + 2 \times [2 + x - 4] = 0$
$+2 + 2 \times [x - 2] = 0$
$2 + 2x - 4 = 0$
$2x - 2 = 0$
$2x = 2$
$x = +1$.
Therefore,the oxidation number of $P$ is $+1$.

(C) Let the oxidation number of $N$ be $x$.
In $N_3H$,the sum of oxidation numbers of all atoms is equal to $0$.
So,$3(x) + 1(+1) = 0$.
$3x + 1 = 0$.
$3x = -1$.
$x = -1/3$.

(A) To find the oxidation state of $N$ in each compound:
$1$. In $N_3H$: Let the oxidation state of $N$ be $x$. $3x + 1 = 0 \implies 3x = -1 \implies x = -1/3 \approx -0.33$.
$2$. In $NH_2OH$: Let the oxidation state of $N$ be $x$. $x + 2(+1) + (-2) + 1 = 0 \implies x + 2 - 2 + 1 = 0 \implies x + 1 = 0 \implies x = -1$.
$3$. In $N_2H_4$: Let the oxidation state of $N$ be $x$. $2x + 4(+1) = 0 \implies 2x + 4 = 0 \implies 2x = -4 \implies x = -2$.
$4$. In $NH_3$: Let the oxidation state of $N$ be $x$. $x + 3(+1) = 0 \implies x + 3 = 0 \implies x = -3$.
Comparing the values: $-0.33 > -1 > -2 > -3$. The highest oxidation state is in $N_3H$.

(A) To find the difference in oxidation numbers,we calculate the oxidation state of the underlined element in each compound:
$A$: In $\underline{N}O_2$,$x + 2(-2) = 0 \implies x = +4$. In $\underline{N}_2H_4$,$2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$. Difference = $|4 - (-2)| = 6$.
$B$: In $\underline{P}_2O_5$,$2x + 5(-2) = 0 \implies 2x = 10 \implies x = +5$. In $\underline{P}_4O_{10}$,$4x + 10(-2) = 0 \implies 4x = 20 \implies x = +5$. Difference = $|5 - 5| = 0$.
$C$: In $\underline{N}_2O$,$2x + (-2) = 0 \implies 2x = 2 \implies x = +1$. In $\underline{N}O$,$x + (-2) = 0 \implies x = +2$. Difference = $|2 - 1| = 1$.
$D$: In $\underline{S}O_2$,$x + 2(-2) = 0 \implies x = +4$. In $\underline{S}O_3$,$x + 3(-2) = 0 \implies x = +6$. Difference = $|6 - 4| = 2$.
Comparing the differences $(6, 0, 1, 2)$,the greatest difference is $6$ in option $A$.

(D) $1$. For option $A$: In $\underline{N}O_2$,$x + 2(-2) = 0 \implies x = +4$. In $\underline{N}_2H_4$,$2x + 4(+1) = 0 \implies x = -2$. Difference = $|4 - (-2)| = 6$.
$2$. For option $B$: In $\underline{S}O_3^{2-}$,$x + 3(-2) = -2 \implies x = +4$. In $\underline{S}O_4^{2-}$,$x + 4(-2) = -2 \implies x = +6$. Difference = $|6 - 4| = 2$.
$3$. For option $C$: In $\underline{S}^{2-}$,$x = -2$. In $\underline{S}O_3^{2-}$,$x = +4$. Difference = $|4 - (-2)| = 6$.
$4$. For option $D$: In $\underline{S}^{2-}$,$x = -2$. In $\underline{S}O_4^{2-}$,$x = +6$. Difference = $|6 - (-2)| = 8$.
$5$. Comparing the differences,the greatest difference is $8$ in option $D$.

(B) Caro's acid is peroxymonosulfuric acid,with the chemical formula $H_2SO_5$.
In $H_2SO_5$,there is one peroxy linkage $(-O-O-)$ where the oxidation state of each oxygen atom is $-1$.
The other three oxygen atoms have an oxidation state of $-2$.
Let the oxidation number of $S$ be $x$.
Applying the sum of oxidation states: $2(+1) + x + 3(-2) + 2(-1) = 0$.
$2 + x - 6 - 2 = 0$.
$x - 6 = 0$.
$x = +6$.

(A) Let the oxidation number of carbon be $x$.
In $CH_2Cl_2$,the oxidation number of hydrogen $(H)$ is $+1$ and the oxidation number of chlorine $(Cl)$ is $-1$.
The sum of oxidation numbers of all atoms in a neutral molecule is $0$.
Therefore,$x + 2(+1) + 2(-1) = 0$.
$x + 2 - 2 = 0$.
$x = 0$.
Thus,the oxidation number of carbon in $CH_2Cl_2$ is $0$.

(B) Let the oxidation number of $S$ be $x$.
In $Na_2S_4O_6$,the oxidation number of $Na$ is $+1$ and $O$ is $-2$.
The sum of oxidation numbers in a neutral molecule is $0$.
$2(+1) + 4(x) + 6(-2) = 0$
$2 + 4x - 12 = 0$
$4x - 10 = 0$
$4x = 10$
$x = 2.5$
Thus,the average oxidation number of $S$ is $+2.5$.

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

(B) Let the oxidation state of $Fe$ be $x$.
For the compound $Fe_{0.94}O$,the sum of oxidation states must be zero.
$0.94(x) + 1(-2) = 0$
$0.94x = 2$
$x = 2 / 0.94$
$x = 200 / 94$
Thus,the oxidation state of $Fe$ is $+200/94$.

(D) The structure of carbon suboxide $(C_3O_2)$ is $O=C=C=C=O$.
In this structure,the terminal oxygen atoms have an oxidation state of $-2$.
The terminal carbon atoms are bonded to one oxygen and one carbon,resulting in an oxidation state of $+2$.
The central carbon atom is bonded to two other carbon atoms,resulting in an oxidation state of $0$.
Therefore,the average oxidation state is $\frac{(+2) + 0 + (+2)}{3} = +4/3$.
Option $D$ is incorrect because the carbon atoms do not have the same oxidation state.

(C) In $(NH_4)_2S_2O_8$ (peroxydisulfuric acid salt),the structure contains a peroxide linkage $(-O-O-)$. Calculating the oxidation state of $S$: $2(+1) + 2(x) + 8(-2) = 0$,but accounting for the peroxide bond,$2(+1) + 2(x) + 6(-2) + 2(-1) = 0$,which gives $2 + 2x - 12 - 2 = 0$,so $2x = 12$,$x = +6$. This is correct.
In $OsO_4$,$x + 4(-2) = 0$,so $x = +8$. This is correct.
In $H_2SO_5$ (Caro's acid),the structure contains a peroxide linkage. Calculating: $2(+1) + x + 5(-2) = 0$ gives $x = +8$,but this is incorrect because the actual oxidation state is $2(+1) + x + 3(-2) + 2(-1) = 0$,which gives $2 + x - 6 - 2 = 0$,so $x = +6$. Thus,the statement that it is $+8$ is incorrect.
In $KO_2$ (potassium superoxide),$O$ is in the superoxide ion $(O_2^-)$,so the oxidation number is $-1/2$. This is correct.