Oxidation number and Oxidation state Questions in English

(A-4, B-5, C-3, D-1) The correct matching is $A-4, B-5, C-3, D-1$.
$(A)$ For $Cr_2O_7^{2-}$: $2(Cr) + 7(-2) = -2 \implies 2(Cr) = +12 \implies Cr = +6$.
$(B)$ For $MnO_4^{-}$: $Mn + 4(-2) = -1 \implies Mn = +7$.
$(C)$ For $VO_3^{-}$: $V + 3(-2) = -1 \implies V = +5$.
$(D)$ For $FeF_6^{3-}$: $Fe + 6(-1) = -3 \implies Fe = +3$.

(A-5, B-4, C-3, D-2, E-6) The correct matches are:
$A-5$ (Positively charged ion is a Cation)
$B-4$ (Oxidation number of all neutral atoms is $0$)
$C-3$ (Oxidation number of $H^+$ is $+1$)
$D-2$ (Oxidation number of $F$ in $NaF$ is $-1$)
$E-6$ (Negatively charged ion is an Anion)
Thus,the correct sequence is: $A-5, B-4, C-3, D-2, E-6$.

(N/A) The oxidation numbers are calculated as follows:
$NaCl: Cl = -1$
$Cl_2: Cl = 0$
$NaClO: +1 + Cl - 2 = 0 \implies Cl = +1$
$Cl_2O: 2Cl - 2 = 0 \implies Cl = +1$
$KClO_2: +1 + Cl + 2(-2) = 0 \implies Cl = +3$
$ClO_2: Cl + 2(-2) = 0 \implies Cl = +4$
$NaClO_3: +1 + Cl + 3(-2) = 0 \implies Cl = +5$
$ClO_3: Cl + 3(-2) = 0 \implies Cl = +6$
$NaClO_4: +1 + Cl + 4(-2) = 0 \implies Cl = +7$
$Cl_2O_7: 2Cl + 7(-2) = 0 \implies Cl = +7$
Increasing order of oxidation number of $Cl$:
$NaCl (-1) < Cl_2 (0) < NaClO/Cl_2O (+1) < KClO_2 (+3) < ClO_2 (+4) < NaClO_3 (+5) < ClO_3 (+6) < NaClO_4/Cl_2O_7 (+7)$
The oxidation state $+2$ is not present in any of the above compounds.

(D) In $CH_4$,the oxidation number of $H$ is $+1$. Let the oxidation number of $C$ be $x$. So,$x + 4(+1) = 0$,which gives $x = -4$.
In $CCl_4$,the oxidation number of $Cl$ is $-1$. Let the oxidation number of $C$ be $y$. So,$y + 4(-1) = 0$,which gives $y = +4$.
Therefore,the oxidation number of carbon changes from $-4$ to $+4$.

(B) $(1) \ HAuCl_4 \Rightarrow (+1) + x + 4(-1) = 0, x = +3$
$(2) \ Cu_2\underline{O} \Rightarrow 2(+1) + x = 0, x = -2$
$(3) \ \underline{Cl}O_3^{-} \Rightarrow x + 3(-2) = -1, x = +5$
$(4) \ K_2\underline{Cr_2O_7} \Rightarrow 2(+1) + 2x + 7(-2) = 0, x = +6$
The oxidation number of oxygen in $Cu_2O$ is $-2$,not $-1$. Therefore,option $B$ is incorrect.

(A) To find the oxidation state of nitrogen $(N)$ in each species,we use the rule that the oxidation state of oxygen $(O)$ is $-2$.
$1$. In $NO_{3}^{-}$: $x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5$.
$2$. In $NO_{2}$: $x + 2(-2) = 0 \implies x - 4 = 0 \implies x = +4$.
$3$. In $NO$: $x + (-2) = 0 \implies x = +2$.
$4$. In $N_{2}O$: $2x + (-2) = 0 \implies 2x = 2 \implies x = +1$.
Comparing the values: $+5 (NO_{3}^{-}) > +4 (NO_{2}) > +2 (NO) > +1 (N_{2}O)$.
Thus,the correct order is $NO_{3}^{-} > NO_{2} > NO > N_{2}O$.

(D) The reaction of dichromate ion with a base is given by: $Cr_{2}O_{7}^{2-} + 2OH^{-} \longrightarrow 2CrO_{4}^{2-} + H_{2}O$.
In the product chromate ion $(CrO_{4}^{2-})$,let the oxidation state of $Cr$ be $x$.
$x + 4(-2) = -2$
$x - 8 = -2$
$x = +6$.
Thus,the oxidation number of $Cr$ in the product is $6$.

(D) The structure of polythionic acid $H_{2}S_{x}O_{6}$ consists of two terminal sulfur atoms bonded to three oxygen atoms each (one double-bonded,one double-bonded,and one hydroxyl group),and a chain of $(x-2)$ sulfur atoms in the middle.
In the terminal sulfur atoms,the oxidation state is $+5$ due to bonding with three oxygen atoms and one sulfur atom.
The central sulfur atoms are bonded only to other sulfur atoms,so their oxidation state is $0$.
Therefore,the oxidation states present are $0$ and $+5$.

(D) To find the change in oxidation state,we calculate the oxidation number of the central atom in both the reactant and the product.
$A$: In $C_{2}O_{4}^{2-}$,$2x + 4(-2) = -2$ $\Rightarrow 2x = 6$ $\Rightarrow x = +3$. In $CO_{2}$,$x + 2(-2) = 0 \Rightarrow x = +4$. Change = $|4 - 3| = 1$.
$B$: In $CrO_{4}^{2-}$,$x + 4(-2) = -2 \Rightarrow x = +6$. In $Cr^{3+}$,$x = +3$. Change = $|3 - 6| = 3$.
$C$: In $Cr_{2}O_{7}^{2-}$,$2x + 7(-2) = -2$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$. In $Cr^{3+}$,$x = +3$. Change per $Cr$ atom = $|3 - 6| = 3$.
$D$: In $MnO_{4}^{-}$,$x + 4(-2) = -1 \Rightarrow x = +7$. In $Mn^{2+}$,$x = +2$. Change = $|2 - 7| = 5$.
Thus,the process with a change of $5$ is $MnO_{4}^{-} \rightarrow Mn^{2+}$.

(A) To find the oxidation state of $P$ (let it be $x$):
For $H_{4}P_{2}O_{7}$: $4(+1) + 2x + 7(-2) = 0 \implies 4 + 2x - 14 = 0 \implies 2x = 10 \implies x = +5$.
For $H_{4}P_{2}O_{5}$: $4(+1) + 2x + 5(-2) = 0 \implies 4 + 2x - 10 = 0 \implies 2x = 6 \implies x = +3$.
For $H_{4}P_{2}O_{6}$: $4(+1) + 2x + 6(-2) = 0 \implies 4 + 2x - 12 = 0 \implies 2x = 8 \implies x = +4$.
Thus,the oxidation states are $5, 3$ and $4$ respectively.

(B) In chromate ion $(CrO_4^{2-})$,let the oxidation state of $Cr$ be $x$.
$x + 4(-2) = -2 \implies x - 8 = -2 \implies x = +6$.
In dichromate ion $(Cr_2O_7^{2-})$,let the oxidation state of $Cr$ be $y$.
$2y + 7(-2) = -2 \implies 2y - 14 = -2 \implies 2y = +12 \implies y = +6$.
The difference in oxidation state is $|+6 - (+6)| = 0$.

(A) To determine the oxidation states of sulphur in the given oxoacids,we examine their structures:
$1$. $H_{2}S_{2}O_{3}$ (Thiosulphuric acid): The structure contains one central $S$ atom bonded to another $S$ atom. The central $S$ atom is in the $+6$ oxidation state,while the terminal $S$ atom is in the $-2$ oxidation state.
$2$. $H_{2}S_{2}O_{6}$ (Dithionic acid): Both $S$ atoms are in the $+5$ oxidation state.
$3$. $H_{2}S_{2}O_{7}$ (Pyrosulphuric acid): Both $S$ atoms are in the $+6$ oxidation state.
$4$. $H_{2}S_{2}O_{8}$ (Peroxodisulphuric acid): Both $S$ atoms are in the $+6$ oxidation state.
Thus,$H_{2}S_{2}O_{3}$ is the only acid among the options that contains sulphur in two different oxidation states ($-2$ and $+6$).

(A) For $Cr_2O_7^{2-}$: Let the oxidation state of $Cr$ be $x$.
$2(x) + 7(-2) = -2$
$2x - 14 = -2$
$2x = 12$
$x = +6$.
For $ClO_3^{-}$: Let the oxidation state of $Cl$ be $x$.
$1(x) + 3(-2) = -1$
$x - 6 = -1$
$x = +5$.
Thus,the oxidation numbers are $+6$ and $+5$ respectively. The correct option is $A$.

(A) The correct option is $A$.
Let the oxidation state of $P$ atom be $x$.
$(I)$ For $POCl_3$: $x + 1(-2) + 3(-1) = 0 \implies x - 2 - 3 = 0 \implies x = +5$.
$(II)$ For $H_3PO_3$: $3(1) + x + 3(-2) = 0 \implies 3 + x - 6 = 0 \implies x = +3$.
$(III)$ For $H_4P_2O_6$: $4(1) + 2x + 6(-2) = 0 \implies 4 + 2x - 12 = 0 \implies 2x = 8 \implies x = +4$.
Thus,the oxidation states are $+5, +3, +4$.

(D) The correct option is $D$.
Let us calculate the oxidation number of $S$ in the given compounds:
$(i)$ In $H_2S$:
Let the oxidation state of $S$ be $x$.
$2(+1) + x = 0$
$x = -2$
$(ii)$ In $CS_2$:
Let the oxidation state of $S$ be $x$.
$C$ is $+4$ (as $C$ is more electronegative than $S$ is not true,but $C$ is $+4$ in $CS_2$ based on standard rules).
$+4 + 2(x) = 0$
$2x = -4$
$x = -2$
$(iii)$ In $Na_2SO_4$:
$2(+1) + x + 4(-2) = 0$
$2 + x - 8 = 0$
$x = +6$
Since none of the compounds have an oxidation state of $-4$ for sulphur,the correct option is $D$.

(D) The chemical formula for hypophosphoric acid is $H_4P_2O_6$.
To calculate the oxidation state of phosphorus $(P)$,let its oxidation state be $x$.
The oxidation state of hydrogen $(H)$ is $+1$ and oxygen $(O)$ is $-2$.
Applying the rule: $4(+1) + 2(x) + 6(-2) = 0$
$4 + 2x - 12 = 0$
$2x - 8 = 0$
$2x = 8$
$x = +4$
Therefore,the oxidation state of phosphorus in hypophosphoric acid is $+4$.

(A) In $HBrO_3$ (bromic acid),let the oxidation state of $Br$ be $x$.
$1 + x + 3(-2) = 0 \implies x - 5 = 0 \implies x = +5$.
In $HBrO_4$ (perbromic acid),let the oxidation state of $Br$ be $y$.
$1 + y + 4(-2) = 0 \implies y - 7 = 0 \implies y = +7$.
The sum of the oxidation states is $5 + 7 = 12$.

(A) The chemical formula for Chromyl chloride is $CrO_2Cl_2$.
Let the oxidation state of $Cr$ be $x$.
$x + 2(-2) + 2(-1) = 0$
$x - 4 - 2 = 0$
$x = +6$
Therefore,the oxidation state of chromium in $CrO_2Cl_2$ is $+6$.

(B) To find the oxidation state of Sulphur $(S)$ in each compound:
$1$. $SO_3$: $x + 3(-2) = 0 \implies x = +6$
$2$. $H_2SO_3$: $2(+1) + x + 3(-2) = 0 \implies x = +4$
$3$. $SOCl_2$: $x + (-2) + 2(-1) = 0 \implies x = +4$
$4$. $SF_4$: $x + 4(-1) = 0 \implies x = +4$
$5$. $BaSO_4$: $(+2) + x + 4(-2) = 0 \implies x = +6$
$6$. $H_2S_2O_7$: $2(+1) + 2x + 7(-2) = 0 \implies 2x = +12 \implies x = +6$
The compounds with $+4$ oxidation state are $H_2SO_3, SOCl_2$,and $SF_4$.
Therefore,the total number of such compounds is $3$.

(A) In the chromyl chloride test,$Cl^{-}$ reacts to form chromyl chloride $(CrO_2Cl_2)$,which dissolves in a basic medium to form a yellow solution containing chromate ions $(CrO_4^{2-})$.
When this yellow solution is acidified and treated with $10\% H_2O_2$ in the presence of amyl alcohol,it forms a blue-colored peroxo compound,chromium pentoxide $(CrO_5)$.
The structure of $CrO_5$ (butterfly structure) contains one double-bonded oxygen atom (oxidation state $-2$) and four peroxo oxygen atoms (each with oxidation state $-1$).
Let the oxidation state of $Cr$ be $x$.
$x + 1(-2) + 4(-1) = 0$
$x - 2 - 4 = 0$
$x = +6$
Thus,the oxidation state of chromium in $CrO_5$ is $+6$.

(B) The reaction is: $K_2Cr_2O_7 + 4KCl + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 6KHSO_4 + 3H_2O$.
This reaction is known as the chromyl chloride test.
In the product chromyl chloride $(CrO_2Cl_2)$,let the oxidation state of $Cr$ be $x$.
$x + 2(-2) + 2(-1) = 0$
$x - 4 - 2 = 0$
$x = +6$.
Thus,the oxidation state of chromium in the product is $+6$.

(D) $1$. For $Br_3O_8$: The structure shows bromine atoms in $+6$ and $+4$ oxidation states. Number of atoms with zero oxidation state = $0$.
$2$. For $F_2O$: The oxygen atom is in $+2$ oxidation state. Number of atoms with zero oxidation state = $0$.
$3$. For $H_2S_4O_6$: The structure contains two central sulfur atoms in $0$ oxidation state. Number of atoms with zero oxidation state = $2$.
$4$. For $H_2S_5O_6$: The structure contains three central sulfur atoms in $0$ oxidation state. Number of atoms with zero oxidation state = $3$.
$5$. For $C_3O_2$: The central carbon atom is in $0$ oxidation state. Number of atoms with zero oxidation state = $1$.
$6$. Total sum = $0 + 0 + 2 + 3 + 1 = 6$.

(C) To find the oxidation state of $P$ in each species,let the oxidation state be $x$:
$1$. In $H_3PO_4$: $3(+1) + x + 4(-2) = 0 \implies 3 + x - 8 = 0 \implies x = +5$.
$2$. In $H_4P_2O_6$: $4(+1) + 2x + 6(-2) = 0 \implies 4 + 2x - 12 = 0 \implies 2x = 8 \implies x = +4$.
$3$. In $H_3PO_3$: $3(+1) + x + 3(-2) = 0 \implies 3 + x - 6 = 0 \implies x = +3$.
$4$. In $H_3PO_2$: $3(+1) + x + 2(-2) = 0 \implies 3 + x - 4 = 0 \implies x = +1$.
Comparing the oxidation states: $+5 (H_3PO_4) > +4 (H_4P_2O_6) > +3 (H_3PO_3) > +1 (H_3PO_2)$.

(A) The structure of $Na_2S_4O_6$ is shown below:
$Na^+O^--S(=O)_2-S-S-S(=O)_2-O^-Na^+$
In $Na_2S_4O_6$ (sodium tetrathionate),there are two types of sulphur atoms:
$1$. The two terminal sulphur atoms are bonded to three oxygen atoms (two via double bonds and one via a single bond) and one sulphur atom. Their oxidation number is calculated as $+5$.
$2$. The two central sulphur atoms are bonded only to two other sulphur atoms. Since the electronegativity of identical atoms is the same,their oxidation number is $0$.
The difference in the oxidation numbers of the two types of sulphur atoms is $|5 - 0| = 5$.

(B) To determine the order,we calculate the oxidation state of nitrogen $(N)$ in each compound:
$HNO_3$: $1 + x + 3(-2) = 0 \implies x = +5$
$NO$: $x + (-2) = 0 \implies x = +2$
$N_2$: The oxidation state of an element in its standard state is $0$.
$NH_4Cl$: $x + 4(1) + (-1) = 0 \implies x = -3$
Comparing the values: $+5 > +2 > 0 > -3$.
Thus,the decreasing order is $HNO_3, NO, N_2, NH_4Cl$.

(B) The reaction for the ozonolysis of $ClO_2$ is:
$2 ClO_2 + 2 O_3 \longrightarrow Cl_2O_6 + 2 O_2$
In the resulting oxide $Cl_2O_6$,let the oxidation state of $Cl$ be $x$.
For $Cl_2O_6$:
$2x + 6(-2) = 0$
$2x - 12 = 0$
$2x = 12$
$x = +6$
Therefore,the average oxidation state of $Cl$ in $Cl_2O_6$ is $6$.

(A) $1$. For $\underline{K}O_2$: Potassium $(K)$ is an alkali metal and always exhibits an oxidation state of $+1$ in its compounds.
$2$. For $H_2\underline{O}_2$: Let the oxidation state of oxygen be $x$. Since hydrogen is $+1$,we have $2(+1) + 2x = 0$,which gives $2x = -2$,so $x = -1$.
$3$. For $H_2\underline{S}O_4$: Let the oxidation state of sulfur be $x$. Hydrogen is $+1$ and oxygen is $-2$. Thus,$2(+1) + x + 4(-2) = 0$,which simplifies to $2 + x - 8 = 0$,so $x = +6$.
Therefore,the oxidation states are $+1, -1$ and $+6$.

(A) To find the oxidation state of nitrogen $(N)$:
$1$. In $\underline{N}H_4 Cl$: $x + 4(+1) = +1$ (since $Cl$ is $-1$),so $x + 4 = 1$,$x = -3$ $(A-R)$.
$2$. In $H\underline{N}O_3$: $1 + x + 3(-2) = 0$,so $1 + x - 6 = 0$,$x = +5$ $(B-Q)$.
$3$. In $\underline{N}_2 O$: $2x + (-2) = 0$,so $2x = 2$,$x = +1$ $(C-P)$.
$4$. In $\underline{N}H_2 OH$: $x + 2(+1) + (-2) + 1 = 0$,so $x + 2 - 2 + 1 = 0$,$x = -1$ $(D-S)$.
Thus,the correct match is $A-R, B-Q, C-P, D-S$.

(C) Let the oxidation number of phosphorus be $x$.
In $Ba(H_2PO_2)_2$,the oxidation state of $Ba$ is $+2$,$H$ is $+1$,and $O$ is $-2$.
The sum of oxidation numbers in a neutral compound is $0$.
$1(+2) + 2[2(+1) + x + 2(-2)] = 0$
$2 + 2[2 + x - 4] = 0$
$2 + 2[x - 2] = 0$
$2 + 2x - 4 = 0$
$2x - 2 = 0$
$2x = 2$
$x = +1$

(D) The molecular formula of glucose is $C_6H_{12}O_6$. Let the oxidation number of carbon be $x$.
$6(x) + 12(+1) + 6(-2) = 0$
$6x + 12 - 12 = 0$
$6x = 0$
$x = 0$. Thus,the oxidation number of carbon in glucose is zero.

(B) For $PF_6^{-}$: Let the oxidation number of $P$ be $x$.
$x + (6 \times -1) = -1$
$x - 6 = -1$
$x = +5$
For $V_2 O_7^{4-}$: Let the oxidation number of $V$ be $y$.
$2y + (7 \times -2) = -4$
$2y - 14 = -4$
$2y = +10$
$y = +5$
Thus,the oxidation numbers are $+5$ and $+5$ respectively.

(D) Let the oxidation number of $Mn$ be $x$.
In $MnO_4^{-}$,the sum of oxidation numbers of all atoms equals the charge on the ion.
$x + (4 \times -2) = -1$
$x - 8 = -1$
$x = +7$
Therefore,the oxidation number of $Mn$ is $+7$.

(C) In $CrO_4^{2-}$,let the oxidation state of $Cr$ be $x$.
$x + 4(-2) = -2$ $\Rightarrow x - 8 = -2$ $\Rightarrow x = +6$.
In $K_2Cr_2O_7$,let the oxidation state of $Cr$ be $x$.
$2(+1) + 2x + 7(-2) = 0$ $\Rightarrow 2 + 2x - 14 = 0$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$.

(C) The chemical formula for the manganite ion is $MnO_4^{2-}$ and for the permanganate ion is $MnO_4^-$.
Thus,the ionic charge of the manganite ion is $-2$ and the ionic charge of the permanganate ion is $-1$.

(C) For $K_{2}Cr_{2}O_{7}$,let the oxidation state of $Cr$ be $x$.
The sum of oxidation states of all atoms in a neutral molecule is $0$.
$2(+1) + 2x + 7(-2) = 0$
$2 + 2x - 14 = 0$
$2x - 12 = 0$
$2x = 12$
$x = +6$
Therefore,the oxidation state of $Cr$ in $K_{2}Cr_{2}O_{7}$ is $+6$.

(D) The oxidation number of nitrogen in $HNO_3$ is calculated as follows:
Let the oxidation number of $N$ be $x$.
For $HNO_3$: $1 + x + 3(-2) = 0$
$1 + x - 6 = 0$
$x - 5 = 0$
$x = +5$
Therefore,the oxidation number of nitrogen is $ +5 $ in $HNO_3$.

(A) The chemical formula for nitric oxide is $NO$.
Let the oxidation state of nitrogen be $x$.
The oxidation state of oxygen in oxides is generally $-2$.
For the neutral molecule $NO$,the sum of oxidation states must be zero:
$x + (-2) = 0$
$x = +2$.
Therefore,the oxidation state of nitrogen in nitric oxide is $+2$.

(B) The oxidation states of nitrogen in the given substances are calculated as follows:
$N_2$: $2x = 0 \Rightarrow x = 0$
$NH_3$: $x + 3(+1) = 0 \Rightarrow x = -3$
$N_2O$: $2x + (-2) = 0$ $\Rightarrow 2x = 2$ $\Rightarrow x = +1$
$NO$: $x + (-2) = 0 \Rightarrow x = +2$
Comparing these values $(0, -3, +1, +2)$,the lowest oxidation state is $-3$,which corresponds to ammonia $(NH_3)$.

(B) Let the oxidation state of sulphur be $x$. The oxidation state of hydrogen is $+1$ and oxygen is $-2$. In $H_2S_2O_7$,the sum of oxidation states is $2(+1) + 2(x) + 7(-2) = 0$. Solving for $x$: $2 + 2x - 14 = 0$,$2x = 12$,$x = +6$. Thus,the oxidation state of sulphur in $H_2S_2O_7$ is $+6$.

(B) The oxidation number of an element in its elemental form or in a homoatomic molecule is always $0$.
Since $S_8$ is a homoatomic molecule consisting only of sulphur atoms,the oxidation number of sulphur in $S_8$ is $0$.

(C) To find the oxidation state of chlorine $(Cl)$ in the given compounds,we use the rule that the sum of oxidation states of all atoms in a neutral molecule is $0$. Let the oxidation state of $Cl$ be $x$. The oxidation state of hydrogen $(H)$ is $+1$ and oxygen $(O)$ is $-2$.
$1.$ For $HClO_4$: $1 + x + 4(-2) = 0 \implies x - 7 = 0 \implies x = +7$.
$2.$ For $HClO_2$: $1 + x + 2(-2) = 0 \implies x - 3 = 0 \implies x = +3$.
$3.$ For $HClO_3$: $1 + x + 3(-2) = 0 \implies x - 5 = 0 \implies x = +5$.
$4.$ For $HCl$: $1 + x = 0 \implies x = -1$.
Thus,$HClO_3$ contains chlorine in the $+5$ oxidation state.

(D) The chemical formula for hypochlorous acid is $HOCl$.
Let the oxidation state of $Cl$ be $x$.
The oxidation state of $H$ is $+1$ and $O$ is $-2$.
Sum of oxidation states in a neutral molecule is $0$:
$(+1) + (-2) + x = 0$
$-1 + x = 0$
$x = +1$.
Therefore,the oxidation state of chlorine in $HOCl$ is $+1$.

(B) In the tetrathionate ion $(S_4O_6^{2-})$,the structure consists of a chain of four sulfur atoms.
For the terminal sulfur atoms (numbered $1$ and $4$): Each is bonded to three oxygen atoms (two double-bonded and one single-bonded) and one sulfur atom. The oxidation state is calculated as: $x + 2(-2) + 1(-2) + 1(-1) = -1$ (considering the charge contribution). More simply,using electronegativity: $S$ bonded to $O$ gains $+2$ for each double bond and $+1$ for the single bond,totaling $+5$.
For the central sulfur atoms (numbered $2$ and $3$): Each is bonded to two other sulfur atoms. Since the atoms are identical,the bond between them contributes $0$ to the oxidation state. Thus,the oxidation state of the central sulfur atoms is $0$.
Therefore,the oxidation states for sulfur atoms $1, 2, 3, 4$ are $+5, 0, 0, +5$ respectively.

(D) The oxidation state of a metal in a compound is considered fractional if it represents the average oxidation state of multiple atoms of the same element in different oxidation environments within the same molecule.
$Fe_3O_4$ is a mixed oxide of $FeO$ and $Fe_2O_3$,where $Fe$ exists in $+2$ and $+3$ states. The average oxidation state is $+8/3$.
$Mn_3O_4$ is a mixed oxide of $MnO$ and $Mn_2O_3$,where $Mn$ exists in $+2$ and $+3$ states. The average oxidation state is $+8/3$.
$Pb_3O_4$ is a mixed oxide of $2PbO$ and $PbO_2$,where $Pb$ exists in $+2$ and $+4$ states. The average oxidation state is $+8/3$.
In $Na_2S_4O_6$ (sodium tetrathionate),the sulfur atoms are not metals. The question asks for a metal. However,if we evaluate the oxidation states of the metals provided in the options,all $Fe$,$Mn$,and $Pb$ in these specific mixed oxides show fractional average oxidation states. Given the options,$Na_2S_4O_6$ is the only compound where the central element $(S)$ is a non-metal,and the metal $(Na)$ is in a fixed $+1$ oxidation state.

(C) For a compound to be electrically neutral,the sum of the oxidation numbers of all atoms must be zero.
Let the formula be $X_a(Y_bZ_c)_d$.
Checking option $C$: $X_3(YZ_4)_3$.
Oxidation state of $X = +3$,$Y = +5$,$Z = -2$.
Total charge $= 3(+3) + 3(+5 + 4(-2)) = 9 + 3(5 - 8) = 9 + 3(-3) = 9 - 9 = 0$.
Since the sum of oxidation numbers is zero,$X_3(YZ_4)_3$ is a possible formula.

(D) Stock notation represents the oxidation state of a metal in a compound using Roman numerals in parentheses.
In $Hg_2Cl_2$,the oxidation state of $Hg$ is $+1$,so it is $Hg(I)Cl$. In $HgCl_2$,the oxidation state of $Hg$ is $+2$,so it is $Hg(II)Cl_2$.
Option $A$ and $B$ are correct because $Hg(I)$ is the reduced form of $Hg(II)$.
Option $C$ is correct as stock notation explicitly states the oxidation state.
Option $D$ is incorrect because 'Aurous' refers to $Au(I)$,whereas $Au(III)Cl_3$ is 'Auric chloride'. Therefore,$Au(III)Cl_3$ is not Aurous chloride.

(B) In $KMnO_4$,let the oxidation number of $Mn$ be $x$. The sum of oxidation numbers is $1 + x + 4(-2) = 0$,so $x - 7 = 0$,which gives $x = +7$.
In $MnO_2$,let the oxidation number of $Mn$ be $y$. The sum of oxidation numbers is $y + 2(-2) = 0$,so $y - 4 = 0$,which gives $y = +4$.
The difference in oxidation number is $|7 - 4| = 3$.

(D) The atomic ratio of $Xe : F$ is given as $0.4 : 2.4$.
Dividing both by $0.4$,we get the molar ratio as $1 : 6$.
Thus,the empirical formula of the compound is $XeF_6$.
In $XeF_6$,let the oxidation number of $Xe$ be $x$.
Since the oxidation number of $F$ is $-1$,we have:
$x + 6(-1) = 0$
$x - 6 = 0$
$x = +6$.
Therefore,the oxidation number of $Xe$ is $+6$.

(C) The chemical formula for calcium phosphate is $Ca_3(PO_4)_2$.
Let the oxidation number of phosphorus be $x$.
The oxidation number of calcium $(Ca)$ is $+2$ and oxygen $(O)$ is $-2$.
For the neutral compound $Ca_3(PO_4)_2$,the sum of oxidation numbers must be zero:
$3(+2) + 2[x + 4(-2)] = 0$
$6 + 2[x - 8] = 0$
$6 + 2x - 16 = 0$
$2x - 10 = 0$
$2x = 10$
$x = +5$
Therefore,the oxidation number of phosphorus is $+5$.

(C) Let the oxidation number of sulphur $(S)$ be $x$.
In $SO_3$,the oxidation number of oxygen $(O)$ is $-2$.
The sum of oxidation numbers of all atoms in a neutral molecule is $0$.
Therefore,$x + 3(-2) = 0$.
$x - 6 = 0$.
$x = +6$.
Thus,the oxidation number of sulphur in $SO_3$ is $+6$.