Oxidation number and Oxidation state Questions in English

(C) Let the oxidation state of sulphur be $x$.
In the sulphate ion $(SO_4^{2-})$,the sum of oxidation states of all atoms must equal the charge on the ion.
$x + 4 \times (-2) = -2$
$x - 8 = -2$
$x = +6$
Therefore,the oxidation state of sulphur in $SO_4^{2-}$ is $+6$.

(D) The correct option is $(D)$.
Let the oxidation number of $S$ be $x$.
In $Na_2SO_4$,the oxidation number of $Na$ is $+1$ and the oxidation number of $O$ is $-2$.
Applying the sum rule: $2 \times (+1) + x + 4 \times (-2) = 0$.
$2 + x - 8 = 0$.
$x - 6 = 0$.
$x = +6$.

(D) The initial oxidation state of the metal ion is $+3$.
When an atom or ion loses electrons,its oxidation number increases.
Since the ion $M^{3+}$ loses $3$ more electrons,the new oxidation number is calculated as:
$3 + 3 = +6$.
The reaction is represented as: $M^{3+} \to M^{6+} + 3e^{-}$.
Therefore,the final oxidation number is $6$.

(D) The oxidation number of an element in its elemental form or in a homonuclear molecule is always $0$.
Since ozone $(O_3)$ is a homonuclear molecule consisting only of oxygen atoms,the oxidation number of oxygen in $O_3$ is $0$.

(B) To find the oxidation state of sulphur $(S)$ in each anion,we use the rule that the sum of oxidation states equals the charge of the ion.
$1$. For $S_2O_4^{2-}$: $2x + 4(-2) = -2 \implies 2x - 8 = -2 \implies 2x = +6 \implies x = +3$.
$2$. For $SO_3^{2-}$: $x + 3(-2) = -2 \implies x - 6 = -2 \implies x = +4$.
$3$. For $S_2O_6^{2-}$: $2x + 6(-2) = -2 \implies 2x - 12 = -2 \implies 2x = +10 \implies x = +5$.
Comparing the values: $+3 < +4 < +5$.
Therefore,the order is $S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$.

(B) In metal hydrides like $LiH$,the alkali metal $Li$ is in the $+1$ oxidation state.
Since the sum of oxidation numbers in a neutral compound is $0$,we have:
$x + (+1) = 0$
$x = -1$
Therefore,the oxidation number of hydrogen in $LiH$ is $-1$.

(A) The molecular mass of $KMnO_4$ is $39 + 55 + (4 \times 16) = 158 \ g/mol$.
In the given reaction,the oxidation state of $Mn$ changes from $+7$ in $KMnO_4$ to $+6$ in $K_2MnO_4$.
The change in oxidation state is $7 - 6 = 1$.
Therefore,the $n$-factor for $KMnO_4$ is $1$.
The equivalent weight is calculated as $E = \frac{\text{Molecular Mass}}{n\text{-factor}} = \frac{158}{1} = 158$.

(A) For $K_2MnO_4$: Let the oxidation number of $Mn$ be $x$. The sum of oxidation states is $2(+1) + x + 4(-2) = 0$. Solving this,$2 + x - 8 = 0$,so $x = +6$.
For $KMnO_4$: Let the oxidation number of $Mn$ be $x$. The sum of oxidation states is $1(+1) + x + 4(-2) = 0$. Solving this,$1 + x - 8 = 0$,so $x = +7$.
Therefore,the oxidation numbers are $+6$ and $+7$ respectively.

(C) When $H_2O_2$ is added to an acidified solution of a dichromate $Cr_2O_7^{2-}$,a deep blue coloured complex,chromic peroxide $CrO_5$ (or $CrO(O_2)_2$) is formed.
$Cr_2O_7^{2-} + 2H^{+} + 4H_2O_2 \rightarrow 2CrO(O_2)_2 + 5H_2O$
This deep blue coloured complex has a butterfly structure with two peroxide linkages $(-O-O-)$ and one double-bonded oxygen.
Let the oxidation state of $Cr$ be $x$.
In $CrO_5$,there is one oxygen atom with oxidation state $-2$ and four oxygen atoms in two peroxide linkages with oxidation state $-1$ each.
Thus,$x + 1(-2) + 4(-1) = 0$.
$x - 2 - 4 = 0$.
$x = +6$.

(C) To find the oxidation state of nitrogen $(N)$ in each compound,we assign hydrogen $(H)$ a value of $+1$ and oxygen $(O)$ a value of $-2$:
$1$. In $N_2H_4$: $2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$.
$2$. In $NH_3$: $x + 3(+1) = 0 \implies x = -3$.
$3$. In $N_3H$: $3x + 1(+1) = 0 \implies 3x = -1 \implies x = -1/3$.
$4$. In $NH_2OH$: $x + 2(+1) + (-2) + 1(+1) = 0 \implies x + 1 = 0 \implies x = -1$.
Comparing the values $-2, -3, -1/3, -1$,the highest value is $-1/3$.

(D) The oxidation state of $P$ $(x)$ is calculated as follows:
For $H_{4}P_{2}O_{5}: 4(+1) + 2x + 5(-2) = 0$ $\Rightarrow 2x - 6 = 0$ $\Rightarrow x = +3$
For $H_{4}P_{2}O_{6}: 4(+1) + 2x + 6(-2) = 0$ $\Rightarrow 2x - 8 = 0$ $\Rightarrow x = +4$
For $H_{4}P_{2}O_{7}: 4(+1) + 2x + 7(-2) = 0$ $\Rightarrow 2x - 10 = 0$ $\Rightarrow x = +5$
Thus,the oxidation states are $+3, +4, +5$.

(D) $CaOCl_2$ is represented as $Ca(OCl)Cl$,which dissociates into $Ca^{2+}$,$OCl^{-}$,and $Cl^{-}$.
The oxidation state of $Cl$ in $OCl^{-}$ is $x + (-2) = -1$,so $x = +1$.
The oxidation state of $Cl$ in $Cl^{-}$ is $-1$.
Thus,the oxidation states are $+1$ and $-1$.

(A) In the given reaction: $3K_2MnO_4 + 2H_2O + 4CO_2 \rightarrow 2KMnO_4 + MnO_2 + 4KHCO_3$.
The products containing $Mn$ are $KMnO_4$ and $MnO_2$.
$1$. In $KMnO_4$: Let the oxidation state of $Mn$ be $x$. Then,$1 + x + 4(-2) = 0$ $\Rightarrow x - 7 = 0$ $\Rightarrow x = +7$.
$2$. In $MnO_2$: Let the oxidation state of $Mn$ be $y$. Then,$y + 2(-2) = 0$ $\Rightarrow y - 4 = 0$ $\Rightarrow y = +4$.
Therefore,the oxidation states of $Mn$ in the products are $+7$ and $+4$.

(C) To determine the order,calculate the oxidation number $(ON)$ of nitrogen in each species:
$NH_4^+$: $x + 4(+1) = +1 \implies x = -3$
$N_2H_4$: $2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$
$NH_2OH$: $x + 2(+1) + (-2) + (+1) = 0 \implies x + 1 = 0 \implies x = -1$
$N_2O$: $2x + (-2) = 0 \implies 2x = +2 \implies x = +1$
Thus,the increasing order is: $\mathop {NH_4^+}\limits_{(-3)} < \mathop {N_2H_4}\limits_{(-2)} < \mathop {NH_2OH}\limits_{(-1)} < \mathop {N_2O}\limits_{(+1)}$.

(A) In $Na_2S_4O_6$ (sodium tetrathionate),the structure contains two terminal sulphur atoms bonded to three oxygen atoms each and one hydroxyl group,and two central sulphur atoms bonded to each other.
The oxidation number of each terminal sulphur atom is $+5$.
The oxidation number of each central sulphur atom is $0$.
The difference between the oxidation numbers of the two types of sulphur atoms is $5 - 0 = 5$.
The structure is: $O_3S-S-S-SO_3^{2-}$ (or more accurately,$^-O_3S-S-S-SO_3^-$).

(A) $1$. For $S_8$: Since it is a homoatomic molecule,the oxidation number of $S$ is $0$.
$2$. For $S_2F_2$: Let the oxidation number of $S$ be $x$. Since the oxidation number of $F$ is $-1$,we have $2x + 2(-1) = 0$,which gives $2x = 2$,so $x = +1$.
$3$. For $H_2S$: Let the oxidation number of $S$ be $x$. Since the oxidation number of $H$ is $+1$,we have $2(+1) + x = 0$,which gives $x = -2$.
Thus,the oxidation numbers are $0, +1, -2$.

(C) The structure of tetrathionic acid $(H_2S_4O_6)$ is $HO-SO_2-S-S-SO_2-OH$.
In this molecule,the two terminal sulphur atoms are in the $+5$ oxidation state,while the two central sulphur atoms are in the $0$ oxidation state.
Therefore,the oxidation states of all sulphur atoms are not the same.
There are three $S-S$ linkages in the molecule ($S-S-S-S$ chain).
The oxygen atoms are present in two environments: terminal $OH$ groups and double-bonded $O$ atoms,so their oxidation states are not all the same.
The oxidation state of hydrogen is $+1$,not $-1$.

(C) The structure of $CrO_5$ (butterfly structure) contains one double-bonded oxygen atom (oxidation state $-2$) and four oxygen atoms involved in two peroxy linkages (each oxygen atom in a peroxy linkage has an oxidation state of $-1$).
Let the oxidation state of $Cr$ be $x$.
$x + 1(-2) + 4(-1) = 0$
$x - 2 - 4 = 0$
$x - 6 = 0$
$x = +6$
Therefore,the oxidation state of $Cr$ in $CrO_5$ is $+6$.

(B) For $PO_4^{3-}$: Let the oxidation number of $P$ be $x$. $x + 4(-2) = -3 \implies x - 8 = -3 \implies x = +5$.
For $SO_4^{2-}$: Let the oxidation number of $S$ be $y$. $y + 4(-2) = -2 \implies y - 8 = -2 \implies y = +6$.
For $Cr_2O_7^{2-}$: Let the oxidation number of $Cr$ be $z$. $2z + 7(-2) = -2 \implies 2z - 14 = -2 \implies 2z = +12 \implies z = +6$.
Thus,the oxidation numbers are $+5, +6$ and $+6$ respectively.

(A) In $HNC$,the structure is $H-N \rightrightarrows C$.
The bond between $N$ and $C$ consists of one covalent bond and one coordinate bond.
The coordinate bond is directed from $N$ to $C$.
Assigning oxidation states:
$H$ is $+1$.
$N$ is more electronegative than $H$,so $N$ takes $-1$ from the $H-N$ bond.
For the $N \rightrightarrows C$ bond,the coordinate bond does not contribute to the oxidation state calculation in the standard convention where electrons are assigned to the more electronegative atom.
Thus,$O.N. \text{ of } H = +1$,$O.N. \text{ of } N = -3$,and $O.N. \text{ of } C = +2$.

(B) To find the oxidation number of $Cl$ in each species:
$1$. In $ClO^-$,$x + (-2) = -1 \implies x = +1$.
$2$. In $HClO_3$,$1 + x + 3(-2) = 0 \implies x = +5$.
$3$. In $ICl$,$I$ is more electropositive than $Cl$,so $Cl$ has an oxidation state of $-1$.
$4$. In $NaCl$,$Na$ is $+1$,so $Cl$ has an oxidation state of $-1$.
$5$. In $ClF_3$,$F$ is $-1$,so $x + 3(-1) = 0 \implies x = +3$.
Comparing the pairs,in the pair $ICl$ and $NaCl$,the oxidation number of $Cl$ is $-1$ in both.

(A) In metal hydrides like $KH$ and $MgH_2$,hydrogen is bonded to a less electronegative metal,so the oxidation state of $H$ is $-1$.
In $NaOH$,hydrogen is part of the hydroxide ion $(OH^-)$,where it is bonded to oxygen. Since oxygen is more electronegative than hydrogen,the oxidation state of $H$ is $+1$.
Therefore,the oxidation numbers are $-1, -1, +1$.

(D) The oxidation number of $Na$ is $+1$ and $O$ is $-2$.
Let the oxidation number of $S$ be $x$.
For the compound $Na_2S_4O_6$,the sum of oxidation numbers of all atoms is $0$.
$2 \times (+1) + 4 \times x + 6 \times (-2) = 0$
$2 + 4x - 12 = 0$
$4x - 10 = 0$
$4x = 10$
$x = +2.5$

(A) The reaction of zinc with dilute $HNO_3$ is given by: $4Zn + 10HNO_3 \to 4Zn(NO_3)_2 + N_2O + 5H_2O$.
In $HNO_3$,the oxidation state of nitrogen is $+5$.
In $N_2O$,the oxidation state of nitrogen is $+1$.
Therefore,the oxidation state of nitrogen changes from $+5$ to $+1$.

(A) In $K_2Cr_2O_7$,the oxidation state of $Cr$ is calculated as: $2(+1) + 2(x) + 7(-2) = 0 \implies 2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +6$.
In $K_2CrO_4$,the oxidation state of $Cr$ is calculated as: $2(+1) + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies x = +6$.
Since the oxidation state of chromium remains $+6$ in both compounds,the change in the oxidation state is $6 - 6 = 0$.

(A) To find the oxidation number of nitrogen $(N)$ in each species:
$1$. In $NO_2^-$: $x + 2(-2) = -1 \implies x - 4 = -1 \implies x = +3$.
$2$. In $N_2O_3$: $2x + 3(-2) = 0 \implies 2x - 6 = 0 \implies 2x = 6 \implies x = +3$.
Since both $NO_2^-$ and $N_2O_3$ have nitrogen in the $+3$ oxidation state,the correct pair is $NO_2^-, N_2O_3$.

(C) To find the oxidation number of sulphur $(S)$ in each species:
$1$. In $SO_3^{2-}$: $x + 3(-2) = -2 \implies x - 6 = -2 \implies x = +4$.
$2$. In $H_2S_2O_8$ (Marshall's acid): This contains a peroxide linkage. $2(+1) + 2(x) + 6(-2) + 2(-1) = 0 \implies 2 + 2x - 12 - 2 = 0 \implies 2x = 12 \implies x = +6$.
$3$. In $S_2O_3^{2-}$: $2x + 3(-2) = -2 \implies 2x - 6 = -2 \implies 2x = 4 \implies x = +2$.
$4$. In $SO_2$: $x + 2(-2) = 0 \implies x - 4 = 0 \implies x = +4$.
Comparing the oxidation states: $+4, +6, +2, +4$. The minimum oxidation number is $+2$ in $S_2O_3^{2-}$.

(A) $FeCr_2O_4$ is a mixed oxide,which can be represented as $FeO \cdot Cr_2O_3$.
In $FeO$,the oxidation state of $Fe$ is $+2$ because oxygen is $-2$.
In $Cr_2O_3$,let the oxidation state of $Cr$ be $x$:
$2(x) + 3(-2) = 0$
$2x - 6 = 0$
$2x = 6$
$x = +3$
Therefore,the oxidation number of $Fe$ is $+2$ and $Cr$ is $+3$.

(D) Bleaching powder is represented as $Ca(OCl)Cl$.
It consists of a calcium ion $(Ca^{2+})$,a hypochlorite ion $(OCl^-)$,and a chloride ion $(Cl^-)$.
In the hypochlorite ion $(OCl^-)$,the oxygen atom has an oxidation state of $-2$. Let the oxidation state of chlorine be $x$.
$x + (-2) = -1 \implies x = +1$.
In the chloride ion $(Cl^-)$,the oxidation state of chlorine is $-1$.
Therefore,the oxidation states of the two chlorine atoms are $+1$ and $-1$.

(D) The initial oxidation state of $N$ in $N_2H_4$ is $x$,where $2x + 4(+1) = 0$,so $x = -2$.
Total electrons lost by $3 \ mole$ of $N_2H_4$ is $30 \ mole$,so electrons lost per mole of $N_2H_4$ is $30 / 3 = 10 \ mole$.
Let the new oxidation state of $N$ be $y$. The change in oxidation state per mole of $N_2H_4$ is $2(y - (-2)) = 10$.
$2(y + 2) = 10 \implies y + 2 = 5 \implies y = +3$.
Thus,the oxidation state of $N$ in the new compound is $+3$.

(C) The oxidation number of $Mg$ is $+2$ and $O$ is $-2$.
Let the oxidation number of $P$ be $x$.
For the compound $Mg_2P_2O_7$,the sum of oxidation numbers is $0$.
$2(+2) + 2(x) + 7(-2) = 0$
$4 + 2x - 14 = 0$
$2x - 10 = 0$
$2x = 10$
$x = +5$

(D) The compound $NH_4NO_2$ is an ionic salt consisting of the ammonium ion $(NH_4^+)$ and the nitrite ion $(NO_2^-)$.
In $NH_4^+$,let the oxidation number of $N$ be $x$. Thus,$x + 4(+1) = +1$,which gives $x = -3$.
In $NO_2^-$,let the oxidation number of $N$ be $y$. Thus,$y + 2(-2) = -1$,which gives $y - 4 = -1$,so $y = +3$.
Therefore,the oxidation numbers of $N$ are $-3$ and $+3$.

(A) Let the oxidation state of iodine be $x$.
For the ion $H_4IO_6^-$,the sum of oxidation states of all atoms equals the charge on the ion.
$4(+1) + x + 6(-2) = -1$
$4 + x - 12 = -1$
$x - 8 = -1$
$x = +7$

(B) For a neutral compound,the sum of the oxidation numbers of all atoms must be equal to $0$.
Checking the options:
$(1)$ $xyz$: $(+3) + (-5) + (+1) = -1 \neq 0$
$(2)$ $xyz_2$: $(+3) + (-5) + 2 \times (+1) = +3 - 5 + 2 = 0$
$(3)$ $x_2yz$: $2 \times (+3) + (-5) + (+1) = +6 - 5 + 1 = +2 \neq 0$
$(4)$ $(xy)_2z$: $2 \times ((+3) + (-5)) + (+1) = 2 \times (-2) + 1 = -3 \neq 0$
Therefore,the correct formula is $xyz_2$.

(B) The structure of $Br_3O_8$ consists of three $Br$ atoms linked together with oxygen atoms attached to them.
Each terminal $Br$ atom is bonded to three oxygen atoms via double bonds and one $Br$ atom via a single bond.
For the terminal $Br$ atoms: Each forms three double bonds with oxygen (contributing $+2$ each) and one single bond with the central $Br$ (contributing $0$ as it is a homonuclear bond). Thus,the oxidation number is $3 \times 2 = +6$.
For the central $Br$ atom: It is bonded to two oxygen atoms via double bonds and two terminal $Br$ atoms via single bonds. The oxidation number is $2 \times 2 = +4$.
Therefore,the oxidation states of the three $Br$ atoms are $6, 4, 6$.

(A) In peroxodisulphuric acid $(H_{2}S_{2}O_{8})$,the structure contains a peroxide linkage $(-O-O-)$. The oxidation state of both sulfur atoms is $+6$.
In sodium tetrathionate $(Na_{2}S_{4}O_{6})$,the structure is $O_{3}S-S-S-SO_{3}^{2-}$. The two terminal sulfur atoms are in the $+5$ oxidation state,while the two central sulfur atoms are in the $0$ oxidation state.
Therefore,the oxidation states are $+6$ for $H_{2}S_{2}O_{8}$ and $+5, 0$ for $Na_{2}S_{4}O_{6}$.

(A) In the molecule $HNC$ (hydrogen isocyanide),the electronegativity values are $C > N > H$.
Assigning oxidation states based on electronegativity:
$H$ is bonded to $N$,so $H$ has an oxidation state of $+1$.
$N$ is bonded to $H$ and $C$. Since $N$ is more electronegative than $H$ but less than $C$,the net charge on $N$ is $-3$.
$C$ is bonded to $N$ with a triple bond,and since $C$ is more electronegative than $N$,$C$ has an oxidation state of $+2$.
Thus,the oxidation states are $H = +1$,$N = -3$,and $C = +2$.

(A) To find the average oxidation number of sulphur $(S)$ in each species:
$1$. $S_8$: The oxidation state of an element in its standard state is $0$.
$2$. $S_2O_3^{2-}$: $2x + 3(-2) = -2 \implies 2x - 6 = -2 \implies 2x = 4 \implies x = +2$.
$3$. $S_4O_6^{2-}$: $4x + 6(-2) = -2 \implies 4x - 12 = -2 \implies 4x = 10 \implies x = +2.5$.
$4$. $SO_3^{2-}$: $x + 3(-2) = -2 \implies x - 6 = -2 \implies x = +4$.
$5$. $S_2O_7^{2-}$: $2x + 7(-2) = -2 \implies 2x - 14 = -2 \implies 2x = 12 \implies x = +6$.
Comparing the values: $0 < +2 < +2.5 < +4 < +6$.
The correct order is $S_8 < S_2O_3^{2-} < S_4O_6^{2-} < SO_3^{2-} < S_2O_7^{2-}$.

(B) In $KO_2$,the nature of oxygen species is superoxide (superoxide ion is $O_2^-$).
To calculate the oxidation state of oxygen,let $x$ be the oxidation state of oxygen.
The oxidation state of $K$ is $+1$.
Applying the rule for the sum of oxidation states in a neutral molecule:
$+1 + 2(x) = 0$
$2x = -1$
$x = -\frac{1}{2}$
Thus,the oxidation state of oxygen in $KO_2$ is $-\frac{1}{2}$.

(A) To find the oxidation state of the central atom in $CrO_2Cl_2$:
Let the oxidation state of $Cr$ be $x$.
Oxygen $(O)$ has an oxidation state of $-2$ and Chlorine $(Cl)$ has an oxidation state of $-1$.
Setting the sum of oxidation states to zero: $x + 2(-2) + 2(-1) = 0$.
$x - 4 - 2 = 0$.
$x - 6 = 0$.
$x = +6$.
Thus,$CrO_2Cl_2$ contains $Cr$ in the $+6$ oxidation state.

(C) For $SO_3^{2-}$: $x + 3(-2) = -2 \implies x = +4$.
For $S_2O_4^{2-}$: $2x + 4(-2) = -2 \implies 2x = +6 \implies x = +3$.
For $S_2O_6^{2-}$: $2x + 6(-2) = -2 \implies 2x = +10 \implies x = +5$.
Comparing the oxidation states: $+3 < +4 < +5$.
Thus,the increasing order is $S_2O_4^{2-} < SO_3^{2-} < S_2O_6^{2-}$.

(A) The chemical reaction between iodine and concentrated nitric acid is given by:
$I_2 + 10HNO_3 \to 2HIO_3 + 10NO_2 + 4H_2O$.
Here,the product $Y$ is $HIO_3$ (iodic acid).
To find the oxidation state of iodine $(I)$ in $HIO_3$:
Let the oxidation state of $I$ be $x$.
$1 + x + 3(-2) = 0$
$1 + x - 6 = 0$
$x - 5 = 0$
$x = +5$.
Thus,the oxidation state of iodine in $Y$ is $+5$.

(D) To find the oxidation state of nitrogen $(N)$ in each compound:
$1$. In $N_2O$: $2x + (-2) = 0 \implies x = +1$
$2$. In $NO$: $x + (-2) = 0 \implies x = +2$
$3$. In $N_2O_3$: $2x + 3(-2) = 0 \implies 2x = 6 \implies x = +3$
$4$. In $NO_2$: $x + 2(-2) = 0 \implies x = +4$
Comparing the oxidation states: $+1 < +2 < +3 < +4$.
Therefore,the correct order is $N_2O < NO < N_2O_3 < NO_2$.

(C) In the given structure $H-N \rightrightarrows C$,the bond between $N$ and $C$ is a coordinate bond where $N$ acts as the donor and $C$ as the acceptor.
Assigning oxidation states:
$H$ is $+1$.
$N$ is $-3$ (as it is more electronegative than $H$ and $C$).
Let the oxidation state of $C$ be $x$.
The sum of oxidation states in a neutral molecule is $0$.
$x + (-3) + (+1) = 0$
$x - 2 = 0$
$x = +2$
Therefore,the oxidation number of carbon is $+2$.

(C) To find the maximum oxidation state,we calculate the oxidation number of the central metal atom in each species:
$(a)$ In $MnO_2$: $x + 2(-2) = 0 \Rightarrow x = +4$
$(b)$ In $MnO$: $x + (-2) = 0 \Rightarrow x = +2$
$(c)$ In $MnO_4^-$: $x + 4(-2) = -1$ $\Rightarrow x - 8 = -1$ $\Rightarrow x = +7$
$(d)$ In $[Fe(CN)_6]^{4-}$: $x + 6(-1) = -4$ $\Rightarrow x - 6 = -4$ $\Rightarrow x = +2$
Comparing the values,the maximum oxidation state is $+7$,which is present in $MnO_4^-$.

(D) The chromyl chloride test is used for the detection of chloride ions $(Cl^-)$.
In this test,a chloride salt is heated with potassium dichromate $(K_2Cr_2O_7)$ and concentrated sulfuric acid $(H_2SO_4)$.
The reaction produces chromyl chloride $(CrO_2Cl_2)$ vapors.
The chemical equation is: $K_2Cr_2O_7 + 4NaCl + 6H_2SO_4 \rightarrow 2KHSO_4 + 4NaHSO_4 + 2CrO_2Cl_2 + 3H_2O$.
In $K_2Cr_2O_7$,the oxidation state of $Cr$ is $+6$.
In the product $CrO_2Cl_2$,the oxidation state of $Cr$ is calculated as: $x + 2(-2) + 2(-1) = 0$,which gives $x - 4 - 2 = 0$,so $x = +6$.
Since the oxidation state of $Cr$ is $+6$ in both the reactant $(K_2Cr_2O_7)$ and the product $(CrO_2Cl_2)$,it remains constant.

(B) The sum of oxidation numbers of all atoms in a neutral molecule must be zero.
For $A_3(B_4C)_2$: $(+2 \times 3) + 2 \times [(+5 \times 4) + (-2)] = 6 + 36 = 42 \neq 0$.
For $A_3(BC_4)_2$: $(+2 \times 3) + 2 \times [(+5) + (-2 \times 4)] = 6 + 2 \times (-3) = 6 - 6 = 0$.
For $A_2(BC_3)_2$: $(+2 \times 2) + 2 \times [(+5) + (-2 \times 3)] = 4 + 2 \times (-1) = 2 \neq 0$.
For $ABC_2$: $(+2) + (+5) + (-2 \times 2) = 7 - 4 = 3 \neq 0$.
Thus,the correct formula is $A_3(BC_4)_2$.

(A) Peroxodisulphuric acid $(H_2S_2O_8)$ contains two sulphur atoms in the $+6$ oxidation state.
In the tetrathionate anion $(S_4O_6^{2-})$,the structure consists of a chain of four sulphur atoms. The two terminal sulphur atoms are in the $+5$ oxidation state,while the two central sulphur atoms are in the $0$ oxidation state.
Therefore,the oxidation states are $+6$ for peroxodisulphuric acid,and $+5$ and $0$ for sodium tetrathionate.

(B) When zinc reacts with very dilute $HNO_3$,it acts as a strong reducing agent and reduces the nitrogen in $HNO_3$ from its $+5$ oxidation state to $-3$ in ammonium nitrate $(NH_4NO_3)$.
The balanced chemical equation is:
$4Zn_{(s)} + 10HNO_{3(aq)} \rightarrow 4Zn(NO_3)_{2(aq)} + NH_4NO_{3(aq)} + 3H_2O_{(l)}$
In $HNO_3$,the oxidation state of $N$ is $+5$.
In $NH_4NO_3$,the oxidation state of $N$ in the $NH_4^+$ ion is $-3$.