Oxidation number and Oxidation state Questions in English

(C) The chemical formula for methanal is $HCHO$.
Let the oxidation number of carbon be $x$.
The oxidation number of hydrogen is $+1$ and that of oxygen is $-2$.
The sum of oxidation numbers in a neutral molecule is $0$.
So,$1(+1) + x + 1(-2) + 1(+1) = 0$.
$1 + x - 2 + 1 = 0$.
$x = 0$.
Therefore,the oxidation number of carbon in methanal is $0$.

(C) To find the oxidation state of carbon in each compound,we use the rule that the sum of oxidation states of all atoms in a neutral molecule is $0$. Let the oxidation state of carbon be $x$.
$1$. In $HCHO$: $2(+1) + x + (-2) = 0 \implies x = 0$.
$2$. In $CH_3OH$: $x + 3(+1) + (-2) + (+1) = 0 \implies x + 2 = 0 \implies x = -2$.
$3$. In $CHCl_3$: $x + 1(+1) + 3(-1) = 0 \implies x - 2 = 0 \implies x = +2$.
$4$. In $C_{12}H_{22}O_{11}$ (Sucrose): $12x + 22(+1) + 11(-2) = 0 \implies 12x + 22 - 22 = 0 \implies 12x = 0 \implies x = 0$.
Comparing the values $(0, -2, +2, 0)$,the maximum oxidation state is $+2$ in $CHCl_3$.

(D) To find the oxidation state of chlorine $(Cl)$ in each compound,we use the rules for oxidation numbers:
$1$. In $KCl$,the oxidation state of $K$ is $+1$. Let the oxidation state of $Cl$ be $x$. So,$1 + x = 0$,which gives $x = -1$.
$2$. In $HClO$,the oxidation state of $H$ is $+1$ and $O$ is $-2$. Let the oxidation state of $Cl$ be $x$. So,$1 + x - 2 = 0$,which gives $x = +1$.
$3$. In $HClO_2$,the oxidation state of $H$ is $+1$ and $O$ is $-2$. Let the oxidation state of $Cl$ be $x$. So,$1 + x + 2(-2) = 0$,which gives $1 + x - 4 = 0$,so $x = +3$.
$4$. In $HClO_4$,the oxidation state of $H$ is $+1$ and $O$ is $-2$. Let the oxidation state of $Cl$ be $x$. So,$1 + x + 4(-2) = 0$,which gives $1 + x - 8 = 0$,so $x = +7$.
Comparing the values: $-1, +1, +3, +7$. The highest oxidation state is $+7$ in $HClO_4$.

(D) To find the oxidation number of iodine $(I)$ in each compound:
$1$. In $KIO_3$: $1 + x + 3(-2) = 0 \implies x = +5$
$2$. In $KI$: $1 + x = 0 \implies x = -1$
$3$. In $IF_5$: $x + 5(-1) = 0 \implies x = +5$
$4$. In $KIO_4$: $1 + x + 4(-2) = 0 \implies x = +7$
Comparing the values: $+5, -1, +5, +7$. The highest oxidation number is $+7$ in $KIO_4$.

(A) Let the oxidation state of $S$ in $SO_4^{2-}$ be $x$.
The sum of oxidation states of all atoms in an ion is equal to the charge on the ion.
$x + 4(-2) = -2$
$x - 8 = -2$
$x = -2 + 8$
$x = +6$
Therefore,the oxidation state of $S$ in $SO_4^{2-}$ is $+6$.

(C) Let $x$ be the oxidation state of $I$ in $I_2Cl_6$.
Since the oxidation state of chlorine $(Cl)$ in this compound is $-1$,we can write the equation:
$2x + 6 \times (-1) = 0$
$2x - 6 = 0$
$2x = 6$
$x = +3$
Therefore,the oxidation state of iodine in $I_2Cl_6$ is $+3$.

(C) Let the oxidation state of $C$ be $x$ in $K_2C_2O_4$.
The sum of oxidation states of all atoms in a neutral molecule is $0$.
$2(+1) + 2(x) + 4(-2) = 0$
$2 + 2x - 8 = 0$
$2x - 6 = 0$
$2x = 6$
$x = +3$
Therefore,the oxidation number of carbon is $+3$.

(A) In the reactant $H_2S$,the oxidation number of $S$ is calculated as follows: $2(+1) + x = 0$,which gives $x = -2$.
In the product $S$ (elemental state),the oxidation number of $S$ is $0$.
Therefore,the change in the oxidation number of $S$ is from $-2$ to $0$.

(A) Let the oxidation state of $S$ be $x$ in $SO_3^{2-}$.
The sum of oxidation states of all atoms in an ion is equal to the charge on the ion.
$x + 3(-2) = -2$
$x - 6 = -2$
$x = -2 + 6$
$x = +4$
Therefore,the oxidation number of $S$ in $SO_3^{2-}$ is $+4$.

(A) In $NO_3^{-}$,let the oxidation number of $N$ be $x$. Since the oxidation number of $O$ is $-2$,we have: $x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5$.
In $NH_4^{+}$,let the oxidation number of $N$ be $y$. Since the oxidation number of $H$ is $+1$,we have: $y + 4(+1) = +1 \implies y + 4 = +1 \implies y = -3$.
Therefore,the oxidation number of nitrogen changes from $+5$ to $-3$.

(C) Let the oxidation state of phosphorus be $x$.
In the phosphate ion $(PO_4^{3-})$,the sum of oxidation states of all atoms equals the charge on the ion.
$x + 4 \times (-2) = -3$
$x - 8 = -3$
$x = -3 + 8$
$x = +5$
Therefore,the oxidation state of phosphorus is $+5$.

(B) For $CaC_2$: The oxidation state of $Ca$ is $+2$. Let the oxidation state of $C$ be $x$. Thus,$+2 + 2x = 0$,which gives $x = -1$.
For $K_2C_2O_4$: The oxidation state of $K$ is $+1$ and $O$ is $-2$. Let the oxidation state of $C$ be $x$. Thus,$2(+1) + 2x + 4(-2) = 0$,which simplifies to $2 + 2x - 8 = 0$,so $2x = 6$ or $x = +3$.
Therefore,the oxidation states are $-1$ and $+3$ respectively.

(B) The chemical formula for potassium permanganate is $KMnO_4$.
Oxidation number of $K = +1$.
Oxidation number of $O = -2$.
Since $KMnO_4$ is a neutral molecule,the sum of the oxidation numbers of all atoms must be $0$.
Let the oxidation number of $Mn$ be $x$.
$(+1) + x + 4 \times (-2) = 0$.
$1 + x - 8 = 0$.
$x - 7 = 0$.
$\therefore x = +7$.
Thus,the oxidation state of manganese in $KMnO_4$ is $+7$.

(B) $H_2SO_5$ is Peroxymonosulfuric acid (Caro's acid).
It contains one peroxide linkage $(-O-O-)$.
Let the oxidation number of sulfur be $x$.
The oxidation number of $H$ is $+1$,and for oxygen,there are three atoms with $-2$ and two atoms in the peroxide linkage with $-1$.
$(2 \times (+1)) + x + (3 \times (-2)) + (2 \times (-1)) = 0$
$2 + x - 6 - 2 = 0$
$x - 6 = 0$
$x = +6$

(C) The oxidation number of $F$ is $-1$ in all of its compounds.
Let the oxidation state of $O$ be $x$.
In $OF_2$,the sum of oxidation states is $x + 2(-1) = 0$.
Therefore,$x - 2 = 0$,which gives $x = +2$.

(D) In the given reaction,we need to find the change in the oxidation number of $Cr$ from the reactant side to the product side.
In the dichromate ion,$Cr_2O_7^{2-}$,let the oxidation number of $Cr$ be $x$.
$2x + 7(-2) = -2$
$2x - 14 = -2$
$2x = 12$
$x = +6$
In the product,the chromium ion is $Cr^{3+}$,so the oxidation number of $Cr$ is $+3$.
Therefore,the oxidation number of $Cr$ changes from $+6$ to $+3$.

(C) To find the change in oxidation number of selenium $(Se)$,we calculate the oxidation state of $Se$ in both the reactant and product species.

Reactant: $SeO_3^{2-}$	Product: $SeO_4^{2-}$
Let the oxidation state of $Se$ be $x$.	Let the oxidation state of $Se$ be $x$.
$x + 3(-2) = -2$	$x + 4(-2) = -2$
$x - 6 = -2$	$x - 8 = -2$
$x = +4$	$x = +6$

The oxidation number of selenium changes from $+4$ to $+6$.

(A) In $OF_2$,the electronegativity of fluorine is higher than that of oxygen.
Let the oxidation number of oxygen be $x$.
The oxidation number of fluorine is $-1$.
Applying the rule for the sum of oxidation numbers in a neutral molecule:
$x + 2(-1) = 0$
$x - 2 = 0$
$x = +2$
Therefore,the oxidation number of oxygen in $OF_2$ is $+2$.

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

(D) In the $O_2^{2-}$ ion (peroxide ion),let the oxidation state of oxygen be $x$.
Since the sum of oxidation states of all atoms in an ion is equal to the charge on the ion:
$2x = -2$
$x = -1$
Therefore,the oxidation state of oxygen in $O_2^{2-}$ is $-1$.

(A) In the molecule $N_3H$ (hydrazoic acid),the sum of the oxidation states of all atoms must be equal to zero.
Let the oxidation state of nitrogen be $x$.
Since there are $3$ nitrogen atoms and $1$ hydrogen atom (which has an oxidation state of $+1$),the equation is:
$3x + 1 = 0$
$3x = -1$
$x = -\frac{1}{3}$
Therefore,the oxidation state of nitrogen in $N_3H$ is $-\frac{1}{3}$.

(A) Let the oxidation state of $Cl$ be $x$.
The sum of oxidation states of all atoms in a neutral molecule is $0$.
For $KClO_3$: $(+1) + x + 3(-2) = 0$.
$1 + x - 6 = 0$.
$x - 5 = 0$.
$x = +5$.

(C) In a superoxide ion $(O_2^-)$,the total charge on the ion is $-1$.
Let the oxidation number of oxygen be $x$.
Since there are two oxygen atoms,the sum of their oxidation numbers must equal the charge of the ion:
$2x = -1$
$x = -\frac{1}{2}$
Therefore,the oxidation number of oxygen in superoxide is $-\frac{1}{2}$.

(B) The chemical formula for potassium dichromate is $K_2Cr_2O_7$.
Let the oxidation state of $Cr$ be $x$.
The sum of oxidation states of all atoms in a neutral molecule is $0$.
$2(+1) + 2(x) + 7(-2) = 0$
$2 + 2x - 14 = 0$
$2x - 12 = 0$
$2x = 12$
$x = +6$

(A) The ion $I_3^{-}$ is composed of three iodine atoms.
Let the average oxidation state of each iodine atom be $x$.
The sum of the oxidation states of all atoms in the ion must equal the charge of the ion.
Therefore,$3x = -1$.
Solving for $x$,we get $x = -(1/3)$.

(C) For $NO_3^{-}$: Let the oxidation number of $N$ be $x$.
$x + 3(-2) = -1$
$x - 6 = -1$
$x = +5$
For $NO_2$: Let the oxidation number of $N$ be $y$.
$y + 2(-2) = 0$
$y - 4 = 0$
$y = +4$
Thus,the oxidation number of nitrogen changes from $+5$ to $+4$.

(C) The sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge on the ion.
For the $S_{2}O_{3}^{2-}$ ion,the total charge is $-2$.
Therefore,the sum of the oxidation numbers of all atoms ($2$ sulfur atoms and $3$ oxygen atoms) is equal to $-2$.

(B) Let the oxidation state of $As$ be $x$.
For $H_{3}AsO_{4}$,the sum of oxidation states of all atoms is $0$.
$(3 \times (+1)) + x + (4 \times (-2)) = 0$
$3 + x - 8 = 0$
$x - 5 = 0$
$x = +5$
Therefore,the oxidation state of $As$ in $H_{3}AsO_{4}$ is $+5$.

(B) The sum of the oxidation states of all atoms in a polyatomic ion is equal to the net charge on the ion.
For the $Cr_{2}O_{7}^{2-}$ ion,the net charge is $-2$.
Therefore,the sum of the oxidation states of all atoms $(2 \times Cr + 7 \times O)$ is equal to $-2$.

(D) Let the oxidation number of $Cr$ be $x$.
For $K_{2}Cr_{2}O_{7}$,the sum of oxidation numbers of all atoms is $0$.
$(+1 \times 2) + (x \times 2) + (-2 \times 7) = 0$
$2 + 2x - 14 = 0$
$2x - 12 = 0$
$2x = 12$
$x = +6$
Therefore,the oxidation number of $Cr$ is $+6$.

(C) The chemical formula for chloric acid is $HClO_3$.
Let the oxidation state of chlorine be $x$.
The oxidation states of hydrogen $(H)$ and oxygen $(O)$ are $+1$ and $-2$ respectively.
Applying the rule for the sum of oxidation states in a neutral molecule:
$(+1) + x + 3(-2) = 0$
$1 + x - 6 = 0$
$x - 5 = 0$
$x = +5$
Therefore,the oxidation state of chlorine in chloric acid is $+5$.

(D) Let the oxidation number of $As$ be $x$.
The sum of oxidation numbers of all atoms in a neutral molecule is $0$.
For $H_{3}AsO_{3}$: $(3 \times (+1)) + x + (3 \times (-2)) = 0$
$3 + x - 6 = 0$
$x - 3 = 0$
$x = +3$
Therefore,the oxidation number of $As$ is $+3$.

(B) The chemical name for oil of vitriol is sulphuric acid,which has the formula $H_2SO_4$.
To find the oxidation state of sulphur $(S)$,let it be $x$.
The oxidation state of hydrogen $(H)$ is $+1$ and oxygen $(O)$ is $-2$.
Setting the sum of oxidation states to zero: $2(+1) + x + 4(-2) = 0$.
$2 + x - 8 = 0$.
$x - 6 = 0$.
$x = +6$.
Therefore,the oxidation state of sulphur in $H_2SO_4$ is $+6$.

(B)
Let the oxidation number of $Mn$ be $x$.
In $MnO_{4}^{2-}$,the oxidation number of oxygen is $-2$.
$x + 4 \times (-2) = -2$
$x - 8 = -2$
$x = +6$
Therefore,the oxidation number of $Mn$ is $+6$.

(A) Let the oxidation number of carbon be $x$.
For $CO_{3}^{2-}$: $x + 3(-2) = -2 \implies x - 6 = -2 \implies x = +4$.
For $CaC_{2}$: $Ca^{2+} + 2x = 0 \implies +2 + 2x = 0 \implies x = -1$.
For $CO$: $x + (-2) = 0 \implies x = +2$.
For $C_{2}O_{4}^{2-}$: $2x + 4(-2) = -2 \implies 2x - 8 = -2 \implies 2x = +6 \implies x = +3$.
Comparing the values,the highest oxidation number of carbon is $+4$,which is present in $CO_{3}^{2-}$.

(A) The chemical formula is $K_{2}C_{2}O_{4}$.
Let the oxidation number of $C$ be $x$.
The oxidation number of $K$ is $+1$ and $O$ is $-2$.
Sum of oxidation numbers in a neutral molecule is $0$.
$(2 \times (+1)) + (2 \times x) + (4 \times (-2)) = 0$
$2 + 2x - 8 = 0$
$2x - 6 = 0$
$2x = 6$
$x = +3$
Therefore,the oxidation number of $C$ is $+3$.

(D) The given reaction is $MnO_{4(aq)}^{-} + Br_{(aq)}^{-} \rightarrow MnO_{2(s)} + BrO_{3(aq)}^{-}$.
In $MnO_{4}^{-}$,the oxidation state of $Mn$ is $x + 4(-2) = -1$,so $x = +7$.
In $MnO_{2}$,the oxidation state of $Mn$ is $x + 2(-2) = 0$,so $x = +4$.
In $Br^{-}$,the oxidation state of $Br$ is $-1$.
In $BrO_{3}^{-}$,the oxidation state of $Br$ is $x + 3(-2) = -1$,so $x = +5$.
Therefore,the oxidation number of bromine changes from $Br^{-1}$ to $Br^{+5}$.

(C) Let the oxidation number of $C$ be $x$.
For $CH_{2}Cl_{2}$:
$x + 2(+1) + 2(-1) = 0$
$x + 2 - 2 = 0$
$x = 0$
For $CCl_{4}$:
$x + 4(-1) = 0$
$x - 4 = 0$
$x = 4$
Thus,the oxidation numbers are $0$ and $4$ respectively.

(C) The manganate ion is represented by the chemical formula $MnO_4^{2-}$.
In this ion,manganese is in the $+6$ oxidation state.
In contrast,the permanganate ion is represented by the formula $MnO_4^-$,where manganese is in the $+7$ oxidation state.
Therefore,the correct option is $C$.

(D) Chromite ore is $FeCr_2O_4$,which can be represented as $FeO \cdot Cr_2O_3$.
In this compound,the oxidation number of $Fe$ is $+2$ and the oxidation number of $Cr$ is $+3$.

(D) The structure of $H_2S_2O_8$ (peroxodisulphuric acid) contains a peroxide linkage $(-O-O-)$.
In this structure,there are two sulphur atoms,two hydroxyl groups $(-OH)$,two peroxide oxygen atoms (oxidation state $-1$ each),and six terminal oxygen atoms (oxidation state $-2$ each).
Let the oxidation state of sulphur be $x$.
The sum of oxidation states in a neutral molecule is zero:
$2(+1) + 2(x) + 2(-1) + 6(-2) = 0$
$2 + 2x - 2 - 12 = 0$
$2x - 12 = 0$
$2x = 12$
$x = +6$
Thus,the oxidation state of $S$ in $H_2S_2O_8$ is $+6$.

(A) In $NH_4NO_3$,the nitrogen element exhibits two different oxidation states.
$NH_4NO_3$ is an ionic compound consisting of $NH_4^+$ and $NO_3^-$ ions.
For $NH_4^+$: Let the oxidation state of $N$ be $x$. Then $x + 4(+1) = +1$,which gives $x = -3$.
For $NO_3^-$: Let the oxidation state of $N$ be $y$. Then $y + 3(-2) = -1$,which gives $y = +5$.
Thus,nitrogen exists in $-3$ and $+5$ oxidation states in $NH_4NO_3$.

(B) $NH_4NO_3 \longrightarrow NH_4^+ + NO_3^-$
In $NH_4^+$,let the oxidation number of $N$ be $x$.
$x + 4(+1) = +1 \implies x = -3$.
In $NO_3^-$,let the oxidation number of $N$ be $y$.
$y + 3(-2) = -1 \implies y - 6 = -1 \implies y = +5$.
Thus,the oxidation numbers of nitrogen atoms in $NH_4NO_3$ are $-3$ and $+5$.

(A) Oleum is formed by the reaction of $H_2SO_4$ $(X)$ and $SO_3$ $(Y)$:
$H_2\stackrel{+6}{S}O_4 + \stackrel{+6}{S}O_3 \rightarrow H_2S_2O_7$ (oleum).
In $H_2SO_4$,the oxidation state of sulfur $(S)$ is $+6$.
In $SO_3$,the oxidation state of sulfur $(S)$ is $+6$.
Sum of oxidation states $= 6 + 6 = 12$.

(B) The structure of the tetrathionate ion $(S_4O_6^{2-})$ consists of a chain of four sulphur atoms.
In this structure,the two central sulphur atoms are bonded to each other and to other sulphur atoms,so their oxidation state is $0$.
The two terminal sulphur atoms are each bonded to three oxygen atoms (two via double bonds and one via a single bond) and one sulphur atom.
For each terminal sulphur atom: $x + 2(-2) + 1(-1) = -1$ (since the terminal group is $-SO_3^-$),which simplifies to $x - 5 = -1$,so $x = +5$.
Thus,the oxidation numbers of the four $S$ atoms are $+5, 0, 0, +5$.

(B) The structure of $Br_3O_8$ consists of three bromine atoms linked in a chain,with oxygen atoms bonded to them.
Terminal bromine atoms ($I$ and $III$) are each bonded to three oxygen atoms via double bonds. Since oxygen is more electronegative than bromine,each $Br-O$ bond contributes $+2$ to the oxidation state of bromine. Thus,for terminal bromine atoms: $3 \times (+2) = +6$.
The central bromine atom $(II)$ is bonded to two oxygen atoms via double bonds. Thus,its oxidation state is $2 \times (+2) = +4$.
Therefore,the oxidation states of the three $Br$ atoms are $+6, +4, +6$.

(A) This is a redox reaction in which sulphur is oxidized and bromine is reduced according to the following balanced equation:
$SO_3^{2-}{_{\text{(aq)}}} + Br_{2\text{(l)}} + H_2O_{\text{(l)}} \rightarrow SO_4^{2-}{_{\text{(aq)}}} + 2Br^{-}{_{\text{(aq)}}} + 2H^{+}{_{\text{(aq)}}}$
The sulphur-containing product is the sulphate ion,$SO_4^{2-}$.
To find the oxidation state of $S$ in $SO_4^{2-}$,let it be $x$.
The sum of oxidation states of all atoms in an ion is equal to the charge on the ion:
$x + (4 \times -2) = -2$
$x - 8 = -2$
$x = +6$
Therefore,the oxidation state of $S$ in the product is $+6$.

(A) The oxidation number of hydrogen is usually $+1$ in compounds with more electronegative atoms (as in $H_2O$).
However,in metal hydrides like lithium hydride $(LiH)$,the oxidation state of the hydrogen atom is $-1$.
Therefore,the statement that the oxidation number of hydrogen is always $+1$ is incorrect.

(B) To find the oxidation state of $S$ in the given compounds:
$1$. $H_2S_2O_8$ (Peroxodisulphuric acid): $2(+1) + 2(x) + 8(-2) = 0$ $\Rightarrow 2x = 14$ $\Rightarrow x = +6$.
$2$. $H_2SO_5$ (Peroxomonosulphuric acid): $2(+1) + x + 5(-2) = 0 \Rightarrow x = +8$ (Incorrect,structure shows $S$ is $+6$ due to peroxide linkage).
$3$. $H_2SO_3$ (Sulphurous acid): $2(+1) + x + 3(-2) = 0 \Rightarrow x = +4$.
$4$. $H_2SO_4$ (Sulphuric acid): $2(+1) + x + 4(-2) = 0 \Rightarrow x = +6$.
$5$. $H_2S_2O_7$ (Pyrosulphuric acid): $2(+1) + 2(x) + 7(-2) = 0$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$.
$6$. $SO_2Cl_2$ (Sulphuryl chloride): $x + 2(-2) + 2(-1) = 0 \Rightarrow x = +6$.
$7$. $SOCl_2$ (Thionyl chloride): $x + (-2) + 2(-1) = 0 \Rightarrow x = +4$.
Thus,the compounds with $S$ in $+6$ oxidation state are $H_2S_2O_8, H_2SO_5, H_2SO_4, H_2S_2O_7, SO_2Cl_2$. The total count is $5$.