Oxidation number and Oxidation state Questions in English

(B) The balanced chemical equation for the reaction is:
$K_2Cr_2O_7 + 4H_2SO_4 + 3H_2S \to K_2SO_4 + Cr_2(SO_4)_3 + 7H_2O + 3S$
In $K_2Cr_2O_7$,the oxidation state of $Cr$ is $+6$.
In $Cr_2(SO_4)_3$,the oxidation state of $Cr$ is $+3$.
Therefore,the oxidation number of chromium decreases from $+6$ to $+3$.

(D) The equivalent weight of an oxidizing agent is calculated as: $\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}}$.
In an acidic medium,$KMnO_4$ undergoes the following reduction reaction: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$.
Here,the change in oxidation state of $Mn$ is from $+7$ to $+2$,which involves a transfer of $5$ electrons.
Therefore,the n-factor is $5$.
Thus,the equivalent weight of $KMnO_4$ is $\frac{1}{5} \times \text{Molecular weight}$.

(C) In $KMnO_4$,the oxidation state of $Mn$ is calculated as: $x + 1 + 4(-2) = 0$,so $x = +7$.
In acidic medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $Mn^{2+}$ ions.
The balanced chemical equation is: $2KMnO_4 + 3H_2SO_4 + 5H_2C_2O_4 \to K_2SO_4 + 2MnSO_4 + 8H_2O + 10CO_2$.
Thus,the oxidation number of $Mn$ changes from $+7$ to $+2$.

(A) The chemical reaction for the conversion of $KMnO_4$ to $K_2MnO_4$ is:
$KMnO_4 + e^- \rightarrow K_2MnO_4$ (in basic medium).
In $KMnO_4$,the oxidation state of $Mn$ is $+7$.
In $K_2MnO_4$,the oxidation state of $Mn$ is $+6$.
The change in oxidation state is $|7 - 6| = 1$.
Thus,the number of electrons gained per molecule is $1$.
Equivalent weight = $\frac{\text{Molecular weight}}{\text{n-factor}} = \frac{M}{1} = M$.

(C) $1. [Co(NH_3)_5Cl]Cl_2$: In $NH_3$,the oxidation state of $N$ is $-3$. The given value $0$ is for the neutral molecule $NH_3$,not for $N$ itself. However,checking other options for the most incorrect value.
$2. NH_2OH$: $x + 3(+1) - 2 = 0 \implies x = -1$. This is correct.
$3. (N_2H_5)_2SO_4$: $2(2x + 5) - 2 = 0 \implies 4x + 10 - 2 = 0 \implies 4x = -8 \implies x = -2$. The given value $+2$ is incorrect.
$4. Mg_3N_2$: $3(+2) + 2x = 0 \implies 6 + 2x = 0 \implies x = -3$. This is correct.
Thus,option $C$ is the incorrectly stated one.

(B) In an alkaline medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $MnO_2$.
The chemical reaction is: $2KMnO_4 + H_2O \rightarrow 2MnO_2 + 2KOH + 3[O]$.
The change in the oxidation state of $Mn$ is from $+7$ to $+4$,so the change in oxidation number is $7 - 4 = 3$.
The equivalent weight is calculated as: $\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} = \frac{\text{Molecular weight}}{3}$.

(B) In an alkaline medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $MnO_4^{2-}$ (manganate ion).
The chemical reaction is: $MnO_4^- + e^- \rightarrow MnO_4^{2-}$.
The change in oxidation state of $Mn$ is from $+7$ to $+6$,so the change in oxidation number is $1$.
The equivalent weight is calculated as: $\text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}}$.
The molar mass of $KMnO_4$ is $158 \ g/mol$.
The n-factor is $1$.
Therefore,$\text{Equivalent weight} = \frac{158}{1} = 158$.

(D) Let us calculate the oxidation state $(X)$ of nitrogen in each compound:
$1$. In $[Co(NH_3)_5Cl]Cl_2$,$NH_3$ is a neutral ligand. The oxidation state of $N$ in $NH_3$ is $X + 3(+1) = 0$,so $X = -3$.
$2$. In $NH_2OH$,$X + 2(+1) + 1(-2) + 1(+1) = 0$,so $X + 1 = 0$,which means $X = -1$.
$3$. In $N_2H_5^+$,$2(X) + 5(+1) = +1$,so $2X = -4$,which means $X = -2$.
$4$. In $Mg_3N_2$,$3(+2) + 2(X) = 0$,so $6 + 2X = 0$,which means $X = -3$.
Thus,the correct representation is in $Mg_3N_2$ where the oxidation state of $N$ is $-3$.

(A) In $F_2O$,fluorine is more electronegative than oxygen.
According to the rules for assigning oxidation numbers,fluorine is assigned an oxidation state of $-1$ in its compounds.
Let the oxidation number of oxygen be $x$.
$2(-1) + x = 0$
$x - 2 = 0$
$x = +2$.
Thus,the oxidation number of fluorine is $-1$.

(D) In the molecule $N_3H$,the oxidation number of hydrogen is $+1$.
Let the oxidation number of nitrogen be $X$.
Since the total oxidation state of a neutral molecule is $0$,we have:
$3(X) + 1(+1) = 0$
$3X + 1 = 0$
$3X = -1$
$X = -\frac{1}{3}$
Therefore,the oxidation number of nitrogen in $N_3H$ is $-\frac{1}{3}$.

(D) Let the oxidation number of $P$ be $x$.
In $KH_2PO_2$,the oxidation numbers of other elements are:
$K = +1$
$H = +1$
$O = -2$
Sum of oxidation numbers = $0$
$(+1) + 2(+1) + x + 2(-2) = 0$
$1 + 2 + x - 4 = 0$
$x - 1 = 0$
$x = +1$
Therefore,the oxidation number of $P$ is $+1$.

(B) The oxidation state of phosphorus $(X)$ in phosphorous acid $(H_3PO_3)$ is calculated as follows:
$3(+1) + X + 3(-2) = 0$
$3 + X - 6 = 0$
$X - 3 = 0$
$X = +3$
Therefore,phosphorus has an oxidation state of $+3$ in phosphorous acid $(H_3PO_3)$.

(C) The correct formula is $A_3(BC_4)_2$.
For a neutral compound,the sum of the oxidation states of all atoms must be zero.
Let the formula be $A_x(BC_4)_y$. The oxidation state of the group $(BC_4)$ is $(+5) + 4(-2) = 5 - 8 = -3$.
To make the compound neutral,the total charge must be $x(+2) + y(-3) = 0$.
For $x=3$ and $y=2$,we get $3(+2) + 2(-3) = 6 - 6 = 0$.
Thus,the formula is $A_3(BC_4)_2$.

(C) Let the oxidation number of carbon be $X$.
In $CH_2O$,the oxidation number of hydrogen $(H)$ is $+1$ and oxygen $(O)$ is $-2$.
The sum of oxidation numbers in a neutral molecule is $0$.
$X + 2(+1) + 1(-2) = 0$
$X + 2 - 2 = 0$
$X = 0$
Therefore,the oxidation number of carbon in $CH_2O$ is $0$.

(C) The chemical formula is $Mg_3(PO_4)_2$. Let the oxidation number of $P$ be $X$.
Assigning oxidation states: $Mg = +2$,$O = -2$.
Sum of oxidation numbers in a neutral compound is $0$.
$3(+2) + 2(X + 4(-2)) = 0$
$6 + 2(X - 8) = 0$
$6 + 2X - 16 = 0$
$2X - 10 = 0$
$2X = 10$
$X = +5$

(D) The oxidation number of $Cl$ in the given compounds is calculated as follows:
For $HCl$: $1(+1) + X = 0 \Rightarrow X = -1$
For $HClO_4$: $1(+1) + X + 4(-2) = 0 \Rightarrow X = +7$
For $ICl$: $1(+1) + X = 0 \Rightarrow X = -1$
For $Cl_2O$: $2(X) + 1(-2) = 0$ $\Rightarrow 2X = +2$ $\Rightarrow X = +1$
Thus,in $Cl_2O$,the oxidation state of chlorine is $+1$.

(B) The oxidation state of $Na$ is $+1$ and $O$ is $-2$.
Let the oxidation number of $S$ be $x$.
For the molecule $Na_2S_4O_6$,the sum of oxidation numbers is $0$.
$2(+1) + 4(x) + 6(-2) = 0$
$2 + 4x - 12 = 0$
$4x = 10$
$x = \frac{10}{4} = 2.5$
Thus,the average oxidation number of sulfur is $2.5$.

(A) Let the oxidation number of carbon be $X$.
In $CH_2Cl_2$,the oxidation number of hydrogen $(H)$ is $+1$ and chlorine $(Cl)$ is $-1$.
Applying the rule for the sum of oxidation numbers in a neutral molecule:
$X + 2(+1) + 2(-1) = 0$
$X + 2 - 2 = 0$
$X = 0$
Therefore,the oxidation number of carbon in $CH_2Cl_2$ is $0$.

(B) The oxidation state of $S$ is calculated as follows:
For $SO_3^{2-}$: $x + 3(-2) = -2 \implies x = +4$
For ${S_2}O_4^{2-}$: $2x + 4(-2) = -2 \implies 2x = +6 \implies x = +3$
For ${S_2}O_6^{2-}$: $2x + 6(-2) = -2 \implies 2x = +10 \implies x = +5$
Comparing the oxidation states: $+3 < +4 < +5$.
Thus,the order is ${S_2}O_4^{2-} < SO_3^{2-} < {S_2}O_6^{2-}$.

(A) The ferrous ion is represented as $Fe^{2+}$.
Therefore,the charge on a ferrous ion is $+2$.

(C) The reaction for the oxidation of hydrazine is: $N_2H_4 \rightarrow N_2^X + n e^-$.
In $N_2H_4$,the oxidation state of $N$ is $-2$.
Let the oxidation state of $N$ in compound $X$ be $X$.
The change in oxidation state per mole of $N_2H_4$ is: $n = 2 \times (X - (-2)) = 2(X + 2)$.
Total electrons lost by $2.5 \ mol$ of $N_2H_4$ is: $2.5 \times 2(X + 2) = 5(X + 2)$.
Given that the total electrons lost is $25 \ mol$,we have: $5(X + 2) = 25$.
$X + 2 = 5$.
$X = +3$.
Thus,the oxidation state of $N$ in compound $X$ is $+3$.

(B) Let the oxidation number of $N$ be $x$.
In hydrazine $(NH_2-NH_2)$,the oxidation number of $H$ is $+1$.
The sum of oxidation numbers of all atoms in a neutral molecule is $0$.
$2x + 4(+1) = 0$
$2x + 4 = 0$
$2x = -4$
$x = -2$
Therefore,the oxidation number of $N$ is $-2$.

(D) Let the oxidation state of $Fe$ be $x$.
In the compound $Fe_{0.94}O$,the total charge must be zero.
The oxidation state of oxygen $(O)$ is $-2$.
Therefore,$0.94(x) + 1(-2) = 0$.
$0.94x = 2$.
$x = \frac{2}{0.94}$.

(D) The oxidation numbers of $Mn$ in the given compounds are calculated as follows:
$MnCl_2$: $x + 2(-1) = 0 \Rightarrow x = +2$
$MnO_2$: $x + 2(-2) = 0 \Rightarrow x = +4$
$Mn(OH)_3$: $x + 3(-1) = 0 \Rightarrow x = +3$
$KMnO_4$: $1(+1) + x + 4(-2) = 0 \Rightarrow x = +7$
Thus,the increasing order of the oxidation number of $Mn$ is: $MnCl_2 (+2) < Mn(OH)_3 (+3) < MnO_2 (+4) < KMnO_4 (+7)$.

(A) Let the oxidation number of $Xe$ be $X$.
In $Ba_2XeO_6$,the oxidation number of $Ba$ is $+2$ and the oxidation number of $O$ is $-2$.
Applying the rule that the sum of oxidation numbers in a neutral compound is $0$:
$2(+2) + X + 6(-2) = 0$
$4 + X - 12 = 0$
$X - 8 = 0$
$X = +8$
Therefore,the oxidation number of $Xe$ is $+8$.

(C) To find the oxidation number $(x)$ of iodine in each compound:
$ICl$: $x + (-1) = 0 \Rightarrow x = +1$
$HI$: $(+1) + x = 0 \Rightarrow x = -1$
$I_2$: $2x = 0 \Rightarrow x = 0$
$HIO_4$: $(+1) + x + 4(-2) = 0$ $\Rightarrow x - 7 = 0$ $\Rightarrow x = +7$
Comparing the oxidation numbers: $+7 (HIO_4) > +1 (ICl) > 0 (I_2) > -1 (HI)$.
Thus,the decreasing order is $HIO_4 > ICl > I_2 > HI$.

(A) Let the oxidation number of $Cl$ be $X$.
For the ion $ClO_3^-$,the sum of oxidation numbers of all atoms is equal to the charge on the ion.
$X + 3 \times (-2) = -1$
$X - 6 = -1$
$X = -1 + 6$
$X = +5$
Therefore,the oxidation number of chlorine in $ClO_3^-$ is $+5$.

(C) The structure of $H_2S_2O_8$ (peroxodisulfuric acid) is $HO-SO_2-O-O-SO_2-OH$.
In this structure,there is a peroxide linkage $(-O-O-)$ where the oxidation state of each oxygen atom is $-1$.
The remaining six oxygen atoms are bonded to sulfur,each having an oxidation state of $-2$.
Let the oxidation number of sulfur be $x$.
Applying the sum of oxidation states equal to zero:
$2(+1) + 2(x) + 2(-1) + 6(-2) = 0$
$2 + 2x - 2 - 12 = 0$
$2x - 12 = 0$
$2x = 12$
$x = +6$

(C) Let the oxidation number of $Fe$ be $x$.
In $Fe_3O_4$,the oxidation number of $O$ is $-2$.
Applying the rule for the sum of oxidation numbers in a neutral molecule:
$3(x) + 4(-2) = 0$
$3x - 8 = 0$
$3x = 8$
$x = \frac{8}{3}$
Thus,the average oxidation number of $Fe$ in $Fe_3O_4$ is $\frac{8}{3}$.
Actually,$Fe_3O_4$ is a mixed oxide of $FeO$ and $Fe_2O_3$,where $Fe$ exhibits oxidation states of $+2$ (in $FeO$) and $+3$ (in $Fe_2O_3$).

(B) The reaction is: $BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2$.
The products are $BaSO_4$ and $H_2O_2$.
The most electronegative element in these products is oxygen $(O)$.
In $BaSO_4$,the oxidation state of $O$ is calculated as: $(+2) + (+6) + 4(x) = 0$,which gives $4x = -8$,so $x = -2$.
In $H_2O_2$,the oxidation state of $O$ is calculated as: $2(+1) + 2(x) = 0$,which gives $2x = -2$,so $x = -1$.
Thus,the oxidation states of oxygen are $-2$ and $-1$.

(A) $1$. For $S_8$: Since it is an elemental form of sulfur,its oxidation number 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 $2(x) + 2(-1) = 0$,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 $x = -2$.
Therefore,the oxidation numbers are $0, +1, -2$ respectively.

(D) To find the oxidation state of the central atom in $CrO_2Cl_2$,let the oxidation state of $Cr$ be $x$.
Given that the oxidation state of oxygen $(O)$ is $-2$ and chlorine $(Cl)$ is $-1$:
$x + 2(-2) + 2(-1) = 0$
$x - 4 - 2 = 0$
$x - 6 = 0$
$x = +6$
Therefore,in $CrO_2Cl_2$,the oxidation state of $Cr$ is $+6$.

(D) The chemical formula $CaOCl_2$ represents calcium oxychloride,which is an ionic compound containing $Ca^{2+}$,$Cl^-$,and $ClO^-$ ions.
In the $Cl^-$ ion,the oxidation number of $Cl$ is $-1$.
In the $ClO^-$ ion,the oxidation number of $Cl$ is $+1$ (since oxygen is $-2$,$x + (-2) = -1$,so $x = +1$).
Therefore,the oxidation numbers of chlorine in $CaOCl_2$ are $+1$ and $-1$.

(B) For $SO_3^{2-}$: Let the oxidation state of $S$ be $x$. Then $x + 3(-2) = -2$,so $x = +4$.
For $S_2O_4^{2-}$: Let the oxidation state of $S$ be $x$. Then $2x + 4(-2) = -2$,so $2x = +6$,$x = +3$.
For $S_2O_6^{2-}$: Let the oxidation state of $S$ be $x$. Then $2x + 6(-2) = -2$,so $2x = +10$,$x = +5$.
The oxidation states are $+4$,$+3$,and $+5$ respectively.
Thus,the correct order is $S_2O_4^{2-} (+3) < SO_3^{2-} (+4) < S_2O_6^{2-} (+5)$.

(D) For $PO_4^{3-}$: $P + 4(-2) = -3 \implies P - 8 = -3 \implies P = +5$.
For $SO_4^{2-}$: $S + 4(-2) = -2 \implies S - 8 = -2 \implies S = +6$.
For $Cr_2O_7^{2-}$: $2Cr + 7(-2) = -2 \implies 2Cr - 14 = -2 \implies 2Cr = +12 \implies Cr = +6$.
Thus,the oxidation states are $+5, +6, +6$.

(D) In the peroxide ion $O_2^{2-}$,the sum of the oxidation numbers of all atoms must equal the total charge of the ion.
Let the oxidation number of each oxygen atom be $x$.
Then,$2x = -2$.
Solving for $x$,we get $x = -1$.
Therefore,the oxidation number of each oxygen atom in the $O_2^{2-}$ ion is $-1$.

(C) In metal hydrides,hydrogen exists as a hydride ion $(H^-)$.
Since $K$,$Mg$,and $Na$ are metals,$KH$,$MgH_2$,and $NaH$ are ionic metal hydrides.
Therefore,the oxidation number of hydrogen in all these compounds is $-1$.

(D) 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 $0$.
$2(K) + 2(Cr) + 7(O) = 0$
$2(+1) + 2(x) + 7(-2) = 0$
$2 + 2x - 14 = 0$
$2x - 12 = 0$
$2x = 12$
$x = +6$
Therefore,the oxidation state of $Cr$ is $6$.

(D) Let the oxidation number of $S$ in $Na_2S_4O_6$ be $x$.
Applying the rule for the sum of oxidation states in a neutral molecule:
$2 \times (+1) + 4x + 6 \times (-2) = 0$
$2 + 4x - 12 = 0$
$4x = 10$
$x = \frac{10}{4} = 2.5$

(D) Let the oxidation state of $Os$ in $OsO_4$ be $x$.
The oxidation state of oxygen $(O)$ is $-2$.
Setting up the equation: $x + 4(-2) = 0$.
$x - 8 = 0$.
$x = +8$.

(D) In $S_8$,the oxidation state of $S$ is $0$ because it is in its elemental form.
In $H_2S$,let the oxidation state of $S$ be $x$. Then,$2(+1) + x = 0$,which gives $x = -2$.
In $S_2F_2$,let the oxidation state of $S$ be $x$. Since $F$ has an oxidation state of $-1$,we have $2x + 2(-1) = 0$,which gives $2x = 2$,so $x = +1$.
Therefore,the oxidation states are $0, -2, +1$.

(D) Let the oxidation state of $Fe$ in $Fe_3O_4$ be $x$.
Since the oxidation state of oxygen is $-2$,the sum of oxidation states in a neutral molecule is $0$.
$3(x) + 4(-2) = 0$
$3x - 8 = 0$
$3x = 8$
$x = +8/3$
Thus,the oxidation state of $Fe$ in $Fe_3O_4$ is $+8/3$.

(A) Oxygen is the second most electronegative element after fluorine.
Therefore,in compounds of oxygen with fluorine,oxygen exhibits a positive oxidation state.
Since the electronegativity of fluorine is higher than that of oxygen,fluorine pulls the electron density towards itself,forcing oxygen into a positive oxidation state.
For example,in $OF_2$,the oxidation state of oxygen is $+2$.

(C) The structure of sodium tetrathionate $(Na_2S_4O_6)$ is $Na^+[O_3S-S-S-SO_3]^-Na^+$.
In this structure,the two terminal sulfur atoms have an oxidation state of $+5$,while the two central sulfur atoms have an oxidation state of $0$.
The average oxidation number is calculated as $\frac{5 + 0 + 0 + 5}{4} = \frac{10}{4} = 2.5$.
The reason is false because the two central sulfur atoms are not directly bonded to oxygen atoms,but the assertion is true.

(A) Graphite is an allotrope of carbon,which consists of only carbon atoms in their elemental state.
According to the rules for assigning oxidation numbers,the oxidation number of an element in its free or uncombined state is always $0$.

(B) The equivalent weight $(E)$ is given by $E = \frac{M}{n}$,where $M$ is the molecular weight and $n$ is the change in oxidation state per molecule.
Given $E = \frac{M}{2}$,it implies $n = 2$.
In $MnSO_4$,the oxidation state of $Mn$ is $+2$.
For $MnO_2$,the oxidation state of $Mn$ is $+4$.
Change in oxidation state = $|4 - 2| = 2$.
Since the change in oxidation state is $2$,the equivalent weight is half of the molecular weight.

(C) To determine the paramagnetic nature,we look at the oxidation state of $Mn$ in the given acids.
$HMnO_4$: $Mn$ is in $+7$ oxidation state ($d^0$ configuration),which is diamagnetic.
$H_2MnO_4$: $Mn$ is in $+6$ oxidation state ($d^1$ configuration). Since it has one unpaired electron,it is paramagnetic.
Therefore,$H_2MnO_4$ is the paramagnetic acid.

(C) Let the oxidation state of $Cr$ be $x$.
In $K_2Cr_2O_7$,the oxidation state of $K$ is $+1$ and $O$ is $-2$.
The sum of oxidation states in a neutral molecule is zero.
$2(+1) + 2(x) + 7(-2) = 0$
$2 + 2x - 14 = 0$
$2x - 12 = 0$
$2x = 12$
$x = +6$
Therefore,the oxidation state of $Cr$ is $+6$.

(A) The chemical reaction for the conversion is: $\mathop {KMnO_4}\limits^{+7} \to \mathop {K_2MnO_4}\limits^{+6}$.
The change in the oxidation state of Manganese $(Mn)$ per atom is: $7 - 6 = 1$.
The equivalent weight is calculated as: $\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}}$.
Here,the n-factor is the change in oxidation state,which is $1$.
Therefore,the equivalent weight $= \frac{M}{1} = M$.

(C) In $Na_{2}O_{2}$,sodium $(Na)$ is in the $+1$ oxidation state.
Let the oxidation state of oxygen be $x$.
Since the molecule is neutral,the sum of oxidation states is zero: $2(+1) + 2(x) = 0$.
$2 + 2x = 0$
$2x = -2$
$x = -1$.
Therefore,each oxygen atom in $Na_{2}O_{2}$ has an oxidation state of $-1$.