Oxidation number and Oxidation state Questions in English

(B) The equivalent weight of a substance is given by the formula: $\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}}$.
For the equivalent weight to be half the molecular weight,the n-factor must be $2$.
The n-factor represents the change in oxidation state of the central metal atom.
In $MnSO_4$,the oxidation state of $Mn$ is $+2$.
In $MnO_2$,the oxidation state of $Mn$ is $+4$.
The change in oxidation state is $|4 - 2| = 2$.
Thus,for the conversion $MnSO_4 \to MnO_2$,the n-factor is $2$,making the equivalent weight $\frac{M}{2}$.

(D) In an acidic medium,$KMnO_4$ acts as an oxidizing agent and undergoes reduction as follows:
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
Here,the oxidation state of $Mn$ changes from $+7$ to $+2$,which corresponds to a change of $5$ electrons per molecule of $KMnO_4$.
Equivalent weight is calculated as:
$\text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}}$
Since the n-factor is $5$,the equivalent weight is $\frac{1}{5}$ of its molecular weight.
Therefore,the correct option is $D$.

(A) For a compound to be neutral,the sum of the oxidation numbers of all atoms must be $0$.
Check option $A$: $A_3(BC_4)_2$
Sum $= 3 \times (+2) + 2 \times [1 \times (+5) + 4 \times (-2)]$
Sum $= 6 + 2 \times [5 - 8]$
Sum $= 6 + 2 \times (-3) = 6 - 6 = 0$.
Since the sum is $0$,the formula $A_3(BC_4)_2$ is correct.

(A) To calculate the average oxidation state of sulphur $(x)$ in $Na_2S_2O_3$:
$2(+1) + 2(x) + 3(-2) = 0$
$2 + 2x - 6 = 0$
$2x - 4 = 0$
$2x = 4$
$x = +2$
Therefore,the average oxidation state of sulphur is $+2$.

(C) The net charge on the compound must be zero,meaning the sum of the oxidation states of all atoms must be $0$.
Given: Oxidation number of $X = +2$,$Y = +5$,and $Z = -2$.
Let us check each option:
$A) XYZ_2: (+2) + (+5) + 2 \times (-2) = 2 + 5 - 4 = +3 \neq 0$
$B) X_2(YZ_3)_2: 2 \times (+2) + 2 \times (5 + 3 \times (-2)) = 4 + 2 \times (5 - 6) = 4 - 2 = +2 \neq 0$
$C) X_3(YZ_4)_2: 3 \times (+2) + 2 \times (5 + 4 \times (-2)) = 6 + 2 \times (5 - 8) = 6 + 2 \times (-3) = 6 - 6 = 0$
$D) X_3(Y_4Z)_2: 3 \times (+2) + 2 \times (4 \times 5 - 2) = 6 + 2 \times (18) = 42 \neq 0$
Thus,the correct formula is $X_3(YZ_4)_2$.

(C) In the reaction $CrO_4^{2-} \to Cr_2O_7^{2-}$,we calculate the oxidation state of $Cr$ in both species:
For $CrO_4^{2-}$: $x + 4(-2) = -2 \implies x = +6$.
For $Cr_2O_7^{2-}$: $2x + 7(-2) = -2 \implies 2x = 12 \implies x = +6$.
Since the oxidation state of $Cr$ remains $+6$ in both reactants and products,there is no change in the oxidation number.
Therefore,this reaction involves neither oxidation nor reduction.

(D) Let the oxidation number of $C$ be $x$.
In $CO_2$,the oxidation number of $O$ is $-2$.
The sum of oxidation numbers in a neutral molecule is $0$.
$x + 2(-2) = 0$
$x - 4 = 0$
$x = +4$
Therefore,the oxidation number of $C$ is $+4$.

(B) Arsenic $(As)$ belongs to group $15$ of the periodic table.
Elements in group $15$ have a valence shell electronic configuration of $ns^2 np^3$.
They can lose three electrons to show a $+3$ oxidation state or lose all five valence electrons to show a $+5$ oxidation state.
Therefore,$As$ exhibits both $+3$ and $+5$ oxidation states.

(B) Barium peroxide is represented as $BaO_2$.
In peroxides,the oxidation state of oxygen is $-1$.
Let the oxidation number of $Ba$ be $x$.
For the neutral molecule $BaO_2$,the sum of oxidation numbers is $0$:
$x + 2(-1) = 0$
$x - 2 = 0$
$x = +2$.
Thus,the oxidation number of $Ba$ is $+2$.

(D) To find the oxidation state of $Cl$ in $Cl_2O$:
Let the oxidation state of $Cl$ be $x$.
Since the oxidation state of $O$ is $-2$,we have:
$2x + (-2) = 0$
$2x = 2$
$x = +1$
Therefore,in $Cl_2O$,chlorine is in the $+1$ oxidation state.

(C) In the elemental form $Br_2$,the oxidation state of bromine is $0$.
In the ion $BrO_3^-$,let the oxidation state of $Br$ be $x$.
$x + 3(-2) = -1$
$x - 6 = -1$
$x = +5$.
Therefore,the oxidation state of bromine changes from $0$ to $+5$.

(C) In the reaction $Cl_2 + H_2S \to 2HCl + S$,the oxidation state of sulphur in $H_2S$ is calculated as $2(+1) + x = 0$,which gives $x = -2$.
In the product side,sulphur is in its elemental form $S$,so its oxidation state is $0$.
Thus,the oxidation number of sulphur changes from $-2$ to $0$.

(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 $Cr$ remains $+6$ in both compounds,the change in the oxidation state is $6 - 6 = 0$.

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

(A) For a monatomic ion,the oxidation number is equal to the charge on the ion.
In the case of the sulfide ion $(S^{2-})$,the charge is $-2$.
Therefore,the oxidation number of $S$ in $S^{2-}$ is $-2$.

(D) In $(NH_4)_2SO_4$,the compound dissociates into $2NH_4^+$ and $SO_4^{2-}$ ions.
We need to find the oxidation number of $N$ in the ammonium ion,$NH_4^+$.
Let the oxidation number of $N$ be $x$.
The oxidation number of $H$ is $+1$.
For the $NH_4^+$ ion,the sum of oxidation numbers equals the charge on the ion:
$x + 4(+1) = +1$
$x + 4 = +1$
$x = 1 - 4 = -3$
Therefore,the oxidation number of $N$ is $-3$.

(B) Let the oxidation state of nitrogen be $x$.
In $N_2O$,$2x + (-2) = 0$,so $2x = 2$,which gives $x = +1$.
In $NO$,$x + (-2) = 0$,so $x = +2$.
In $NH_2OH$,$x + 2(+1) + (-2) + (+1) = 0$,so $x + 1 = 0$,which gives $x = -1$.
In $N_2H_4$,$2x + 4(+1) = 0$,so $2x = -4$,which gives $x = -2$.
Therefore,the correct option is $(B)$.

(D) The oxidation number of $H$ is $+1$ and $O$ is $-2$.
Let the oxidation number of $S$ be $x$.
For the molecule $H_2SO_4$,the sum of oxidation numbers is $0$.
$2(+1) + x + 4(-2) = 0$
$2 + x - 8 = 0$
$x - 6 = 0$
$x = +6$
Therefore,the oxidation number of sulphur in $H_2SO_4$ is $+6$.

(D) The chemical formula for perchloric acid is $HClO_4$.
Let the oxidation state of chlorine $(Cl)$ be $x$.
The oxidation state of hydrogen $(H)$ is $+1$ and oxygen $(O)$ is $-2$.
Applying the sum of oxidation states equal to zero:
$1 + x + (4 \times -2) = 0$
$1 + x - 8 = 0$
$x - 7 = 0$
$x = +7$.
Therefore,the oxidation state of chlorine in $HClO_4$ is $+7$.

(D) Let the oxidation number of $N$ be $x$.
The oxidation number of $H$ is $+1$ and that of $O$ is $-2$.
For the neutral molecule $HNO_3$,the sum of oxidation numbers is $0$.
$(+1) + x + 3(-2) = 0$
$1 + x - 6 = 0$
$x - 5 = 0$
$x = +5$
Thus,the oxidation number of $N$ in $HNO_3$ is $+5$.

(A) The oxidation number of $Mn$ in $MnO_4^{-}$ is calculated as follows:
Let the oxidation number of $Mn$ be $x$.
For oxygen,the oxidation number is $-2$.
$x + (4 \times -2) = -1$
$x - 8 = -1$
$x = +7$
Therefore,the oxidation number of $Mn$ is $+7$.

(C) The oxidation number of an element increases when it loses electrons (oxidation).
Initially,the oxidation number of $Sn$ is $+2$.
When $Sn^{2+}$ loses two electrons,the reaction is: $Sn^{2+} \to Sn^{4+} + 2e^-$.
Therefore,the final oxidation number of tin is $+4$.

(D) To find the oxidation state of $Mn$ in $K_2MnO_4$,let the oxidation state of $Mn$ be $x$.
The oxidation state of $K$ is $+1$ and the oxidation state of $O$ is $-2$.
$2(+1) + x + 4(-2) = 0$
$2 + x - 8 = 0$
$x - 6 = 0$
$x = +6$.
Therefore,the oxidation state of $Mn$ in $K_2MnO_4$ is $+6$.

(B) In any elemental form or homonuclear diatomic molecule,the atoms are bonded to identical atoms,resulting in no net transfer of electrons.
Therefore,the oxidation number of each oxygen atom in an $O_2$ molecule is $0$.

(A) To determine the oxidation state of carbon in each compound:
$1$. In $CH_4$,let the oxidation state of $C$ be $x$. Then $x + 4(+1) = 0$,so $x = -4$.
$2$. In $CCl_4$,$x + 4(-1) = 0$,so $x = +4$.
$3$. In $CF_4$,$x + 4(-1) = 0$,so $x = +4$.
$4$. In $CO_2$,$x + 2(-2) = 0$,so $x = +4$.
Comparing these values,the lowest oxidation state is $-4$,which corresponds to $CH_4$.

(B) Let the oxidation number of carbon be $x$.
In $H_2C_2O_4$,the oxidation number of $H$ is $+1$ and $O$ is $-2$.
The sum of oxidation numbers of all atoms in a neutral molecule is $0$.
$2(+1) + 2(x) + 4(-2) = 0$
$2 + 2x - 8 = 0$
$2x - 6 = 0$
$2x = 6$
$x = +3$.

(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 the oxidation numbers of all atoms in a neutral molecule is $0$.
Therefore,$x + 2(+1) + 2(-1) = 0$.
$x + 2 - 2 = 0$.
$x = 0$.

(C) $H_2S_2O_8$ is known as peroxydisulfuric acid.
Its structure contains a peroxide linkage $(-O-O-)$.
The oxidation state of $H$ is $+1$,oxygen in peroxide linkage is $-1$,and other oxygen atoms are $-2$.
Let the oxidation number of $S$ be $x$.
$2(+1) + 2(x) + 6(-2) + 2(-1) = 0$
$2 + 2x - 12 - 2 = 0$
$2x - 12 = 0$
$2x = 12$
$x = +6$.

(D) The chemical formula for hydrazoic acid is $N_3H$.
Let the oxidation state of nitrogen be $x$.
The oxidation state of hydrogen in this compound is $+1$.
Since the molecule is neutral,the sum of oxidation states must be zero:
$3x + 1 = 0$
$3x = -1$
$x = -\frac{1}{3}$
Therefore,the oxidation state of nitrogen in $N_3H$ is $-\frac{1}{3}$.

(B) To find the oxidation state of sulphur $(S)$ in each compound:
$1$. In $SO_2$: $x + 2(-2) = 0 \implies x = +4$.
$2$. In $H_2SO_4$: $2(+1) + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies x = +6$.
$3$. In $Na_2S_2O_3$: $2(+1) + 2x + 3(-2) = 0 \implies 2 + 2x - 6 = 0 \implies 2x = 4 \implies x = +2$.
$4$. In $Na_2S_4O_6$: $2(+1) + 4x + 6(-2) = 0 \implies 2 + 4x - 12 = 0 \implies 4x = 10 \implies x = +2.5$.
Comparing the values $(+4, +6, +2, +2.5)$,the highest oxidation state is $+6$ in $H_2SO_4$.

(A) Iron pyrites is $FeS_2$.
Let the oxidation number of $Fe$ be $x$ and $S$ be $y$.
For $FeS_2$,the compound is composed of $Fe^{2+}$ and $S_2^{2-}$ ions.
In the disulfide ion $S_2^{2-}$,the oxidation state of each $S$ atom is $-1$.
Substituting this into the equation: $x + 2(-1) = 0$.
$x - 2 = 0 \Rightarrow x = +2$.
Thus,the oxidation number of $Fe$ is $+2$ and $S$ is $-1$.

(D) Let the oxidation number of nitrogen be $x$.
In the nitrate ion,$NO_3^-$,the sum of the oxidation numbers of all atoms equals the charge on the ion,which is $-1$.
The oxidation number of oxygen is $-2$.
Therefore,$x + 3(-2) = -1$.
$x - 6 = -1$.
$x = +5$.

(A) The correct answer is $(A)$.
By definition,the oxidation state of an element in its elemental form (uncombined state) is always $0$.

(B) In $C_6H_5CHO$ (benzaldehyde),the structure consists of a benzene ring attached to an aldehyde group $(-CHO)$.
For the benzene ring $(C_6H_5-)$,the average oxidation state of each carbon is $-1/6$,so the sum for $6$ carbons is $-1$.
The carbonyl carbon in the $-CHO$ group has an oxidation state of $+1$.
Therefore,the sum of the oxidation numbers of all $7$ carbon atoms is $(-1) + (+1) = 0$.

(D) To find the oxidation number of iodine $(I)$ in each compound:
$A) \ KI_3$: $1 + 3x = 0 \implies x = -1/3$
$B) \ KI$: $1 + x = 0 \implies x = -1$
$C) \ IF_5$: $x + 5(-1) = 0 \implies x = +5$
$D) \ KIO_4$: $1 + x + 4(-2) = 0 \implies x - 7 = 0 \implies x = +7$
Comparing the values,$KIO_4$ has the highest oxidation number of iodine,which is $+7$.

(B) Let the oxidation number of $N$ be $x$.
For the ion ${N_2}H_5^+$,the sum of oxidation numbers of all atoms equals the charge on the ion.
$2x + 5(+1) = +1$
$2x + 5 = +1$
$2x = -4$
$x = -2$
Therefore,the oxidation number of $N$ is $-2$.

(B) To find the oxidation number of $C$,let it be $x$:
$1.$ For $HCHO$: $x + 1(+1) + 1(-2) + 1(+1) = 0 \implies x + 2 - 2 = 0 \implies x = 0$.
$2.$ For $CHCl_3$: $x + 1(+1) + 3(-1) = 0 \implies x + 1 - 3 = 0 \implies x = +2$.
$3.$ For $CH_3OH$: $x + 3(+1) + 1(-2) + 1(+1) = 0 \implies x + 3 - 2 + 1 = 0 \implies x = -2$.
$4.$ For $C_{12}H_{22}O_{11}$: $12x + 22(+1) + 11(-2) = 0 \implies 12x + 22 - 22 = 0 \implies 12x = 0 \implies x = 0$.
Comparing the values,the maximum oxidation number is $+2$ in $CHCl_3$.

(C) Let the oxidation state of chlorine be $x$.
The oxidation state of potassium $(K)$ is $+1$ and oxygen $(O)$ is $-2$.
For the compound $KClO_4$,the sum of oxidation states is $0$:
$(+1) + x + 4 \times (-2) = 0$
$1 + x - 8 = 0$
$x - 7 = 0$
$x = +7$

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

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

(D) All free metals in their elemental state always show a $zero$ oxidation state because there is no difference in electronegativity between identical atoms.

(A) To determine the number of electrons transferred,we calculate the change in the oxidation state of the central metal atom.
$A$. In $MnO_4^-$,the oxidation state of $Mn$ is $+7$. In $Mn^{2+}$,it is $+2$. The change is $7 - 2 = 5$ electrons.
$B$. In $CrO_4^{2-}$,the oxidation state of $Cr$ is $+6$. In $Cr^{3+}$,it is $+3$. The change is $6 - 3 = 3$ electrons.
$C$. In $MnO_4^{2-}$,the oxidation state of $Mn$ is $+6$. In $MnO_2$,it is $+4$. The change is $6 - 4 = 2$ electrons.
$D$. In $Cr_2O_7^{2-}$,the oxidation state of $Cr$ is $+6$. In $2Cr^{3+}$,the total change for two $Cr$ atoms is $2 \times (6 - 3) = 6$ electrons.
Therefore,the correct change involving the transfer of five electrons is $MnO_4^- \to Mn^{2+}$.

(C) Let the oxidation number of $C$ be $x$.
For $C_6H_{12}O_6$,the sum of oxidation numbers of all atoms is equal to $0$.
$6(x) + 12(+1) + 6(-2) = 0$
$6x + 12 - 12 = 0$
$6x = 0$
$x = 0$
Therefore,the oxidation number of $C$ in glucose is $0$.

(B) Iodine is a halogen belonging to group $17$ of the periodic table.
Its valence shell configuration is $ns^2 np^5$.
It can gain one electron to achieve a stable noble gas configuration (oxidation state $-1$) or lose electrons from its valence shell (up to $7$ electrons) to exhibit positive oxidation states up to $+7$ in compounds with more electronegative elements like fluorine or oxygen (e.g.,$IF_7$ or $HIO_4$).
Thus,the range of oxidation states for iodine is $-1$ to $+7$.

(C) The balanced chemical equation for the reaction is:
$K_2Cr_2O_7 + 3SO_2 + H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O$
In $K_2Cr_2O_7$,the oxidation state of chromium $(Cr)$ is calculated as:
$2(+1) + 2(x) + 7(-2) = 0$ $\Rightarrow 2 + 2x - 14 = 0$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$
In $Cr_2(SO_4)_3$,the oxidation state of chromium $(Cr)$ is $+3$ because the sulphate ion $(SO_4^{2-})$ has a charge of $-2$.
Thus,the oxidation state of chromium changes from $+6$ to $+3$.

(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 $0$.
$x + 2(+1) + (-2) = 0$
$x + 2 - 2 = 0$
$x = 0$
Therefore,the oxidation number of carbon is $0$.

(A) Let the oxidation number of $Cr$ be $x$.
In $K_2Cr_2O_7$,the oxidation number of $K$ is $+1$ and $O$ is $-2$.
The sum of oxidation numbers of all atoms in a neutral molecule is $0$.
$2(+1) + 2(x) + 7(-2) = 0$
$2 + 2x - 14 = 0$
$2x - 12 = 0$
$2x = 12$
$x = +6$
Therefore,the oxidation number of $Cr$ is $+6$.

(D) Let the oxidation number of $Os$ be $x$.
In $OsO_4$,the oxidation number of oxygen is $-2$.
Applying the rule that the sum of oxidation numbers in a neutral molecule is zero:
$x + 4(-2) = 0$
$x - 8 = 0$
$x = +8$
Therefore,the oxidation number of $Os$ in $OsO_4$ is $+8$.

(B) The oxidation state of oxygen in $F_2O$ is calculated as follows:
Let the oxidation state of oxygen be $x$.
The oxidation state of fluorine $(F)$ is $-1$ because it is the most electronegative element.
In $F_2O$,the sum of oxidation states is $x + 2(-1) = 0$.
$x - 2 = 0$,so $x = +2$.
Therefore,oxygen shows a $+2$ oxidation state in $F_2O$.

(A) To find the oxidation state of nitrogen $(x)$ in each compound:
$1$. In $N_3H$: $3x + 1 = 0$ $\Rightarrow 3x = -1$ $\Rightarrow x = -\frac{1}{3} \approx -0.33$
$2$. In $NH_2OH$: $x + 2(1) + (-2) + 1 = 0$ $\Rightarrow x + 1 = 0$ $\Rightarrow x = -1$
$3$. In $N_2H_4$: $2x + 4(1) = 0$ $\Rightarrow 2x = -4$ $\Rightarrow x = -2$
$4$. In $NH_3$: $x + 3(1) = 0 \Rightarrow x = -3$
Comparing the values: $-0.33 > -1 > -2 > -3$. The highest oxidation state is in $N_3H$.