Oxidation number and Oxidation state Questions in English

(D) The structure of $CrO_5$ is a butterfly structure where one oxygen atom is bonded to $Cr$ via a double bond (oxidation state $-2$) and four oxygen atoms are bonded via single bonds forming two peroxide linkages (each oxygen atom has an oxidation state of $-1$).
Let the oxidation state of $Cr$ be $x$.
The sum of oxidation states in a neutral molecule is zero.
$x + 1(-2) + 4(-1) = 0$
$x - 2 - 4 = 0$
$x - 6 = 0$
$x = +6$

(C) In an acidic medium,the permanganate ion $(MnO_4^{-})$ acts as a strong oxidizing agent and reacts with hydrogen peroxide $(H_2O_2)$.
The balanced chemical equation for this reaction is: $2MnO_4^{-} + 5H_2O_2 + 6H^{+} \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O$.
In $MnO_4^{-}$,the oxidation state of $Mn$ is calculated as: $x + 4(-2) = -1$,which gives $x = +7$.
In the product $Mn^{2+}$,the oxidation state of $Mn$ is $+2$.
Therefore,the oxidation state of $Mn$ changes from $+7$ to $+2$.

(C) The chemical formula of Peroxydisulfuric acid (Marshall's acid) is $H_2S_2O_8$.
Its structure is $HO-S(=O)_2-O-O-S(=O)_2-OH$.
From the structure,each sulfur atom is bonded to two double-bonded oxygen atoms,one oxygen atom of the peroxide linkage $(-O-O-)$,and one hydroxyl group $(-OH)$.
The oxidation state of each sulfur atom is $+6$.
There are two $S-OH$ bonds in the molecule (one on each sulfur atom).

(B) In the structure of $H_2S_2O_8$ (peroxodisulphuric acid),there is a peroxide linkage $(-O-O-)$.
Two oxygen atoms are involved in the peroxide linkage,each having an oxidation state of $-1$.
The remaining $6$ oxygen atoms have an oxidation state of $-2$.
Let the oxidation state of $S$ be $x$.
Applying the oxidation state rule: $2(+1) + 2(x) + 2(-1) + 6(-2) = 0$.
$2 + 2x - 2 - 12 = 0$.
$2x = 12$.
$x = +6$.

(B) To find the oxidation state of nitrogen $(N)$ in each compound,we assign the oxidation states of other elements (Hydrogen = $+1$,Oxygen = $-2$,Chlorine = $-1$):
$1$. In $NH_2OH$: $x + 2(+1) + (-2) + (+1) = 0 \implies x + 1 = 0 \implies x = -1$.
$2$. In $NH_4Cl$: $x + 4(+1) + (-1) = 0 \implies x + 3 = 0 \implies x = -3$.
$3$. In $N_2H_4$: $2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$.
$4$. In $HNO_2$: $(+1) + x + 2(-2) = 0 \implies x - 3 = 0 \implies x = +3$.
Comparing the values: $-1, -3, -2, +3$. The lowest oxidation state is $-3$ in $NH_4Cl$.

(A) The structure of carbon suboxide $(C_3O_2)$ is $O=C=C=C=O$.
In this structure,the two terminal carbon atoms are bonded to oxygen atoms.
Since the oxidation state of oxygen is $-2$,each terminal carbon atom must have an oxidation state of $+2$ to balance the bond.
The central carbon atom is bonded to two other carbon atoms,and since there is no electronegativity difference between identical atoms,its oxidation state is $0$.
Thus,the oxidation states are $+2, 0, +2$.

(C) $Pb_3O_4$ is a mixed oxide,which can be represented as $2PbO \cdot PbO_2$.
In $PbO$,the oxidation state of $Pb$ is $+2$.
In $PbO_2$,the oxidation state of $Pb$ is $+4$.
Calculation for $PbO$: $x + (-2) = 0 \Rightarrow x = +2$.
Calculation for $PbO_2$: $x + 2(-2) = 0$ $\Rightarrow x - 4 = 0$ $\Rightarrow x = +4$.
Thus,the oxidation states of $Pb$ in $Pb_3O_4$ are $+2$ and $+4$.

(C) In an aqueous solution,potassium dichromate $(K_2Cr_2O_7)$ exists in equilibrium with potassium chromate $(K_2CrO_4)$ depending on the $pH$ of the solution.
The equilibrium reaction is: $Cr_2O_7^{2-} + 2OH^- \rightleftharpoons 2CrO_4^{2-} + H_2O$.
In $K_2Cr_2O_7$,the oxidation state of $Cr$ is calculated as: $2(+1) + 2(x) + 7(-2) = 0$ $\Rightarrow 2 + 2x - 14 = 0$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$.
In $K_2CrO_4$,the oxidation state of $Cr$ is calculated as: $2(+1) + x + 4(-2) = 0$ $\Rightarrow 2 + x - 8 = 0$ $\Rightarrow x = +6$.
Since the oxidation state of $Cr$ remains $+6$ in both acidic $(Cr_2O_7^{2-})$ and basic $(CrO_4^{2-})$ media,the correct answer is $6$.

(A) $H_2S_2O_7$ is known as oleum or pyrosulphuric acid.
In the structure of $H_2S_2O_7$,each sulphur atom is bonded to three oxygen atoms via double bonds (contributing $+2$ each) and one oxygen atom via a single bond (contributing $+1$),and one hydroxyl group (contributing $+1$).
Therefore,the oxidation state for each sulphur atom is $+2 + 2 + 1 + 1 = +6$.
Thus,the oxidation states of both sulphur atoms are $VI$ and $VI$.

(C) The initial oxidation state of $N$ in $N_2H_4$ is calculated as: $2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$.
Since $1 \ mole$ of $N_2H_4$ loses $10 \ moles$ of electrons,the total increase in oxidation number is $10$.
Let the oxidation state of $N$ in $Z$ be $y$. Since there are $2$ nitrogen atoms,the total oxidation state of $N$ in $Z$ is $2y$.
The change in oxidation state is $2y - (-4) = +10$.
$2y + 4 = 10 \implies 2y = 6 \implies y = +3$.
Therefore,the oxidation state of nitrogen in $Z$ is $+3$.

(B) $NH_4NO_3$ consists of $NH_4^+$ and $NO_3^-$ ions.
For $NH_4^+$: Let the oxidation number of $N$ be $x$. $x + 4(+1) = +1$,so $x = -3$.
For $NO_3^-$: Let the oxidation number of $N$ be $x$. $x + 3(-2) = -1$,so $x - 6 = -1$,which gives $x = +5$.
Thus,the oxidation numbers of nitrogen are $-3$ and $+5$.

(B) In the initial compound $LiCoO_2$,the oxidation state of $Li$ is $+1$ and oxygen is $-2$.
Let the oxidation state of $Co$ be $x$.
$1(1) + x + 2(-2) = 0$
$x - 3 = 0 \implies x = +3$.
After $50\%$ lithium extraction,the formula becomes $Li_{0.5}CoO_2$.
Let the new oxidation state of $Co$ be $x'$.
$0.5(1) + x' + 2(-2) = 0$
$x' + 0.5 - 4 = 0 \implies x' = +3.5$.
The change in oxidation state of $Co$ is $3.5 - 3.0 = +0.5$.
The percentage increase in the oxidation state is $\frac{0.5}{3} \times 100 = 16.66\%$.

(C) For $K_2Cr_2O_7$: Let the oxidation state of $Cr$ be $x$.
$2(+1) + 2x + 7(-2) = 0$
$2 + 2x - 14 = 0$
$2x = 12$
$x = +6$
For $CrO_5$: This compound has a butterfly structure containing four peroxy oxygen atoms (each with $-1$ charge) and one oxo oxygen atom (with $-2$ charge). Let the oxidation state of $Cr$ be $x$.
$x + 4(-1) + 1(-2) = 0$
$x - 4 - 2 = 0$
$x - 6 = 0$
$x = +6$
Thus,the oxidation states of $Cr$ in both compounds are $+6$.

(C) In acidic medium,the reduction reaction is:
$Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \longrightarrow 2Cr^{3+} + 7H_{2}O$
Here,the oxidation state of $Cr$ changes from $+6$ to $+3$.
Since there are two $Cr$ atoms in $K_{2}Cr_{2}O_{7}$,the total change in oxidation state is $2 \times (6 - 3) = 6$.
Therefore,the n-factor is $6$.
Equivalent weight = $\frac{\text{Molecular weight}}{\text{n-factor}} = \frac{M}{6}$.