Oxidation number and Oxidation state Questions in English

(A) Let the oxidation number of $P$ be $x$.
In $KH_2PO_2$,the oxidation numbers of $K$,$H$,and $O$ are $+1$,$+1$,and $-2$ respectively.
The sum of oxidation numbers of all atoms in a neutral molecule is $0$.
Therefore,$(+1) + 2(+1) + x + 2(-2) = 0$.
$1 + 2 + x - 4 = 0$.
$3 + x - 4 = 0$.
$x - 1 = 0$.
$x = +1$.

(D) To find the oxidation number of sulphur $(S)$ in each compound:
$1$. In $H_2SO_3$: $2(+1) + x + 3(-2) = 0$ $\Rightarrow 2 + x - 6 = 0$ $\Rightarrow x = +4$.
$2$. In $SO_2$: $x + 2(-2) = 0$ $\Rightarrow x - 4 = 0$ $\Rightarrow x = +4$.
$3$. In $H_2SO_4$: $2(+1) + x + 4(-2) = 0$ $\Rightarrow 2 + x - 8 = 0$ $\Rightarrow x = +6$.
$4$. In $H_2S$: $2(+1) + x = 0$ $\Rightarrow 2 + x = 0$ $\Rightarrow x = -2$.
Comparing the values $(+4, +4, +6, -2)$,the lowest oxidation number is $-2$,which corresponds to $H_2S$.

(A) In the $S_8$ molecule,all sulphur atoms are in their elemental state,so the oxidation number of each sulphur atom is $0$.
Each sulphur atom is bonded to two other sulphur atoms by single covalent bonds,so the covalency of each sulphur atom is $2$.

(B) Ferrous ammonium sulphate is also known as Mohr's salt,with the chemical formula $(NH_4)_2Fe(SO_4)_2 \cdot 6H_2O$.
In this compound,the iron is present as a ferrous ion,which is $Fe^{2+}$.
Therefore,the oxidation state of $Fe$ is $+2$.

(B) Let the oxidation number of nitrogen be $x$.
In $NH_2OH$,the oxidation number of hydrogen is $+1$ and oxygen is $-2$.
The sum of oxidation numbers of all atoms in a neutral molecule is $0$.
$x + 2(+1) + (-2) + 1 = 0$
$x + 2 - 2 + 1 = 0$
$x + 1 = 0$
$x = -1$.

(B) The chemical formula is $Ba(H_2PO_2)_2$.
Let the oxidation number of $P$ be $x$.
The oxidation number of $Ba$ is $+2$,$H$ is $+1$,and $O$ is $-2$.
Sum of oxidation numbers $= 0$.
$2 + 2 \times [2 \times (+1) + x + 2 \times (-2)] = 0$
$2 + 2 \times [2 + x - 4] = 0$
$2 + 2 \times [x - 2] = 0$
$2 + 2x - 4 = 0$
$2x - 2 = 0$
$2x = 2$
$x = +1$.

(D) To find the oxidation number of sulphur $(S)$ in $H_2SO_4$:
Let the oxidation number of $S$ be $x$.
$2 \times (+1) + x + 4 \times (-2) = 0$
$2 + x - 8 = 0$
$x = +6$
In $H_2SO_4$,sulphur is in its $+6$ oxidation state,meaning it has lost all its valence electrons $(3s^2 3p^4)$.
The electronic configuration of neutral sulphur $(S)$ is $1s^2 2s^2 2p^6 3s^2 3p^4$.
After losing $6$ electrons to reach the $+6$ state,the configuration becomes $1s^2 2s^2 2p^6$.

(A) Let the oxidation number of $Mn$ be $x$.
In $KMnO_4$,the oxidation number of $K$ is $+1$ and the oxidation number of $O$ is $-2$.
The sum of oxidation numbers in a neutral molecule is $0$.
Therefore,$(+1) + x + 4 \times (-2) = 0$.
$1 + x - 8 = 0$.
$x - 7 = 0$.
$x = +7$.

(D) The oxidation number of $H$ is $+1$ and $O$ is $-2$.
Let the oxidation number of $As$ be $x$.
In $H_3AsO_4$,the sum of oxidation numbers is $0$:
$3(+1) + x + 4(-2) = 0$
$3 + x - 8 = 0$
$x - 5 = 0$
$x = +5$.

(B) To find the oxidation number of carbon $(x)$ in $CH_3Cl$:
Assign oxidation states to other atoms: Hydrogen $(H)$ is $+1$ and Chlorine $(Cl)$ is $-1$.
The sum of oxidation numbers in a neutral molecule is $0$.
$x + 3(+1) + (-1) = 0$
$x + 3 - 1 = 0$
$x + 2 = 0$
$x = -2$
Therefore,the oxidation number of carbon is $-2$.

(C) Let the oxidation state of $Cr$ be $x$.
In the ion $Cr_2O_7^{2-}$,the sum of oxidation states of all atoms must equal the charge on the ion.
$2x + 7(-2) = -2$
$2x - 14 = -2$
$2x = 12$
$x = +6$
Therefore,the oxidation state of $Cr$ is $+6$.

(C) To find the oxidation state of $S$ in $H_2SO_3$,let the oxidation state of $S$ be $x$.
The oxidation state of $H$ is $+1$ and $O$ is $-2$.
Applying the sum of oxidation states equal to zero:
$2(+1) + x + 3(-2) = 0$
$2 + x - 6 = 0$
$x - 4 = 0$
$x = +4$
Therefore,the oxidation state of $S$ is $+4$.

(B) Bleaching powder is represented as $Ca(OCl)Cl$.
In this compound,the calcium ion is $Ca^{2+}$.
The two chlorine atoms are present in different environments:
$1$. One $Cl$ atom is part of the hypochlorite ion $(OCl^-)$,where the oxidation state of $O$ is $-2$,so $x + (-2) = -1$,which gives $x = +1$.
$2$. The other $Cl$ atom exists as a chloride ion $(Cl^-)$,which has an oxidation state of $-1$.
Therefore,the oxidation numbers of the two $Cl$ atoms are $+1$ and $-1$.

(C) To find the oxidation number of chlorine $(Cl)$ in each compound,we use the rule that the sum of oxidation numbers in a neutral molecule is $0$. Let the oxidation number of $Cl$ be $x$. The oxidation number of $H$ is $+1$ and $O$ is $-2$.
For $HClO_3$:
$1 + x + 3(-2) = 0$
$1 + x - 6 = 0$
$x - 5 = 0$
$x = +5$.
Therefore,in $HClO_3$,the oxidation number of chlorine is $+5$.

(C) The balanced chemical equation for the reaction between $KMnO_4$ and oxalic acid in an acidic medium is:
$5(COOH)_2 + 2KMnO_4 + 3H_2SO_4 \to K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O$
In $KMnO_4$,the oxidation state of $Mn$ is calculated as: $x + 1 + 4(-2) = 0$,which gives $x = +7$.
In $MnSO_4$,the oxidation state of $Mn$ is $+2$ because the sulfate ion $(SO_4^{2-})$ has a charge of $-2$.
Therefore,the oxidation number of $Mn$ changes from $+7$ to $+2$.

(D) In $OF_2$,fluorine is more electronegative than oxygen.
Since the oxidation state of fluorine is $-1$,let the oxidation state of oxygen be $x$.
$x + 2(-1) = 0$
$x - 2 = 0$
$x = +2$.
Therefore,oxygen has an oxidation state of $+2$ in $OF_2$.

(B) Let the oxidation number of sulphur be $x$.
In the thiosulphate ion,$S_2O_3^{2-}$,the sum of the oxidation numbers of all atoms equals the charge on the ion.
$2x + 3(-2) = -2$
$2x - 6 = -2$
$2x = +4$
$x = +2$
Therefore,the average oxidation number of sulphur in $S_2O_3^{2-}$ is $+2$.

(C) To find the oxidation number of carbon $(x)$ in $CH_2Cl_2$:
$x + 2 \times (+1) + 2 \times (-1) = 0$
$x + 2 - 2 = 0$
$x = 0$
Therefore,carbon has an oxidation number of $0$ in $CH_2Cl_2$.

(C) In potassium superoxide $(KO_2)$,the sum of oxidation states of all atoms must be $0$.
Let the oxidation state of oxygen be $x$.
The oxidation state of potassium $(K)$ is $+1$.
Thus,$1 + 2x = 0$.
$2x = -1$.
$x = -\frac{1}{2} = -0.5$.
Therefore,the correct option is $(C)$.

(A) Let the oxidation number of $S$ be $x$.
In $S_2Cl_2$,the oxidation number of $Cl$ is $-1$.
The sum of oxidation numbers in a neutral molecule is $0$.
$2x + 2(-1) = 0$
$2x - 2 = 0$
$2x = 2$
$x = +1$
Therefore,the oxidation number of $S$ in $S_2Cl_2$ is $+1$.

(D) For the molecule $Na_2S_4O_6$,let the oxidation number of sulphur be $x$.
Assigning oxidation states: $Na = +1$,$O = -2$.
The sum of oxidation states in a neutral molecule is $0$.
$2(+1) + 4(x) + 6(-2) = 0$
$2 + 4x - 12 = 0$
$4x - 10 = 0$
$4x = 10$
$x = 10/4 = 5/2$
Therefore,the average oxidation number of sulphur is $5/2$.

(D) The compound $NH_4Cl$ dissociates as: $NH_4Cl \rightarrow NH_4^+ + Cl^-$
In the ammonium ion $(NH_4^+)$,let the oxidation number of $N$ be $x$.
The sum of oxidation numbers of all atoms in the ion is equal to its charge:
$x + 4 \times (+1) = +1$
$x + 4 = +1$
$x = 1 - 4 = -3$
Therefore,the oxidation number of $N$ in $NH_4Cl$ is $-3$.

(D) Let the average oxidation state of $Fe$ be $x$.
In $Fe_3O_4$,the oxidation state of oxygen is $-2$.
The sum of oxidation states of all atoms in a neutral molecule is $0$.
$3x + 4(-2) = 0$
$3x - 8 = 0$
$3x = 8$
$x = \frac{8}{3}$

(B) 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 gives $2 + x - 8 = 0$,so $x = +6$.
For $MnSO_4$: Let the oxidation number of $Mn$ be $x$. The sulfate ion $SO_4^{2-}$ has an oxidation state of $-2$. Thus,$x + (-2) = 0$,so $x = +2$.
Therefore,the oxidation numbers are $+6$ and $+2$ respectively.

(B) In $F_2O$,fluorine is more electronegative than oxygen.
Oxygen is assigned an oxidation state of $-2$.
Let the oxidation number of fluorine be $x$.
$2x + (-2) = 0$
$2x = 2$
$x = +1$.
Therefore,the oxidation number of fluorine in $F_2O$ is $+1$.

(A) Let the oxidation number of $N$ be $x$.
In $NH_3$,the oxidation number of each $H$ atom is $+1$.
The sum of oxidation numbers in a neutral molecule is $0$.
Therefore,$x + 3(+1) = 0$.
$x + 3 = 0$.
$x = -3$.
Thus,the oxidation number of $N$ in $NH_3$ is $-3$.

(A) The oxidation number of $H$ in $H_2S$ is $+1$.
Let the oxidation number of $S$ be $x$.
For the neutral molecule $H_2S$,the sum of oxidation numbers is $0$.
$(+1) \times 2 + x = 0$
$2 + x = 0$
$x = -2$
Therefore,the oxidation number of sulphur in $H_2S$ is $-2$.

(B) Let the oxidation number of $N$ in $NaNO_2$ be $x$.
$Na$ has an oxidation number of $+1$ and $O$ has an oxidation number of $-2$.
Sum of oxidation numbers in a neutral compound is $0$.
$(+1) + x + 2 \times (-2) = 0$
$1 + x - 4 = 0$
$x - 3 = 0$
$x = +3$

(A) Let the oxidation number of $S$ be $x$.
For the sulfate ion $SO_4^{2-}$,the sum of the oxidation numbers of all atoms equals the charge on the ion.
$x + 4 \times (-2) = -2$
$x - 8 = -2$
$x = +6$
Therefore,the oxidation number of $S$ in $SO_4^{2-}$ is $+6$.

(C) In $KMnO_4$,the oxidation state of $Mn$ is $+7$.
The number of electrons transferred is equal to the change in the oxidation state of $Mn$.
$1$. For $[MnO_4]^{2-}$,$Mn$ is $+6$. Change $= |7 - 6| = 1$.
$2$. For $MnO_2$,$Mn$ is $+4$. Change $= |7 - 4| = 3$.
$3$. For $Mn_2O_3$,$Mn$ is $+3$. Change per $Mn$ atom $= |7 - 3| = 4$.
$4$. For $Mn^{2+}$,$Mn$ is $+2$. Change $= |7 - 2| = 5$.
Thus,the number of electrons transferred is $1, 3, 4, 5$.

(C) To determine the oxidation state of $N$ in each compound:
$(A)$ In $NO_2$: $x + 2(-2) = 0 \Rightarrow x = +4$. Thus,$(A) \rightarrow (3)$.
$(B)$ In $HNO$: $+1 + x + (-2) = 0 \Rightarrow x = +1$. Thus,$(B) \rightarrow (4)$.
$(C)$ In $NH_3$: $x + 3(+1) = 0 \Rightarrow x = -3$. Thus,$(C) \rightarrow (2)$.
$(D)$ In $N_2O_5$: $2x + 5(-2) = 0$ $\Rightarrow 2x = 10$ $\Rightarrow x = +5$. Thus,$(D) \rightarrow (1)$.
The correct sequence is $3, 4, 2, 1$.

(C) The initial oxidation state of the ion is $+3$.
When an atom or ion loses electrons,its oxidation number increases by the number of electrons lost.
Final oxidation number = Initial oxidation number + (Number of electrons lost)
Final oxidation number = $+3 + 3 = +6$.

(C) In potassium superoxide $(KO_2)$,the oxidation number of potassium $(K)$ is $+1$.
Let the oxidation number of oxygen be $x$.
Since the molecule is neutral,the sum of oxidation numbers is $0$.
$1 + 2x = 0$
$2x = -1$
$x = -1/2$
Therefore,the oxidation number of oxygen in $KO_2$ is $-1/2$.

(A) In $N_2H_4$,the oxidation state of $H$ is $+1$. Let the oxidation state of $N$ be $x$.
$2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$.
When $1 \ mol$ of $N_2H_4$ loses $10 \ mol$ of electrons,the total oxidation state increases by $10$.
Let the oxidation state of $N$ in $Y$ be $a$.
Since there are $2$ atoms of $N$,the total change is $2(a - (-2)) = 10$.
$2(a + 2) = 10
$a + $2$ = $5$
$a = +3$.

(D) To find the oxidation state of the central atom,we assign oxidation numbers to the ligands:
In $MnO_4^-$,$Mn + 4(-2) = -1 \implies Mn = +7$.
In $Cr(CN)_6^{3-}$,$Cr + 6(-1) = -3 \implies Cr = +3$.
In $NiF_6^{2-}$,$Ni + 6(-1) = -2 \implies Ni = +4$.
In $CrO_2Cl_2$,let the oxidation state of $Cr$ be $x$. Since $O$ is $-2$ and $Cl$ is $-1$,we have $x + 2(-2) + 2(-1) = 0 \implies x - 4 - 2 = 0 \implies x = +6$.
Therefore,the species with an atom in $+6$ oxidation state is $CrO_2Cl_2$.

(C) In $I_3^-$,the total charge on the triiodide ion is $-1$. Let the oxidation number of iodine be $x$. Then,$3x = -1$,which gives $x = -1/3$. Since $-1/3$ is a fractional value,the oxidation number of iodine in $I_3^-$ is fractional.

(B) Let the oxidation state of $Cu$ be $x$.
Given that $Y$ is in $+3$ state and $Ba$ is in $+2$ state (alkaline earth metal),and $O$ is in $-2$ state.
The sum of oxidation states in a neutral compound is $0$.
$3 + (2 \times 2) + 3x + (7 \times -2) = 0$
$3 + 4 + 3x - 14 = 0$
$7 + 3x - 14 = 0$
$3x - 7 = 0$
$3x = 7$
$x = 7/3$

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

(C) The compound $NH_4NO_2$ dissociates into ions as $NH_4NO_2 \rightleftharpoons NH_4^+ + NO_2^-$.
For the ammonium ion $(NH_4^+)$,let the oxidation number of $N$ be $x$. Thus,$x + 4(+1) = +1$,which gives $x = -3$.
For the nitrite ion $(NO_2^-)$,let the oxidation number of $N$ be $x$. Thus,$x + 2(-2) = -1$,which gives $x = +3$.
Therefore,the oxidation numbers of nitrogen in $NH_4NO_2$ are $-3$ and $+3$.

(B) Let the oxidation state of phosphorus be $x$.
For $P_4O_8$:
$4x + 8(-2) = 0$
$4x - 16 = 0$
$4x = 16$
$x = +4$
Therefore,the oxidation state of phosphorus in $P_4O_8$ is $+4$.

(C) Let the oxidation state of nitrogen be $x$.
In $NH_2OH$,the oxidation states of $H$ is $+1$ and $O$ is $-2$.
Applying the sum of oxidation states equal to zero:
$x + 2(+1) + (-2) + 1 = 0$
$x + 2 - 2 + 1 = 0$
$x + 1 = 0$
$x = -1$
Thus,the oxidation state of nitrogen in $NH_2OH$ is $-1$.

(C) Let the oxidation number of $As$ be $x$.
The oxidation number of $H$ is $+1$ and that of $O$ is $-2$.
For the ion $H_2AsO_4^-$,the sum of oxidation numbers equals the charge on the ion:
$2(+1) + x + 4(-2) = -1$
$2 + x - 8 = -1$
$x - 6 = -1$
$x = +5$
Therefore,the oxidation number of $As$ is $5$.

(C) Dithionous acid $(H_2S_2O_4)$ has sulphur in $+3$ oxidation state.
Structure: $HO-S(=O)-S(=O)-OH$
Calculation: $2(+1) + 2x + 4(-2) = 0$
$2x = 8 - 2 = 6$
$x = +3$

(A) Oxygen is the second most electronegative element after fluorine. Therefore,it usually exhibits a negative oxidation state.
In $F_2O$,fluorine is the most electronegative element and has an oxidation state of $-1$. Let the oxidation state of oxygen be $x$. Then,$x + 2(-1) = 0$,which gives $x = +2$.
In $Cl_2O$,chlorine is less electronegative than oxygen,so oxygen has an oxidation state of $-2$.
In $Na_2O_2$ (sodium peroxide),oxygen is in the peroxide state with an oxidation state of $-1$.
In $Na_2O$ (sodium oxide),oxygen has an oxidation state of $-2$.

(D) To determine the correct sequence,calculate the oxidation number of iodine in each compound:
$1$. In $HI$,the oxidation state of $I$ is $-1$.
$2$. In $I_2$,the oxidation state of $I$ is $0$.
$3$. In $ICl$,since $Cl$ is more electronegative than $I$,the oxidation state of $I$ is $+1$.
$4$. In $HIO_4$,let the oxidation state of $I$ be $x$. Then $1 + x + 4(-2) = 0$,which gives $x = +7$.
Comparing these values: $-1 < 0 < +1 < +7$.
Therefore,the correct sequence is $HI < I_2 < ICl < HIO_4$.

(D) In an alkaline medium,$KMnO_4$ acts as an oxidizing agent and is reduced to $K_2MnO_4$.
The reaction is: $MnO_4^- + e^- \to MnO_4^{2-}$.
The change in oxidation state of $Mn$ is from $+7$ to $+6$,so the $n$-factor is $1$.
Equivalent weight = $\frac{\text{Molecular weight}}{n\text{-factor}} = \frac{158}{1} = 158$.

(C) The equivalent weight is calculated as: $\text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}}$.
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 (n-factor) is $|7 - 6| = 1$.
The molar mass of $KMnO_4$ is $39 + 55 + (4 \times 16) = 158 \ g/mol$.
Therefore,the equivalent weight = $\frac{158}{1} = 158$.

(B) In a neutral medium,the reduction reaction of $KMnO_4$ is: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$.
Here,the oxidation state of $Mn$ changes from $+7$ in $KMnO_4$ to $+4$ in $MnO_2$.
The change in oxidation state is $7 - 4 = 3$.
Thus,the number of electrons involved ($n$-factor) is $3$.
The equivalent weight is calculated as: $\text{Equivalent weight} = \frac{\text{Molecular weight}}{n\text{-factor}} = \frac{M}{3}$.

(A) Permanganic acid is a strong acid with the chemical formula $HMnO_4$.
In this compound,manganese is in the $+7$ oxidation state.