NEET PYQs

Hint

(d) \(\mathrm{CaF}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{CaSO}_{4}+2 \mathrm{HF}\)
Here, the oxidation state of every atom remains the same so, it is not a redox reaction.

Hint

(c)Both \(\mathrm{FeCl}_{2}\) and \(\mathrm{SnCl}_{2}\) are reducing agents with low oxidation numbers

Hint

(c)

\(\mathrm{H}_{2} \mathrm{O}_{2}\) acts as reducing agent in all those reactions in which \(\mathrm{O}_{2}\) is evolved

Hint

(c) \(\mathrm{CrO}_{5}\) has butterfly structure having two peroxo bonds.
Peroxo oxygen has \(-1\) oxidation state.
Let oxidation state of \(\mathrm{Cr}\) be ‘ \(x\) ‘
\(
\mathrm{CrO}_{5}: x+4(-1)+1(-2)=0 \Rightarrow x=+6
\)

Hint

(b)
\(
\stackrel{0}{3 \mathrm{Cl}_{2}}+\underset{\text { (hot and conc.) }}{6 \mathrm{NaOH}} \longrightarrow 5 \mathrm{Na} \stackrel{-1}{\mathrm{Cl}}+\stackrel{+5}{\mathrm{NaCl} \mathrm{O}_3}+3 \mathrm{H}_2 \mathrm{O}
\)
This is an example of disproportionation reaction and oxidation state of chlorine changes from 0 to \(-1\) and \(+5\)

Hint

(c) \(
{\mathrm{~K}} \stackrel{+5}{\mathrm{Cl}} \mathrm{O}_3+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+\mathrm{H}_2 \stackrel{+6}{\mathrm{S}} \mathrm{{O}}_4 \rightarrow {\mathrm{~K_2}} \stackrel{+6}{\mathrm{~S}} \mathrm{{O}}_4+\stackrel{-1}{\mathrm{KCl}}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}
\)
i.e. maximum change in oxidation number is observed in \(\mathrm{Cl}\) ( +5 to -1).

Hint

(d) Let oxidation number of \(\mathrm{P}\) in \(\mathrm{PO}_{4}^{3-}\) be \(x\).
\(
\therefore \quad x+4(-2)=-3 \quad \Rightarrow x=+5
\)
Let oxidation number of \(\mathrm{S}\) in \(\mathrm{SO}_{4}^{2-}\) be \(y\).
\(
\therefore \quad y+4(-2)=-2 \quad \Rightarrow \quad y=+6
\)

Let oxidation number of \(\mathrm{Cr}\) in \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) be \(z\).
\(
\therefore \quad 2 z+7(-2)=-2 \Rightarrow z=+6
\)

Hint

(d) \(\left[5 e^{-}+\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}\right.\)..(i) \(] \times 2\)
\(\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 2 e^{-}+2 \mathrm{CO}_{2} \ldots\right.\) (ii) \(] \times 5\)
\(2 \mathrm{MnO}_{4}^{-}+16 \mathrm{H}^{+}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_{2}\)
2 moles of \(\mathrm{MnO}_{4}^{-}\)required to oxidise 5 moles of oxalate.
\(\therefore\) Number of moles of \(\mathrm{MnO}_{4}^{-}\)required to oxidise 1 mole of oxalate \(=2 / 5=0.4\)

Hint

(b) \(
\mathrm{H}_2 \mathrm{O}+\stackrel{0}{\mathrm{B{r_2}}} \rightarrow \mathrm{HO}\stackrel{+1}{\mathrm{Br}}+\mathrm{H} \stackrel{-1}{\mathrm{Br}}
\)
In the above reaction the oxidation number of \(\mathrm{Br}_{2}\) increases from zero (in \(\mathrm{Br}_{2}\) ) to \(+1\) (in \(\mathrm{HOBr}\) ) and decreases from zero (in \(\mathrm{Br}_{2}\) ) to \(-1\) (in \(\mathrm{HBr}\) ). Thus \(\mathrm{Br}_{2}\) is oxidised as well as reduced and hence it is a redox reaction.

Hint

(a) \(\mathrm{SO}_{3}{ }^{2-}: x+(-2) 3=-2\)
or \(x-6=-2\) or \(x=+4\)
\(\mathrm{S}_{2} \mathrm{O}_{4}{ }^{2-}: 2 x+(-2) 4=-2\)
or \(2 x-8=-2\) or \(2 x=+6 \therefore x=+3\)
\(\mathrm{S}_{2} \mathrm{O}_{6}^{2-}: 2 x+(-2) 6=-2\)
or, \(2 x-12=-2\) or \(2 x=+10 \therefore x=+5\)
Oxidation states follow the order:
\(
\mathrm{S}_{2} \mathrm{O}_{4}{ }^{2-} \lt \mathrm{SO}_{3}{ }^{2-} \lt \mathrm{S}_{2} \mathrm{O}_{6}{ }^{2-}
\)

Hint

\(
\text { (d) } \mathrm{Fe}_{3} \mathrm{O}_{4} \rightarrow 3 x+4(-2)=0 \Rightarrow x=+\frac{8}{3}
\)

Hint

 (b) Redox reactions are those chemical reactions which involve transfer of electrons from one chemical species to another.

Hint

(a) Since carbon is in maximum state of \(+4\), therefore carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) cannot act as a reducing agent

Hint

 (d)  Since the oxidation number of Ni increases from 0 to 2 , therefore it acts as a reducing agent.

Hint

(c) Let \(x=\) Oxidation state of \(\mathrm{I}\). Since oxidation state of \(\mathrm{H}=+1\) and oxidation state of \(\mathrm{O}=-2\), therefore for \(\mathrm{H}_{4} \mathrm{IO}_{6}^{-}\), we get
\(
(4 \times 1)+x+(6 \times-2)=-1
\)

\(
\text { or } x=+7
\)

Hint

(c)

Oxidation Number of \(\mathrm{Cr}\) in \(\mathrm{CrO}_5\)

• The oxidation number of \(\mathrm{Cr}\) be \(\mathrm{x}\).
• Four oxygen atoms are attached to \(\mathrm{Cr}\) through a single bond. The oxygen is in peroxide form. Hence oxidation state of the oxygen atom, in this case, is -1 .
• Another oxygen atom is attached through a double bond and this has an oxidation number -2 .
  \(
  \begin{array}{c}
  x+(-2)+4(-1)=0 \\
  x=+6
  \end{array}
  \)

Hence, the oxidation number of \(\mathrm{Cr}\) in \(\mathrm{CrO}_5\) is +6 .

Hint

(c) Redox couple is both the reduced and oxidised form involve same element.

Hint

(d) Reaction has to be balanced in acidic medium ‘O’ atoms are balanced by adding \(\mathrm{H}_2 \mathrm{O}\) and then \(\mathrm{H}\)-atom is balanced by adding \(\mathrm{H}^{+}\)ions and charge is balanced by \(e^{\ominus}\).
Oxidation: \(\left.\mathrm{SO}_3^{2-}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{SO}_4^{2-}+2 \mathrm{H}^{+}+2 e^{\ominus}\right] \times 3\)
Reduction: \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e^{\ominus} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)
\(
\mathrm{Cr}_2 \mathrm{O}_7^{2-}+3 \mathrm{SO}_3^{2-}+8 \mathrm{H}^{\oplus} \rightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{SO}_4^{2-}+4 \mathrm{H}_2 \mathrm{O}
\)
Comparing with given equation
\(
\begin{array}{l}
\mathrm{a}=1 \\
\mathrm{~b}=3 \\
\mathrm{c}=8
\end{array}
\)

Hint

(b) Decomposition redox reaction leads to breakdown of a compound into two or more compounds at least one of which must be in the elemental state with change in oxidation number.
\(
2 \mathrm{~Pb}\left(\mathrm{NO}_3^{2-}\right)_2(\mathrm{~s}) \rightarrow 2 \mathrm{PbO}(\mathrm{s})+4 \mathrm{NO}_2(\mathrm{~g})+\stackrel{0}{\mathrm{O}_2}(\mathrm{~g})
\)

Hint

(b)

\(
\stackrel{-4}{\mathrm{CH}_4}(\mathrm{~g})+4 \mathrm{Cl}_2(\mathrm{~g}) \rightarrow \stackrel{+4}{\mathrm{CCl}_4}(l)+4 \mathrm{HCl}(\mathrm{g})
\)
Change in oxidation state of carbon is from -4 to +4.

Hint

(b) In a disproportionation reaction, same substance undergoes oxidation (increase in oxidation number) and reduction (decrease in oxidation number forming two different products.
\(
\text { (a) } 2 \mathrm{Cu}^{+1} \longrightarrow \mathrm{Cu}^{+2}+\mathrm{Cu}^0 \text { (Disproportionation) }
\)
\(
\text { (b) } 3 \stackrel{+6}{\mathrm{Mn}} \mathrm{O}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2\stackrel{+7}{\mathrm{M}} \mathrm{nO}_4^{-}+\stackrel{+4}{\mathrm{M}} \mathrm{nO}_2+2 \mathrm{H}_2 \mathrm{O} \text { (Disproportionation) }
\)
\(
\text { (c) } 2 \mathrm{K} \stackrel{+7}{\mathrm{M}}{nO}_4 \stackrel{\Delta}{\longrightarrow} \mathrm{K}_2 \stackrel{+6}{\mathrm{M}} \mathrm{nO}_4+\stackrel{+4}{\mathrm{M}} \mathrm{nO}_2+\stackrel{0}{\mathrm{O_2}}
\)
\(
\text { (d) } 2 \mathrm{K} \stackrel{+7}{\mathrm{M}} {\mathrm{nO}}_4^{-}+3 \stackrel{+2}{\mathrm{M}}\mathrm{n}^{2+}+2 \mathrm{H}_2 \mathrm{O}\longrightarrow \stackrel{+4}{\mathrm{M}}\mathrm{nO}_2+4 \mathrm{H}^{+}
\)

Hint

(b) The correct balanced equation is
\(
2 \mathrm{MnO}_4{ }^{-}+5 \mathrm{C}_2 \mathrm{O}_4{ }^{2-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}
\)

Hint

(a)

\(
\stackrel{+5}{\mathrm{~N}} \mathrm{O}_3, \stackrel{+2}{\mathrm{~N}} \mathrm{O}, \stackrel{0}{\mathrm{~N}}, \stackrel{-3}{\mathrm{~N}} \mathrm{H}_4 \mathrm{Cl}
\)

Hint

(d) Zinc gives \(\mathrm{H}_2\) gas with dil \(\mathrm{H}_2 \mathrm{SO}_4 / \mathrm{HCl}\) but not with \(\mathrm{HNO}_3\) because in \(\mathrm{HNO}_3, \mathrm{NO}_3^{-}\)ion is reduced and give \(\mathrm{NH}_4 \mathrm{NO}_3, \mathrm{~N}_2 \mathrm{O}, \mathrm{NO}\) and \(\mathrm{NO}_2\)
Note:
\(
\begin{array}{r}
[\mathrm{Zn}+\underset{(\text { nearly } 6 \%)}{2 \mathrm{HNO}_3}\left.\longrightarrow \mathrm{Zn}\left(\mathrm{NO}_3\right)_3+2 \mathrm{H}\right] \times 4
\end{array}
\)
\(
\begin{array}{l}
\mathrm{HNO}_3+8 \mathrm{H} \longrightarrow \mathrm{NH}_3+3 \mathrm{H}_2 \mathrm{O} \\
\mathrm{NH}_3+\mathrm{HNO}_3 \longrightarrow \mathrm{NH}_4 \mathrm{NO}_3 \\
4 \mathrm{Zn}+10 \mathrm{HNO}_3 \longrightarrow 4 \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+\mathrm{NH}_4 \mathrm{NO}_3+3 \mathrm{H}_2 \mathrm{O}
\end{array}
\)
\(\mathrm{Zn}\) is above of hydrogen in electrochemical series. So, \(\mathrm{Zn}\) displaces \(\mathrm{H}_2\) from dilute \(\mathrm{H}_2 \mathrm{SO}_4\) and \(\mathrm{HCl}\) with liberation of \(\mathrm{H}_2\).
\(
\mathrm{Zn}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{ZnSO}_4+\mathrm{H}_2
\)

Hint

(c)

\(
\text { (a) } 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \underset{\text { (neutralization) }}{\mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}}
\)
\(
\text { (b) } \quad \stackrel{0}{3 \mathrm{O}_2} \xrightarrow{\text { Light }} \stackrel{0}2{\mathrm{O}_3} \text { (not redox reaction) }
\)
\(
\text { (c) } \underset{0}{\mathrm{N}_2}+\underset{0}{\mathrm{O}_2} \xrightarrow{\text { Light }} \underset{+2-2}{2 \mathrm{NO}} \text { (redox reaction) }
\)
here oxidation of \(\mathrm{N}_2\) \& reduction of \(\mathrm{O}_2\) is taking place
\(
\text { (d) } \mathrm{H}_2 \mathrm{O} \text { (l) } \xrightarrow{\Delta} \mathrm{H}_2 \mathrm{O} \text { (g) (not redox reaction) }
\)

Hint

(a) Losing of electron is called oxidation.

Hint

(b) Oxidation number of a compound must be 0 . Using the values for \(A, B\) and \(C\) in the four options we find that \(A_3\left(B C_4\right)_2\) is the answer.
Check: \((+2) 3+[(+5)+4(-2)] 2=6+(5-8) 2=0\)

Hint

(d) Pyrophosphoric acid \(\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7\)
Let oxidation state of phosphorus is \(x\)
\(
\begin{array}{l}
(4 \times 1+(-2) \times 7+2 x)=0 \\
\therefore 2 x=10 \text { or } x=+5
\end{array}
\)

Hint

(a) Let \(x=\) oxidation no. of \(\mathrm{Cr}\) in \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\).
\(
\therefore(2 \times 1)+(2 \times x)+7(-2)=0
\)
or \(2+2 x-14=0\) or \(x=+6\).

Hint

(a) O.N. of \(\mathrm{P}\) in \(\mathrm{H}_3 \mathrm{PO}_3\) (phosphorous acid)
\(3 \times 1+x+3 \times(-2)=0\) or \(x=+3\)
In orthophosphoric acid \(\left(\mathrm{H}_3 \mathrm{PO}_4\right)\) O.N. of \(\mathrm{P}\) is +5 , in hypophosphorous acid \(\left(\mathrm{H}_3 \mathrm{PO}_2\right)\) it is +1 while in metaphosphoric acid \(\left(\mathrm{HPO}_3\right)\), it is +5 .

Hint

(c) Calculate \(E_{\text {cell }}^{\circ}\) corresponding to each compound undergoing disproportionation reaction. The reaction for which \(E_{\text {cell }}^{\circ}\) comes out + ve is spontaneous.
\(
\begin{array}{l}
\mathrm{HBrO} \longrightarrow \mathrm{Br}_2; E^{\circ}=1.595 \mathrm{~V}, \mathrm{SRP} \text { (cathode) } \\
\mathrm{HBrO} \longrightarrow \mathrm{BrO}_3^{-}; E^{\circ}=-1.5 \mathrm{~V}, \mathrm{SOP} \text { (anode) }
\end{array}
\)
\(
2 \mathrm{HBrO} \longrightarrow \mathrm{Br}_2+\mathrm{BrO}_3^{-}
\)
\(
\begin{aligned}
E_{\text {cell }}^{\circ} & =\mathrm{SRP}(\text { cathode) }-\mathrm{SRP} \text { (anode) } \\
& =1.595-1.5 \\
& =0.095 \mathrm{~V} \\
E_{\text {cell }}^{\circ}>0 \Rightarrow & \Delta G^{\circ}<0 \text { [spontaneous] }
\end{aligned}
\)
Note: Reaction in (d) involves comproportionation or synproportionation. When two reactants, each containing the same element but with a different oxidation number, form a product in which the element involved reach the same oxidation number.

It is opposite to disproportionation.

Hint

(d)

\(
\mathrm{Zn} \rightarrow \mathrm{Zn}^{+2}+2 \mathrm{e}^{-} \dots(1)
\)
\(
8 \mathrm{e}^{-}+10 \mathrm{H}^{+}+\mathrm{NO}_3^{-} \rightarrow \mathrm{NH}_4^{+}+3 \mathrm{H}_2 \mathrm{O} \dots(2)
\)
operate eq. (1) \(\times 4+\) eq. (2) \(\times 1\)
\(
4 \mathrm{Zn}+10 \mathrm{H}^{+}+\mathrm{NO}_3^{-} \rightarrow 4 \mathrm{Zn}^{2+}+\mathrm{NH}_4^{+}+3 \mathrm{H}_2 \mathrm{O}
\)

Hint

(c)

In this reaction, none of the elements undergoes a change in oxidation number or valency.

Hint

(c) Lesser is the reduction potential greater is the reducing power.
Reducing power : \(\mathrm{K}>\mathrm{Al}>\mathrm{Cr}>\mathrm{Ag}\)

Hint

(a) \(\mathrm{F}_2\) is the strongest oxidising agent as it has highest reduction potential while \(\mathrm{I}^{-}\)is the strongest reducing agent since it has lowest reduction potential.
Higher the value of reduction potential, higher will be the oxidising power whereas the lower the value of reduction potential higher will be the reducing power.

Hint

(d)

\(
\begin{aligned}
&\text { In the following reaction }\\
&3 MnO _4^{2-}+4 H ^{+} \longrightarrow 2 MnO _4^{-}+ MnO _2+2 H _2 O
\end{aligned}
\)
\(
\begin{array}{|c|l|}
\hline \text { Oxidation state of } Mn & \text { Species } \\
\hline+6 & MnO _4^{2-} \\
\hline+7 & MnO _4^{-} \\
\hline+4 & MnO _2 \\
\hline
\end{array}
\)
\(
\text { So }+2 \text { and }+3 \text { oxidation state is not shown by } Mn \text {. }
\)

Hint

(d)