3 Marks Question

Question 13 Marks

Permanganate ion $\left( MnO _4^{-}\right)$gives different reduction products at different concentration of hydrogen ions, write their equations.

Answer

The products formed in the reduction reaction of permanganate ion depend on the concentration of hydrogen ion. The reactions occurring at different concentration of $H ^{+}$ are as follows:
$
\begin{aligned}
MnO_{+7}^{-}+e^{-} & \rightarrow \underset{+6}{MnO_4^{2-}} \\
MnO_4^{-}+4 H^{+}+3 e^{-} & \rightarrow \underset{+4}{MnO_2}+2 H_2 O \\
MnO_4^{-}+8 H^{+}+5 e^{-} & \rightarrow Mn^{2+}+4 H_2 O
\end{aligned}
$
Water should be oxidized by permanganate ion at $\left[ H ^{+}\right]$ $=1$ but this reaction happens very slowly but in this. The rate of reaction increases by using $Mn ^{2+}$ ion as a catalyst or by increasing the temperature.

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Question 23 Marks

(a) Which is the salt of Mn which is isostructural to $KClO _4$ ?
(b) Explain the structure and magnetic properties of manganese and permanganate ions.

Answer

(a) $KMnO _4$ (potassium permanganate) is isostructural to $KClO _4$ (potassium chlorate), that is, both have the same structure.
(b) Manganate and permanganate ions are tetrahedral in which $\pi$-bond is formed due to overlap between the $p$ orbital of oxygen and $d$-orbitals of Mn and $sp ^3$ hybridization occurs on Mn . Their structure is as follows :

Permanganate ion is diamagnetic, when heated at 513 K it gets converted into manganese ion which is paramagnetic because it has one unpaired electron.

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Question 33 Marks

(a) $Cu ^{+} X ^{-}$is unstable, why?
(b) Explain the structure of $Mn _2 O _7$ and $MnO _4^{-}$.

Answer

(a) $Cu ^{+} X ^{-}$is unstable because the disproportionation of Cu in aqueous solution occurs.
$
2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+}+Cu_{(s)}
$
Therefore, $Cu ^{2+}$ is more stable than $Cu ^{+}$in aqueous solution because the value of hydration enthalpy of $Cu ^{2+}$ is much more negative than that of $Cu ^{+}$, which can easily balance the energy (second ionization enthalpy) required to form $Cu ^{2+}$ from $Cu ^{+}$.
(b) In $Mn _2 O _7$, each Mn atom is tetrahedrally surrounded by oxygen atoms and an $Mn - O - Mn$ bridge is also found in it and the structure of $MnO _4^{-}$is also tetrahedral.

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Question 43 Marks

Write down the number of 3d electrons in each of the following ions :
$Ti ^{2+}, V ^{2+}, Cr ^{3+}, Mn ^{2+}, Fe ^{2+}, Fe ^{3+}, Co ^{2+}, Ni ^{2+}, Cu ^{2+}$.
Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).

Answer

In hydrated ions, water H₂O acts as a liquid due to which the five 3d orbitals of equal energy get divided into two parts- t_2g and e_g. The energy of t_2g orbitals is less : than that of t_2g orbitals. The energy difference between t_2g orbitals is less because H₂O is a weak ligand. Electrons of different ions are filled in these orbitals in the following manner :
(i) $ Ti ^{2+}:[ Ar ] 3 d^2=\left( t _{2 g}{ }^2\right)$
(ii) $ V ^{2+}:[ Ar ] 3 d^3=\left( t _{2 g}\right)^3$
(iii) $ Cr ^{3+}:[ Ar ] 3 d^3=\left( t _{2 g}\right)^3$
(iv) $ Mn ^{2+}:[ Ar ] 3 d^5=\left( t _{2 g}\right)^3\left( e _{ g }\right)^2$
(v) $ Fe ^{2+}:[ Ar ] 3 d^6=\left( t _{2 g}\right)^4\left( e _{ g }\right)^2$
(vi) $ Fe ^{3+}:[ Ar ] 3 d^5=\left( t _{2 g}\right)^3\left( e _{ g }\right)^2$
(vii) $ Co ^{2+}:[ Ar ] 3 d^7=\left( t _{2 g}\right)^5\left( e _{ g }\right)^2$
(viii) $ Ni ^{2+}:[ Ar ] 3 d^8=\left( t _{2 g}\right)^6\left( e _{ g }\right)^2$
(ix) $ Cu ^{2+}:[ Ar ] 3 d^9=\left( t _{2 g}\right)^6\left( e _{ g }\right)^3$

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Question 53 Marks

For M²⁺/M and M³⁺/M²⁺ systems the $E ^{\circ}$ values for some metals are as follows :
$\begin{array}{llll} Cr ^{2+} / Cr & -0.9 V & Cr ^3 / Cr ^{2+} & -0.4 V \\ Mn ^{2+} / Mn & -1.2 V & Mn ^{3+} / Mn ^{2+} & +1.5 V \\ Fe ^{2+} / Fe & -0.4 V & Fe ^{3+} / Fe ^{2+} & +0.8 V\end{array}$
Use this data to comment upon :
(i) the stability of Fe³⁺ in acid solution are compared to that of Cr³⁺or Mn³⁺ and
(ii) the case with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Answer

(i) When the reduction potential (electrode potential) of a species is high, then its tendency to be reduced is high. Reduction potential of Mn⁺³ is maximum hence it gets easily reduced to Mn²⁺ hence Mn⁺³ is less stable than Fe⁺³ But Cr⁺³ is much lower than the reduction potential of Fe⁺³.
(ii) When the value of electrode potential (reduction potential) of a metal ion is low, then the tendency of that metal atom to be oxidized will the more, hence Mn will have the highest tendency to be oxidized to Mn⁺² and Fe will have the highest tendency to be oxidized to Fe⁺². The tendency for oxidation will be minimum in +2. Therefore the order of their oxidation is as follows - $Mn > Cr > Fe$.

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Question 63 Marks

Describe the oxidising action of potassium dichromate and write the ionic equation for its reaction with :
(i) Iodide, (ii) Iron, (II) solution and (iii) H₂S

Answer

Potassium dichromate is a strong oxidant. The oxidation action of dichromate ion in acidic medium is shown as follows, in this Cr⁺⁶ changes into Cr⁺³.
$Cr _2 O _7^{2-}+14 \stackrel{+}{ H }+6 e ^{-} \rightarrow 2 Cr ^{3+}+7 H _2 O \left( E ^{\circ}=1.33 V\right)$
(i) Iodide ion : $K _2 Cr _2 O _7$, Oxidizes iodide ion to iodine.
$Cr _2 O _7^{2-}+14 \stackrel{+}{ H }+6 I ^{-} \rightarrow 2 Cr ^{3+}+7 H _2 O +3 I _2$
(ii) Iron(II) solution : $K _2 Cr _2 O _7$, oxidises $Fe ^{2+}$ to $Fe ^{3+}$.
$Cr _2 O _7^{2-}+14 \stackrel{+}{ H }+6 Fe ^{2+} \rightarrow 2 Cr ^{3+}+6 Fe ^{+3}+7 H _2 O$
(iii) H₂S : Dichromate oxidizes H₂S to sulphur.
$Cr _2 O _7^{2-}+8 H ^{+}+3 H _2 S \rightarrow 2 Cr ^{3+}+7 H _2 O +3 S$

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Question 73 Marks

Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Answer

Making potassium dichromate from iron chromite ore : When iron chromite [chromite ore (FeCr₂O₄)] is fused with sodium carbonate in the presence of air, sodium chromate is obtained. The reaction of chromite with sodium carbonate is as follows :
$ 4 FeCr _2 O _4+8 Na _2 CO _3+7 O _2 \rightarrow 8 Na _2 CrO _4 +2 Fe _2 O _3+8 CO _2$
The sodium chromate solution is filtered and made acidic by sulphuric acid from which orange sodium dichromate (Na₂Cr₂O7. 2H₂O) is crystallized.
$2 Na _2 CrO _4+2 H ^{+} \rightarrow Na _2 Cr _2 O _7+2 Na ^{+}+ H _2 O$
The solubility of sodium dichromate is more than that of potassium dichromate. Therefore, when potassium chloride is added to sodium dichromate solution, potassium dichromate is obtained.
$Na _2 Cr _2 O _7+2 KCl \rightarrow K _2 Cr _2 O _7+2 NaCl$
Orange coloured crystals crystallize from solution. On increasing the pH of potassium dichromate solution, i.e. on making alkaline medium, it gets converted into chromate.
$Cr _2 O _7^{2-}+2 \overline{ O } H \rightarrow 2 CrO _4^{2-}+ H _2 O$

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