d and f-Block Elements — JEE Main practice

1. Find the oxidation state of Mn in $\mathrm{MnO_4^{2-}}$.

   Let the oxidation state of Mn be $\mathit{x}$.

   $$\mathit{x}+4(-2)=-2$$ $$\mathit{x}-8=-2$$ $$\mathit{x}=+6$$

   Thus, manganese is in the $+6$ oxidation state.

2. Understand disproportionation.

   A disproportionation reaction is one in which the same element in a single oxidation state undergoes both oxidation and reduction simultaneously.

   Here, Mn in the $+6$ state changes into:

   • $+7$ by oxidation, forming $\mathrm{MnO_4^-}$ (permanganate).
   • $+4$ by reduction, forming $\mathrm{MnO_2}$.

3. Balanced reaction in acidic medium.

   $$3\mathrm{MnO_4^{2-}}+4\mathrm{H^+} \rightarrow 2\mathrm{MnO_4^-}+\mathrm{MnO_2}+2\mathrm{H_2O}$$

4. Why the other options are incorrect.

   • Option A: MnO contains Mn in the $+2$ oxidation state and is not formed in this disproportionation.
   • Option C: $\mathrm{Mn_2O_7}$ is not produced during the reaction.
   • Option D: Neither $\mathrm{Mn_2O_7}$ nor MnO are the products of this disproportionation.

5. Therefore, in acidic medium, $\mathrm{MnO_4^{2-}}$ disproportionates to $\mathrm{MnO_4^-}$ and $\mathrm{MnO_2}$.

Answer: D

1. Statement A is correct.

   Determine the oxidation state of Mn in $\mathrm{Mn_2O_7}$.

   $$2\left(\mathit{x}\right)+7(-2)=0$$ $$2\mathit{x}=14$$ $$\mathit{x}=+7$$

   Since manganese belongs to Group 7, its highest oxidation state is $+7$. Therefore, Mn is present in its maximum oxidation state in $\mathrm{Mn_2O_7}$.

2. Statement B is correct.

   Oxygen stabilizes transition metals in very high oxidation states by forming strong covalent multiple bonds such as $\mathrm{Mn=O}$.

   These multiple bonds help distribute electron density efficiently, making highly oxidized manganese compounds more stable.

3. Statement C is incorrect.

   $\mathrm{Mn_2O_7}$ is not an ionic oxide.

   Because Mn is in the very high oxidation state of $+7$, the Mn–O bonds have strong covalent character. Its physical properties, such as being a dark green molecular liquid with a low melting point, also indicate that it is a covalent compound rather than an ionic solid.

4. Statement D is correct.

   The structure of $\mathrm{Mn_2O_7}$ is represented as:

   $$\mathrm{O_3Mn-O-MnO_3}$$

   It consists of two $\mathrm{MnO_4}$ tetrahedra connected through one bridging oxygen atom, while the remaining oxygen atoms are terminal.

5. Hence, the correct statements are A, B and D, whereas C is false.

Answer: B

1. On heating potassium dichromate with sodium chloride (common salt) and concentrated sulphuric acid, chromyl chloride is formed.

   $K_2Cr_2O_7 + 4NaCl + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 2KHSO_4 + 4NaHSO_4 + 3H_2O$

   Therefore, the gas evolved is:

   $$CrO_2Cl_2$$

2. Let the oxidation state of chromium in (CrO_2Cl_2) be (x).

   Oxygen contributes:

   $$2(-2)=-4$$

   Chlorine contributes:

   $$2(-1)=-2$$

   Since the molecule is neutral:

   $$x-4-2=0$$ $$x=+6$$

3. Hence, the evolved gas is chromyl chloride ((CrO_2Cl_2)) and chromium is in the +6 oxidation state.

Answer: C