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Question #44319

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The Question

When solid $KMnO_4$ is treated with cold concentrated $H_2SO_4$ at $298\text{ K}$ , a highly explosive, green oily liquid $\mathbf{X}$ is formed. Separately, the thermal decomposition of manganese(II) oxalate in an inert atmosphere yields a solid oxide $\mathbf{Y}$ . Which of the following correctly identifies $\mathbf{X}$ and $\mathbf{Y}$ , and accurately explains the fundamental origin of their contrasting acid-base behavior?

Options

A.
$\mathbf{X}$ is $Mn_2O_7$ , $\mathbf{Y}$ is $MnO$ . The high $Z_{eff}$ and charge density of $Mn^{+7}$ in $\mathbf{X}$ cause extreme polarization of the $O^{2-}$ cloud (Fajans' rule), yielding a covalent acidic anhydride. $\mathbf{Y}$ has low polarizing power, forming an ionic basic oxide.
B.
$\mathbf{X}$ is $Mn_2O_7$ , $\mathbf{Y}$ is $MnO_2$ . The acidity of $\mathbf{X}$ is due to unpaired d-electrons in $Mn^{+7}$ accepting lone pairs from water, whereas $\mathbf{Y}$ is basic due to its polymeric lattice.
C.
$\mathbf{X}$ is $MnO_3$ , $\mathbf{Y}$ is $MnO$ . The basicity of $\mathbf{Y}$ is due to the inert pair effect stabilizing the $+2$ state, while $\mathbf{X}$ is acidic because of $p\pi-d\pi$ back-bonding.
D.
$\mathbf{X}$ is $Mn_2O_7$ , $\mathbf{Y}$ is $Mn_3O_4$ . The acidic nature of $\mathbf{X}$ arises from the low $Z_{eff}$ of $Mn^{+7}$ allowing easy release of protons in water, while $\mathbf{Y}$ is a mixed basic oxide.

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Acid-Base Nature of Oxides|Variable Oxidation State Stability|Effective Nuclear Charge (Zeff)

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental reason behind the transition of transition metal oxides from basic to acidic as the oxidation state increases.

Students often memorize the trend but fail to connect it to Fajans' Rules and Effective Nuclear Charge ( $Z_{eff}$ ).

The trap lies in the distractors: Option B incorrectly attributes acidity to unpaired d-electrons (but $Mn^{+7}$ is $d^0$ ), Option C misapplies the inert pair effect (which is strictly for heavier p-block elements, not d-block), and Option D incorrectly states that $Mn^{+7}$ has a low $Z_{eff}$ .

The correct logic requires understanding that high $Z_{eff}$ in $Mn^{+7}$ leads to extreme polarization of the oxide ion, shifting the bond character from ionic (basic) to covalent (acidic).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Classification of Elements and Periodicity in Properties - Class 11 Part 1
Concept	Acid-Base Nature of Oxides
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free