Q31 / 180 · COMEDK 2024 (Evening Shift) · Chemistry
Identify the correct statement from the following.
- A$\mathrm{Cr}(\mathrm{VI})$ is more stable than $\mathrm{W}(\mathrm{VI})$ and hence acts as a good oxidising agent
- BInterstitial compounds of transition metals have lower melting points than that of pure transition metals and their compounds are chemically reactive
- Correct answerCCerium is a lanthanoid metal which exists in a stable oxidation state of +4 , besides exhibiting an oxidation state of +3
- DThe green manganate ion shows diamagnetic nature but the permanganate ion exhibits paramagnetic nature
Solution
Correct answer: (3)
(1) Incorrect: Higher oxidation states become more stable down the group, so W(VI) is more stable than Cr(VI). Cr(VI) is less stable and hence acts as a strong oxidising agent.
(2) Incorrect: Interstitial compounds have higher melting points than pure metals (lattice becomes more rigid), and they are chemically inert.
(3) Correct: Cerium (Ce) is a lanthanoid showing +3 state normally, but also exists in a stable +4 state since $Ce^{4+}$ achieves noble gas configuration $[Xe]4f^0$.
(4) Incorrect: $MnO_4^{2-}$ has $Mn^{6+}$ ($3d^1$, one unpaired electron) $\to$ paramagnetic. $MnO_4^{-}$ has $Mn^{7+}$ ($3d^0$, no unpaired electron) $\to$ diamagnetic. The statement reverses these.
(1) Incorrect: Higher oxidation states become more stable down the group, so W(VI) is more stable than Cr(VI). Cr(VI) is less stable and hence acts as a strong oxidising agent.
(2) Incorrect: Interstitial compounds have higher melting points than pure metals (lattice becomes more rigid), and they are chemically inert.
(3) Correct: Cerium (Ce) is a lanthanoid showing +3 state normally, but also exists in a stable +4 state since $Ce^{4+}$ achieves noble gas configuration $[Xe]4f^0$.
(4) Incorrect: $MnO_4^{2-}$ has $Mn^{6+}$ ($3d^1$, one unpaired electron) $\to$ paramagnetic. $MnO_4^{-}$ has $Mn^{7+}$ ($3d^0$, no unpaired electron) $\to$ diamagnetic. The statement reverses these.
Hence, the correct answer is option: 'C'.