The correct answer is Option (3) → (A), (B), (C) and (D)
- (A) Yttrium does not exhibit variable oxidation states, yet it is a transition element: Correct. A transition element is defined as an element having a partially filled d subshell in its ground state or in any of its common oxidation states. Yttrium ($Y, Z=39$) has the configuration $[Kr] 4d^1 5s^2$. Because it has a partially filled $4d$ orbital in its ground state, it is a transition metal, even though it almost exclusively shows the $+3$ oxidation state.
- (B) Many compounds of transition elements act as catalysts in various chemical reactions: Correct. This is one of the most famous properties of transition metals. Their ability to adopt variable oxidation states and provide a large surface area for adsorption allows them to form intermediate complexes that lower the activation energy of reactions.
- Examples: $V_2O_5$ in the Contact Process, $Fe$ in the Haber Process, and $Ni$ in hydrogenation.
- (C) Technetium is an artificially synthesized element of 4d transition series: Correct. Technetium ($Tc, Z=43$) was the first element to be produced artificially (in 1937). While trace amounts exist in nature from uranium fission, almost all technetium used today is synthetic. It sits in the middle of the $4d$ (second) transition series.
- (D) The $E^\circ$ value for the $Mn^{3+}/Mn^{2+}$ couple is much more positive than that for $Cr^{3+}/Cr^{2+}$: Correct. * For $Mn^{3+}/Mn^{2+}$, $E^\circ \approx +1.57\text{ V}$.
- For $Cr^{3+}/Cr^{2+}$, $E^\circ \approx -0.41\text{ V}$.
- Reason: The $Mn^{2+}$ ion has a $d^5$ configuration, which is exceptionally stable because it is half-filled. Thus, $Mn^{3+}$ ($d^4$) has a very strong tendency to gain an electron to reach this stable state, resulting in a high positive reduction potential.
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