Q37 / 180 · COMEDK 2025 (10 May Shift 3) · Chemistry
Choose the correct metal/ ion from the brackets which ..........
A. has chemical reactivity similar to that of the first few members of the Lanthanoids $(\mathrm{Zn}, \mathrm{Ca}, \mathrm{Fe}, \mathrm{Cu})$.
B. has stable $4 \mathrm{f}^7$ electronic configuration, but acts as a strong reducing agent and converts to $\mathrm{M}^{3+}$ state. $\left(\mathrm{Eu}^{2+}, \mathrm{Ce}^{2+}, \mathrm{Pr}^{2+}, \mathrm{Dy}^{2+}\right)$
C. is a colorless ion $\left(\mathrm{Tm}^{3+}, \mathrm{Lu}^{3+}, \mathrm{Gd}^{3+}, \mathrm{Sm}^{3+}\right)$.
D. shows stable +2 oxidation state and is diamagnetic $(\mathrm{Ce}, \mathrm{Sm}, \mathrm{Ho}, \mathrm{Yb})$
- A$\mathrm{A}: \mathrm{Zn} \quad \mathrm{B}: \mathrm{Ce}^{2+}, \quad \mathrm{C}:\mathrm{Gd}^{3+} \quad \mathrm{D}: \mathrm{Sm}$
- Correct answerB$\mathrm{A}: \mathrm{Ca} \quad \mathrm{B}: \mathrm{Eu}^{2+} \quad \mathrm{C}: \mathrm{Lu}^{3+} \quad \mathrm{D}: ~ \mathrm{Yb}$
- C$\mathrm{A}: \mathrm{Fe} \quad \mathrm{B}: \mathrm{Pr}^{2+} \quad \mathrm{C}: \mathrm{Tm}^{3+} \quad \mathrm{D}: ~ \mathrm{Ce}$
- D$\mathrm{A}: \mathrm{Cu} \quad \mathrm{B}: \mathrm{Dy}^{2+} \quad \mathrm{C}: \mathrm{Sm}^{3+} \quad \mathrm{D}: ~ \mathrm{Ho}$
Solution
A. The lanthanoids are highly reactive metals, similar to calcium. Among the choices, calcium (Ca) exhibits chemical reactivity comparable to the early lanthanoids.
B. Europium (Eu) has the electronic configuration $[Xe] 4f^7 6s^2$. The ion $\mathrm{Eu}^{2+}$ has a stable half-filled $4f^7$ configuration. However, it is a strong reducing agent because it readily loses an electron to form the more stable $\mathrm{Eu}^{3+}$ state ($4f^6$).
C. $\mathrm{Lu}^{3+}$ has the electronic configuration $[Xe] 4f^{14}$. Since the $4f$ subshell is completely filled, there are no unpaired electrons, making the ion colorless.
D. Ytterbium (Yb) in the $+2$ oxidation state has the configuration $[Xe] 4f^{14}$. Since all electrons are paired in the $4f$ subshell, $\mathrm{Yb}^{2+}$ is diamagnetic and stable.
Answer: $\mathrm{A}: \mathrm{Ca} \quad \mathrm{B}: \mathrm{Eu}^{2+} \quad \mathrm{C}: \mathrm{Lu}^{3+} \quad \mathrm{D}: ~ \mathrm{Yb}$
B. Europium (Eu) has the electronic configuration $[Xe] 4f^7 6s^2$. The ion $\mathrm{Eu}^{2+}$ has a stable half-filled $4f^7$ configuration. However, it is a strong reducing agent because it readily loses an electron to form the more stable $\mathrm{Eu}^{3+}$ state ($4f^6$).
C. $\mathrm{Lu}^{3+}$ has the electronic configuration $[Xe] 4f^{14}$. Since the $4f$ subshell is completely filled, there are no unpaired electrons, making the ion colorless.
D. Ytterbium (Yb) in the $+2$ oxidation state has the configuration $[Xe] 4f^{14}$. Since all electrons are paired in the $4f$ subshell, $\mathrm{Yb}^{2+}$ is diamagnetic and stable.
Answer: $\mathrm{A}: \mathrm{Ca} \quad \mathrm{B}: \mathrm{Eu}^{2+} \quad \mathrm{C}: \mathrm{Lu}^{3+} \quad \mathrm{D}: ~ \mathrm{Yb}$
Hence, the correct answer is option: 'B'.