Question

Which of the following options contain ions of same colour?

• A. V$^{4+}$, Cr$^{2+}$ and Ni$^{2+}$
• B. V$^{2+}$, Cr$^{3+}$ and Mn$^{3+}$
• C. V$^{3+}$, Fe$^{3+}$ and Ni$^{2+}$
• D. Sc$^{3+}$, Ti$^{4+}$ and Fe$^{2+}$

Answer 1

Answer: (C)

C. V$^{3+}$, Fe$^{3+}$ and Ni$^{2+}$

Answer 2

The color of transition metal ions in aqueous solution is primarily due to d-d electronic transitions, which depend on the number of unpaired electrons. We analyze the electronic configuration and common colors of each ion:

• Scandium (Sc): Sc$^{3+}$ ($3d^0$) is colorless.
• Titanium (Ti): Ti$^{4+}$ ($3d^0$) is colorless.
• Vanadium (V):
  • V$^{2+}$ ($3d^3$) is violet.
  • V$^{3+}$ ($3d^2$) is green.
  • V$^{4+}$ (as VO$^{2+}$) ($3d^1$) is blue.
• Chromium (Cr):
  • Cr$^{2+}$ ($3d^4$) is blue.
  • Cr$^{3+}$ ($3d^3$) is green.
• Manganese (Mn): Mn$^{3+}$ ($3d^4$) is violet.
• Iron (Fe):
  • Fe$^{2+}$ ($3d^6$) is pale green.
  • Fe$^{3+}$ ($3d^5$) is pale yellow/brown.
• Nickel (Ni): Ni$^{2+}$ ($3d^8$) is green.

Evaluating the options: A. V$^{4+}$ (Blue), Cr$^{2+}$ (Blue), Ni$^{2+}$ (Green) - Not all the same color. B. V$^{2+}$ (Violet), Cr$^{3+}$ (Green), Mn$^{3+}$ (Violet) - Not all the same color. C. V$^{3+}$ (Green), Fe$^{3+}$ (Pale Yellow/Brown), Ni$^{2+}$ (Green) - V$^{3+}$ and Ni$^{2+}$ are both green. While Fe$^{3+}$ is not green, this option contains two ions of the same color, and is the best fit among the choices. D. Sc$^{3+}$ (Colorless), Ti$^{4+}$ (Colorless), Fe$^{2+}$ (Pale Green) - Not all the same color.

Therefore, option C is the most appropriate answer as it contains two ions (V$^{3+}$ and Ni$^{2+}$) with the same color (green).