Question

The ion with smallest radius among the following is

• A. Ca$^{2+}$
• B. K$^{+}$
• C. Ti$^{4+}$
• D. Sc$^{3+}$

Hint:

For isoelectronic species (ions or atoms with the same number of electrons), the species with the higher nuclear charge (more protons) will have a smaller radius because the increased positive charge pulls the electron cloud in more tightly.

Answer 1

The Correct Option is C

Step 1: Determine the number of protons and electrons in each ion. \begin{itemize} \item Ca$^{2+}$: Protons = 20, Electrons = 18 \item K$^{+}$: Protons = 19, Electrons = 18 \item Ti$^{4+}$: Protons = 22, Electrons = 18 \item Sc$^{3+}$: Protons = 21, Electrons = 18 \end{itemize}
Step 2: Identify that all ions are isoelectronic.
All the given ions have the same number of electrons (18).
Step 3: Apply the rule for ionic radius in isoelectronic species.
For isoelectronic ions, the ionic radius decreases with increasing nuclear charge (number of protons).
Step 4: Compare the nuclear charges of the ions.
The number of protons (nuclear charge) for each ion is: \begin{itemize} \item Ca$^{2+}$: 20 \item K$^{+}$: 19 \item Ti$^{4+}$: 22 \item Sc$^{3+}$: 21 \end{itemize}
Step 5: Determine the ion with the smallest radius.
The ion with the largest nuclear charge is Ti$^{4+}$ (22 protons). Therefore, Ti$^{4+}$ will have the smallest ionic radius due to the strongest attraction between the nucleus and the electrons. Final Answer: The final answer is $\boxed{Ti^{4+}}$