Periodic Classification and Periodic Properties 2 Question 36
36. $\mathrm{Ca}^{2+}$ has a smaller ionic radius than $\mathrm{K}^{+}$because it has
(1993, 1M
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Answer:
Correct Answer: 36. $\mathrm{F}$
Solution:
- Higher effective nuclear charge due to greater $p / e$ ratio.
