Question

which of the following series of isoelectronic ions (mg²⁺, n³⁻, f⁻, si⁴⁺) has the ionic radii in order of largest to smallest? a: mg²⁺ > n³⁻ > f⁻ > si⁴⁺; b: mg²⁺ > si⁴⁺ > f⁻ > n³⁻; c: n³⁻ > f⁻ > si⁴⁺ > mg²⁺; d: n³⁻ > f⁻ > mg²⁺ > si⁴⁺; e: f⁻ > n³⁻ > si⁴⁺ > mg²⁺. answer:

Explanation:

Step1: Recall ionic - radius rule for isoelectronic ions

For isoelectronic ions, the ion with the highest nuclear charge (number of protons) has the smallest ionic radius.

Step2: Determine the number of electrons and protons for each ion

• $N^{3 - }$: Nitrogen has atomic number $Z = 7$. $N^{3 - }$ has $7+3=10$ electrons and 7 protons.
• $F^{-}$: Fluorine has atomic number $Z = 9$. $F^{-}$ has $9 + 1=10$ electrons and 9 protons.
• $Mg^{2+}$: Magnesium has atomic number $Z = 12$. $Mg^{2+}$ has $12-2 = 10$ electrons and 12 protons.
• $Si^{4+}$: Silicon has atomic number $Z = 14$. $Si^{4+}$ has $14 - 4=10$ electrons and 14 protons.

Step3: Compare ionic radii based on nuclear charge

As the nuclear charge increases, the electrons are pulled in more tightly, reducing the ionic radius. So, the order of ionic radii from largest to smallest is $N^{3 - }>F^{-}>Mg^{2+}>Si^{4+}$.

Answer:

C. $N^{3 - }>F^{-}>Si^{4+}>Mg^{2+}$