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 species

For isoelectronic species, the ionic radius decreases with an increase in the nuclear charge (number of protons).

Step2: Determine the number of protons in each ion

$N^{3 - }$ has 7 protons, $F^{-}$ has 9 protons, $Mg^{2+}$ has 12 protons and $Si^{4+}$ has 14 protons.

Step3: Arrange ions based on ionic - radius

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 for isoelectronic species is $N^{3 - }>F^{-}>Mg^{2+}>Si^{4+}$.

Answer:

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