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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⁺

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Answer:

Option (d) is correct

N³⁻ > F⁻ > Mg²⁺ > Si⁴⁺

Explanation:

Total electrons for all the species = 10

So these are iso electronic with each other.

We know

Ionic radii ∝ [tex]\frac{Magnitude of Negative Charge}{Magnitude of Positive Charge}[/tex]

  • Si⁴⁺ has 14 protons and 10 electrons
  • Mg²⁺ has 12 protons and 10 electrons
  • N³⁻ has 7 protons and 10 electrons
  • F⁻ has 9 protons and 10 electrons
  • Iso electronic species with greatest number of protons have small size and vice versa.
  • So Si⁺⁴ have smallest size and N³⁻ have largest in size
pstnonsonjoku

The correct arrangement of the isoelectronic ions  in order of largest to smallest ionic radii is; N³⁻ > F⁻ > Mg²⁺ > Si⁴⁺

Isoelectronic ions are ions that have the same number of electrons, All the ions listed have ten electrons.

We must know that ionic radius is a periodic trend that increases across the period.

The ionic radii of the ions are listed as follows;

  • F⁻    =   147 pm
  • N³⁻  =   146 pm
  • Mg²⁺ =  72 pm
  • Si⁴⁺  =  54 pm

Hence, the correct order of ionic radii from largest to smallest is N³⁻ > F⁻ > Mg²⁺ > Si⁴⁺.

Learn more; https://brainly.com/question/17029830

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