What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
Ions of different elements which have the same number of electrons but different magnitude of the nuclear charge are called isoelectronic ions.
To view explanation, please take trial in the course below. NEET 2023 - Target Batch - Aryan Raj Singh To view explanation, please take trial in the course below. NEET 2023 - Target Batch - Aryan Raj Singh Please attempt this question first. Add Note 
Tardigrade - CET NEET JEE Exam App
© 2022 Tardigrade®. All rights reserved Answer Hint: The given ions are all isoelectronic species. For isoelectronic species, the radii depend upon their ${\text{Z/e}}$ ratio. Lesser is the ${\text{Z/e}}$ ratio, more is the ionic size of the ion, i.e., the ionic size of an isoelectronic species is inversely proportional to its ${\text{Z/e}}$ ratio. Complete step by step answer: Those species which have the same number of electrons and belong to different elements and possess different magnitude of nuclear charge are called isoelectronic species. The sodium ion carries a positive charge of 1. This means it has 10 electrons as a neutral sodium atom has 11 electrons. The magnesium ion carries a positive charge of 2 which means it also has 10 electrons as a neutral magnesium atom has 12 electrons. The fluoride ion has negative charge of 1 and so it also has 10 electrons as a neutral fluorine atom has 9 electrons. Similarly, the oxide ion carries a negative charge of 2 and so it also has 10 electrons as a neutral oxygen atom has 8 electrons. Hence, all these four species are isoelectronic. The ${\text{Z/e}}$ ratio on which the radii of these species depend is actually the ratio of the atomic number ‘Z’ and the number of electrons ‘n’ in the ionic species. Let us determine the ${\text{Z/e}}$ ratio of these four given species.For the ${\text{N}}{{\text{a}}^{\text{ + }}}$ ion, the atomic number ‘Z’ is 11 and the number of electrons in the ion ‘n’ is 10. So, the ${\text{Z/e}}$ ratio is equal to $11/10 = 1.1$ . For the ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ ion, the atomic number ‘Z’ is 12 and the number of electrons in the ion ‘n’ is 10. So, the ${\text{Z/e}}$ ratio is equal to $12/10 = 1.2$ . For the ${{\text{F}}^{\text{ - }}}$ ion, the atomic number ‘Z’ is 9 and the number of electrons in the ion ‘n’ is 10. So, the ${\text{Z/e}}$ ratio is equal to $9/10 = 0.9$ . For the ${{\text{O}}^{{\text{2 - }}}}$ ion, the atomic number ‘Z’ is 8 and the number of electrons in the ion ‘n’ is 10. So, the ${\text{Z/e}}$ ratio is equal to $8/10 = 0.8$ . Comparing the ${\text{Z/e}}$ ratios of all the species, we can see that ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ ion has the highest ${\text{Z/e}}$ ratio, followed by ${\text{N}}{{\text{a}}^{\text{ + }}}$ , ${{\text{F}}^{\text{ - }}}$ and ${{\text{O}}^{{\text{2 - }}}}$ . Since ionic radii is inversely proportional to ${\text{Z/e}}$ ratio, so, ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ ion has the least ionic radii and increases from ${\text{N}}{{\text{a}}^{\text{ + }}}$ to ${{\text{F}}^{\text{ - }}}$ to ${{\text{O}}^{{\text{2 - }}}}$ . Hence, the correct increasing order is ${\text{M}}{{\text{g}}^{{\text{2 + }}}}{\text{ < N}}{{\text{a}}^{\text{ + }}}{\text{ < }}{{\text{F}}^{\text{ - }}}{\text{ < }}{{\text{O}}^{{\text{2 - }}}}$ .Hence option B is correct. Note: Cations have a lesser number of electrons than their corresponding neutral atoms. As a result, the effective nuclear charge of cations increases and the electron cloud shrinks. Hence, the size of cations is less than the size of their corresponding neutral atoms. On the other hand, anions have more electrons than their corresponding neutral atoms. As a result, the effective nuclear charge of anions decreases and the electron cloud is held less tightly. Hence, the size of anions is larger than the size of their corresponding neutral atoms. |
What do the ions O2 F Na+ and Mg2+ are isoelectronic with?
Postagens relacionadas
direito autoral © 2024 estudarpara Inc.
