
For ${\text{N}}{{\text{a}}^{\text{ + }}}$ , ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ , ${{\text{F}}^{\text{ - }}}$ and ${{\text{O}}^{{\text{2 - }}}}$ the correct order of increasing ionic radii is:
A.${{\text{O}}^{{\text{2 - }}}}{\text{ < }}{{\text{F}}^{\text{ - }}}{\text{ < N}}{{\text{a}}^{\text{ + }}}{\text{ < M}}{{\text{g}}^{{\text{2 + }}}}$
B.${\text{M}}{{\text{g}}^{{\text{2 + }}}}{\text{ < N}}{{\text{a}}^{\text{ + }}}{\text{ < }}{{\text{F}}^{\text{ - }}}{\text{ < }}{{\text{O}}^{{\text{2 - }}}}$
C.${\text{M}}{{\text{g}}^{{\text{2 + }}}}{\text{ < }}{{\text{O}}^{{\text{2 - }}}}{\text{ < N}}{{\text{a}}^{\text{ + }}}{\text{ < }}{{\text{F}}^{\text{ - }}}$
D.${\text{N}}{{\text{a}}^{\text{ + }}}{\text{ < M}}{{\text{g}}^{{\text{2 + }}}}{\text{ < }}{{\text{F}}^{\text{ - }}}{\text{ < }}{{\text{O}}^{{\text{2 - }}}}$
Answer
567.3k+ views
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.
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.
Recently Updated Pages
1Pentyne and 2Pentyne can be distinguished by ASilver class 11 chemistry CBSE

The twisting couple per unit twist for a solid cylinder class 11 physics CBSE

The sum of two point charges is7mu C They repel each class 11 physics CBSE

Least count of ammeter and voltmeter are class 11 physics CBSE

Let n be the number of ways in 5 boys and 5 girls -class-11-maths-CBSE

Three rods made of the same material and having the class 11 physics CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

