How would you rank the atomic radius of the following elements from largest to smallest: ${\text{N}}{{\text{a}}^ + },{{\text{F}}^ - },{{\text{O}}^{2 - }},{\text{A}}{{\text{l}}^{3 + }},{\text{M}}{{\text{g}}^{2 + }},{{\text{N}}^{3 - }}$.
Answer
580.8k+ views
Hint: We know that the atomic size is described by the atomic radius. Atomic radius is defined as the distance from the centre of the nucleus to the outermost shell containing the electrons.
Complete step by step answer:
Let’s first discuss isoelectronic species in detail. They are the atoms or ions of different elements but the number of electrons in them is equal. For example, ${\text{N}}{{\text{a}}^ + }$
and Helium are isoelectronic species because the number of electrons in both the species is 10.
Now, come to the question. Here, we have to count the number of protons and electrons of all the chemical species.
In ${\text{N}}{{\text{a}}^ + }$, Number of protons=11, number of electrons=10
In ${{\text{F}}^ - }$, Number of protons=9, number of electrons=10
In ${{\text{O}}^{2 - }}$, Number of protons=8, number of electrons=10
In ${\text{A}}{{\text{l}}^{3 + }}$, Number of protons=13, number of electrons=10
In ${\text{M}}{{\text{g}}^{2 + }}$, Number of protons=12, number of electrons=10
In ${{\text{N}}^{3 - }}$, Number of protons=7, number of electrons=10
As all the above species have the same number of electrons, they are called isoelectronic species.
Let’s discuss how the atomic radius is affected by the Number of protons and number of electrons. All isoelectronic species have the same number of electrons but their nuclear charges differ as all of them have different numbers of protons. If the number of protons increases in the nucleus, the electrons are more attracted towards the nucleus and this results in the decrease of ionic radii.
So, the decreasing order of atomic radius of the isoelectronic species is, ${{\text{N}}^{ - 3}} > {{\text{O}}^{2 - }} > {{\text{F}}^ - } > {\text{N}}{{\text{a}}^ + } > {\text{M}}{{\text{g}}^{2 + }} > {\text{A}}{{\text{l}}^{3 + }}$
Note: It is to be remembered that Along a period (moving from left to right of the periodic table), the atomic radii of the elements generally decreases and down a group, the atomic radius of elements increases.
Complete step by step answer:
Let’s first discuss isoelectronic species in detail. They are the atoms or ions of different elements but the number of electrons in them is equal. For example, ${\text{N}}{{\text{a}}^ + }$
and Helium are isoelectronic species because the number of electrons in both the species is 10.
Now, come to the question. Here, we have to count the number of protons and electrons of all the chemical species.
In ${\text{N}}{{\text{a}}^ + }$, Number of protons=11, number of electrons=10
In ${{\text{F}}^ - }$, Number of protons=9, number of electrons=10
In ${{\text{O}}^{2 - }}$, Number of protons=8, number of electrons=10
In ${\text{A}}{{\text{l}}^{3 + }}$, Number of protons=13, number of electrons=10
In ${\text{M}}{{\text{g}}^{2 + }}$, Number of protons=12, number of electrons=10
In ${{\text{N}}^{3 - }}$, Number of protons=7, number of electrons=10
As all the above species have the same number of electrons, they are called isoelectronic species.
Let’s discuss how the atomic radius is affected by the Number of protons and number of electrons. All isoelectronic species have the same number of electrons but their nuclear charges differ as all of them have different numbers of protons. If the number of protons increases in the nucleus, the electrons are more attracted towards the nucleus and this results in the decrease of ionic radii.
So, the decreasing order of atomic radius of the isoelectronic species is, ${{\text{N}}^{ - 3}} > {{\text{O}}^{2 - }} > {{\text{F}}^ - } > {\text{N}}{{\text{a}}^ + } > {\text{M}}{{\text{g}}^{2 + }} > {\text{A}}{{\text{l}}^{3 + }}$
Note: It is to be remembered that Along a period (moving from left to right of the periodic table), the atomic radii of the elements generally decreases and down a group, the atomic radius of elements increases.
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