
Identify the correct order in which the ionic radius of the following ion increases.
I. ${{F}^{-}}$ II. $N{{a}^{+}}$ III. ${{N}^{3-}}$
(A) III, I, II
(B) I, II, III
(C) II, III, I
(D) II, I, III
Answer
555.3k+ views
Hint: Ionic radius the radius of a monatomic ion of an ionic compound. Remember that the ionic radius of a cation is less than that of an anion. Use this to answer the given question.
Complete step by step solution:
Before answering this, let us discuss the meaning of ionic radii.
The radius of a monatomic ion of an ionic compound is known as the ionic radius. We calculate the distance between the ions in a crystal lattice by taking the sum of the cationic and the anionic radii.
We know that the ions can be smaller or larger depending upon their charge. We know that when an atom loses an electron it gains a positive charge and thus gets smaller in size. Also, when an atom accepts an electron, it becomes negative and gets larger in size.
So we can understand that cations are smaller in size so they will have smaller ionic radii and anions are larger in size so they will give a larger ionic radius.
However, the ionic radius is not fixed for an atom and depends on various other factors other than charges like spin state and coordination number. But, in the given question we can determine the ion with the smallest and largest ionic radii based upon their charge.
Now, let us see the ions given to us.
We can see that $N{{a}^{+}}$ is a cation so it will have the lowest ionic radius.
Then, among ${{F}^{-}}$ and ${{N}^{3-}}$, we can see that nitrogen has a negative charge of -3. Thus, it will have a much larger ionic radius than fluorine.
So, the ionic radius increases from $N{{a}^{+}}$ to ${{F}^{-}}$ to ${{N}^{3-}}$.
Therefore, the correct answer is option [D] II, I, III.
Note: There are various types of radii that we can calculate depending upon the type of bond prevailing between atoms or ions. We have Van der Waals radii, metallic radii and ionic radii. We can find the radii of an atom by measuring the distance between the nuclei of the two touching atoms and then taking its half.
Complete step by step solution:
Before answering this, let us discuss the meaning of ionic radii.
The radius of a monatomic ion of an ionic compound is known as the ionic radius. We calculate the distance between the ions in a crystal lattice by taking the sum of the cationic and the anionic radii.
We know that the ions can be smaller or larger depending upon their charge. We know that when an atom loses an electron it gains a positive charge and thus gets smaller in size. Also, when an atom accepts an electron, it becomes negative and gets larger in size.
So we can understand that cations are smaller in size so they will have smaller ionic radii and anions are larger in size so they will give a larger ionic radius.
However, the ionic radius is not fixed for an atom and depends on various other factors other than charges like spin state and coordination number. But, in the given question we can determine the ion with the smallest and largest ionic radii based upon their charge.
Now, let us see the ions given to us.
We can see that $N{{a}^{+}}$ is a cation so it will have the lowest ionic radius.
Then, among ${{F}^{-}}$ and ${{N}^{3-}}$, we can see that nitrogen has a negative charge of -3. Thus, it will have a much larger ionic radius than fluorine.
So, the ionic radius increases from $N{{a}^{+}}$ to ${{F}^{-}}$ to ${{N}^{3-}}$.
Therefore, the correct answer is option [D] II, I, III.
Note: There are various types of radii that we can calculate depending upon the type of bond prevailing between atoms or ions. We have Van der Waals radii, metallic radii and ionic radii. We can find the radii of an atom by measuring the distance between the nuclei of the two touching atoms and then taking its half.
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