The radius of $N{a^ + }$ ion is 95 pm and that of $C{l^ - }$ ions is 181 pm. Predict whether the coordination number of $N{a^ + }$ ion is 6 or 4.
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Hint: Given the radius of sodium and chloride ions. To find the coordination number of Sodium ions, use the limiting radius ratio. Limiting radius ratio is the ratio of the radius of cation over the radius of anion. Find the range where this ratio falls and then we can find the coordination number of sodium ions.
Complete Step by step answer: We are given that the radius of $N{a^ + }$ ion is 95 pm and that of $C{l^ - }$ ions is 181 pm.
We have to predict whether the coordination number of $N{a^ + }$ ion is 6 or 4.
We can solve this question using the limited radius ratio.
Limited radius ratio (L) is the minimum allowed value for the ratio of ionic radii for the structure to be stable. It is calculated by the ratio of radius of cation to the radius of anion.
$L = \dfrac{{{r^ + }}}{{{r^ - }}}$, where ${r^ + }$ is the radius of the cation and ${r^ - }$ is the radius of the anion.
Here cation is the sodium ion and anion is the chloride ion.
Therefore, the limited radius ratio will be
$
{r^ + } = 95pm,{r^ - } = 181pm \\
\Rightarrow L = \dfrac{{95}}{{181}} = 0.524 \\
$
Therefore, the limited radius ratio is 0.524.
When L is 1.0, then the coordination number will be 12, and the shape of the compound is hexagonal.
When L is between 0.732 and 1.0, then the coordination number will be 8, and the shape of the compound is cubical.
When L is between 0.414 and 0.732, then the coordination number will be 6, and the shape of the compound is octahedral.
As we can see, the obtained L value is 0.524 which is between 0.414 and 0.732.
Therefore, the coordination number of the Sodium ($N{a^ + }$) ion is 6 not 8.
Note: Do not confuse coordination number with oxidation number. As oxidation number is the charge of the central atom when we remove all the ligands and electron pairs which are present around it and coordination number is the total number of ligands attached to the central metal atom in a coordination compound.
Complete Step by step answer: We are given that the radius of $N{a^ + }$ ion is 95 pm and that of $C{l^ - }$ ions is 181 pm.
We have to predict whether the coordination number of $N{a^ + }$ ion is 6 or 4.
We can solve this question using the limited radius ratio.
Limited radius ratio (L) is the minimum allowed value for the ratio of ionic radii for the structure to be stable. It is calculated by the ratio of radius of cation to the radius of anion.
$L = \dfrac{{{r^ + }}}{{{r^ - }}}$, where ${r^ + }$ is the radius of the cation and ${r^ - }$ is the radius of the anion.
Here cation is the sodium ion and anion is the chloride ion.
Therefore, the limited radius ratio will be
$
{r^ + } = 95pm,{r^ - } = 181pm \\
\Rightarrow L = \dfrac{{95}}{{181}} = 0.524 \\
$
Therefore, the limited radius ratio is 0.524.
When L is 1.0, then the coordination number will be 12, and the shape of the compound is hexagonal.
When L is between 0.732 and 1.0, then the coordination number will be 8, and the shape of the compound is cubical.
When L is between 0.414 and 0.732, then the coordination number will be 6, and the shape of the compound is octahedral.
As we can see, the obtained L value is 0.524 which is between 0.414 and 0.732.
Therefore, the coordination number of the Sodium ($N{a^ + }$) ion is 6 not 8.
Note: Do not confuse coordination number with oxidation number. As oxidation number is the charge of the central atom when we remove all the ligands and electron pairs which are present around it and coordination number is the total number of ligands attached to the central metal atom in a coordination compound.
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