Answer
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Hint: In a unit cell, atoms of adjacent corners are in contact with each other, so the edge length will be equal to two atomic radii or can also be said as one atomic radii.
Complete step by step answer:
The distance between the cation and anion in NaCl unit cell = \[\frac{a}{2}\]
\[{{r}_{+}}+{{r}_{-}}=\frac{a}{2}\]
Here is the edge length. It is given that the edge length of NaCl a = 562 pm.
\[{{r}_{N{{a}^{+}}}}+{{r}_{C{{l}^{-}}}}=\frac{a}{2}=\frac{562}{2}=281pm\]
Here \[{{r}_{N{{a}^{+}}}}\] is the radius of sodium ion and \[{{r}_{C{{l}^{-}}}}\] is the radius of chloride ions.
Distance between two chloride ion is given as = \[\frac{a}{\sqrt{2}}=\frac{562}{\sqrt{2}} = 397.39 pm\]
We know, \[2{{r}_{C{{l}^{-}}}} = 397.39\]
\[{{r}_{C{{l}^{-}}}} = 198.695pm\]
Therefore substituting the values in \[{{r}_{C{{l}^{-}}}}+{{r}_{N{{a}^{+}}}}\], we can get the value of the radius of sodium ion.
\[{{r}_{N{{a}^{+}}}}=281-198.695=82.35pm\]
Now the nearest distance between two \[N{{a}^{+}}\] ions= \[\frac{a}{\sqrt{2}}\]
\[=\frac{562}{\sqrt{2}}=281\sqrt{2}pm\]
Hence, the Edge length of the NaCl unit cell is 562pm. Then distance between \[N{{a}^{+}}\] and \[C{{l}^{-}}\] ions is 281pm and nearest distance between two \[N{{a}^{+}}\] ions is \[281\sqrt{2}\]pm.
Hence, The correct options are (a) and (c).
Additional Information: NaCl has a face-centered cubic lattice. It has an fcc array of chloride anions with an interpenetrating fcc sodium cation lattice or vice versa. Each ion in NaCl crystal lattice is 6 coordinated, that is, each atom is surrounded by 6 other atoms. Sodium cation is surrounded by 6 chloride anions and a chloride anion is surrounded by 2 sodium cations.
Note: Edge length and the distance between two chloride ions and two sodium ions are different. Edge length is the distance between the chloride ion and a sodium ion.
Complete step by step answer:
The distance between the cation and anion in NaCl unit cell = \[\frac{a}{2}\]
\[{{r}_{+}}+{{r}_{-}}=\frac{a}{2}\]
Here is the edge length. It is given that the edge length of NaCl a = 562 pm.
\[{{r}_{N{{a}^{+}}}}+{{r}_{C{{l}^{-}}}}=\frac{a}{2}=\frac{562}{2}=281pm\]
Here \[{{r}_{N{{a}^{+}}}}\] is the radius of sodium ion and \[{{r}_{C{{l}^{-}}}}\] is the radius of chloride ions.
Distance between two chloride ion is given as = \[\frac{a}{\sqrt{2}}=\frac{562}{\sqrt{2}} = 397.39 pm\]
We know, \[2{{r}_{C{{l}^{-}}}} = 397.39\]
\[{{r}_{C{{l}^{-}}}} = 198.695pm\]
Therefore substituting the values in \[{{r}_{C{{l}^{-}}}}+{{r}_{N{{a}^{+}}}}\], we can get the value of the radius of sodium ion.
\[{{r}_{N{{a}^{+}}}}=281-198.695=82.35pm\]
Now the nearest distance between two \[N{{a}^{+}}\] ions= \[\frac{a}{\sqrt{2}}\]
\[=\frac{562}{\sqrt{2}}=281\sqrt{2}pm\]
Hence, the Edge length of the NaCl unit cell is 562pm. Then distance between \[N{{a}^{+}}\] and \[C{{l}^{-}}\] ions is 281pm and nearest distance between two \[N{{a}^{+}}\] ions is \[281\sqrt{2}\]pm.
Hence, The correct options are (a) and (c).
Additional Information: NaCl has a face-centered cubic lattice. It has an fcc array of chloride anions with an interpenetrating fcc sodium cation lattice or vice versa. Each ion in NaCl crystal lattice is 6 coordinated, that is, each atom is surrounded by 6 other atoms. Sodium cation is surrounded by 6 chloride anions and a chloride anion is surrounded by 2 sodium cations.
Note: Edge length and the distance between two chloride ions and two sodium ions are different. Edge length is the distance between the chloride ion and a sodium ion.
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