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(this question has multiple correct options)

(a) distance between \[N{{a}^{+}}\]and \[C{{l}^{-}}\]ions is 281pm.

(b) radii of \[N{{a}^{+}}\]and \[C{{l}^{-}}\]ions will be 281pm and 95pm.

(c) nearest distance between two \[N{{a}^{+}}\]ions is \[281\sqrt{2}\]pm

(d) nearest distance between two \[C{{l}^{-}}\]ions is \[95\sqrt{2}\]pm

Answer
Verified

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.