What is the minimum distance between two Na+ ions which are present along the edges in an ideal NaCl type crystal? Let, rNa+= radius of Na + and rCl−= radius of Cl−.
A
2(rNa+)
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B
2×(√2+1)rNa+
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C
2rCl−
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D
√2×(rNa++rCl−)
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Solution
The correct option is D√2×(rNa++rCl−)
One face of NaCl crystal lattice looks like,
Na+ ions are situated at the centre of every edge in octahedral voids.
The minimum distance between two Na+ ions is AC
(AC)2=(AB)2+(BC)2 d=√2(rCl−+rNa+)
The minimum distance between two Na+ ions which are present along the edges is √2×(rNa++rCl−).