Coordination number of $ N{a^ + } $ in $ NaCl $ is:
A: $ 3 $
B: $ 4 $
C: $ 5 $
D: $ 6 $
Answer
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Hint: In chemistry, coordination number, also known as legacy, refers to the number of ions, atoms or molecules which a central metal ion or atom holds as its nearest neighbours in a molecule. The ion or molecule or atom surrounding the central atom or ion or molecule is known as a ligand.
Complete step by step solution:
$ NaCl $ possess a fcc arrangement in which ions are located at the octahedral voids. The fcc (i.e. face-centered cubic) lattice possesses a coordination number of 12 and comprises 4 atoms per unit cell. Equal number of cations and anions are required to balance the charge. In fcc, tetrahedral voids are located on the body diagonals while octahedral voids are located at the edge centre and at the body centre. The total number of octahedral voids in fcc is equal to the total number of closed packed particles. Tetrahedral voids are two times the number of octahedral voids.
In fcc, two tetrahedral voids are present on each body diagonal at a distance of $ \left( {1/4} \right)th $ from every corner. And, there are 4 octahedral voids in fcc that means we have 4 chloride ions and to balance charge we require 4 $ N{a^ + } $ ions. If we check their coordination with other atoms, we know:
$ 1{\text{ }}\left( {for{\text{ }}body{\text{ }}centre} \right) + 1/4*12\left( {for{\text{ }}edge{\text{ }}centres} \right) $
It is clear that each $ N{a^ + } $ and $ C{l^ - } $ are coordinated to 6 other atoms. That means each $ N{a^ + } $ ion is surrounded by 6 $ C{l^ - } $ ions while each $ C{l^ - } $ ions is surrounded by 6 $ N{a^ + } $ ions. As a result, the coordination number of $ N{a^ + } $ in $ NaCl $ is 6.
Hence, the correct answer is Option D.
Note:
FCC crystal structure exhibits more ductility (deforms more readily) in comparison to BCC crystal structure. Actually, bcc lattice, though cubic, is not closely packed and thus, forms strong metals. On the other hand, fcc lattice is both cubic as well as closely packed and thus, forms more ductile materials. Hence, $ NaCl $ crystal is ductile.
Complete step by step solution:
$ NaCl $ possess a fcc arrangement in which ions are located at the octahedral voids. The fcc (i.e. face-centered cubic) lattice possesses a coordination number of 12 and comprises 4 atoms per unit cell. Equal number of cations and anions are required to balance the charge. In fcc, tetrahedral voids are located on the body diagonals while octahedral voids are located at the edge centre and at the body centre. The total number of octahedral voids in fcc is equal to the total number of closed packed particles. Tetrahedral voids are two times the number of octahedral voids.
In fcc, two tetrahedral voids are present on each body diagonal at a distance of $ \left( {1/4} \right)th $ from every corner. And, there are 4 octahedral voids in fcc that means we have 4 chloride ions and to balance charge we require 4 $ N{a^ + } $ ions. If we check their coordination with other atoms, we know:
$ 1{\text{ }}\left( {for{\text{ }}body{\text{ }}centre} \right) + 1/4*12\left( {for{\text{ }}edge{\text{ }}centres} \right) $
It is clear that each $ N{a^ + } $ and $ C{l^ - } $ are coordinated to 6 other atoms. That means each $ N{a^ + } $ ion is surrounded by 6 $ C{l^ - } $ ions while each $ C{l^ - } $ ions is surrounded by 6 $ N{a^ + } $ ions. As a result, the coordination number of $ N{a^ + } $ in $ NaCl $ is 6.
Hence, the correct answer is Option D.
Note:
FCC crystal structure exhibits more ductility (deforms more readily) in comparison to BCC crystal structure. Actually, bcc lattice, though cubic, is not closely packed and thus, forms strong metals. On the other hand, fcc lattice is both cubic as well as closely packed and thus, forms more ductile materials. Hence, $ NaCl $ crystal is ductile.
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