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In NaCl lattice structure if one sodium ion removed from the corner,then find the formula of resulting compound?

In NaCl lattice structure if one sodium ion removed from the corner,then find the formula of resulting compound?

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2 Answers

rishabh
27 Points
7 years ago

first of all the na+ion will not be present at the corner while it can be said that na+ion is lost from one of he octahedral voids of one face then

the formula will become Na7Cl8

in nacl sructure na+ ion will occupy the tetrahedral voids while the cl-ion will occupy the fcc latice 

total no. of octahedral voids in fcc latice is =4

given by [1/2*6(of the corner atoms i.e the centre of each face)+ 1(centre octahedral void along body diagonal)]

thus if the na +ion is lost fromone of the corner then the total no.of na+ion is {1/2(6-1)+1}=7/2 and no. of cl-ion =4 therefore the formula is Na7Cl8

approve by clicking yes

Yash Kataria
24 Points
4 years ago
According to the question given , it can be assumed that na+ is present at corners,cl- us present at face centre. Therefore,for Na 1/8(8-1)=7/8For Cl 1/2(6)=3Na7Cl24Actually,In NaCl,Cl- occupy lattice points and Na+ occupy octahedral void.

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