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

rishabh
27 Points
9 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
6 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.