In NaCl lattice structure if one of the sodium atom is removed from corner the formula of resulting compound is
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
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