Question

The edge length of a unit cell of nacl crystal lattice is 552 pm. if the ionic radius of sodium ion is 95 pm. what is the ionic radius of chloride ion? (a)276 pm (b)181 pm (c)368 pm (d)190 pm

Answer 1

answer : option (b) 181 pm

explanation : structure of NaCl crystal is fcc type cubic lattice. see figure, it is clearly shown that, each edge contains two Cl- ions and  one Na+ ion.

so, edge length = $r_{Na^+}+2r_{Cl^-}+r_{Na^+}$

edge length = $2r_{Na^+}+2r_{Cl^-}$

here, given edge length = 552 pm

ionic radius of sodium,$r_{Na^+}$ = 95 pm

now, 552 pm = 2 × 95pm + 2$r_{Cl^-}$

or, 552pm - 190pm = 2$r_{Cl^-}$

or, 362pm = 2$r_{Cl^-}$

hence, $r_{Cl^-}$ = 181 pm

so, ionic radius of chloride ion is 181 pm

Answer 2

hii mate your answer is 181 pm