The ionic radius of Cl- ion is 1.81 mathring A . The interionic distances of NaCl and NaF are 2.79 mathring A and 2.31 mathring A respectively. The ionic radius of F- ion will be

Question

The ionic radius of Cl- ion is 1.81 mathring A . The interionic distances of NaCl and NaF are 2.79 mathring A and 2.31 mathring A respectively. The ionic radius of F- ion will be (A) 0.98 mathring A  (B) 0.80 mathring A  (C) 1.33 mathring A  (D) 2.29 mathring A

Tardigrade Admin · Accepted Answer

d NaCl =r Na ++r Cl - 2.79=r Na+ +1.81 or r Na +=2.79-1.81=0.98 mathringA d NaF =r Na ++r F -, text i.e., 2.31=0.98+r F - or r F -=2.31-0.98=1.33 mathringA

Q. The ionic radius of $C l^{-}$ ion is $1.81 \overset{˚}{A}$ . The interionic distances of $N a Cl$ and $N a F$ are $2.79 \overset{˚}{A}$ and $2.31 \overset{˚}{A}$ respectively. The ionic radius of $F^{-}$ ion will be

0.98 $\overset{˚}{A}$

0.80 $\overset{˚}{A}$

1.33 $\overset{˚}{A}$

2.29 $\overset{˚}{A}$

$d_{N a Cl} = r_{N a} + + r_{Cl}^{-}$

$2.79 = r_{N a^{+}} + 1.81$

or $r_{N a^{+}} = 2.79 - 1.81 = 0.98 \overset{˚}{A}$

$d_{N a F} = r_{N a^{+}} + r_{F^{-}}, i.e., 2.31 = 0.98 + r_{F^{-}}$

or $r_{F^{-}} = 2.31 - 0.98 = 1.33 \overset{˚}{A}$

Q. The ionic radius of Cl− ion is 1.81A˚. The interionic distances of NaCl and NaF are 2.79A˚ and 2.31A˚ respectively. The ionic radius of F− ion will be

0.98 A˚

0.80 A˚

1.33 A˚

2.29 A˚