The SrO has NaCl structure. If the lattice constant a =0.514 nm and ionic radius of Sr 2+=0.113 nm. The radius of O 2- ion is :-

Question

The SrO has NaCl structure. If the lattice constant a =0.514 nm and ionic radius of Sr 2+=0.113 nm. The radius of O 2- ion is :- (A) 1.44 × 10-10 m (B) 1.44 × 10-8 m (C) 1.44 × 10-9 m (D) 1.44 × 10-12 m

Tardigrade Admin · Accepted Answer

2 r Sr ++2 r O -2=0.514 nm ∴ r O -2=0.144 nm =1.44 × 10-10 m

Q. The $S r O$ has $N a Cl$ structure. If the lattice constant $a = 0.514 nm$ and ionic radius of $S r^{2 +} = 0.113 nm$ . The radius of $O^{2 -}$ ion is :-

$1.44 \times 1 0^{- 10} m$

$1.44 \times 1 0^{- 8} m$

$1.44 \times 1 0^{- 9} m$

$1.44 \times 1 0^{- 12} m$

$2 r_{S r^{+}} + 2 r_{O^{- 2}} = 0.514$ nm
$∴ r_{O^{- 2}} = 0.144$ nm
$= 1.44 \times 1 0^{- 10} m$

Q. The SrO has NaCl structure. If the lattice constant a=0.514nm and ionic radius of Sr2+=0.113nm. The radius of O2− ion is :-

1.44×10−10m

1.44×10−8m

1.44×10−9m

1.44×10−12m