Explanation

(b)
$SF_{6}$ has octahedral geometry, $sp^{3}d^{2}$ hybridisation and bond angle is $90{^\circ}$
[image:4_image92.emf]
$\ \ \ \ \ \ \ \ \%\$of d-character$\ = \frac{2\ (no.of\ d - orbitals)}{6(total\ hybridised\ orbitals)} \times 100$
$= 33\%$
So, $SF_{6}$ are bond angle = $90{^\circ}$
and d-character = 33%.