Q. The shape of $I_{3}^{-}$ is

1344 254 Chemical Bonding and Molecular Structure Report Error

A

Linear

32%

B

Bent

26%

C

Pyramidal

18%

D

See saw

24%

Solution:

$I_{3}^{-} = [1^{-} \cdot I_{2}]$
$H = \frac{1}{2} [V + M - C + A]$
$= \frac{1}{2} [7 + 2 - 0 + 1] = \frac{10}{2} = s p^{3} d$

$I^{-}$ have $8 e^{-}$
$∴$ by combining with two $I_{2}, I^{-}$ have $3 L P$ and $2 BP$