Question

In which of the following, the maximum number of lone pairs is present on the central atom?

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

As we already know how to calculate the number of electron pairs.

and we also know that number of lone pairs (out of those ep) = number of ep - number of Ligands

Therefore, firstly we will calculate the number of ep in each compound.

As, the number is not coming in integers, we will calculate using AVSEPR Theory (by not considering oxygen atoms)

$CI{O}_3^{-}\to ep=\frac{(7+1)}{2}=4$
$Xe{F}_4\to ep=\frac{(8+4)}{2}=6$
$S{F}_4\to ep=\frac{(6+4)}{2}=5$
$I_3^{-}\to \left[I\right]I_2^{-}\to ep=\frac{(7+2+1)}{2}=5$
We devided $I_3^{-}$ into I and $I_2^{-}$ and took I as central atom

Now, number of lone pairs in each one will be:

$Cl{O}_3^{-}\to$ number of lone pairs $=4-3=1$

$\downarrow \downarrow$
ep Ligands

$Xe{F}_4\to$ number of lone pairs $=6-4=2$

$S{F}_4\to$ number of lone pairs $=5-4=1$

$I_{3-}\to$ number of lone pairs $=5-2=3$

Therefore, the number of lone pairs are maximum in $I_3^{-}$

So, our answer will be option (d) $I_3^{-}$