# How many bonding pairs and lone pairs surround the central atom in the $\text{I}_{3}^{-}$ ion?

A. 2, 2

B. 2, 3

C. 3, 2

D. 4, 3

Answer

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**Hint:**Hybridisation is the mixing of an atomic orbital to form the chemical bonds according to the valence band theory. It forms a new orbital with the different geometry and shape of the hybridised molecule.

**Complete answer:**

-Firstly, we will calculate the hybridisation by using the formula that is

$\text{H = }\dfrac{1}{2}(\text{V + M +C - A)}$

-Here, H is the hybridisation, V is the valence electron of the central atom, M is the monovalent atom that is linked to the central atom, C is the charge present on the cation and A is the charge present on the anion.

-Now, we know that there are a total of 7 valence electrons in the iodine and it is a monovalent atom also.

-So, the value of V and M for iodine will be 7 and 2 respectively because two 2 molecules of iodine are linked to the central molecule of iodine.

-The value of C will be 1 because there is only a -1 charge on the iodine molecule.

-So, the value of H is:

$\text{H = }\dfrac{1}{2}(7\text{ }+\text{ }2\text{ + 1)}$

$\text{H = }\dfrac{1}{2}\text{ }\cdot \text{ 10 = 5}$

-The hybridisation of the centre iodine will be $\text{s}{{\text{p}}^{3}}\text{d}$ and it will have 3 lone pairs and 2 bond pairs with a linear shape.

-Central iodine makes 2 bonds and the 3 pairs remain unused.

-So, they arrange themselves such that there is minimum repulsion between the lone pairs

**So, the correct answer is “Option B”.**

**Note:**VSEPR or Valence Shell Electron Pair Theory tells us about the geometry of a molecule by arranging the molecules in such a way that there will be less repulsion between them.

Last updated date: 24th Sep 2023

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