The number of lone pairs of electron on the central I atom in I3- is …… .

Q: Why is one Iodine atom considered the 'central' atom?

In a polyatomic ion made of three identical atoms, they cannot all be bonded to each other in a ring without severe strain. One atom must act as the central bridge connecting the other two terminal atoms.

Q: Why do we add an electron for the negative charge?

Anions carry a negative charge because they have gained extra electrons from their environment. To accurately predict the geometry, we assign this extra electron to the central atom's valence shell.

Q: Why are the lone pairs placed in the equatorial positions?

In a trigonal bipyramidal geometry ($sp^3d$), equatorial positions are $120^\circ$ apart, while axial positions are only $90^\circ$ from the equatorial plane. Lone pairs repel more strongly than bond pairs, so they take the equatorial spots to maximize their distance from each other.

Q: What is the hybridization of the central Iodine?

The central Iodine has 2 bond pairs and 3 lone pairs, giving a steric number of 5. This requires 5 hybrid orbitals, which corresponds to $sp^3d$ hybridization.

Q: Why is the shape called 'linear' if the geometry is trigonal bipyramidal?

'Geometry' refers to the arrangement of all electron pairs (both bonds and lone pairs). 'Shape' refers only to the visible arrangement of the atoms. Since the 3 lone pairs form an invisible triangle in the middle, the 3 Iodine atoms form a straight line, hence a linear shape.

Question

The number of lone pairs of electron on the central I atom in $I_3^-$ is …… .

Sigma Percentile · Accepted Answer

### Decoding the Triiodide Ion

The triiodide ion, $I_3^-$, is a classic example in chemical bonding that perfectly illustrates the power of VSEPR (Valence Shell Electron Pair Repulsion) theory. At first glance, a molecule made entirely of iodine atoms might seem confusing. Which one is the central atom? In polyatomic ions made of a single element, one atom acts as the central hub while the others bond to it as terminal atoms. Let's break down the electron math to find out exactly what's happening around that central iodine.

### Counting the Electrons

To determine the geometry, we first need to count the valence electrons on the central iodine atom. Iodine is a halogen, residing in Group 17 of the periodic table, which means it naturally possesses **$7$ valence electrons**.

However, we must look closely at the overall charge of the ion. The $I_3^-$ ion carries a negative charge ($-1$). This indicates the presence of one extra electron in the system. We add this extra electron to the central atom's valence shell, giving us a total of:

$$V = 7 + 1 = 8 	ext{ electrons}$$

### Bonding and Lone Pairs

Now, let's look at the connections. The central iodine atom is bonded to two terminal iodine atoms. Each of these single bonds requires one electron from the central atom. Therefore, **$2$ electrons** are locked up in bonding pairs (BP).

We started with $8$ electrons and used $2$ for bonding. This leaves us with $6$ non-bonding electrons. Since a lone pair consists of two electrons, we simply divide the remaining electrons by two:

$$LP = \frac{8 - 2}{2} = \frac{6}{2} = 3 	ext{ lone pairs}$$

### The Final Geometry

With $2$ bond pairs and $3$ lone pairs, the central iodine atom has a **steric number of $5$**. This corresponds to an **$sp^3d$ hybridization**, which has a base geometry of a trigonal bipyramid.

According to VSEPR theory, lone pairs demand more space and will occupy positions that minimize repulsion. In a trigonal bipyramid, the equatorial positions offer bond angles of $120^\circ$, providing much more room than the crowded $90^\circ$ axial positions. Therefore, all **three lone pairs occupy the equatorial plane**.

This forces the two terminal iodine atoms into the axial positions (top and bottom). If you ignore the invisible lone pairs and only look at the atoms, the resulting shape is perfectly **linear** with a bond angle of $180^\circ$.

So, the number of lone pairs on the central iodine atom is exactly **$3$**.

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The number of lone pairs of electron on the central I atom in $I_{3}^{-}$ is …… .

Enter Numerical Value:

Visualized Solution

$I_{3}^{-}$

$V = 8$

$B P = 2$

$L P = 3$

$s p^{3} d$

$I_{3}^{+}$

The Sigma Insight: Hybridisation and VSEPR Theory

Decoding the Triiodide Ion

Counting the Electrons

Bonding and Lone Pairs

The Final Geometry

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Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of lone pairs of electron on the central I atom in I3−​ is …… .