Be sure to answer all parts. describe the hybrid orbitals used by the central atom and the type(s) of bonds formed in i3−. sp sp2 sp3 sp3d number of σ bonds: number of π bonds:

Triiodide molecule (I₃⁻) has sp³d hybridization

The number of sigma (σ) bonds = 2

and the number of Pi (π) bonds = 0

Explanation:

In a Triiodide molecule (I₃⁻), three iodine atoms combine to form a linear and symmetrical polyhalogen ion. The triiodide ion (I₃⁻) is formed by the chemical reaction of diatomic iodine (I₂) with iodide ion (I⁻).

The involved chemical reaction: I₂ + I⁻ ⇌ I₃⁻

The structure of the triiodide ion: [ I − I − I ]⁻

As we know, that an iodine atom has 7 valence electrons.

To complete its octet, if the central iodine atom accepts an electron to form an iodide ion, thus having 8 valence electrons.

Then this central iodide bonds with two terminal iodine atoms by sigma (σ) bonds. So, the number of non-bonding electrons on the central iodide ion is 6.

Thus, the number of lone pairs on the central iodide atom = 3

Also, the number of sigma (σ) bonds = 2

and the number of Pi (π) bonds = 0

Now, the hybridization of this molecule can be determined by the steric number.

Steric number = number of σ bonds + number of lone pairs = 2 +3 = 5 ⇒ sp³d hybridization

Therefore, the Triiodide molecule (I₃⁻) has sp³d hybridization.

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