Question

Write the electron dot formula and draw the structural formula for each of the following polyatomic ions: (a) IO (b) ${\mathrm{IO}}_2$ (c) ${\mathrm{IO}}_3^{-}$ (d) ${\mathrm{IO}}_4^{-}$

Short answer

Draw Lewis structures and adjust for electrons according to stability: (a) IO has a single I-O bond, (b) IO_2 likely radical as legs allow, (c) IO_3^- optimizes 26 electrons achieving stability, and (d) IO_4^- aims for tetrahedral equivalence distributing 32 electrons in double bonds.

Step-by-step solution

Understand the Requirements

For each polyatomic ion given, we need to determine the total number of valence electrons, draw the Lewis (electron dot) structure, and then provide the corresponding structural formula that shows how atoms are bonded together.

Define Valence Electrons for IO

Calculate the total valence electrons for IO. Iodine (I) has 7 valence electrons, and oxygen (O) has 6. Thus, the total is 13 electrons. This ion doesn't have any charge, so no extra electrons are added or removed.

Draw Lewis Structure for IO

Place iodine in the center and oxygen as its surrounding atom. Distribute electrons to achieve the octet structure. The Lewis structure shows: I with three lone pairs, an O with two lone pairs, and a single bond (represented as a pair of dots) between them.

Draw Structural Formula for IO

The structural formula features an I-O single bond, represented as I-O with a single line indicating a single bond.

Determine Valence Electrons for IO_2

Calculate valence electrons for ${\text{IO}}_2$. Iodine contributes 7, and each oxygen atom contributes 6. Therefore, the total is $7+6\times 2=19$ electrons. No additional charge is present.

Construct Lewis Structure for IO_2

Typically for odd number electron molecules like ${\text{IO}}_2$ you may end up with a radical. Arrange I in the center and attach two O atoms. Distribute electrons considering typical bonding, lone pair completion and acknowledging the molecule holds an odd number of electrons.

Represent Structural Formula for IO_2

Show iodine double-bonded to one oxygen and single-bonded to another: $O=I-O$. This represents the typical arrangement with iodine central and includes formal consideration of stability using resonance.

Calculate Valence Electrons for IO_3^-

Calculate for ${\text{IO}}_3^{-}$ where the charge of $-1$ adds an electron. Total is $7+6\times 3+1=26$ electrons.

Draw Lewis Structure for IO_3^-

Place Iodine in the center and bind to each oxygen with single bonds initially. Distribute the remaining electrons to satisfy the octet rule, placing any extra on iodine and ensuring formal charge optimization to achieve $I{O}_3^{-}$.

Sketch Structural Formula for IO_3^-

Show the arrangement with resonance or equivalent bond-length simplicity, often as $O=I(O)-O^{-}.$

Tally Valence Electrons for IO_4^-

Include extra for the negative charge, yielding $7+6\times 4+1=32$ electrons.

Form Lewis Structure for IO_4^-

Place iodine at the center with oxygen atoms around it. Distribute the electrons to form complete octets, often leading to adding double bonds with oxygens appropriately for correct formal charge.

Create Structural Formula for IO_4^-

Expected bonding symmetry: each bond I=O showcasing the double-bond equivalence found in typical tetra-coordinate geometries.

Key concepts

Valence Electrons

Valence electrons play a crucial role in understanding how atoms bond and form molecules. Each element on the periodic table has a certain number of valence electrons, which are the electrons found in the outermost shell of an atom. These electrons are vital because they are the ones that atoms share, lose, or gain during chemical reactions.

For example, iodine (I) has 7 valence electrons, and oxygen (O) has 6. When considering polyatomic ions like IO, IO₂, IO₃⁻, and IO₄⁻, it's essential to tally up the number of valence electrons from each atom involved.

• For IO: Iodine's 7 electrons plus oxygen's 6, making 13 valence electrons total.
• For IO₂: Iodine's 7 plus oxygen's 12 (6 per oxygen atom) gives 19 valence electrons in total.
• For IO₃⁻: Includes iodine's 7, oxygen's 18, and an extra electron for the ion's negative charge, totaling 26 electrons.
• For IO₄⁻: Adds iodine's 7, oxygen's 24, and one additional electron for the negative charge, culminating in 32 valence electrons.

Understanding the concept of valence electrons is crucial for determining how atoms will bond and for drawing meaningful Lewis structures.

Polyatomic Ions

A polyatomic ion is a charged entity composed of two or more atoms covalently bonded together, or of a metal complex that acts as a single unit. The total charge of a polyatomic ion can be positive or negative. This charge arises because the total number of electrons is not equal to the total number of protons.

In the ions provided in the exercise, IO₃⁻ and IO₄⁻ are examples of negatively charged polyatomic ions. The negative charge indicates an excess of one electron which must be included when counting total valence electrons, influencing the way bonds are drawn.

• IO: This is simply iodine and oxygen bonded, and it's neutral, so no extra electrons are calculated.
• IO₂: Although it sounds like a polyatomic ion, it’s neutral because it doesn't carry any extra charge.
• IO₃⁻: The negative charge signifies one extra electron, resulting in a total of 26 valence electrons to account for.
• IO₄⁻: Similarly, the charge must be considered when drawing Lewis structures, leading up to 32 valence electrons.

Understanding the nature of polyatomic ions and their charge helps in accurately predicting the behavior of these ions during chemical reactions and while drawing Lewis structures.

Structural Formulas

Structural formulas are a way of visually representing the arrangements of atoms within a molecule. This includes the bonds between atoms and often the geometry around each central atom. By looking at a structural formula, you can understand which atoms are bonded and the type of bonds that exist, whether they are single, double, or even triple bonds.

For the ions listed in the problem, creating accurate structural formulas involves several approaches to display possible bonding arrangements.

• IO: Represented as I-O. It's depicted as a simple single line showing a single covalent bond between iodine and oxygen.
• IO₂: Shown with a double bond to one oxygen and a single bond to the other, represented as O=I-O, often indicating that the molecule can exist in multiple resonant forms.
• IO₃⁻: Utilizes resonance structures to best distribute the charge, with one common form being O=I(O)-O⁻, with a mix of single and double bonds.
• IO₄⁻: Typically represented with iodine as the central atom with double bonds to each oxygen, achieving symmetry often attributed to resonance stabilization.

These structural formulas provide a glimpse into the molecular arrangement and can help predict the ion's chemical reactivity and interaction with other chemical species.