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Some chemists believe that satisfaction of the octet rule should be the top criterion for choosing the dominant Lewis structure of a molecule or ion. Other chemists believe that achieving the best formal charges should be the top criterion. Consider the dihydrogen phosphate ion, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-},\) in which the \(\mathrm{H}\) atoms are bonded to \(\mathrm{O}\) atoms. (a) What is the predicted dominant Lewis structure if satisfying the octet rule is the top criterion? (b) What is the predicted dominant Lewis structure if achieving the best formal charges is the top criterion?

Short Answer

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a) The predicted dominant Lewis structure satisfying the octet rule has phosphorus (P) single bonded to four oxygen (O) atoms, with two of the oxygen atoms also bonded to a hydrogen (H) atom. b) The predicted dominant Lewis structure achieving the best formal charges has phosphorus (P) double bonded to one oxygen (O) atom and single bonded to the other three oxygen atoms, with two of the single-bonded oxygen atoms also bonded to a hydrogen (H) atom.

Step by step solution

01

Identifying the central atom

In the dihydrogen phosphate ion, \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\), the central atom is phosphorus (P). This is because it has the lowest electronegativity among the other atoms (excluding H).
02

Counting the total number of valence electrons

In order to draw the Lewis structure, we must first count the total number of valence electrons for this ion. Here is the number of valence electrons for each atom: - Hydrogen (H): 1 valence electron x 2 atoms = 2 electrons - Phosphorus (P): 5 valence electrons - Oxygen (O): 6 valence electrons x 4 atoms = 24 electrons There is also an extra electron added to the species due to the negative charge, so we have a total of 32 electrons.
03

Drawing the dominant Lewis structure based on the octet rule

In this step, we will draw the Lewis structure by satisfying the octet rule. 1. Connect the atoms: Connect the 4 oxygen atoms to the central phosphorus atom. 2. Distribute the remaining electrons: Distribute the remaining electrons to each atom (except hydrogen) to complete their octet. 3. Place the hydrogens: Hydrogen will form a single bond with two of the oxygen atoms. Following these steps, we get the Lewis structure as shown below: P with single bonds to 4 O atoms. Each of the 2 O atoms that form a single bond with H has an extra lone pair, while the other 2 O atoms have 3 lone pairs of electrons. The formal charges are: P = +1, H-bonded O = 0, Non-H-bonded O = -1 - this maintains the overall charge of -1. This Lewis structure satisfies the octet rule.
04

Drawing the dominant Lewis structure based on achieving the best formal charges

In this step, we will draw the Lewis structure by achieving the best formal charges. 1. Connect the atoms: Connect the 4 oxygen atoms to the central phosphorus atom. 2. Distribute the remaining electrons: Distribute the remaining electrons to each atom (except hydrogen) to complete their octet. 3. Place the hydrogens: Hydrogen will form a single bond with two of the oxygen atoms. 4. Adjust the bonds: Convert one of the single bonds between the P and O that has no hydrogen bonded to form a double bond. This will let P to have a formal charge of 0 and satisfy the best formal charges rule. The Lewis structure we get is the following: P with a double bond to one O atom, and single bonds to the other 3 O atoms. The O atom with the double bond has 2 lone pairs; the other 2 single-bonded O atoms have 3 lone pairs each, and the H-bonded O atoms have an extra lone pair of electrons. This structure has formal charges: P = 0, O with the double bond = 0, H-bonded O = 0, Non-H-bonded O = -1 - this maintains the overall charge of -1. The Lewis structure with the best formal charges has a double bond between P and one of the O atoms. a) The predicted dominant Lewis structure satisfying the octet rule has 4 single bonds between phosphorus and oxygen atoms, with hydrogens bonded to two of the oxygen atoms. b) The predicted dominant Lewis structure achieving the best formal charges has a double bond between phosphorus and one oxygen atom, with single bonds to the other 3 oxygen atoms, and hydrogens bonded to two of the oxygen atoms.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Octet Rule
The octet rule is a fundamental concept in chemistry that suggests atoms are most stable when they have eight electrons in their valence shell. This rule is derived from the observation that noble gases, which naturally have complete valence shells, are exceptionally stable. In constructing Lewis structures, the octet rule guides chemists to distribute electrons such that each atom (except hydrogen, which prefers two electrons) achieves a full set of eight around them.
For the dihydrogen phosphate ion, \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-}\), this involves ensuring that the phosphorus and oxygen atoms each complete their octet. In a simplified view, you connect the central phosphorus atom with oxygen atoms, then arrange the remaining electrons to satisfy this rule. With four oxygen atoms each needing electrons to fill their valence shell, the goal is to distribute the electrons so all attain this magic "eight."
  • Distribute single bonds from phosphorus to each oxygen.
  • Adjust lone pairs to achieve the octet structure.
Satisfying the octet rule is a common guideline, but it might not always yield the lowest energy or most accurate depiction of a molecule's Lewis structure. Nonetheless, it's a useful starting point for many chemical species, including the dihydrogen phosphate ion.
Formal Charges
Formal charges are hypothetical charges that help predict the most stable Lewis structure for a molecule. By calculating the formal charge of each atom within a molecule, chemists can find the configuration that minimizes these charges across the structure. A formal charge is calculated using the formula:
\[\text{Formal Charge} = \text{Valence Electrons} - (\text{Non-Bonding Electrons} + 0.5 \times \text{Bonding Electrons})\]
For the dihydrogen phosphate ion, achieving the best formal charges ensures each atom within the structure has charges as close to zero as possible. A zero formal charge indicates the atom is in its most stable state with regard to electron sharing.
  • Ensure phosphorus ideally has a formal charge of zero.
  • Check for minimized formal charges across oxygen and hydrogen atoms.
Optimizing formal charges often leads chemists to adjust bond types, such as adding double bonds between phosphorus and oxygen. This effectively redistributes electrons, allowing the dihydrogen phosphate ion to maintain its structure with the lowest possible energy state, reflecting a more accurate portrayal of electron distribution.
Dihydrogen Phosphate Ion
The dihydrogen phosphate ion, denoted as \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-}\), is a common polyatomic ion that plays a crucial role in several chemical and biological systems. It is part of the phosphate buffering system that helps maintain pH in biological organisms.
The structure consists of a central phosphorus atom surrounded by four oxygen atoms, with two hydrogens bonded to two of these oxygens. This setup means that phosphorus serves as the atom to which all others are connected, making it crucial to how electrons are distributed. Its negative charge arises from the extra electron added to its structure, making it negatively charged.
  • Central phosphorus surrounded by four oxygen atoms.
  • Each hydrogen bonds to an oxygen, contributing to the stability of the ion.
The dihydrogen phosphate ion exemplifies the principles of Lewis structures where balance between the octet rule and formal charge distribution ensures structural stability. Understanding its structure helps in both predicting reactions it may participate in and appreciating its role within larger biochemical pathways.

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Most popular questions from this chapter

(a) Does the lattice energy of an ionic solid increase or decrease (i) as the charges of the ions increase, (ii) as the sizes of the ions increase? (b) Arrange the following substances not listed in Table 8.1 according to their expected lattice energies, listing them from lowest lattice energy to the highest: MgS, KI, GaN, LiBr.

A major challenge in implementing the "hydrogen economy" is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example, \(\mathrm{NaAlH}_{4}\) can release \(5.6 \%\) of its mass as \(\mathrm{H}_{2}\) upon decomposing to \(\mathrm{NaH}(s), \mathrm{Al}(s),\) and \(\mathrm{H}_{2}(g) . \mathrm{NaAlH}_{4}\) possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (a) Write a balanced equation for the decomposition of \(\mathrm{NaAlH}_{4}\). (b) Which element in \(\mathrm{NaAlH}_{4}\) is the most electronegative? Which one is the least electronegative? (c) Based on electronegativity differences, predict the identity of the polyatomic anion. Draw a Lewis structure for this ion. (d) What is the formal charge on hydrogen in the polyatomic ion?

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For Group 13-17 elements in the third row of the periodic table and beyond, the octet rule is often not obeyed. A friend of yours says this is because these heavier elements are more likely to make double or triple bonds. Another friend of yours says that this is because the heavier elements are larger and can make bonds to more than four atoms at a time. Which friend is more correct?

The compound chloral hydrate, known in detective stories as knockout drops, is composed of \(14.52 \% \mathrm{C}, 1.83 \% \mathrm{H}\), \(64.30 \% \mathrm{Cl}\), and \(13.35 \% \mathrm{O}\) by mass, and has a molar mass of \(165.4 \mathrm{~g} / \mathrm{mol} .(\mathbf{a})\) What is the empirical formula of this substance? (b) What is the molecular formula of this substance? (c) Draw the Lewis structure of the molecule, assuming that the Cl atoms bond to a single \(C\) atom and that there are a \(\mathrm{C}-\mathrm{C}\) bond and two \(\mathrm{C}-\mathrm{O}\) bonds in the compound.

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