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Chapter 8: Problem 90

Calculate the formal charge on the indicated atom in each of the following molecules or ions: (a) the central oxygen atom in \(\mathrm{O}_{3},\) (b) phosphorus in \(\mathrm{PF}_{6}^{-},(\mathbf{c})\) nitrogen in \(\mathrm{NO}_{2}\) (d) iodine in ICl_\(\mathrm{ICl}_{3},\) (e) chlorine in \(\mathrm{HClO}_{4}\) (hydrogen is bonded to \(\mathrm{O} )\)

Short Answer

Expert verified
The formal charges on each of the indicated atoms are as follows: a) O3 (central oxygen atom): \(0\) b) PF6- (phosphorus): \(-1\) c) NO2 (nitrogen): \(0\) d) ICl3 (iodine): \(+2\) e) HClO4 (chlorine): \(+3\)

Step by step solution

01

Determine Valence Electrons

Determine the number of valence electrons for each atom of interest: a) O (central oxygen): 6 valence electrons b) P (phosphorus): 5 valence electrons c) N (nitrogen): 5 valence electrons d) I (iodine): 7 valence electrons e) Cl (chlorine): 7 valence electrons
02

Identify Electrons Assigned to Atom

Determine the number of electrons assigned to each atom of interest. a) O3 (central oxygen): 2 lone pairs (4 electrons) + 2 half bonding pairs (2 electrons) = 6 electrons b) PF6- (phosphorus): 6 half bonding pairs (6 electrons) = 6 electrons c) NO2 (nitrogen): 1 lone pair (2 electrons) + 3 half bonding pairs (3 electrons) = 5 electrons d) ICl3 (iodine): 1 lone pair (2 electrons) + 3 half bonding pairs (3 electrons) = 5 electrons e) HClO4 (chlorine): 4 half bonding pairs (4 electrons) = 4 electrons
03

Calculate Formal Charge for each Atom

Calculate the formal charge of each atom by subtracting the assigned electrons from the valence electrons. a) O3 (central oxygen): Formal charge = 6 - 6 = 0 b) PF6- (phosphorus): Formal charge = 5 - 6 = -1 c) NO2 (nitrogen): Formal charge = 5 - 5 = 0 d) ICl3 (iodine): Formal charge = 7 - 5 = +2 e) HClO4 (chlorine): Formal charge = 7 - 4 = +3 The formal charges on each of the indicated atoms are as follows: a) O3 (central oxygen atom): 0 b) PF6- (phosphorus): -1 c) NO2 (nitrogen): 0 d) ICl3 (iodine): +2 e) HClO4 (chlorine): +3

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Key Concepts

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

Valence Electrons
Valence electrons are the outermost electrons of an atom and play a crucial role in forming chemical bonds. These electrons are vital as they determine an atom's chemical properties and reactivity.
An easy way to determine the number of valence electrons is by looking at the group number in the periodic table. For instance:
  • Oxygen (O) 鈥� found in group 16 鈥� has 6 valence electrons.
  • Phosphorus (P) 鈥� located in group 15 鈥� possesses 5 valence electrons.
  • Nitrogen (N) 鈥� also in group 15 鈥� has 5 valence electrons.
  • Iodine (I) 鈥� from group 17 鈥� features 7 valence electrons.
  • Chlorine (Cl) 鈥� similarly in group 17 鈥� has 7 valence electrons.
Understanding valence electrons is crucial because they participate in bond formation and determine how atoms interact together in a molecule or ion.
Lewis Structure
A Lewis structure is a visual way of representing molecules showing both the arrangement of atoms and the distribution of valence electrons around them. This helps predict the geometry, reactivity, and physical properties of molecules.
When drawing a Lewis structure, consider:
  • The total number of valence electrons across all atoms in the molecule or ion.
  • How atoms are connected, typically starting by placing atoms around the central atom.
  • Assigning pairs of electrons as lines (bonds) between atoms, or as lone pairs.
For example, in sulfur hexafluoride (\( ext{SF}_6\)), sulfur is the central atom surrounded by six fluorine atoms with each sharing a pair of electrons, forming strong bonds without lone pairs on sulfur.
This is essential for understanding the bond structure and how molecules such as \( ext{O}_3\) or \( ext{PF}_6^鈭抃) stay together.
Molecular Ions
Molecular ions are charged molecules formed by the gain or loss of electrons, giving them an overall positive or negative charge. Understanding molecular ions is essential in chemistry as they often play significant roles in reactions and processes.
Formation of Molecular Ions:
  • When a molecule gains extra electrons, it becomes negatively charged, forming an anion, like \( ext{PF}_6^-\).
  • Conversely, losing electrons results in a positively charged molecule, known as a cation.
These charged species behave differently compared to neutral molecules, influencing things like solubility and reactivity.
It's important to note how molecular ions retain their overall charge. For instance, in the calculation of the phosphorus atom's formal charge in \( ext{PF}_6^-\), the overall negative charge reflects in balancing every bond and lone pair while sticking to chemical stability rules.

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Most popular questions from this chapter

Write electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Cd}^{2+}\) , \((\mathbf{b}) \mathrm{P}^{3-},(\mathbf{c}) \mathrm{Zr}^{4+},(\mathbf{d}) \mathrm{Ru}^{3+},(\mathbf{e}) \mathrm{As}^{3-},(\mathbf{f}) \mathrm{Ag}^{+}\)

Fill in the blank with the appropriate numbers for both electrons and bonds (considering that single bonds are counted as one, double bonds as two, and triple bonds as three). (a) Fluorine has _____ valence electrons and makes______ bond(s) in compounds. (b) Oxygen has _____ valence electrons and makes ______ bond(s) in compounds. (c) Nitrogen has _____ valence electrons and makes ______ bond(s) in compounds. (d) Carbon has _____ valence electrons and makes ______ bond(s) in compounds.

We can define average bond enthalpies and bond lengths for ionic bonds, just like we have for covalent bonds. Which ionic bond is predicted to have the smaller bond enthalpy, Li-F or \(\mathrm{Cs}-\mathrm{F}\) ?

Consider the lattice energies of the following Group 2 \(\mathrm{A}\) compounds: \(\mathrm{Be} \mathrm{H}_{2}, 3205 \mathrm{kJ} / \mathrm{mol} ; \mathrm{MgH}_{2}, 2791 \mathrm{kJ} / \mathrm{mol}\) \(\mathrm{CaH}_{2}, 2410 \mathrm{kJ} / \mathrm{mol} ; \mathrm{SrH}_{2}, 2250 \mathrm{kJ} / \mathrm{mol} ; \mathrm{BaH}_{2}, 2121 \mathrm{kJ} / \mathrm{mol}\) (a) What is the oxidation number of \(\mathrm{H}\) in these compounds? (b) Assuming that all of these compounds have the same three-dimensional arrangement of ions in the solid, which of these compounds has the shortest cation-anion distance? (c) Consider BeH \(_{2} .\) Does it require 3205 kJ of energy to break one mole of the solid into its ions, or does breaking up one mole of solid into its ions release 3205 \(\mathrm{kJ}\) of energy? (d) The lattice energy of \(\mathrm{ZnH}_{2}\) is 2870 \(\mathrm{kJ} / \mathrm{mol}\) . Considering the trend in lattice enthalpies in the Group 2 \(\mathrm{A}\) compounds, predict which Group 2 \(\mathrm{A}\) element is most similar in ionic radius to the \(\mathrm{Zn}^{2+}\) ion.

Ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl},\) is a very soluble salt in water. (a) Draw the Lewis structures of the ammonium and chloride ions. (b) Is there an \(\mathrm{N}-\) Cl bond in solid ammonium chloride? (c) If you dissolve 14 gof ammonium chloride in 500.0 \(\mathrm{mL}\) of water, what is the molar concentration of the solution? (d) How many grams of silver nitrate do you need to add to the solution in part (c) to precipitate all of the chloride as silver chloride?
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