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Which of these molecules have an odd number of valence electrons: \(\mathrm{NO}_{2}, \mathrm{SCl}_{2}, \mathrm{NH}_{3}, \mathrm{NO}_{3} ?\)

Short Answer

Expert verified
The molecule NO2 has an odd number of valence electrons.

Step by step solution

01

Count valence electrons in NO2

For the molecule NO2 (Nitrogen dioxide), nitrogen (N) has 5 valence electrons and each oxygen (O) has 6 valence electrons. Calculate the total number of valence electrons: \[5 + (2 \times 6) = 17\] Thus, NO2 has 17 valence electrons, which is an odd number.
02

Count valence electrons in SCl2

For the molecule SCl2 (Sulfur dichloride), sulfur (S) has 6 valence electrons and each chlorine (Cl) atom has 7 valence electrons. Calculate the total number of valence electrons:\[6 + (2 \times 7) = 20\]Thus, SCl2 has 20 valence electrons, which is an even number.
03

Count valence electrons in NH3

For the molecule NH3 (Ammonia), nitrogen (N) has 5 valence electrons and each hydrogen (H) has 1 valence electron. Calculate the total number of valence electrons:\[5 + (3 \times 1) = 8\]Thus, NH3 has 8 valence electrons, which is an even number.
04

Count valence electrons in NO3−

For the nitrate ion NO3−, nitrogen (N) has 5 valence electrons, each oxygen (O) has 6 valence electrons, and there is an extra electron due to the negative charge. Calculate the total number of valence electrons:\[5 + (3 \times 6) + 1 = 24\]Thus, the ion NO3− has 24 valence electrons, which is an even number.

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Key Concepts

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

Molecular Structure
Molecular structure refers to the three-dimensional arrangement of atoms within a molecule. This structure determines the physical and chemical properties of a molecule. Understanding molecular structure helps us to predict and explain molecule behavior, reactivity, polarity, phase of matter, color, magnetism, biological activity, and much more.

To analyze the molecular structure, two common methods are used:
  • Lewis Structures: A simple diagram that shows the arrangement of electrons in a molecule. It helps to identify valence electrons and possible bonding and lone pairs.
  • VSEPR Theory (Valence Shell Electron Pair Repulsion): A model used to predict the geometry of individual molecules based on the number of pairs of valence electrons surrounding their central atoms.
Both methods play a significant role in determining the shape and angles of bonds in the molecules involved, which directly influence the molecular structure and its properties.
Chemical Bonding
Chemical bonding is a fundamental concept that refers to the joining of atoms to form molecules and compounds. Atoms bond by sharing or exchanging electrons to achieve a more stable electron configuration usually resembling that of noble gases.

There are several types of chemical bonds:
  • Covalent Bonds: Formed when two atoms share one or more pairs of valence electrons, as in the bonds within a water molecule (H2O).
  • Ionic Bonds: Occur when electrons are transferred from one atom to another, resulting in oppositely charged ions that attract each other, like in sodium chloride (NaCl).
  • Metallic Bonds: Found in metals, where atoms in a metallic lattice share a "sea" of electrons that move freely, providing metals with unique properties such as conductivity.
The type of bond influences a molecule's strength, flexibility, conductibility, and melting and boiling points.
Electron Configuration
Electron configuration describes the arrangement of electrons in an atom or molecule. Understanding electron configuration helps in predicting chemical bonding patterns, reactivity, and molecular structure.

An electron configuration is often depicted using the notation where an element's electrons are arranged among various energy levels and orbitals. For example, the electron configuration for an oxygen atom is written as 1s2 2s2 2p4.

Key aspects of electron configuration include:
  • Principle Energy Levels: Represented by "n" values, where electrons reside in increasing energy levels from the nucleus outward.
  • Sublevels: Composed of "s, p, d," and "f" orbitals, each having a specific shape and number of orientations in space.
  • Pauli Exclusion Principle: States that each orbital can hold a maximum of two electrons with opposite spins.
  • Hund's Rule: Electrons will fill an unoccupied orbital before they pair up in an occupied one.
By understanding electron configurations, chemists can predict atomic interactions in chemical reactions and the formation of various molecules.

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Most popular questions from this chapter

In each case, tell whether cis and trans isomers exist. If they do, write structural formulas for the two isomers and label each cis or trans. For those that cannot have cistrans isomers, explain why. (a) \(\mathrm{Br}_{2} \mathrm{CH}_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CHCH}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}\) (d) \(\mathrm{CH}_{2}=\mathrm{CHCH}_{2} \mathrm{CH}_{3}\)

For each pair of bonds, identify the more polar one and use \(\delta+\) or \(\delta-\) to indicate the partial charge on each atom. (a) \(\mathrm{B}-\mathrm{Cl}\) and \(\mathrm{B}-\mathrm{O}\) (b) \(\mathrm{O}-\mathrm{F}\) and \(\mathrm{O}-\mathrm{Se}\) (c) \(\mathrm{S}-\mathrm{Cl}\) and \(\mathrm{B}-\mathrm{F}\) (d) \(\mathrm{N}-\mathrm{H}\) and \(\mathrm{N}-\mathrm{F}\)

When we estimate \(\Delta_{\mathrm{r}} H^{\circ}\) from bond enthalpies we assume that all bonds of the same type (single, double, triple) between the same two atoms have the same energy, regardless of the molecule in which they occur. The purpose of this problem is to show you that this is only an approximation. You will need these standard enthalpies of formation: $$ \begin{array}{ll} \mathrm{C}(\mathrm{g}) & \Delta_{\mathrm{f}} H^{\circ}=716.7 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{CH}(\mathrm{g}) & \Delta_{\mathrm{f}} H^{\circ}=596.3 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{CH}_{2}(\mathrm{~g}) & \Delta_{\mathrm{f}} H^{\circ}=392.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{CH}_{3}(\mathrm{~g}) & \Delta_{\mathrm{f}} H^{\circ}=146.0 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}(\mathrm{g}) & \Delta_{\mathrm{f}} H^{\circ}=218.0 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ (a) What is the average \(\mathrm{C}-\mathrm{H}\) bond energy in methane, \(\mathrm{CH}_{4} ?\) (b) Using bond enthalpies, estimate \(\Delta_{1} H^{\circ}\) for the reaction $$ \mathrm{CH}_{4}(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) $$ (c) By heating \(\mathrm{CH}_{4}\) in a flame it is possible to produce the reactive gaseous species \(\mathrm{CH}_{3}, \mathrm{CH}_{2}, \mathrm{CH},\) and even carbon atoms, C. Experiments give these values of \(\Delta_{r} H^{\circ}\) for the reactions shown: $$ \begin{aligned} \mathrm{CH}_{3}(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g})+\mathrm{H}(\mathrm{g}) & & \Delta_{\mathrm{r}} H^{\circ} &=788.7 \mathrm{~kJ} \\ \mathrm{CH}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g}) & & \Delta_{\mathrm{r}} H^{\circ} &=324.2 \mathrm{~kJ} \\ \mathrm{CH}(\mathrm{g}) \longrightarrow \mathrm{C}(\mathrm{g})+\mathrm{H}(\mathrm{g}) & & \Delta_{\mathrm{r}} H^{\circ} &=338.3 \mathrm{~kJ} \end{aligned} $$ For each of the reactions in part (c), draw a diagram similar to Figure 6.6 . Then, calculate the average \(\mathrm{C}-\mathrm{H}\) bond energy in \(\mathrm{CH}_{3}, \mathrm{CH}_{2}\), and \(\mathrm{CH}\). Comment on any trends you see.

Write the Lewis structures for (a) \(\left(\mathrm{Cl}_{2} \mathrm{PN}\right)_{3}\) (b) \(\left(\mathrm{Cl}_{2} \mathrm{PN}\right)_{4}\)

Draw resonance structures for each of these ions: \(\mathrm{NSO}^{-}\) and \(\mathrm{SNO}^{-}\). (The atoms are bonded in the order given in each case, that is, \(\mathrm{S}\) is the central atom in \(\mathrm{NSO}^{-} .\).) (a) Use formal charges to determine which ion is likely to be more stable. (b) Explain why the two ions cannot be considered resonance structures of each other.

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