Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 8: Problem 105

Which of the following species are isoelectronic: \(\mathrm{NH}_{4}^{+}, \mathrm{C}_{6} \mathrm{H}_{6}, \mathrm{CO}, \mathrm{CH}_{4}, \mathrm{~N}_{2}, \mathrm{~B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6} ?\)

Short Answer

Expert verified
The isoelectronic pairs are \(\mathrm{NH}_{4}^{+}\) with \(\mathrm{CH}_{4}\), and \(\mathrm{CO}\) with \(\mathrm{N}_{2}\).

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards
Start your free trial Skip

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Understand Isoelectric Species

Isoelectric species are atoms, ions, or molecules that have the same number of electrons. To find which of the given species are isoelectronic, we need to count the total number of electrons in each species.
02

Count Electrons in Each Species

Let's find the total number of electrons for each species:- **\(\mathrm{NH}_{4}^{+}\):** \(\mathrm{N}\) has 7 electrons, and each \(\mathrm{H}\) has 1 electron. Total is \(7 + 4 = 11\). Since it has a \(+1\) charge, subtract 1 electron, resulting in 10 electrons. - **\(\mathrm{C}_{6} \mathrm{H}_{6}\):** \(\mathrm{C}\) has 6 electrons, \(\mathrm{H}\) has 1 electron. Total is \(6\times6 + 6\times1 = 42\) electrons.- **\(\mathrm{CO}\):** \(\mathrm{C}\) has 6 electrons, \(\mathrm{O}\) has 8 electrons. Total is \(6 + 8 = 14\) electrons.- **\(\mathrm{CH}_{4}\):** \(\mathrm{C}\) has 6 electrons, \(\mathrm{H}\) has 1 electron. Total is \(6 + 4 = 10\) electrons.- **\(\mathrm{N}_{2}\):** Each \(\mathrm{N}\) has 7 electrons. Total is \(7 + 7 = 14\) electrons.- **\(\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6}\):** \(\mathrm{B}\) has 5 electrons, \(\mathrm{N}\) has 7 electrons, \(\mathrm{H}\) has 1 electron. Total is \(3\times5 + 3\times7 + 6\times1 = 24\) electrons.
03

Identify Isoelectronic Species

Compare the electron counts. We find that:- \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{CH}_{4}\) both have 10 electrons.- \(\mathrm{CO}\) and \(\mathrm{N}_{2}\) both have 14 electrons.Thus, these pairs are isoelectronic.

Key Concepts

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

Electron Counting
Electron counting is a critical skill for identifying isoelectronic species. By determining the total number of electrons in different atoms, ions, or molecules, we can see if they match up. Finding connections helps us understand chemical properties better.

To count electrons, follow these simple steps:
  • Identify the number of valence electrons for each element, using the periodic table as your guide.
  • Add electrons based on the composition of the molecule or compound.
  • Adjust for any charges by adding electrons for negative charges or subtracting for positive ones.
These steps help you understand how atoms or molecules might interact with others.
Atoms and Ions
Atoms are the fundamental units of matter, consisting of protons, neutrons, and electrons. Ions form when atoms gain or lose electrons, resulting in a charge. Understanding this helps us with electron counting, which is essential for identifying isoelectronic species.

Atoms and ions share this basic structure:
  • Protons determine the atomic number and identity of an element.
  • Neutrons provide stability, and together with protons form the nucleus.
  • Electrons, which orbit the nucleus, are involved in chemical reactions and bonding.
Ions differ from neutral atoms mainly because of their unbalanced charge, but they can still share electron counts with other species.
Valence Electrons
Valence electrons are the outermost electrons of an atom and play key roles in chemical bonding and reactions. Knowing how to identify and count them is crucial for understanding molecular interactions.

Here's how valence electrons influence chemistry:
  • They determine the reactivity and bonding ability of an element.
  • Elements in the same group of the periodic table have similar valence electron configurations, affecting their chemistry.
  • Counting valence electrons helps predict how an atom will bond with others, forming stable compounds.
Mastering this concept makes it easier to see patterns in chemical behavior, like identifying isoelectronic species.
Chemical Species
Chemical species refer to atoms, molecules, ions, or particles that participate in chemical reactions. Understanding them allows us to compare and predict reactivity, polarity, and other chemical properties.

Key points about chemical species:
  • They can be neutral (like atoms or molecules) or charged (like ions).
  • Isoelectronic species are a subset where species have the same number of electrons.
  • Recognizing them helps in predicting similar properties or behaviors in different substances.
By understanding chemical species, students can better grasp complex ideas like isoelectronicity and its significance in chemistry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For each of the following organic molecules draw a Lewis structure in which the carbon atoms are bonded to each other by single bonds: (a) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (b) \(\mathrm{C}_{4} \mathrm{H}_{10}\) (c) \(\mathrm{C}_{5} \mathrm{H}_{12}\) For parts (b) and (c), show only structures in which each \(\mathrm{C}\) atom is bonded to no more than two other \(\mathrm{C}\) atoms.

(a) From the following data calculate the bond enthalpy of the \(\mathrm{F}_{2}^{-}\) ion. $$ \begin{array}{ll} \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{~F}(g) & \Delta H_{\mathrm{rxn}}^{\circ}=156.9 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{F}^{-}(g) \longrightarrow \mathrm{F}(g)+e^{-} & \Delta H_{\mathrm{rxn}}^{\circ}=333 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{F}_{2}^{-}(g) \longrightarrow \mathrm{F}_{2}(g)+e^{-} & \Delta H_{\mathrm{rxn}}^{\circ}=290 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ (b) Explain the difference between the bond enthalpies of \(\mathrm{F}_{2}\) and \(\mathrm{F}_{2}^{-}\).

Give three examples of compounds that do not satisfy the octet rule. Write a Lewis structure for each.

The species \(\mathrm{H}_{3}^{+}\) is the simplest polyatomic ion. The geometry of the ion is that of an equilateral triangle. (a) Draw three resonance structures to represent the ion. (b) Given the following information $$ \begin{aligned} 2 \mathrm{H}+\mathrm{H}^{+} \longrightarrow & \mathrm{H}_{3}^{+} & & \Delta H^{\circ}=-849 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2} \longrightarrow & 2 \mathrm{H} & & \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{H}^{+}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{3}^{+} $$

Draw three resonance structures for the hydrogen sulfite ion \(\left(\mathrm{HSO}_{3}^{-}\right)-\) one that obeys the octet rule for the central atom, and two that expand the octet of the central atom. Calculate the formal charges on all atoms in each structure and determine which, if any, of the resonance structures has formal charges that are inconsistent with the elements' electronegativities.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free