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Chapter 9: Problem 103

Determine the molecular structure and hybridization of the central atom \(\mathrm{X}\) in the polyatomic ion \(\mathrm{XY}_{3}+\) given the following information: A neutral atom of \(\mathrm{X}\) contains 36 electrons, and the element \(Y\) makes an anion with a \(1-\) charge, which has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6}\).

Short Answer

Expert verified
The molecular structure of thepolyatomic ion XY3+ is trigonal pyramidal, with the central atom X (Kr+) having sp3 hybridization.

Step by step solution

01

Determine electron configuration of X and Y

From the given information, a neutral atom of X contains 36 electrons. To determine its electron configuration, we'll use the periodic table to find out which element it is. The atomic number of X (number of protons) is equal to the number of electrons. Since there are 36 electrons, X is element 36, or Krypton (Kr). The electron configuration of Kr is [Ar]\(4s^{2} 3d^{10} 4p^{6}\). Now, let's find out which element Y is. We know that Y makes an anion with a \(1-\) charge that has an electron configuration \(1s^{2} 2s^{2} 2p^{6}\). This is the electron configuration of a complete second shell, with 10 electrons. Therefore, Y must be a period 2 element with 9 electrons (since the anion has one extra electron), which is Fluorine (F). The electron configuration of F- would be that of the neutral Neon atom: \(1s^{2} 2s^{2} 2p^{6}\).
02

Draw Lewis structure of the polyatomic ion

Now, let's draw the Lewis structure for the ion XY3+. Since X is the central atom, we'll place it in the center and surround it with three Y atoms. Keep in mind that X has lost one electron to make a \(1+\) charge, so its electron configuration is now [Ar]\(4s^2 3d^{10} 4p^5\). Each F- has one lone pair and one single bond with X. Therefore, the Lewis structure of XY3+ will have X in the center, bonded to three Y atoms, each with three lone pairs.
03

Determine the hybridization of the central atom

To determine the hybridization of the central atom X, we will count the number of electron domains (bonding and non-bonding electron pairs) around X. In XY3+, X is bonded to three Y atoms, so there are three bonding electron pairs. Since X has one unpaired electron left in its outer shell after forming the three bonds, it has only one non-bonding electron pair. This means X has a total of four electron domains (three bonding pairs and one non-bonding pair). Based on the number of electron domains, we can determine the hybridization of the central atom. In this case, X (Kr+) has four electron domains, which corresponds to sp3 hybridization. So, the molecular structure of XY3+ is trigonal pyramidal, with the central atom X (Kr+) having sp3 hybridization.

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Key Concepts

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

Electron Configuration
Understanding the electron configuration is crucial for determining how an atom interacts with others to form compounds. Electron configuration represents the distribution of electrons in an atom's orbitals. For the central atom X in the ion XY extsubscript{3} extsuperscript{+}, we start by identifying that it contains 36 electrons when neutral.
Using the periodic table, we determine that X is Krypton (Kr), which has an electron configuration of [Ar]\(4s^{2} 3d^{10} 4p^{6}\). For the element Y, we look at the electron configuration provided: \(1s^{2} 2s^{2} 2p^{6}\). This suggests that the anion derived from Y has gained an extra electron, indicating that Y is Fluorine (F). The fluoride ion (F extsuperscript{-}) matches the electron configuration of Neon, a noble gas, thanks to the added electron.
Lewis Structure
The Lewis structure helps visualize the arrangement of atoms in a molecule and how the valence electrons are involved in bonding. In the XY extsubscript{3} extsuperscript{+} ion, X acts as the central atom. We represent it surrounded by three Y (Fluorine) atoms.
When X departs one electron, forming X extsuperscript{+}, its electron configuration alters to [Ar]\(4s^{2} 3d^{10} 4p^{5}\), having one less electron.
Here's how it works:
  • X is connected to three Y atoms through single covalent bonds.
  • Each Y atom has seven valence electrons, needing one more to complete their octet, which they obtain by forming one bond with X.
  • In the Lewis structure, each Y displays three lone pairs.
The structure accounts for the positive charge by the lack of one electron in X.
Hybridization
Hybridization helps us understand the molecular geometry of a molecule by analyzing the mixing of atomic orbitals into new hybrid orbitals. The central atom, X, in XY extsubscript{3} extsuperscript{+}, has a total of four electron domains. This includes three bonds with Y atoms and one unpaired electron due to the ion's overall positive charge.
To determine X's hybridization:
  • Count the electron domains: 3 bonding pairs + 1 non-bonding pair = 4 domains.
  • With four domains, X undergoes sp extsuperscript{3} hybridization.
While Krypton typically exhibits a filled outer shell as a noble gas, it is capable of forming such hybridized bonds under special conditions.
Central Atom
The central atom in a molecule or ion often dictates the overall geometry and stability of the structure. In this context, X (Krypton) serves as the central atom in XY extsubscript{3} extsuperscript{+}. As Krypton picks up the central role:
  • It forms three bonds with Y (Fluorine) atoms, facilitating the molecular structure.
  • The unpaired electron results in a positive charge on X, indicated in the molecular formula by XY extsubscript{3} extsuperscript{+}.
  • Due to its central position and hybridization (sp extsuperscript{3}), the ion adopts a trigonal pyramidal shape.
This role of X as the central atom with a specific hybridization provides a stable yet fascinating molecule structure.

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Most popular questions from this chapter

Carbon monoxide (CO) forms bonds to a variety of metals and metal ions. Its ability to bond to iron in hemoglobin is the reason that \(\mathrm{CO}\) is so toxic. The bond carbon monoxide forms to metals is through the carbon atom: \(\mathrm{M}-\mathrm{C}=\mathrm{O}\) a. On the basis of electronegativities, would you expect the carbon atom or the oxygen atom to form bonds to metals? b. Assign formal charges to the atoms in CO. Which atom would you expect to bond to a metal on this basis? c. In the MO model, bonding MOs place more electron density near the more electronegative atom. (See the HF molecule in Figs. \(9.42\) and \(9.43\).) Antibonding MOs place more. electron density near the less electronegative atom in the diatomic molecule. Use the MO model to predict which atom of carbon monoxide should form bonds to metals.

Use the MO model to determine which of the following has the smallest ionization energy: \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{~N}_{2}{ }^{2-}, \mathrm{N}_{2}^{-}, \mathrm{O}_{2}^{+} .\) EX- plain your answer.

In terms of the molecular orbital model, which species in each of the following two pairs will most likely be the one to gain an electron? Explain. a. CN or \(\mathrm{NO}\) b. \(\mathrm{O}_{2}^{2+}\) or \(\mathrm{N}_{2}^{2+}\)

The diatomic molecule OH exists in the gas phase. The bond length and bond energy have been measured to be \(97.06 \mathrm{pm}\) and \(424.7 \mathrm{~kJ} / \mathrm{mol}\), respectively. Assume that the \(\mathrm{OH}\) molecule is analogous to the HF molecule discussed in the chapter and that molecular orbitals result from the overlap of a lowerenergy \(p_{z}\) orbital from oxygen with the higher-energy \(1 s\) orbital of hydrogen (the \(\mathrm{O}-\mathrm{H}\) bond lies along the \(z\) -axis). a. Which of the two molecular orbitals will have the greater hydrogen \(1 s\) character? b. Can the \(2 p_{x}\) orbital of oxygen form molecular orbitals with the \(1 s\) orbital of hydrogen? Explain. c. Knowing that only the \(2 p\) orbitals of oxygen will interact significantly with the \(1 s\) orbital of hydrogen, complete the molecular orbital energy- level diagram for \(\mathrm{OH}\). Place the correct number of electrons in the energy levels. d. Estimate the bond order for OH. e. Predict whether the bond order of \(\mathrm{OH}^{+}\) will be greater than, less than, or the same as that of OH. Explain.

The atoms in a single bond can rotate about the internuclear axis without breaking the bond. The atoms in a double and triple bond cannot rotate about the internuclear axis unless the bond is broken. Why?
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