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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 \(X\) 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 the XY3+ ion is trigonal planar with sp2 hybridization of the central atom X (Kr), where X is Krypton and Y is Fluorine. The central Krypton atom forms three covalent bonds with three Fluorine atoms, and the bond angles are approximately 120 degrees.

Step by step solution

01

Identify element X

A neutral atom of X contains 36 electrons. To identify element X, count the electrons in each energy level (electron configuration) until you reach 36 electrons. From the periodic table, the electron configuration with 36 electrons is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}\), corresponding to the element Krypton (Kr).
02

Identify element Y

The electron configuration of the Y- ion is given as \(1s^{2} 2s^{2} 2p^{6}\). Since Y forms an anion with a -1 charge, its neutral form will have one less electron. The neutral form of Y would have the electron configuration of \(1s^{2} 2s^{2} 2p^{5}\). This corresponds to the element Fluorine (F).
03

Determine the Valence Electrons and the Bonds formed

The valence electrons of Kr (element X) are 8 (4s2 and 4p6). For F (element Y), the valence electrons are 7 (2s2 and 2p5). Since the overall charge of the ion is +1, one of the electrons from the central atom Kr will be removed. This leaves Kr with one unpaired electron, allowing it to form three covalent bonds with three F atoms, each having one unoccupied electron in its valence shell.
04

Determine Hybridization

The central atom X (Kr) is forming three covalent bonds. To accommodate these bonds, X needs to hybridize its orbitals. The hybridization will involve one s-orbital from the 4s shell and two p-orbitals from the 4p shell. The hybridization will be sp2.
05

Describe the Molecular Structure

In the XY3+ ion, the central atom X (Kr) is sp2 hybridized and forms three covalent bonds with three Y (F) atoms. Since sp2 hybridization results in a trigonal planar geometry, the molecular structure of XY3+ ion will have the central atom Kr bond with the three F atoms in a trigonal planar arrangement with bond angles of approximately 120 degrees. In conclusion, the molecular structure of the XY3+ ion is trigonal planar with sp2 hybridization of the central atom X (Kr).

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Key Concepts

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

Hybridization
Hybridization is a critical concept in chemistry that explains how atoms in a molecule bond together to form its shape. It describes the merging of the atomic orbitals of an atom (usually the central atom in a molecule) to form new hybrid orbitals that can host pairs of electrons involved in chemical bonding.

For instance, in our exercise, the central atom Krypton (Kr) undergoes sp2 hybridization. This means that one s orbital merges with two p orbitals to create three hybrid orbitals. Each of these hybrid orbitals can form a covalent bond with the fluorine atoms, leading to a trigonal planar molecular geometry. The concept of hybridization helps us understand that despite Kr typically being an inert gas, under certain conditions, it can form stable compounds.
Valence Electrons
Valence electrons are the outermost electrons of an atom and play a key role in chemical bonding. They are the electrons involved in forming bonds with other atoms. The number of valence electrons can often predict the chemical properties and bonding behavior of an element.

In our example, Krypton (Kr) has 8 valence electrons initially, but because the molecule has an overall positive charge, one electron is removed, leaving 7 valence electrons for bonding. Fluorine (F), on the other hand, has 7 valence electrons naturally, as reflected in its electron configuration. Knowing the count of valence electrons is essential for determining the types of bonds that will form and the resulting molecular structure.
Covalent Bonds
Covalent bonds are formed when atoms share pairs of electrons, allowing them to stick together and form molecules. These bonds typically occur between non-metal atoms with similar electronegativities. The shared electrons allow each atom to achieve a more stable electronic configuration.

In the molecular structure of the polyatomic ion in our exercise, three covalent bonds are formed between the central Kr atom and three F atoms. The sharing of electrons fills the valence shells of both Kr and F, ultimately stabilizing the system. Covalent bonding is a fundamental concept for understanding molecular structures in chemistry.
Polyatomic Ion
A polyatomic ion is a charged entity composed of two or more covalently bonded atoms. The charge results from the total number of protons not being equal to the total number of electrons in the molecule, creating either a positive or negative charge.

In our exercise, the XY3+ ion is a polyatomic ion consisting of one Kr atom and three F atoms with a surplus positive charge. Such ions are often part of larger chemical compounds and their formation involves electrons from different atoms being shared or transferred, leading to a charge imbalance. Understanding polyatomic ions is crucial in the study of inorganic and organic chemistry, as they frequently occur in various chemical reactions and compounds.
Electron Configuration
Electron configuration refers to the distribution of electrons of an atom or molecule in atomic or molecular orbitals. It is often described using the principles of the Aufbau principle, Hund's rule, and the Pauli exclusion principle. These rules help us predict the electron arrangement and thus the chemical behavior of an element.

Through the example provided, we deduced that Krypton, with the electron configuration that culminates in the 4p6 orbital, initially has 36 electrons. Analogously, Fluorine typically ends in the 2p5 orbital in its neutral state, possessing 9 electrons. Such electron configurations give us insight into the valence electrons and potential chemical bonds an element can form. Recognizing electron configuration patterns is crucial for predicting how atoms will interact and combine to form molecules.

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Most popular questions from this chapter

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 OH molecule is analogous to the HF molecule discussed in the chapter and that molecular orbitals result from the overlap of a lower-energy \(p_{z}\) 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 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 \(\mathrm{OH}\). Explain.

Biacetyl and acetoin are added to margarine to make it taste more like butter. Complete the Lewis structures, predict values for all \(\mathrm{C}-\mathrm{C}-\mathrm{O}\) bond angles, and give the hybridization of the carbon atoms in these two compounds. Must the four carbon atoms and two oxygen atoms in biacetyl lie the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in biacetyl and acetoin?

A flask containing gaseous \(\mathrm{N}_{2}\) is irradiated with \(25-\mathrm{nm}\) light. a. Using the following information, indicate what species can form in the flask during irradiation. $$ \begin{aligned} \mathrm{N}_{2}(g) & \longrightarrow 2 \mathrm{~N}(g) & \Delta H &=941 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{N}_{2}(g) & \longrightarrow \mathrm{N}_{2}^{+}(g)+\mathrm{e}^{-} & \Delta H &=1501 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{N}(g) & \longrightarrow \mathrm{N}^{+}(g)+\mathrm{e}^{-} & \Delta H &=1402 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ b. What range of wavelengths will produce atomic nitrogen in the flask but will not produce any ions? c. Explain why the first ionization energy of \(\mathrm{N}_{2}(1501 \mathrm{~kJ} / \mathrm{mol})\) is greater than the first ionization energy of atomic nitrogen \((1402 \mathrm{~kJ} / \mathrm{mol})\).

The allene molecule has the following Lewis structure: Must all hydrogen atoms lie the same plane? If not, what is their spatial relationship? Explain.

Describe the bonding in the first excited state of \(\mathrm{N}_{2}\) (the one closest in energy to the ground state) using the molecular orbital model. What differences do you expect in the properties of the molecule in the ground state as compared to the first excited state? (An excited state of a molecule corresponds to an electron arrangement other than that giving the lowest possible energy.)

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