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Chapter 4: Problem 15

Why are \(d\) orbitals sometimes used to form hybrid orbitals? Which period of elements does not use \(d\) orbitals for hybridization? If necessary, which \(d\) orbitals \((3 d, 4 d, 5 d, \text { or } 6 d)\) would sulfur use to form hybrid orbitals requiring \(d\) atomic orbitals? Answer the same question for arsenic and for iodine.

Short Answer

Expert verified
D orbitals are sometimes used to form hybrid orbitals in transition elements and heavier elements since they can accommodate more atoms in the molecule and provide necessary energy for bonding. The second period elements do not use d orbitals for hybridization as they only have access to s and p orbitals. If sulfur, a 3rd period element, needs d orbitals for hybridization, it would use 3d orbitals. Likewise, arsenic (a 4th period element) would use 4d orbitals, and iodine (a 5th period element) would use 5d orbitals for hybridization if necessary.

Step by step solution

01

Understanding d Orbital Hybridization

Hybridization is a process in which atomic orbitals of an atom mix to form new hybrid orbitals. The main purpose of hybridization is to allow the atom to form bonds with other atoms by achieving a lower energy state. In general, atoms utilize s, p, and d orbitals for hybridization. The involvement of d orbitals occurs in transition elements and heavier elements, where these orbitals can be easily accessed for bonding. The reason behind the use of d orbitals is to accommodate more number of atoms in the molecule and fulfill the energy requirements for bonding. Sometimes, d orbitals will also be involved in the hybridization process to form d-蟺 bonds with other atoms like transition metals.
02

Period Not Using d Orbitals for Hybridization

The second period elements do not use d orbitals for hybridization. Elements in this period only have access to s and p orbitals because the electron configuration of second period elements ends with the 2s and 2p subshells. Due to this electron configuration restriction, second period elements such as carbon, nitrogen, oxygen, and fluorine cannot form d orbitals for hybridization.
03

Sulfur's d Orbital Hybridization

Sulfur is a 3rd period element and has an electron configuration of \([Ne]3s^{2}3p^{4}\). If sulfur needs to form hybrid orbitals requiring d atomic orbitals, it would use the 3d orbitals. This is because sulfur's 3d orbitals are energetically accessible for hybridization and can accommodate additional electrons to form bonds with other atoms.
04

Arsenic's d Orbital Hybridization

Arsenic is a 4th period element and has an electron configuration of \([Ar]4s^{2}3d^{10}4p^{3}\). If arsenic needs to form hybrid orbitals requiring d atomic orbitals, it would use the 4d orbitals. This is because arsenic's 4d orbitals are energetically accessible for hybridization and can accommodate additional electrons to form bonds with other atoms.
05

Iodine's d Orbital Hybridization

Iodine is a 5th period element and has an electron configuration of \([Kr]5s^{2}4d^{10}5p^{5}\). If iodine needs to form hybrid orbitals requiring d atomic orbitals, it would use the 5d orbitals. This is because iodine's 5d orbitals are energetically accessible for hybridization and can accommodate additional electrons to form bonds with other atoms.

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Key Concepts

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

Hybridization
Hybridization is a key concept in chemistry that explains how atoms mix their orbitals to form new hybrid orbitals. This mixing allows atoms to form bonds and achieve more stable configurations.
In essence, hybridization combines the properties of different orbitals (usually s, p, and sometimes d orbitals). This leads to forms of bonding that can support complex structures like those seen in molecules.
  • Hybridization enables the overlapping of orbitals which is essential for effective bonding.
  • It minimizes energy, favoring more stable arrangements.
  • d orbital hybridization occurs mainly in larger atoms, including transition metals and other heavier elements.
The d orbitals' involvement helps accommodate more atoms or electrons, especially in larger molecules, enhancing the flexibility of molecular geometry.
Atomic Orbitals
Atomic orbitals are regions around an atom's nucleus where electrons are most likely to be found. Each type of atomic orbital has a unique shape and energy level.
There are several types of orbitals: s, p, d, and f. Here鈥檚 a simple breakdown:
  • s orbitals: Spherical shape, each energy level has one s orbital.
  • p orbitals: Dumbbell-shaped, three p orbitals per energy level starting from the second level.
  • d orbitals: More complex shapes, appear starting from the third energy level.
d orbitals can combine with s and p orbitals for hybridization. They bring about unique properties in elements, especially for bond formation in larger, heavier atoms.
The use of d orbitals in hybridization is more prominent in heavier atoms where these orbitals are readily available for bonding.
Transition Elements
Transition elements are metals found in the d-block of the periodic table, known for their unique ability to use d orbitals in bonding.
One hallmark of transition elements is their variable oxidation states facilitated by the occupation of d orbitals.
  • They often exhibit a wide range of colors due to d-d electron transitions.
  • The use of d orbitals allows these elements to form complexes with various elements.
  • Typical transition elements include iron, copper, and nickel.
The adaptability of their d orbitals is crucial for catalysis and bonding, which explains their usefulness in chemical reactions and industrial processes.
Electron Configuration
Electron configuration is the arrangement of electrons in an atom's orbitals. It determines how an atom interacts with other atoms.
Each element has a unique electron configuration that follows the Aufbau principle, Hund鈥檚 rule, and Pauli exclusion principle to fill orbitals.
  • For example, sulfur's electron configuration is \( [Ne]3s^{2}3p^{4} \).
  • Arsenic has \( [Ar]4s^{2}3d^{10}4p^{3} \).
  • Iodine is \( [Kr]5s^{2}4d^{10}5p^{5} \).
Understanding electron configurations helps explain why certain elements use d orbitals in hybridization. For example, elements in the second period do not have d orbitals available, unlike heavier elements from the third period onwards.
This understanding is crucial for predicting chemical reactivity and bond formation.

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Most popular questions from this chapter

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 NO b. \(\mathrm{O}_{2}^{2+}\) or \(\mathrm{N}_{2}^{2+}\)

Which is the more correct statement: "The methane molecule \(\left(\mathrm{CH}_{4}\right)\) is a tetrahedral molecule because it is \(s p^{3}\) hybridized" or "The methane molecule \(\left(\mathrm{CH}_{4}\right)\) is \(s p^{3}\) hybridized because it is a tetrahedral molecule"? What, if anything, is the difference between these two statements?

Using the molecular orbital model to describe the bonding in \(\mathrm{F}_{2}^{+}, \mathrm{F}_{2},\) and \(\mathrm{F}_{2}^{-},\) predict the bond orders and the relative bond lengths for these three species. How many unpaired electrons are present in each species?

Complete a Lewis structure for the compound shown below, then answer the following questions. What are the predicted bond angles about the carbon and nitrogen atoms? How many lone pairs of electrons are present in the Lewis structure? How many double bonds are present?

Cyanamide \(\left(\mathrm{H}_{2} \mathrm{NCN}\right),\) an important industrial chemical, is produced by the following steps: Calcium cyanamide (CaNCN) is used as a direct-application fertilizer, weed killer, and cotton defoliant. It is also used to make cyanamide, dicyandiamide, and melamine plastics: a. Write Lewis structures for \(\mathrm{NCN}^{2-}, \mathrm{H}_{2} \mathrm{NCN}\), dicyandiamide, and melamine, including resonance structures where appropriate. b. Give the hybridization of the \(\mathrm{C}\) and \(\mathrm{N}\) atoms in each species. c. How many \(\sigma\) bonds and how many \(\pi\) bonds are in each species? d. Is the ring in melamine planar? e. There are three different \(\mathrm{C}-\mathrm{N}\) bond distances in dicyandiamide, NCNC(NH_)_2, and the molecule is nonlinear. Of all the resonance structures you drew for this molecule, predict which should be the most important.
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