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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-π 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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Most popular questions from this chapter

Complete the following resonance structures for \(\mathrm{POCl}_{3}\) a. Would you predict the same molecular structure from each resonance structure? b. What is the hybridization of \(P\) in each structure? c. What orbitals can the \(P\) atom use to form the \(\pi\) bond in structure B? d. Which resonance structure would be favored on the basis of formal charges?

Explain the difference between the \(\sigma\) and \(\pi\) MOs for homonuclear diatomic molecules. How are bonding and antibonding orbitals different? Why are there two \(\pi\) MOs and one \(\sigma\) MO? Why are the \(\pi\) MOs degenerate?

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?

Predict the molecular structure, bond angles, and polarity (has a net dipole moment or has no net dipole moment) for each of the following compounds. a. \(\mathrm{SeCl}_{4}\) b. \(\mathrm{SF}_{2}\) c. \(\mathrm{KrF}_{4}\) d. \(C B r_{4}\) e. \(\mathrm{IF}_{3}\) f. \(\mathrm{ClF}_{5}\)

Which of the following statements concerning \(\mathrm{SO}_{2}\) is(are) true? a. The central sulfur atom is \(s p^{2}\) hybridized. b. One of the sulfur-oxygen bonds is longer than the other(s). c. The bond angles about the central sulfur atom are about 120 degrees. d. There are two \(\sigma\) bonds in \(\mathrm{SO}_{2}\) e. There are no resonance structures for \(\mathrm{SO}_{2}\)
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