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Chapter 9: Problem 11

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 \((3 d, 4 d, 5 d\), 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 used in hybridization to satisfy the VSEPR theory, which minimizes electron pair repulsion around a central atom. Elements in the second period (n=2) do not use d orbitals for hybridization. Sulfur, being in the third period (n=3), uses its 3d orbitals for hybridization. Arsenic, in the fourth period (n=4), uses its 4d orbitals. Iodine, in the fifth period (n=5), uses its 5d orbitals for hybridization.

Step by step solution

01

1. Why d orbitals are used in hybridization

: d orbitals are sometimes used to form hybrid orbitals to satisfy the valence shell electron pair repulsion (VSEPR) theory, which states that electron pairs around a central atom will arrange themselves in space to minimize repulsion. In some molecules or ions, the use of d orbitals in hybridization can provide the correct geometry and bonding according to this theory.
02

2. Period of elements that do not use d orbitals

: Elements in the second period (n=2) do not use d orbitals for hybridization since they don't possess d orbitals in their valence shell. Elements in this period include lithium (Li), beryllium (Be), boron (B), carbon (C), nitrogen (N), oxygen (O), and fluorine (F).
03

3. d orbitals used by sulfur to form hybrid orbitals

: Sulfur is in the third period (n=3) and has the electron configuration: \[1s^2 2s^2 2p^6 3s^2 3p^4\] When sulfur forms hybrid orbitals requiring d atomic orbitals, it uses the 3d orbitals since they are in the same principal quantum level (n=3) as its valence shell.
04

4. d orbitals used by arsenic to form hybrid orbitals

: Arsenic is in the fourth period (n=4) and has the electron configuration: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^3\] When arsenic forms hybrid orbitals requiring d atomic orbitals, it uses the 4d orbitals since they are in the same principal quantum level (n=4) as its valence shell.
05

5. d orbitals used by iodine to form hybrid orbitals

: Iodine is in the fifth period (n=5) and has the electron configuration: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^5\] When iodine forms hybrid orbitals requiring d atomic orbitals, it uses the 5d orbitals since they are in the same principal quantum level (n=5) as its valence shell.

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Most popular questions from this chapter

Cyanamide \(\left(\mathrm{H}_{2} \mathrm{NCN}\right)\), an important industrial chemical, is produced by the following steps: $$ \begin{aligned} \mathrm{CaC}_{2}+\mathrm{N}_{2} & \longrightarrow \mathrm{CaNCN}+\mathrm{C} \\\ \mathrm{CaNCN} & \stackrel{\text { Acid }}{\longrightarrow} \mathrm{H}_{2} \mathrm{NCN} \end{aligned} $$ Cyanamid 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, \(\mathrm{NCNC}\left(\mathrm{NH}_{2}\right)_{2}\), and the molecule is nonlinear. Of all the resonance structures you drew for this molecule, predict which should be the most important.

Why does the molecular orbital model do a better job in explaining the bonding in \(\mathrm{NO}^{-}\) and NO than the hybrid orbital model?

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