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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

Show how a \(d_{x z}\) atomic orbital and a \(p_{z}\) atomic orbital combine to form a bonding molecular orbital. Assume the \(x\) -axis is the internuclear axis. Is a \(\sigma\) or a \(\pi\) molecular orbital formed? Explain.

Draw the Lewis structure for HCN. Indicate the hybrid orbitals, and draw a picture showing all the bonds between the atoms, labeling each bond as \(\sigma\) or \(\pi\).

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 \(1 s^{2} 2 s^{2} 2 p^{6}\).

Which of the following would you expect to be more favorable energetically? Explain. a. an \(\mathrm{H}_{2}\) molecule in which enough energy is added to excite one electron from the bonding to the antibonding \(\mathrm{MO}\) b. two separate \(\mathrm{H}\) atoms

In Exercise 89 in Chapter 8, the Lewis structures for benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) were drawn. Using one of the Lewis structures, estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) using bond energies and given that the standard enthalpy of formation of \(\mathrm{C}(g)\) is \(717 \mathrm{~kJ} / \mathrm{mol}\). The experimental \(\Delta H_{\mathrm{f}}^{\circ}\) value of \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) is \(83 \mathrm{~kJ} / \mathrm{mol} .\) Explain the discrepancy between the experimental value and the calculated \(\Delta H_{\mathrm{f}}^{\circ}\) value for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\)
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