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Chapter 11: Problem 95

The hybridization scheme for Xe in \(\mathrm{XeF}_{2}\) is (a) \(s p\) (b) \(s p^{3} ;\) (c) \(s p^{3} d ;\) (d) \(s p^{3} d^{2}\)

Short Answer

Expert verified
The hybridization scheme for \(Xe\) in \(\mathrm{XeF}_{2}\) is \(sp^{3}d\).

Step by step solution

01

Determine number of Valence Electrons

Determine how many valence electrons \(Xe\) has. Xenon (Xe) is a noble gas and is in the 18th group of the periodic table. Therefore it has 8 valence electrons.
02

Determine number of Bonding and Non-Bonding Electrons

Next, draw the Lewis structure for \(XeF_{2}\) to determine the number of bonding and non-bonding electrons (also known as lone pairs). \(XeF_{2}\) is linear with 2 bonding pairs of electrons shared with the Fluorines and 3 lone pairs of electrons.
03

Calculate Hybridization Scheme

Add up the number of bonding pairs and lone pairs. In this case, we get a total of 2+3=5. The hybridization scheme must provide enough orbitals to accommodate all of these electron pairs. The number 5 indicates the involvement of five orbitals (one s, three p and one d) in hybridization. Hence the hybridization scheme for \(Xe\) in \(XeF_{2}\) is \(sp^{3}d\).

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Most popular questions from this chapter

Which of these diatomic molecules do you think has the greater bond energy, \(\mathrm{Li}_{2}\) or \(\mathrm{C}_{2} ?\) Explain.

A molecule in which \(s p^{2}\) hybrid orbitals are used by the central atom in forming covalent bonds is (a) \(\mathrm{PCl}_{5}\) (b) \(\mathrm{N}_{2} ;\) (c) \(\mathrm{SO}_{2} ;\) (d) \(\mathrm{He}_{2}\)

Which of the following substances, when added in trace amounts to germanium, would produce an n-type semiconductor: (a) sulfur, (b) aluminum, (c) tin, (d) cadmium sulfide, (e) arsenic, (f) gallium arsenide? Explain.

From this list of terms-electrical conductor, insulator, semiconductor- -choose the one that best characterizes each of the following materials: (a) stainless steel; (b) solid sodium chloride; (c) sulfur; (d) germanium; (e) seawater; (f) solid iodine.

Construct the molecular orbital diagram for CF. Would you expect the bond length of \(\mathrm{CF}^{+}\) to be longer or shorter than that of CF?
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