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Chapter 10: Problem 14

Which of the following species are tetrahedral? \(\mathrm{SiCl}_{4}, \mathrm{SeF}_{4}, \mathrm{XeF}_{4}, \mathrm{Cl}_{4}, \mathrm{CdCl}_{4}^{2-}\)

Short Answer

Expert verified
The molecules SiCl4 and CdCl42- are tetrahedral in shape according to VSEPR theory.

Step by step solution

01

- Calculation for SiCl4

Silicon (Si) is in Group 4A, so it has 4 valence electrons. Chlorine (Cl) has 7 valence electrons. Since 4 chlorine atoms are present, the total number of valence electrons is \(4 + 4*7 = 32\). Silicon forms 4 bonds with 4 chlorine atoms using these electrons and there's no lone pair on the central atom, so SiCl4 is tetrahedral.
02

- Calculation for SeF4

Selenium (Se) is in Group 6A, so it has 6 valence electrons. Fluorine (F) has 7 valence electrons. Since 4 fluorine atoms are present, the total number of valence electrons is \(6 + 4*7 = 34\). With these electrons, selenium forms 4 bonds with fluorine atoms and also has one lone pair on it, so SeF4 is not tetrahedral.
03

- Calculation for XeF4

Xenon (Xe) is in Group 8A, so it has 8 valence electrons. Each fluorine (F) atom has 7 valence electrons. Since 4 fluorine atoms are present, the total number of valence electrons is \(8 + 4*7 = 36\). Xenon forms 4 bonds with fluorine atoms but also holds 2 lone pairs, so XeF4 isn't tetrahedral.
04

- Calculation for Cl4

In this case, since every Chlorine (Cl) atom has 7 valence electrons. However, chlorine can't act as the central atom forming molecule Cl4. Therefore, the molecule Cl4 doesn't exist.
05

- Calculation for CdCl42-

The cadmium ion (Cd2+) can form a stable complex ion with 4 chlorine ions (Cl-). In this complex, Cd2+ serves as the central ion surrounded by four Cl- ions, so CdCl42- is tetrahedral.

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Most popular questions from this chapter

Sketch the bond moments and resultant dipole moments for the following molecules: \(\mathrm{H}_{2} \mathrm{O}, \mathrm{PCl}_{3},\) \(\mathrm{XeF}_{4}, \mathrm{PCl}_{5}, \mathrm{SF}_{6}\)

How many atoms are directly bonded to the central atom in a tetrahedral molecule, a trigonal bipyramidal molecule, and an octahedral molecule?

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) has a tendency to lose two protons \(\left(\mathrm{H}^{+}\right)\) and form the carbide ion \(\left(\mathrm{C}_{2}^{2-}\right),\) which is present in a number of ionic compounds, such as \(\mathrm{CaC}_{2}\) and \(\mathrm{MgC}_{2}\). Describe the bonding scheme in the \(\mathrm{C}_{2}^{2-}\) ion in terms of molecular orbital theory. Compare the bond order in \(\mathrm{C}_{2}^{2-}\) with that in \(\mathrm{C}_{2}\).

What is the hybridization state of the central \(\mathrm{N}\) atom in the azide ion, \(\mathrm{N}_{3}^{-} ?\) (Arrangement of atoms: NNN.)

Describe the bonding in the nitrate ion \(\mathrm{NO}_{3}^{-}\) in terms of delocalized molecular orbitals.
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