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Chapter 9: Problem 29

(a) Explain why \(\mathrm{BrF}_{4}^{-}\) is square planar, whereas \(\mathrm{BF}_{4}^{-}\) is tetrahedral. (b) Water, \(\mathrm{H}_{2} \mathrm{O}\), is a bent molecule. Predict the shape of the molecular ion formed from the water molecule if you were able to remove four electrons to make \(\left(\mathrm{H}_{2} \mathrm{O}\right)^{4+}\).

Short Answer

Expert verified
The molecular geometry of BrF4- is square planar because it has 4 bond pairs and 2 lone pairs around the central Br atom, which arrange themselves in an octahedral electronic geometry. The molecular geometry of BF4- is tetrahedral due to the presence of 4 bond pairs and no lone pairs around the central B atom. For the (H2O)4+ ion, after losing 4 electrons, it has 2 bond pairs and no lone pairs, so its molecular geometry would be linear.

Step by step solution

01

Determine the electronic geometry for BrF4- and BF4- ions

First, we need to determine the electronic geometry of BrF4- and BF4- ions. Electronic geometry is defined by the arrangement of electron pairs (lone pairs and bonded pairs) around the central atom. For BrF4-, the central atom is Br with 7 valence electrons. Since it gains one electron, it has 8 valence electrons in total. The ion forms 4 bonds with F atoms which results in 4 single bonds. As there are 8-4=4 electrons left, there are 2 lone pairs on the Br atom. For BF4-, the central atom is B with 3 valence electrons. It forms 4 single bonds with F atoms resulting in a total of 4 bond pairs. There are no lone pairs.
02

Determine the molecular geometry of BrF4- and BF4- ions

Now, we will use the VSEPR theory to determine the molecular geometry of the ions. For BrF4-, there are 4 bond pairs and 2 lone pairs around the central Br atom. These six electron pairs arrange themselves in the most stable geometry: an octahedral arrangement. However, we only account for the positions of bond pairs to define the molecular geometry. Since the bond pairs form a square in the middle of the octahedron, the molecular geometry of BrF4- is square planar. For BF4-, there are 4 bond pairs and no lone pairs around the central B atom. These four electron pairs arrange themselves in a tetrahedral geometry, so the molecular geometry of BF4- is also tetrahedral.
03

Determine the electronic geometry for H2O

Now, we will determine the electronic geometry for H2O molecule. The central atom in H2O is O, which has 6 valence electrons. Oxygen forms 2 single bonds with hydrogen atoms, leaving 4 electrons as 2 lone pairs. There are 2 bond pairs and 2 lone pairs around the oxygen atom.
04

Determine the molecular geometry for (H2O)4+

We are asked to predict the shape of the H2O molecule when it loses 4 electrons. As H2O originally has 2 lone pairs and 2 bond pairs, losing 4 electrons would remove both lone pairs, leaving only 2 bond pairs. With 2 bond pairs and no lone pairs, the electron pairs arrange themselves in a linear geometry. Therefore, the shape of (H2O)4+ would be linear.

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Most popular questions from this chapter

Antibonding molecular orbitals can be used to make bonds to other atoms in a molecule. For example, metal atoms can use appropriate \(d\) orbitals to overlap with the \(\pi_{2 p}^{*}\) orbitals of the carbon monoxide molecule. This is called \(d-\pi\) backbonding. (a) Draw a coordinate axis system in which the \(y\) -axis is vertical in the plane of the paper and the \(x\) -axis horizontal. Write \(^{\prime} \mathrm{M}^{\prime \prime}\) at the origin to denote a metal atom. (b) Now, on the \(x\) -axis to the right of \(M\), draw the Lewis structure of a CO molecule, with the carbon nearest the \(\mathrm{M}\). The \(\mathrm{CO}\) bond axis should be on the \(x\) -axis. (c) Draw the CO \(\pi_{2 p}^{*}\) orbital, with phases (see the Closer Look box on phases) in the plane of the paper. Two lobes should be pointing toward \(\mathrm{M}\). (d) Now draw the \(\mathrm{d}_{x y}\) orbital of \(\mathrm{M}\), with phases. Can you see how they will overlap with the \(\pi_{2 p}^{*}\) orbital of CO? (e) What kind of bond is being made with the orbitals between \(\mathrm{M}\) and \(\mathrm{C}, \sigma\) or \(\pi ?\) (f) Predict what will happen to the strength of the CO bond in a metal-CO complex compared to CO alone.

How many nonbonding electron pairs are there in each of the following molecules: (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S} ;\) (b) \(\mathrm{HCN}\); (c) \(\mathrm{H}_{2} \mathrm{C}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{~F}\) ?

Draw a picture that shows all three \(2 p\) orbitals on one atom and all three \(2 p\) orbitals on another atom. (a) Imagine the atoms coming close together to bond. How many \(\sigma\) bonds can the two sets of \(2 p\) orbitals make with each other? (b) How many \(\pi\) bonds can the two sets of \(2 p\) orbitals make with each other? (c) How many antibonding orbitals, and of what type, can be made from the two sets of \(2 p\) orbitals?

From their Lewis structures, determine the number of \(\sigma\) and \(\pi\) bonds in each of the following molecules or ions: (a) \(\mathrm{CO}_{2} ;\) (b) thiocyanate ion, \(\mathrm{NCS}^{-}\) : (c) formaldehyde, \(\mathrm{H}_{2} \mathrm{CO} ;\) (d) formic acid, \(\mathrm{HCOOH}\), which has one \(\mathrm{H}\) and two \(\mathrm{O}\) atoms attached to \(\mathrm{C}\).

Consider the bonding in an \(\mathrm{MgH}_{2}\) molecule. (a) Draw a Lewis structure for the molecule, and predict its molecular geometry. (b) What hybridization scheme is used in \(\mathrm{MgH}_{2} ?\) (c) Sketch one of the two-electron bonds between an \(\mathrm{Mg}\) hybrid orbital and an \(\mathrm{H} 1 \mathrm{~s}\) atomic orbital.
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