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(a) Which geometry and central atom hybridization would you expect in the series \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}^{+}\)? (b) What would you expect for the magnitude and direction of the bond dipoles in this series? (c) Write the formulas for the analogous species of the elements of period 3 ; would you expect them to have the same hybridization at the central atom?

Short Answer

Expert verified
(a) All three molecules, BH₄⁻, CH₄, and NH₄⁺, have a tetrahedral geometry and sp³ hybridization at the central atom. (b) The bond dipoles in all three molecules have small electronegativity differences and are symmetrical, resulting in a net dipole moment of zero. (c) The analogous species for Period 3 elements are AlH₄⁻, SiH₄, and PH₄⁺. They all have the same tetrahedral molecular geometry and sp³ hybridization at the central atom.

Step by step solution

01

Geometry and hybridization of BH₄⁻

To determine the molecular geometry of BH₄⁻, first count the number of electron pairs around the central B atom (1 bond pair for each hydrogen and 1 lone pair). Since there are 4 electron pairs, BH₄⁻ has a tetrahedral geometry. The hybridization of the B atom is therefore sp³.
02

Geometry and hybridization of CH₄

Similar to BH₄⁻, CH₄ has 4 bond pairs around the central C atom, so it also has a tetrahedral geometry and sp³ hybridization.
03

Geometry and hybridization of NH₄⁺

NH₄⁺ has four bond pairs around the central N atom, giving it a tetrahedral geometry and sp³ hybridization like the other two molecules. 2. Bond dipoles:
04

Bonds in BH₄⁻

The electronegativity difference between B (2.04) and H (2.20) is small, and since all bonds are symmetrical around the central atom, the net dipole moment of BH₄⁻ is zero.
05

Bonds in CH₄

The electronegativity difference between C (2.55) and H (2.20) is also small, and the bonds are symmetric, so the net dipole moment of CH₄ is zero as well.
06

Bonds in NH₄⁺

Like the other two molecules, the electronegativity difference between N (3.04) and H (2.20) is small and the bonds are symmetrical, so the net dipole moment of NH₄⁺ is zero. 3. Analogy to Period 3 elements:
07

Analogous species in period 3

The analogous species for the elements of Period 3 would be: AlH₄⁻ (Aluminum Hydride ion), SiH₄ (Silicon Hydride), and PH₄⁺ (Phosphonium ion).
08

Hybridization of analogous species

AlH₄⁻ is similar to BH₄⁻, with the Al atom in the center bonding to four H atoms. It has a tetrahedral geometry and sp³ hybridization. SiH₄ is like CH₄, also with a tetrahedral geometry and sp³ hybridization. PH₄⁺, like NH₄⁺, has a tetrahedral geometry and sp³ hybridization. All of these Period 3 elements have analogous species with the same tetrahedral molecular geometry and sp³ hybridization at the central atom.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tetrahedral Geometry
Tetrahedral geometry is a common molecular shape that results when a central atom forms four bonds directed towards the corners of a geometric tetrahedron. Picture this shape like a three-dimensional pyramid with a triangular base, where each of the three-dimensional lines sticking out from the center represents a bond.
  • This structure results when there are four areas of electron density around the central atom.
  • The angles between each pair of bonds in a perfect tetrahedral geometry are about 109.5 degrees.
  • It's commonly observed in molecules like \( \text{CH}_4 \) (methane), where the central carbon atom is bonded to four hydrogen atoms symmetrically.
In the exercise provided, \( \text{BH}_4^- \), \( \text{CH}_4 \), and \( \text{NH}_4^+ \) all exhibit this tetrahedral geometry. This is due to each central atom having four bonds organized symmetrically in space.
Hybridization
Hybridization is a concept in chemistry used to explain the formation and arrangement of bonds in molecules. Through hybridization, atomic orbitals on the central atom mix to form new hybrid orbitals.
  • For tetrahedral geometry, the hybridization state is \( ext{sp}^3 \). This denotes the mixing of one s and three p orbitals.
  • This hybridization results in four equivalent hybrid orbitals, each participating in the formation of a sigma bond with hydrogen atoms, as seen in \( \text{CH}_4 \), \( \text{BH}_4^- \), and \( \text{NH}_4^+ \).
  • The concept helps explain the observed bond angles and molecular shapes.
In our series, all molecules share this \( ext{sp}^3 \) hybridization at their central atom, resulting in uniform tetrahedral geometries.
Bond Dipoles
Bond dipoles occur when there is an electronegativity difference between two atoms bonded together. Electronegativity is a measure of an atom's ability to attract shared electrons. In molecules like \( \text{CH}_4 \) and \( \text{NH}_4^+ \), the electronegativity differences are minimal:
  • For \( \text{CH}_4 \), carbon's electronegativity is 2.55, and hydrogen's is 2.20.
  • This small difference means each C-H bond is slightly polar.
  • However, the tetrahedral symmetry ensures that these bond dipoles cancel out, leading to a net dipole moment of zero.
  • Similarly, \( \text{NH}_4^+ \) and \( \text{BH}_4^- \) have their individual bond dipoles cancel out due to their symmetrical bond arrangements.
Understanding bond dipoles is crucial in predicting whether a molecule is polar or nonpolar.
Electronegativity
Electronegativity is a fundamental chemical property that describes an atom's ability to attract and bind with electrons. It plays a crucial role in determining a molecule's polarity.
  • A higher electronegativity means stronger pull on electron pairs, leading potentially to more polar bonds.
  • When examining molecules like \( \text{NH}_4^+ \), \( \text{CH}_4 \), and \( \text{BH}_4^- \), slight differences in electronegativity between the central atoms and hydrogen contribute to small bond dipoles.
  • However, the tetrahedral geometry minimizes the overall polarity as bond dipoles cancel out, leading to non-polar molecules.
In this way, electronegativity allows us to predict and explain molecular behavior and interaction.
Period 3 Elements
Period 3 elements in the periodic table include elements from sodium (Na) to argon (Ar). When considering analogs such as \( \text{AlH}_4^- \), \( \text{SiH}_4 \), and \( \text{PH}_4^+ \), these molecules can also adopt tetrahedral geometry and \( ext{sp}^3 \) hybridization:
  • \( \text{SiH}_4 \) (silicon hydride) mirrors \( \text{CH}_4 \) in structure and hybridization.
  • \( \text{AlH}_4^- \) functions similarly to \( \text{BH}_4^- \) with aluminum at the center.
  • \( \text{PH}_4^+ \) bears resemblance to \( \text{NH}_4^+ \), showcasing a similar arrangement.
This demonstrates that period 3 elements can form molecules with geometries and properties akin to their period 2 counterparts, exhibiting analogous chemical behavior. Understanding this helps in predicting and drawing parallels between chemistry across different elemental periods.

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Most popular questions from this chapter

(a) What does the term paramagnetism mean? (b) How can one determine experimentally whether a substance is paramagnetic? (c) Which of the following ions would you expect to be paramagnetic: \(\mathrm{O}_{2}^{+}, \mathrm{N}_{2}{ }^{2-}, \mathrm{Li}_{2}^{+}, \mathrm{O}_{2}{ }^{2-}\) ? For those ions that are paramagnetic, determine the number of unpaired electrons.

There are two compounds of the formula \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\) : The compound on the right, cisplatin, is used in cancer therapy. The compound on the left, transplatin, is ineffective for cancer therapy. Both compounds have a square-planar geometry. (a) Which compound has a nonzero dipole moment? (b) The reason cisplatin is a good anticancer drug is that it binds tightly to DNA. Cancer cells are rapidly dividing, producing a lot of DNA. Consequently, cisplatin kills cancer cells at a faster rate than normal cells. However, since normal cells also are making DNA, cisplatin also attacks healthy cells, which leads to unpleasant side effects. The way both molecules bind to DNA involves the \(\mathrm{Cl}^{-}\)ions leaving the Pt ion, to be replaced by two nitrogens in DNA. Draw a picture in which a long vertical line represents a piece of DNA. Draw the \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}\) fragments of cisplatin and transplatin with the proper shape. Also draw them attaching to your DNA line. Can you explain from your drawing why the shape of the cisplatin causes it to bind to DNA more effectively than transplatin?

In ozone, \(\mathrm{O}_{3}\), the two oxygen atoms on the ends of the molecule are equivalent to one another. (a) What is the best choice of hybridization scheme for the atoms of ozone? (b) For one of the resonance forms of ozone, which of the orbitals are used to make bonds and which are used to hold nonbonding pairs of electrons? (c) Which of the orbitals can be used to delocalize the \(\pi\) electrons? (d) How many electrons are delocalized in the \(\pi\) system of ozone?

Explain the following: (a) The peroxide ion, \(\mathrm{O}_{2}{ }^{2-}\), has a longer bond length than the superoxide ion, \(\mathrm{O}_{2}^{-}\). (b) The magnetic properties of \(\mathrm{B}_{2}\) are consistent with the \(\pi_{2 p}\) MOs being lower in energy than the \(\sigma_{2 p} \mathrm{MO}\). (c) The \(\mathrm{O}_{2}{ }^{2+}\) ion has a stronger \(\mathrm{O}-\mathrm{O}\) bond than \(\mathrm{O}_{2}\) itself.

Place the following molecules and ions in order from smallest to largest bond order: \(\mathrm{H}_{2}^{+}, \mathrm{B}_{2}, \mathrm{~N}_{2}^{+}, \mathrm{F}_{2}^{+}\), and \(\mathrm{Ne}_{2}\).

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