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Chapter 10: Problem 16

The bonds in beryllium hydride \(\left(\mathrm{BeH}_{2}\right)\) molecules are polar, and yet the dipole moment of the molecule is zero. Explain.

Short Answer

Expert verified
Despite the bonds in beryllium hydride (BeH2) being polar due to the difference in electronegativity between beryllium and hydrogen, the molecule has a linear shape. Therefore, the dipole moments of the two bonds are equal and opposite, cancelling each other out and resulting in an overall dipole moment of zero.

Step by step solution

01

Understanding molecular polarity

Polarity in a molecule occurs due to the difference in electronegativity between atoms. In beryllium hydride, the difference in electronegativity between beryllium and hydrogen leads to the bonds being polar.
02

Understanding dipole moment

Dipole moment is the measure of the polarity of a molecule. It is calculated as the product of the charge and the distance between the charges. A dipole moment points from the positive to the negative charge with a magnitude proportional to the difference in charge and the distance between the charges.
03

Analyzing the dipole moment of BeH2

Beryllium hydride has a linear shape due to the electron configuration of beryllium. Therefore, the dipole moments of the Beryllium-Hydrogen bonds, which are polar, point in opposite directions. Because they are equal in magnitude, they cancel each other out. Hence the overall dipole moment of the molecule is zero, even though it contains polar bonds.

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Most popular questions from this chapter

Assume that the third-period element phosphorus forms a diatomic molecule, \(\mathrm{P}_{2}\), in an analogous way as nitrogen does to form \(\mathrm{N}_{2}\). (a) Write the electronic configuration for \(\mathrm{P}_{2}\). Use \(\left[\mathrm{Ne}_{2}\right]\) to represent the electron configuration for the first two periods. (b) Calculate its bond order. (c) What are its magnetic properties (diamagnetic or paramagnetic)?

Specify which hybrid orbitals are used by carbon atoms in these species: (a) \(\mathrm{CO},\) (b) \(\mathrm{CO}_{2},\) (c) \(\mathrm{CN}^{-}\).

What is the hybridization state of \(\mathrm{Si}\) in \(\mathrm{SiH}_{4}\) and in \(\mathrm{H}_{3} \mathrm{Si}-\mathrm{SiH}_{3} ?\)

Briefly compare the VSEPR and hybridization approaches to the study of molecular geometry.

Consider a \(\mathrm{N}_{2}\) molecule in its first excited electronic state; that is, when an electron in the highest occupied molecular orbital is promoted to the lowest empty molecular obital. (a) Identify the molecular orbitals involved and sketch a diagram to show the transition. (b) Compare the bond order and bond length of \(\mathrm{N}_{2}{ }^{*}\) with \(\mathrm{N}_{2}\), where the asterisk denotes the excited molecule. (c) Is \(\mathrm{N}_{2}{ }^{*}\) diamagnetic or paramagnetic? (d) When \(\mathrm{N}_{2} *\) loses its excess energy and converts to the ground state \(\mathrm{N}_{2}\), it emits a photon of wavelength \(470 \mathrm{nm}\), which makes up part of the auroras lights. Calculate the energy difference between these levels.
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