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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

How would you distinguish between a sigma bond and a pi bond?

The formation of \(\mathrm{H}^{+}\) from two \(\mathrm{H}\) atoms is an energetically favorable process. Yet statistically there is less than a 100 percent chance that any two \(\mathrm{H}\) atoms will undergo the reaction. Apart from energy considerations, how would you account for this observation based on the electron spins in the two \(\mathrm{H}\) atoms?

The ionic character of the bond in a diatomic molecule can be estimated by the formula $$ \frac{\mu}{e d} \times 100 \% $$ where \(\mu\) is the experimentally measured dipole moment (in \(\mathrm{C} \mathrm{m}\) ), \(e\) is the electronic charge \((1.6022 \times\) \(10^{-19} \mathrm{C}\) ), and \(d\) is the bond length in meters. (The quantity \(e d\) is the hypothetical dipole moment for the case in which the transfer of an electron from the less electronegative to the more electronegative atom is complete.) Given that the dipole moment and bond length of \(\mathrm{HF}\) are \(1.92 \mathrm{D}\) and \(91.7 \mathrm{pm},\) respectively, calculate the percent ionic character of the molecule.

Draw the Lewis structure of ketene \(\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{O}\right)\) and describe the hybridization states of the \(\mathrm{C}\) atoms. The molecule does not contain \(\mathrm{O}-\mathrm{H}\) bonds. On separate diagrams, sketch the formation of sigma and pi bonds.

Which of these pairs of atomic orbitals of adjacent nuclei can overlap to form a sigma bond? Which overlap to form a pi bond? Which cannot overlap (no bond)? Consider the \(x\) -axis to be the internuclear axis, that is, the line joining the nuclei of the two atoms. (a) \(1 s\) and \(1 s,\) (b) \(1 s\) and \(2 p_{x},\) (c) \(2 p_{x}\) and \(2 p_{y}\) (d) \(3 p_{y}\) and \(3 p_{y},\) (e) \(2 p_{x}\) and \(2 p_{x}\), (f) 1 s and 2 s.
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