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Chapter 10: Problem 91

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.

Short Answer

Expert verified
The Lewis structure of ketene is H2C=C=O. Carbon 1 is 'sp2' hybridized, and Carbon 2 is 'sp' hybridized. There are five '蟽' bonds formed and two '蟺' bonds formed in ketene.

Step by step solution

01

Draw the Lewis structure of Ketene (C2H2O)

The following relates to drawing Lewis structure: (1) Start by counting the valence electrons on each atom. Oxygen (O) has 6 valence electrons, and each Carbon (C) atom has 4 valence electrons and Hydrogen (H) has 1, so the total number of valence electrons is 10. (2) Ketene has a linear structure with the hydrogen atoms bonded to the first carbon atom and the oxygen atom bonded to the second carbon atom. (3) Start by joining the atoms with single bonds (each consuming 2 electrons) and then use the remaining electrons to fill up the octets (8 electrons structure) of each atom starting from the ones on the outskirts towards the central atom. (4) However, sometimes, like in this case, a double bond might be necessary to ensure every atom other than Hydrogen has 8 electrons in its outer shell. The resulting Lewis structure is H2C=C=O.
02

Determine the Hybridization of the Carbon Atoms

The hybridization depends on how many '蟽' bonds and 'Lone pairs' are around the atom. The first Carbon (from the Left, let鈥檚 call this Carbon 1) is connected to two Hydrogens through '蟽' bonds and to the other Carbon through a '蟽' bond and a '蟺' bond. Any Double bond consists of a Sigma (蟽) bond and a Pi (蟺) bond. So, Carbon 1 has three '蟽' bonds and no lone pair, which means it is 'sp2' hybridized. The second Carbon (Carbon 2) is bonded to Carbon 1 and the Oxygen atom through a '蟽' bond and a '蟺' bond. It has a total of two '蟽' bonds, implying it is 'sp' hybridized.
03

Sketch the Sigma and Pi bond formation

In Sigma (蟽) bond formation: (1) The 's' orbital of the Hydrogen combines with the 'sp2' and 'sp' hybrid orbitals of the Carbon 1 and Carbon 2 respectively to form three '蟽' bonds. (2) Also, the 'sp2' orbital of Carbon 1 combines with an 'sp' orbital of Carbon 2 and 'sp' orbital of Carbon 2 combines with a 'p' orbital of Oxygen to form two more '蟽' bonds. Thus, we have a total of five '蟽' bonds. For the Pi (蟺) bond formation: (1) The remaining 'p' orbitals of Carbon 1 and Carbon 2 overlap sideways to form one '蟺' bond. (2) The remaining 'p' orbital of Carbon 2 also overlaps sideways with a 'p' orbital of Oxygen to form the second '蟺' bond. Thus, we have two '蟺' bonds in Ketene.

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Most popular questions from this chapter

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 mercury(II) bromide. Is this molecule linear or bent? How would you establish its geometry?

Compare the relative stability of these species and indicate their magnetic properties (that is, diamagnetic or paramagnetic): \(\mathrm{O}_{2}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2}^{-}\) (superoxide ion), \(\mathrm{O}_{2}^{2-}\) (peroxide ion).

What are the hybrid orbitals of the carbon atoms in these molecules? (a) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{3}\) (b) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2}\) (c) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{O}\) (e) \(\mathrm{CH}_{3} \mathrm{COOH}\).

What is the hybridization of atomic orbitals? Why is it impossible for an isolated atom to exist in the hybridized state?
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