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Chapter 3: Problem 51

Develop a bonding scheme for the compound \(\mathrm{H}_{2} \mathrm{CO}\) in which both carbon and oxygen are hybridized \(s p^{2}\).

Short Answer

Expert verified
The bonding scheme for \(\text{H}_{2}\text{CO}\) has carbon forming σ bonds with 2 H and 1 O, and a π bond between C and O.

Step by step solution

01

Identify the Atoms and Their Valence Electrons

The compound \(\text{H}_{2}\text{CO}\) consists of 2 hydrogen atoms (H), one carbon atom (C), and one oxygen atom (O). The valence electrons are: H: 1, C: 4, O: 6.
02

Sum the Valence Electrons

The total number of valence electrons is determined by adding the valence electrons of each atom: \[2(1) + 4 + 6 = 12 \text{valence electrons} \]
03

Determine Hybridization of C and O

Both the carbon and oxygen atoms are hybridized as \(sp^2\). In \(sp^2\) hybridization, one s orbital mixes with two p orbitals, resulting in three \(sp^2\) hybrid orbitals.
04

Form Sigma Bonds

The carbon atom forms three sigma (σ) bonds: one with each hydrogen atom and one with the oxygen atom using the \(sp^2\) hybrid orbitals.
05

Form Pi Bond

The unhybridized p orbitals on both carbon and oxygen overlap side-by-side to form a pi (π) bond between them.
06

Lone pairs on Oxygen

Oxygen has one \(sp^2\) hybrid orbital left with a lone pair of electrons. Additionally, it has a lone pair in its unhybridized p orbital.
07

Draw the Bonding Scheme

The bonding scheme includes: \[ \text{H} - \text{C} (\text{sp}^2-\text{sp}^2 \ σ \text{bond}) - \text{O} (\text{sp}^2 \ σ \text{bond} and a \text{p}-\text{p} \ π \text{bond}) - \text{H} \]

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Key Concepts

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

Valence Electrons
In chemistry, valence electrons play a crucial role in the formation of bonds between atoms. These are the electrons found in the outermost shell of an atom and they participate in chemical bonding.
In the molecule \(\text{H}_{2}\text{CO}\), the valence electrons for each atom are as follows:
  • Hydrogen (H): 1 electron
  • Carbon (C): 4 electrons
  • Oxygen (O): 6 electrons
By summing these up, we get a total of \[2(1) + 4 + 6 = 12 \text{ valence electrons\] for the entire molecule. This total count is important because it helps us understand how electrons are shared or transferred to create chemical bonds.
Sigma Bond
A sigma (\[\sigma\] bond is the strongest type of covalent chemical bond. It is formed by the direct overlap of atomic orbitals. In the \(\text{H}_{2}\text{CO}\) molecule, the carbon atom forms three sigma bonds. Here's how it works:
  • The carbon uses \[sp^{2}\] hybrid orbitals to bond with two hydrogen atoms and one oxygen atom.
  • Each of these sigma bonds is stable due to the head-on overlapping of the \[sp^{2}\] orbitals.
This stability is essential for the structural integrity of the molecule. Sigma bonds provide a strong foundation for molecule formation due to their effective overlapping.
Pi Bond
Pi (\[\pi\] bonds arise from the side-by-side overlap of unhybridized p orbitals. Unlike sigma bonds, pi bonds are generally weaker since their overlap is less direct. In the \[H_2CO\] molecule:
  • Both the carbon and oxygen have unhybridized p orbitals.
  • These p orbitals overlap to form a pi bond.
The presence of this pi bond creates a double bond between carbon and oxygen, combining with the sigma bond to form a stronger, more rigid structure. This combination of sigma and pi bonds affects the molecule's shape and reactivity.
Hybrid Orbitals
Hybrid orbitals are a blend of standard atomic orbitals and are formed to explain the geometry and bonding properties of molecules. In \[sp^2\] hybridization:
  • One s orbital mixes with two p orbitals.
  • This results in three \[sp^2\] hybrid orbitals.
In the \[H_2CO\] molecule, both carbon and oxygen undergo \[sp^2\] hybridization. This means that both atoms have three \[sp^2\] hybrid orbitals, which are used as follows:
  • Carbon forms three sigma bonds (two with hydrogen, one with oxygen) using \[sp^2\] orbitals.
  • Oxygen uses one of its \[sp^2\] orbitals to form a sigma bond with carbon.
The remaining \[sp^2\] orbitals on oxygen hold lone pairs, and an unhybridized p orbital on each forms the pi bond. Hybrid orbitals help explain the observed geometry and bonding patterns in complex molecules.

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Most popular questions from this chapter

Draw 2-methyl-2-butene, 1-pentene, \((E)-2\)-pentene, and (Z)2-pentene. Put them in order of least stable to most stable.

A compound containing only carbon, hydrogen, and oxygen was found to contain \(70.58 \%\) carbon and \(5.92 \%\) hydrogen by weight. Calculate the empirical formula for this compound. If the compound has a molecular weight of approximately \(135 \mathrm{~g} / \mathrm{mol}\), what is the molecular formula?

Write the six isomers of \(\mathrm{C}_{5} \mathrm{H}_{10}\) that are alkenes (the pentenes). There are 17 isomers of \(\mathrm{C}_{6} \mathrm{H}_{12}\) that are alkenes (the hexenes). Draw six of them. Be alert for isomerism of the cis/trans [or \((Z / E)]\) kind in these molecules.

Devise syntheses of the following three compounds, starting in each case with any alkene that contains four carbons or fewer. You do not have to write mechanisms, although at this point it may be very helpful for you to do so. $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH} \quad\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH} \quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}_{3} $$

Alkanes combine with oxygen to produce carbon dioxide and water according to the following scheme: $$ \mathrm{C}_{n} \mathrm{H}_{2 n+2}+(3 n+1) / 2 \mathrm{O}_{2} \rightarrow(n+1) \mathrm{H}_{2} \mathrm{O}+n \mathrm{CO}_{2} $$ This process is generally referred to as combustion. An important use of this reaction is the quantitative determination of elemental composition (elemental analysis). Typically, a small sample of the compound is completely burned, and the water and carbon dioxide produced are collected and weighed. From the weight of water, the amount of hydrogen in the original compound can be determined. Similarly, the amount of carbon dioxide formed allows us to determine the amount of carbon in the original compound. Oxygen, if present, is usually determined by difference. The determination of the relative molar proportions of carbon and hydrogen in a compound is the first step in deriving its molecular formula. If combustion of \(5.00 \mathrm{mg}\) of a hydrocarbon gives \(16.90\) \(\mathrm{mg}\) of carbon dioxide and \(3.46 \mathrm{mg}\) of water, what are the weight percents of carbon and hydrogen in the sample?
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