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Draw Lewis structures for the following: (a) \(\operatorname{SiH}_{4},\) (b) \(\mathrm{CO}\) \((\mathbf{c}) \mathrm{SF}_{2},(\mathbf{d}) \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{H}\) is bonded to \(\mathrm{O})\) , \((\mathbf{e}) \mathrm{ClO}_{2}^{-},(\mathbf{f}) \mathrm{NH}_{2} \mathrm{OH}\)

Short Answer

Expert verified
The Lewis structures for the given molecules and ion are as follows: a) SiH4: Si-H | H-Si-H | H b) CO: C≡O c) SF2: F | S - F d) H2SO4: O | O=S-O | O | H e) ClO2-: O | Cl=O | / \ LP LP f) NH2OH: H | H-N-O-H | LP

Step by step solution

01

1. Determine the total number of valence electrons

Count the total number of valence electrons for each molecule/ion based on the valence electron configuration of the individual elements. a) SiH4: Si has 4 valence electrons and each H has 1 valence electron, so for SiH4 we have: 4 (for Si) + 4x1 (for 4 H atoms) = 8 valence electrons b) CO: C has 4 valence electrons and O has 6 valence electrons, so for CO we have: 4 + 6 = 10 valence electrons c) SF2: S has 6 valence electrons and each F has 7 valence electrons, so for SF2 we have: 6 + 2x7 = 20 valence electrons d) H2SO4: S has 6 valence electrons, each O has 6 valence electrons, and each H has 1 valence electron, so for H2SO4 we have: 6 + 4x6 (for 4 O atoms) + 2x1 (for 2 H atoms) = 32 valence electrons e) ClO2-: Cl has 7 valence electrons, each O has 6 valence electrons, and there is an extra electron gained by the ion, so for ClO2- we have: 7 + 2x6 + 1 = 20 valence electrons f) NH2OH: N has 5 valence electrons, each H has 1 valence electron, and O has 6 valence electrons, so for NH2OH we have: 5 + 3x1 + 6 = 14 valence electrons
02

2. Connect atoms using single bonds and complete the octet

Form single bonds between the central atom and the surrounding atoms. Then add electrons to complete the octet for each atom. a) SiH4 Lewis structure: Si is the central atom, forming single bond with hydrogen atoms. Si-H | H-Si-H | H b) CO Lewis structure: C is the central atom, forming a triple bond with oxygen to complete their octet. C≡O c) SF2 Lewis structure: S is the central atom, forming single bond with fluorine atoms. F | S - F d) H2SO4 Lewis structure: S is the central atom, forming single bonds with four oxygen atoms, and two oxygen atoms have H atoms attached to them. O | O-S-O | O | H e) ClO2- Lewis structure: Cl is the central atom, forming single bonds with two oxygen atoms. O | Cl-O f) NH2OH Lewis structure: N is the central atom, forming single bonds with two hydrogen atoms and the oxygen atom. Oxygen is then attached to a hydrogen atom. H | H-N-O-H
03

3. Add extra electrons and refine the structure

For any remaining electrons, distribute them to the outer atoms (except hydrogen) to fulfill the octet rule. In some cases, refine the structure by forming double or triple bonds, if needed. a) SiH4 - no need to modify; all atoms have a full outer shell. b) CO - triple bond already formed; no modification needed. c) SF2: To complete the octet rule for sulfur, we need to add two non-bonding electron pairs: F | S - F | / \ LP LP d) H2SO4: Add two non-bonding electron pairs to the oxygen atoms that are not bonded to hydrogen atoms, forming double bonds: O | O=S-O | O | H e) ClO2-: Add three non-bonding electron pairs to the outer oxygen atoms, and complete the octet rule for chlorine by adding two non-bonding electron pairs. O | Cl=O | / \ LP LP f) NH2OH: Add a non-bonding electron pair to nitrogen to complete the octet rule. Hydrogen and oxygen are already complete. H | H-N-O-H | LP

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Key Concepts

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

Valence Electrons
Valence electrons are the outermost electrons of an atom. They play a crucial role in chemical bonding because they can be gained, lost, or shared to form different types of bonds. For each element, the number of valence electrons is determined by its group number in the periodic table. For instance, carbon, which belongs to group 14, has four valence electrons. Similarly, oxygen, from group 16, has six valence electrons.

To draw Lewis structures, we first have to determine the total number of valence electrons for a molecule or ion. This helps us understand how these electrons might be shared or transferred among the atoms in the structure. For example, in the molecule \( ext{SiH}_4\), silicon has four valence electrons, and each hydrogen has one, resulting in a total of eight valence electrons.

Understanding valence electrons allows chemists to predict how atoms will interact in a chemical reaction. It also helps in determining molecular geometry and the resulting properties of the compound.
Octet Rule
The octet rule is a guiding principle in chemistry that suggests atoms tend to bond in order to have eight electrons in their valence shell, mimicking the electron configuration of a noble gas. Noble gases are stable due to their complete valence shell. This rule is especially applicable to main group elements.

In the step-by-step solution for Lewis structures, after forming initial bonds between atoms, it is important to check if each atom satisfies the octet rule. For some molecules, such as \( ext{CO}\), a triple bond between carbon and oxygen helps both atoms achieve a stable octet configuration.

However, there are exceptions to the octet rule. Elements like hydrogen are content with two valence electrons, akin to the noble gas helium. Compounds like \( ext{SF}_2\) have sulfur as the central atom, which can have more than eight electrons due to its ability to expand its octet using d-orbitals.

Ensuring that atoms follow the octet rule, or understanding when the rule doesn't apply, is fundamental in predicting the stability and reactivity of molecules.
Covalent Bonds
Covalent bonds arise from the sharing of valence electrons between atoms. This sharing allows each atom to obtain a complete octet, achieving greater stability. At the core of covalent bonding is the mutual attraction of each atom's nucleus to the shared electron pair.

When drawing Lewis structures, covalent bonds are typically depicted as lines connecting atoms. Each line represents a pair of shared electrons. For example, in \( ext{NH}_2 ext{OH}\), nitrogen forms single covalent bonds with its hydrogen atoms and the hydroxyl group (OH).

Depending on the molecule, covalent bonds can be single, double, or even triple, with each type involving different numbers of shared electron pairs. Single bonds share one pair of electrons, double bonds share two pairs, and triple bonds share three. These bonding types affect the molecule's length and strength; generally, triple bonds are shorter and stronger than double or single bonds.

Understanding covalent bonding helps in predicting how molecules form, interact, and react. It also influences a molecule's physical characteristics, such as melting and boiling points, as well as its solubility and electrical conductivity.

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Most popular questions from this chapter

Predict the chemical formula of the ionic compound formed between the following pairs of elements: (a) Al and F, (b) K and \(S,(\mathbf{c}) \mathrm{Y}\) and \(\mathrm{O},(\mathbf{d}) \mathrm{Mg}\) and \(\mathrm{N} .\)

True or false: (a) The \(\mathrm{C}-\) Cbonds in benzene are all the same length and correspond to typical single \(\mathrm{C}-\mathrm{Cbond}\) lengths. (b) The \(\mathrm{C}-\mathrm{C}\) bond in acetylene, HCCH, is longer than the average \(\mathrm{C}-\mathrm{C}\) bond length in benzene.

A common form of elemental phosphorus is the tetrahedral \(\mathrm{P}_{4}\) molecule, where all four phosphorus atoms are equivalent: At room temperature phosphorus is a solid. (a) Are there any lone pairs of electrons in the \(\mathrm{P}_{4}\) molecule? (b) How many \(\mathrm{p}-\mathrm{p}\) bonds are there in the molecule? (c) Draw a Lewis structure for a linear \(P_{4}\) molecule that satisfies the octet rule. Does this molecule have resonance structures? (d) On the basis of formal charges, which is more stable, the linear molecule or the tetrahedral molecule?

Consider the formate ion, \(\mathrm{HCO}_{2}^{-},\) which is the anion formed when formic acid loses an \(\mathrm{H}^{+}\) ton. The \(\mathrm{H}\) and the two O atoms are bonded to the central \(\mathrm{C}\) atom. (a) Draw the best Lewis structure(s) for this ion. (b) Are resonance structures needed to describe the structure? (c) Would you predict that the \(\mathrm{C}-\mathrm{O}\) bond lengths in the formate ion would be longer or shorter relative to those in \(\mathrm{CO}_{2} ?\)

The Ti \(\mathrm{Ti}^{2+}\) ion is isoelectronic with the Ca atom. (a) Write the electron configurations of \(\mathrm{Ti}^{2+}\) and (b) Calculate the number of unpaired electrons for Ca and for \(\mathrm{Ti}^{2+} .\) (c) What charge would Ti have to be isoelectronic with \(\mathrm{Ca}^{2}+?\)

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