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Draw the Lewis structures for each of the following molecules or ions. Identify instances where the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state how many electrons surround these atoms: (a) \(\mathrm{NO},(\mathbf{b}) \mathrm{BF}_{3},(\mathbf{c}) \mathrm{ICl}_{2}^{-},(\mathbf{d}) \mathrm{OPBr}_{3}(\) the \(\mathrm{P}\) is the central atom), (e) XeF.

Short Answer

Expert verified
In summary: - NO does not obey the octet rule as nitrogen is surrounded by six electrons. - BF3 does not obey the octet rule as boron is surrounded by six electrons. - ICl2- does not obey the octet rule as iodine is surrounded by ten electrons. - OPBr3 does not obey the octet rule as phosphorus is surrounded by ten electrons. - XeF does not obey the octet rule as xenon is surrounded by ten electrons.

Step by step solution

01

(a) Drawing Lewis Structure for NO

To draw the Lewis structure for the NO molecule, find the total number of valence electrons. Nitrogen has five valence electrons, and oxygen has six. The total count is 11 valence electrons. Next, place nitrogen and oxygen atoms next to each other since nitrogen has fewer lone pairs. Connect them with a single bond and distribute remaining electrons as lone pairs. NO Lewis Structure: O=N:\( \,\,\,\) The nitrogen atom is surrounded by six electrons, violating the octet rule. Therefore, NO does not obey the octet rule.
02

(b) Drawing Lewis Structure for BF3

Boron has three valence electrons, and each fluorine atom has seven valence electrons. Overall, there are 24 valence electrons. Place boron at the center and surround it with three fluorine atoms. Connect boron and each fluorine atom with single bonds and distribute the remaining electrons. BF3 Lewis Structure: F | B - F | F The boron atom is surrounded by six electrons, violating the octet rule. Therefore, BF3 does not obey the octet rule.
03

(c) Drawing Lewis Structure for ICl2-

Iodine has seven valence electrons, chlorine has seven valence electrons, and there is one extra electron from the negative charge, so the total count is 22 valence electrons. Place iodine at the center and surround it with two chlorine atoms. Connect iodine and each chlorine atom with single bonds and distribute the remaining electrons as lone pairs. ICl2- Lewis Structure: Cl | I - Cl | - The iodine atom is surrounded by ten electrons, violating the octet rule. Therefore, ICl2- does not obey the octet rule.
04

(d) Drawing Lewis Structure for OPBr3

Oxygen has six valence electrons, phosphorus has five valence electrons, and each bromine atom has seven valence electrons. Overall, there are 32 valence electrons. Place phosphorus at the center, with oxygen above and three bromine atoms surrounding it. Connect each atom with single bonds and distribute the remaining electrons as lone pairs. OPBr3 Lewis Structure: O | Br - P - Br | Br The phosphorus atom is surrounded by ten electrons, violating the octet rule. Therefore, OPBr3 does not obey the octet rule.
05

(e) Drawing Lewis Structure for XeF

Xenon has eight valence electrons, and fluorine has seven valence electrons. The total count is 15 valence electrons. Place xenon at the center and connect it to the fluorine atom with single bonds and distribute the remaining electrons as lone pairs. XeF Lewis Structure: F | Xe The xenon atom is surrounded by ten electrons, violating the octet rule. Therefore, XeF does not obey the octet rule.

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Key Concepts

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

Octet Rule
The octet rule is a fundamental principle in chemistry that suggests most atoms seek to have eight electrons in their outermost shell. This configuration is similar to that of noble gases, which are stable and unreactive. The rule is based on the tendency for atoms to complete their valence shell, achieving a stable octet. However, there are exceptions to the octet rule. For example, molecules or ions such as NO, BF₃, and XeF, as mentioned in the exercise, do not adhere to this rule.
NO has an odd number of valence electrons, resulting in a stable molecule without a complete octet for nitrogen. The molecule BF₃ has only six electrons around boron, making it electron deficient. For XeF, xenon expands its octet to accommodate ten electrons, utilizing its d-orbitals. Understanding these exceptions is crucial for mastering molecular structures and predicting the behavior of certain compounds.
Valence Electrons
Valence electrons are the electrons found in the outermost shell of an atom and are crucial for forming chemical bonds. These electrons participate in bonding with other atoms and determine elements' chemical properties. To find the number of valence electrons for an element, look at its group number in the periodic table. For instance:
  • Nitrogen has five valence electrons,
  • Oxygen has six,
  • Fluorine has seven,
  • Boron has three,
  • Iodine and chlorine each have seven,
  • Xenon has eight.
The counting of total valence electrons is essential in creating Lewis structures. In the exercises provided, nitrogen and oxygen together have 11 valence electrons in NO, leading to an odd number that results in a stable yet unfulfilled octet for nitrogen. Proper distribution of these electrons in compounds is key to accurately depicting molecular structures.
Molecular Geometry
Molecular geometry describes the three-dimensional arrangement of atoms in a molecule, dictating the molecule's properties and reactions. The specific geometry depends on the number of bonding pairs and lone pairs around the central atom. This arrangement is guided by the VSEPR (Valence Shell Electron Pair Repulsion) theory, aiming to minimize repulsion between electron pairs.
For example, in BF₃, the geometry is trigonal planar because three bonding pairs around boron spread out evenly. In ICl₂⁻, the geometry is linear due to three lone pairs on iodine. These geometric distinctions help to predict molecule behavior and interactions. Visualizing these shapes can be challenging, but knowing the types of bonds and lone pairs allows for a more precise understanding of molecular geometry.
Electron Distribution
Electron distribution refers to how electrons are allocated around atoms within a molecule. The aim is to fulfill the octet rule whenever possible and depict accurate bonds, both single and multiple, between atoms.
In constructing Lewis structures, electrons are distributed to satisfy the valence requirements of each participating element. Lone pairs often indicate regions of electrons that are not involved in bonding. For example, the molecule NO has a total of 11 valence electrons. While the distribution forms a double bond, the nitrogen atom does not achieve an octet. Similarly, in OPBr₃, phosphorus has ten electrons around it due to bonding with oxygen and three bromine atoms.
Electron distribution can also show when elements expand their octet, which happens in compounds like XeF, where xenon accommodates more than eight electrons. Correctly distributing electrons aids in a better understanding of molecular stability and reactivity.

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Most popular questions from this chapter

A classmate of yours is convinced that he knows everything about electronegativity. (a) In the case of atoms \(X\) and \(Y\) having different electronegativities, he says, the diatomic molecule \(X-Y\) must be polar. Is your classmate correct? (b) Your classmate says that the farther the two atoms are apart in a bond, the larger the dipole moment will be. Is your classmate correct?

Arrange the bonds in each of the following sets in order of increasing polarity: (a) \(\mathrm{C}-\mathrm{F}, \mathrm{O}-\mathrm{F}, \mathrm{Be}-\mathrm{F}\) ; (b) \(\mathrm{O}-\mathrm{Cl}, \mathrm{S}-\mathrm{Br}, \mathrm{C}-\mathrm{P} ;(\mathbf{c}) \mathrm{C}-\mathrm{S}, \mathrm{B}-\mathrm{F}, \mathrm{N}-\mathrm{O}\)

By referring only to the periodic table, select (a) the most electronegative element in group \(6 \mathrm{A} ;(\mathbf{b})\) the least electronegative element in the group Al, Si, P; (c) the most electronegative element in the group \(\mathrm{Ga}, \mathrm{P}, \mathrm{Cl}, \mathrm{Na} ;(\mathbf{d})\) the element in the group \(\mathrm{K}\) \(\mathrm{C}, \mathrm{Zn}, \mathrm{F}\) that is most likely to form an ionic compound with Ba.

Although \(\mathrm{I}_{3}\) is a known ion, \(\mathrm{F}_{3}^{-}\) is not. (a) Draw the Lewis structure for \(\mathrm{I}_{3}^{-}\) (it is linear, not a triangle). (b) One of your classmates says that \(\mathrm{F}_{3}^{-}\) does not exist because \(\mathrm{Fis}\) too electronegative to make bonds with another atom. Give an example that proves your classmate is wrong. (c) Another classmate says \(\mathrm{F}_{3}^{-}\) does not exist because it would violate the octet rule. Is this classmate possibly correct? (d) Yet another classmate says \(\mathrm{F}_{3}^{-}\) does not exist because \(\mathrm{F}\) is too small to make bonds to more than one atom. Is this classmate possibly correct?

(a) Which of these compounds is an exception to the octet rule: carbon dioxide, water, ammonia, phosphorus trifluoride, or arsenic pentafluoride? (b) Which of these compounds or ions is an exception to the octet rule: borohydride \(\left(\mathrm{BH}_{4}^{-}\right),\) borazine \(\left(\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6},\) which is analogous \right. to benzene with alternating \(\mathrm{B}\) and \(\mathrm{N}\) in the ring \(),\) or boron trichloride?

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