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Draw the Lewis structures for each of the following ions or molecules. Identify those in which the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state, for those atoms, how many electrons surround these atoms: (a) \(\mathrm{PH}_{3},\) (b) AlH_ \(_{3},(\mathbf{c}) \mathrm{N}_{3}^{-}\) (d) \(\mathrm{CH}_{2} \mathrm{Cl}_{2},(\mathbf{e}) \mathrm{SnF}_{6}^{2-}\)

Short Answer

Expert verified
In summary, the Lewis structures for the given ions/molecules are as follows: (a) PH鈧 obeys the octet rule: H | H--P--H (b) AlH鈧 does not obey the octet rule; the Aluminum atom has 6 electrons: H | H--Al--H (c) N鈧冣伝 obeys the octet rule: N鈮--N (-) (d) CH鈧侰l鈧 obeys the octet rule: Cl | H--C--H | Cl (e) SnF鈧喡测伝 does not obey the octet rule; the Tin atom has 12 electrons: F | F --Sn-- F | F --Sn-- F | F

Step by step solution

01

(a) Drawing Lewis structure for PH鈧 and checking the Octet rule:

First, count the total number of valence electrons: Phosphorus has 5 valence electrons, and each of the 3 Hydrogens has 1 valence electron. So the total number of valence electrons is: \(5 + 3 * 1 = 5 + 3 = 8\) Now, let's draw the Lewis structure for PH鈧 by putting Phosphorus in the center and Hydrogens around it, and distribute the 8 valence electrons surrounding them: H | H--P--H This molecule obeys the octet rule for all atoms, having 8 electrons surrounding P (2 from each bond) and 2 electrons surrounding each H (1 from the bond).
02

(b) Drawing Lewis structure for AlH鈧 and checking the Octet rule:

Aluminum has 3 valence electrons and each of the 3 Hydrogens has 1 valence electron. The total number of valence electrons is: \(3 + 3*1 = 3 + 3 = 6\) Now, let's draw the Lewis structure for AlH鈧 by putting Aluminum in the center and Hydrogens around it: H | H--Al--H The octet rule is not obeyed for the Aluminum atom, which only has 6 electrons surrounding it (2 from each bond). Hydrogen atoms obey the rule with 2 electrons each (1 from the bond).
03

(c) Drawing Lewis structure for N3鈦 and checking the Octet rule:

Two nitrogen atoms have 5 valence electrons each, and the extra electron from the negative charge (-1) adds one more electron. The total number of valence electrons is: \(5 + 5 + 1 = 11\) Now, let's draw the Lewis structure for N3鈦 by placing N atoms connected by single and triple bonds, and add lone pair(s) to satisfy the octet rule: N鈮--N (-) Note the extra electron in brackets added to the last Nitrogen atom. This ion obeys the octet rule for all nitrogen atoms, having 8 electrons surrounding each nitrogen atom.
04

(d) Drawing Lewis structure for CH鈧侰l鈧 and checking the Octet rule:

Carbon has 4 valence electrons, 2 Hydrogens with 1 valence electron each and 2 Chlorines with 7 valence electrons each. The total number of valence electrons is: \(4 + 2*1 + 2*7 = 4 + 2 + 14 = 20\) Now, draw the Lewis structure for CH鈧侰l鈧 by placing Carbon in the center, Hydrogens and Chlorines around it, and distribute the 20 valence electrons: Cl | H--C--H | Cl This molecule obeys the octet rule for all atoms, having 8 electrons surrounding C and each Cl (2 from each bond plus lone pairs), and 2 electrons surrounding each H (1 from the bond).
05

(e) Drawing Lewis structure for SnF6虏鈦 and checking the Octet rule:

Tin (Sn) has 4 valence electrons, each of the 6 fluorines has 7, and the extra 2 electrons from the negative charge (-2) contribute: \(4 + 6 * 7 + 2 = 4 + 42 + 2 = 48\) Now, let's draw the Lewis structure for SnF6虏鈦 by placing Sn in the center, the Fluorines around it, and distribute the 48 valence electrons: F | F --Sn-- F | F --Sn-- F | F The octet rule is not obeyed for the Tin atom (Sn), which has 12 electrons (2 from each bond) surrounding it. However, the Fluorine atoms all obey the octet rule with 8 electrons (2 from the bond and 6 lone pair electrons).

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Key Concepts

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

Understanding the Octet Rule
The octet rule is a fundamental principle in chemistry, particularly when discussing the stability of atoms. According to this rule, atoms are most stable when they have eight electrons in their valence or outermost electron shell.

This stems from the desire of atoms to reach a noble gas configuration, which is recognized for being particularly stable due to the full valence shell. In simple molecules and ions, atoms form bonds to complete their octet, sharing, gaining, or losing electrons as necessary.

However, there are exceptions where atoms do not follow the octet rule. This is often seen in molecules with an odd number of electrons, those containing atoms that can have more than eight electrons like phosphorus and sulfur, and elements in the third period and beyond on the periodic table because they have available d orbitals.

For instance, in PH鈧, the phosphorus atom is surrounded by only 8 electrons, complying with the octet rule. On the contrary, in AlH鈧, aluminum breaks the octet rule since it is only able to bond with 6 electrons due to its three valence electrons. In the exercise, recognizing such deviation from the octet rule is crucial for accurately assessing chemical stability and reactivity.
Valence Electrons and Their Importance
Valence electrons play a pivotal role in chemical reactivity and the formation of bonds. These are the electrons found in an atom's outermost shell and are involved in bond formation with other atoms.

The number of valence electrons an atom has determines how it will interact with other atoms and how many bonds it can form. For example, in the Lewis structures exercise, elements like hydrogen have one valence electron and typically form one bond, while carbon, with its four valence electrons, is known for forming four bonds.

Examining Valence Electrons in Exercise Compounds

When drawing Lewis structures for molecules like CH鈧侰l鈧, we first count the valence electrons 鈥 for carbon (4), hydrogen (1 each), and chlorine (7 each) 鈥 yielding a total that then guides the distribution of electrons around the central atom to satisfy the octet rule. Tin in SnF鈧喡测伝 has four valence electrons, but being a heavier element, it can exceed the octet and form more bonds, as it does with six fluorine atoms in the exercise.
Electron Dot Structures, aka Lewis Structures
Electron dot structures, also known as Lewis structures, are a visual representation of the valence electron arrangement in molecules and ions. These diagrams showcase how electrons are distributed around atoms and how the atoms are bonded together.

Lewis structures give us insight into molecular shapes, potential bond types, and whether the octet rule is obeyed or not. Drawing them correctly is foundational for understanding chemical bonding and molecule reactivity.

In the exercise where we draw the electron dot structures for various molecules, we must account for lone pairs of electrons that do not participate in bonding. For example, with N鈧冣伝, it's important to not only form bonds between the nitrogen atoms but also to allocate lone pairs correctly so each nitrogen ends up with eight electrons.

Overall, successfully interpreting and drawing electron dot structures require a solid grasp of both the octet rule and the concept of valence electrons, as seen in the compounds provided in the textbook exercise.

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Most popular questions from this chapter

(a) True or false: An element's number of valence electrons is the same as its atomic number. (b) How many valence electrons does a nitrogen atom possess? (c) An atom has the electron configuration 1\(s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2}\) . How many valence electrons does the atom have?

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?

Which of the following bonds are polar? (a) \(\mathrm{B}-\mathrm{F}\) (b) Cl-Cl, (c) Se-O, (d) H-I. Which is the more electronegative atom in each polar bond?

Using only the periodic table as your guide, select the most electronegative atom in each of the following sets: (a) Na, \(\mathrm{Mg}, \mathrm{K}, \mathrm{Ca} ;(\mathbf{b}) \mathrm{P}, \mathrm{S},\) As, \(\mathrm{Se} ;(\mathbf{c}) \mathrm{Be}, \mathrm{B}, \mathrm{C}, \mathrm{Si} ;(\mathbf{d}) \mathrm{Zn}, \mathrm{Ge}, \mathrm{Ga}, \mathrm{Gs}\)

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