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In which of these bonding patterns does X obey the octet rule?

Short Answer

Expert verified
  1. X obeys the octet rule.
  2. X obeys the octet rule.
  3. X does not obey the octet rule.
  4. X obeys the octet rule.
  5. X obeys the octet rule.
  6. X obeys the octet rule.
  7. X does not obey the octet rule.
  8. X does not obey the octet rule.

Step by step solution

01

Octet Rule

The octet rule is important to attain noble gas configuration i.e., the atom should have 8 electrons in the valence shell.

It is the requirement of an electron by the molecule (by gaining an electron from another molecule or by losing its electron to another atom) or having the electron in the valence shell of the atom.

02

Subpart (a) Explanation

  • The central atom X in this bonding pattern obeys the octet rule.
  • It has attained the noble gas configuration as its valence shell of the central atom has 8 electrons in the valence shell after the formation of a bond with 4 different elements.
  • In one bond there are two electrons. Hence, 4 bonds contain 8 electrons. Example:CH4
03

Subpart (b) Explanation

  • The central atom X in this bonding pattern obeys the octet rule.
  • It has attained the noble gas configuration.
  • Here, 2 electrons are present in one bond pair and 6 electrons are present in three bond pairs. Example; PH3
04

Subpart (c) Explanation

  • Here, it does not follow the noble gas configuration as it has more than 8 electrons in its valence shell.
  • The existence of these type of molecule is very rare as it does not follow the basic noble gas configuration rule. Example:PF5
05

Subpart (d) Explanation

  • The central atom X in this bonding pattern obeys the octet rule.
  • It has attained the noble gas configuration.
  • The two atoms that form three bonds and have one lone pair that does not participate in bonding. So, the total of 8electrons in valence shell. Example:N2
06

Subpart (e) Explanation

  • The central atom X in this bonding pattern obeys the octet rule.
  • It has attained the noble gas configuration.
  • The central atom forms two bonds with one atom and two single bonds with two other atoms.
  • All the 8 electrons are bond pairs. Example:COCl2 phosgene.
07

Subpart (f) Explanation

  • The central atom X in this bonding pattern obeys the octet rule.
  • It has attained the noble gas configuration.
  • The central atom has 3 bond pairs and 1 lone pair. Therefore, a total of 8 electrons. Example:O=NF
08

Subpart (g) Explanation

  • Here, it does not follow the noble gas configuration as it has less than 8 electrons in its valence shell.
  • The compound has only 6 electrons in its outermost shell.
  • The central atom has three valence electrons and forms a bond with three atoms. A total of 6 electrons are present in the valence shell. Example:BeF3
09

Subpart (h) Explanation

  • Here, it does not follow the noble gas configuration as it has more than 8 electrons in its valence shell.
  • The X atom contains 8 electrons already in its outermost shell and accepts two-electron more, making the total of 10 electrons in the outermost shell.
  • This kind of compound exists very rarely.
  • The atom has 8electrons in its valence shell and accepts two electrons more in bonding. Example;Xe2

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Most popular questions from this chapter

A gaseous compound has a composition by mass of 24.8% carbon, 2.08% hydrogen, and 73.1% chlorine. At STP, the gas has a density of 4.3 g/L. Draw a Lewis structure that satisfies these facts. Would another structure also satisfy them? Explain

Dinitrogen monoxide (N2O ) supports combustion in a manner similar to oxygen, with the nitrogen atoms forming N2. Draw three resonance structures forN2O (one N is central), and use formal charges to decide the relative importance of each. What correlation can you suggest between the most important structure and the observation thatN2O supports combustion?

These species do not obey the octet rule. Draw a Lewis structure for each, and state the type of octet-rule exception:

(a)PF6-

(b)ClO3

(c) H3PO3 (one P卤H bond)

Ethanol ( CH3CH2OH ) is being used as a gasoline additive or alternative in many parts of the world.

  1. Use bond energies to find role="math" localid="1658829887119" 螖贬rxnfor the combustion of gaseous ethanol. (Assumerole="math" localid="1658829903069" H2O forms as a gas.)
  2. In its standard state at25C , ethanol is a liquid. Its vaporization requires 40.5 kJ/mol. Correct the value from part (a) to find the heat of reaction for the combustion of liquid ethanol.
  3. How does the value from part (b) compare with the value you calculate from standard heats of formation (Appendix B)?
  4. 鈥淕reener鈥 methods produce ethanol from corn and other plant material, but the main industrial method involves hydrating ethylene from petroleum. Use Lewis structures and bond energies to calculate螖贬rxn for the formation of gaseous ethanol from ethylene gas with water vapor.

Consider the following molecular shapes.

a.Which has the most electron pairs (both bonding and lone pairs) around the central atom?

b.Which has the most lone pairs around the central atom?

c.Do any have only bonding pairs around the central atom? A B C

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