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Fill in the blank with the appropriate numbers for both electrons and bonds (considering that single bonds are counted as one, double bonds as two, and triple bonds as three). (a) Fluorine has _____ valence electrons and makes______ bond(s) in compounds. (b) Oxygen has _____ valence electrons and makes ______ bond(s) in compounds. (c) Nitrogen has _____ valence electrons and makes ______ bond(s) in compounds. (d) Carbon has _____ valence electrons and makes ______ bond(s) in compounds.

Short Answer

Expert verified
(a) Fluorine has \(7\) valence electrons and makes \(1\) bond(s) in compounds. (b) Oxygen has \(6\) valence electrons and makes \(2\) bond(s) in compounds. (c) Nitrogen has \(5\) valence electrons and makes \(3\) bond(s) in compounds. (d) Carbon has \(4\) valence electrons and makes \(4\) bond(s) in compounds.

Step by step solution

01

Valence Electrons for Fluorine (F)

Fluorine is in Group 17 of the periodic table and has an atomic number of 9. Its electronic configuration is 1s²2s²2p⁵. Since the electrons in the last energy level contribute to the chemical bonding, Fluorine has 7 valence electrons.
02

Fluorine (F) Bonding Behavior

To achieve a full outer electron shell (the octet rule), Fluorine needs to gain one electron. Thus, it forms one single bond with other elements. (a) Fluorine has 7 valence electrons and makes 1 bond(s) in compounds.
03

Valence Electrons for Oxygen (O)

Oxygen is in Group 16 of the periodic table and has an atomic number of 8. Its electronic configuration is 1s²2s²2p⁴. Just like in the case of Fluorine, the last energy level electrons contribute to the chemical bonding. Oxygen has 6 valence electrons.
04

Oxygen (O) Bonding Behavior

To achieve a full outer electron shell, Oxygen needs to gain two electrons. Thus, it can either form two single bonds or one double bond with other elements. (b) Oxygen has 6 valence electrons and makes 2 bond(s) in compounds.
05

Valence Electrons for Nitrogen (N)

Nitrogen is in Group 15 of the periodic table and has an atomic number of 7. Its electronic configuration is 1s²2s²2p³. The last energy level electrons contribute to the chemical bonding. Nitrogen has 5 valence electrons.
06

Nitrogen (N) Bonding Behavior

To achieve a full outer electron shell, Nitrogen needs to gain three electrons. So, it can either form three single bonds, one single bond and one double bond, or one triple bond with other elements. (c) Nitrogen has 5 valence electrons and makes 3 bond(s) in compounds.
07

Valence Electrons for Carbon (C)

Carbon is in Group 14 of the periodic table and has an atomic number of 6. Its electronic configuration is 1s²2s²2p². The last energy level electrons contribute to the chemical bonding. Carbon has 4 valence electrons.
08

Carbon (C) Bonding Behavior

To achieve a full outer electron shell, Carbon needs to gain four electrons. So, it can either form four single bonds, two single bonds and one double bond, or two double bonds with other elements. (d) Carbon has 4 valence electrons and makes 4 bond(s) in compounds.

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Key Concepts

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

Bonding Behavior
Atoms bond with one another to achieve stability. They aim to fill their outermost electron shell. This is often referred to as achieving an "octet."
  • Fluorine, with 7 valence electrons, forms one single bond to gain an eighth electron. This fills its outer shell and satisfies the octet rule.
  • Oxygen, having 6 valence electrons, needs two more to achieve a full shell. It forms two single bonds or one double bond with other atoms.
  • Nitrogen has 5 valence electrons and typically forms three bonds. It can form a triple bond to fill the necessary 3 electrons.
  • Carbon, with 4 valence electrons, forms four bonds. This can be four single bonds, two double bonds, or any variation adding up to four electron pair bonds.
These bonds ensure each atom reaches a stable electronic configuration.
Periodic Table Groups
The periodic table is a systematic way to display the elements, organized by their atomic structure. One important structure is the column, or group, which tells us how many valence electrons an element has. The number of valence electrons affects bonding behavior.
  • Group 17 elements, like Fluorine, typically have 7 valence electrons.
  • Group 16 elements, such as Oxygen, consist of atoms with 6 valence electrons.
  • Group 15 elements, including Nitrogen, possess 5 valence electrons.
  • Group 14 elements, like Carbon, have 4 valence electrons.
The group number thus predicts bonding capability, linked to the concept of filling the outer electron shell.
Octet Rule
The octet rule is a fundamental concept in understanding chemical bonding. It states that atoms strive to have eight electrons in their valence shell, mimicking the electron configuration of the noble gases.
  • Fluorine achieves its octet by forming a single bond.
  • Oxygen reaches an octet through forming two bonds, which could be double or two singles.
  • Nitrogen needs three additional electrons, forming three bonds in a variety of arrangements like one triple bond.
  • Carbon, needing four electrons, easily makes four bonds in diverse configurations.
The octet rule is a good rule of thumb for predicting the bonding patterns of many elements, ensuring a stable and low-energy state.

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Most popular questions from this chapter

(a) Using Lewis symbols, diagram the reaction between magnesium and oxygen atoms to give the ionic substance MgO. (b) How many electrons are transferred? (c) Which atom loses electrons in the reaction?

Formic acid has the chemical formula HCOOH. It is a colorless liquid that has a density of 1.220 \(\mathrm{g} / \mathrm{mL}\) . (a) The carbon atom in formic acid is bound to one \(\mathrm{H}\) and both \(\mathrm{O}^{\prime}\) 's. Draw the Lewis structure for formic acid, showing resonance if present. (b) Formic acid can react with NaOH in aqueous solution to produce the formate ion, HCOO- . Write the balanced chemical equation for this reaction. (c) Draw the Lewis structure of the formate ion, showing resonance if present. (d) How many milliliters of a 0.100 M solution of NaOH would it take to completely react with 0.785 \(\mathrm{mL}\) of formic acid?

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