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Chapter 7: Q16E (page 391)

From its position in the periodic table, determine which atom in each pair is more electronegative: (a)\({\rm{N or P}}\)(b)\({\rm{N or Ge}}\)(c)\({\rm{S or F}}\)(d)\({\rm{Cl or S}}\)(e)\({\rm{H or C}}\)(f)\({\rm{Se or P}}\)(g)\({\rm{C or Si}}\).

Short Answer

Expert verified
  1. Nitrogen is more electronegative.
  2. Nitrogen is more electronegative.
  3. Fluorine is more electronegative.
  4. Chlorine is more electronegative.
  5. Carbon is more electronegative.
  6. Phosphorus is more electronegative.
  7. Carbon is more electronegative.

Step by step solution

01

Define Chemical Bonding

A chemical bond is a long-term attraction between atoms, ions, or molecules that allows chemical compounds to form.

02

Explanation

As, the addition of a proton has a greater effect than the addition of an electron, the size of the atom decreases as you travel from left to right in the periodic table. Because of the strong force of the nucleus, it will be able to receive electrons more easily as its size decreases.

According to the foregoing argument, electronegativity rises as one moves from left to right in the periodic table.

As you progress from top to bottom in a group, the size of the group grows due to the addition of fresh energy shells. Because the force of attraction of the nucleus is no longer able to attract additional electrons as the size of the nucleus rises, electronegativity will drop as the group size decreases.

03

Determining which atom in each pair is more electronegative?

a. Due to nitrogen is heavier than phosphorus, it is more electronegative.

Therefore, nitrogen is more electronegative.

04

Determining which atom in each pair is more electronegative?

b. Nitrogen is more electronegative than Germanium since it is located to the right of it.

Therefore, nitrogen is more electronegative.

05

Determining which atom in each pair is more electronegative?

c. As, Fluorine is heavier than Sulfur, it is more electronegative.

Therefore, Fluorine is more electronegative.

06

Determining which atom in each pair is more electronegative?

d. The Chlorine is in front of Sulfur, it is more electronegative.

Therefore, Chlorine is more electronegative.

07

Determining which atom in each pair is more electronegative?

e. As, Carbon is located above and to the right of Hydrogen, it is more electronegative.

08

Determining which atom in each pair is more electronegative?

f. Due to phosphorus is located to the right and above \({\rm{Se}}\), \({\rm{P}}\) is more electronegative.

Therefore, phosphorus is more electronegative.

09

Determining which atom in each pair is more electronegative?

g. As, the value of carbon is to the right and above sillicon, it is more electronegative.

Therefore, carbon is more electronegative.

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Most popular questions from this chapter

Identify the atoms that correspond to each of the following electron configurations. Then, write the Lewis symbol for the common ion formed from each atom: (a)\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{5}}}\)(b)\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}\)(c)\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{d}}^{{\text{10}}}}\)(d)\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{d}}^{{\text{10}}}}{\text{4}}{{\text{p}}^{\text{4}}}\)(e)\({\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{d}}^{{\text{10}}}}{\text{4}}{{\text{p}}^{\text{1}}}\).

In the Lewis structures listed here, M and X represent various elements in the third period of the periodic table. Write the formula of each compound using the chemical symbols of each element:

Draw the Lewis structures and predict the shape of each compound or ion:

(a) \({\rm{C}}{{\rm{O}}_{\rm{2}}}\)

(b) \({\rm{NO}}_{\rm{2}}^{\rm{ - }}\)

(c) \({\rm{S}}{{\rm{O}}_{\rm{3}}}\)

(d) \({\rm{S}}{{\rm{O}}_{\rm{3}}}^{{\rm{2 - }}}\)

Use the simulation (http://openstaxcollege.org/l/16MolecPolarity) to perform the following exercises for a two-atom molecule: (a) Adjust the electronegativity value so the bond dipole is pointing toward B. Then determine what the electronegativity values must be to switch the dipole so that it points toward A. (b) With a partial positive charge on A, turn on the electric field and describe what happens. (c) With a small partial negative charge on A, turn on the electric field and describe what happens. (d) Reset all, and then with a large partial negative charge on A, turn on the electric field and describe what happens.

From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity: (a)\({\rm{C, F, H, N, O}}\)(b)\({\rm{Br, Cl, F, H, I }}\)(c)\({\rm{F, H, O, P, S }}\)(d)\({\rm{Al, H, Na, O, P}}\)(e)\({\rm{Ba, H, N, O, As}}\).
See all solutions

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