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Andru [333]
3 years ago
11

The image compares the arrangement of electrons in two different neutral atoms.

Chemistry
1 answer:
jekas [21]3 years ago
8 0

Answer:

  • <u>Option b. Atom P has an estimated Zeff of 7 and is therefore to the right of Atom Q, which has a Zeff of 6.</u>

Explanation:

Please, find attached the figures of both atom Q and atom P corresponding to this question.

The <u>features of atom Q are</u>:

  • Each <em>black sphere</em> represents an electron
  • In total this atom has 8 electrons: 2 in the inner shell and 6 in the outermost shell.
  • Since it is assumed that the atom is neutral, it has 8 protons: one positive charge of a proton balances one negative charge of an electron. Thus, the atomic number of this atom is 8.
  • Since only two shells are ocuppied, you can assert that the atom belongs to the period 2 (which is confirmed looking into a periodic table with the atomic number 8).
  • <em>Zeff </em>is the effective nuclear charge of the atom. It accounts for the  net positive charge the valence electrons experience. And may, in a very roughly way, be estimated as the number of protons less the number of electrons in the inner shells. Thus, for this atom, an estimated  Z eff = 8 - 2 = 6.

The <u>features of atom P</u> are:

  • Again, each black sphere represents an electron
  • In total this atom has 9 electrons: 2 in the inner shell and 7 in the outermost shell.
  • Since it is assumed that the atom is neutral, it has 9 protons.
  • The atomic number of this atom is 9.
  • Using the same reasoning used for atom Q, this atom is also in the period 2.
  • Estimated Z eff = 9 - 2 = 7.

Then, since atom P has a greater Z eff than atom Q (an estimated Zeff of 7 for atom P against an estimated Z eff of 6 for atom Q),  and both atoms are in the same period, you can affirm that <em>atom P</em> has a greater atomic number and<em> is therefore to the right of atom Q</em>.

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The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.

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