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Chapter 4: Problem 178

What is the total number of \(p\) -subshell electrons for each of the following atoms: \(\mathrm{P}, \mathrm{Mg}\), Se, \(\mathrm{Zn} ?\)

Short Answer

Expert verified
Phosphorus (P) has 9 p-subshell electrons, Magnesium (Mg) has 6 p-subshell electrons, Selenium (Se) has 16 p-subshell electrons, and Zinc (Zn) has 12 p-subshell electrons.

Step by step solution

01

Find the atomic numbers of the given elements

First, we need to determine the atomic numbers (number of protons) of the given elements. This information can be found on the periodic table. For Phosphorus (P), the atomic number is 15. For Magnesium (Mg), the atomic number is 12. For Selenium (Se), the atomic number is 34. For Zinc (Zn), the atomic number is 30.
02

Determine the electron configurations

Now that we have the atomic numbers of the given elements, we can determine their electron configurations. - P: (Atomic number = 15) => 1s^2 2s^2 2p^6 3s^2 3p^3 - Mg: (Atomic number = 12) => 1s^2 2s^2 2p^6 3s^2 - Se: (Atomic number = 34) => 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^4 - Zn: (Atomic number = 30) => 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10
03

Count the number of p-subshell electrons

Now, we can count the number of p-subshell electrons by looking for p-orbitals in the electron configurations we've found in Step 2. - P: has a 2p^6 and 3p^3 => 6 (from 2p) + 3 (from 3p) = 9 p-subshell electrons - Mg: has a 2p^6 => 6 p-subshell electrons - Se: has a 2p^6, 3p^6, and 4p^4 => 6 (from 2p) + 6 (from 3p) + 4 (from 4p) = 16 p-subshell electrons - Zn: has a 2p^6 and 3p^6 => 6 (from 2p) + 6 (from 3p) = 12 p-subshell electrons
04

Present the final results

Now that we have the total number of p-subshell electrons for each given element, we can present our final results. - Phosphorus (P) has 9 p-subshell electrons - Magnesium (Mg) has 6 p-subshell electrons - Selenium (Se) has 16 p-subshell electrons - Zinc (Zn) has 12 p-subshell electrons

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Most popular questions from this chapter

The second ionization energy of an atom is the minimum energy required to remove an electron from the \(+1\) cation of the atom, and it is always larger than the first ionization energy. Why is this so? (Hint: Think of atomic electrons as clouds, and each electron must "look" through every other electron cloud in the atom to "see" (feel) the nucleus.)

Why are roman-numeral group numbers for the representative elements useful in predicting how many electrons an atom will gain or lose in a chemical reaction?

What is the implication of thinking about atomic electrons as clouds that got Einstein so upset? What did Schrödinger call these clouds?

Predict the formula for lithium nitride, made from lithium, \(\mathrm{Li}\), and nitrogen, \(\mathrm{N}\). Explain your reasoning.

Below are data on the first four ionization energies for a fictitious element \(X\). First ionization energy \(=500 \mathrm{~kJ} / \mathrm{mol}\) Second ionization energy \(=2000 \mathrm{~kJ} / \mathrm{mol}\) Third ionization energy \(=3500 \mathrm{~kJ} / \mathrm{mol}\) Fourth ionization energy \(=25,000 \mathrm{~kJ} / \mathrm{mol}\) From the data, which of the following statements is(are) incorrect? (a) \(\mathrm{X}\) could belong to group IIIA. (b) The fourth ionization energy is so much greater than the third ionization energy because \(\mathrm{X}^{3+}\) consists of a noble-gas core or a pseudo- noble-gas core. (c) The third ionization energy is greater than the second ionization energy because \(\mathrm{X}^{2+}\) has a bigger charge than \(\mathrm{X}^{+}\). (d) \(\mathrm{X}\) could belong to group IIIB. (e) \(X\) could belong to group VA.
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