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Chapter 4: Problem 21

Write the ground-state electron configuration of radium (Ra), using both the full notation and the noble gas abbreviated notation.

Short Answer

Expert verified
The ground-state electron configuration of radium (Ra) is: Full notation: \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 \) Noble gas abbreviated notation: \( [Rn] 7s^2 \)

Step by step solution

01

Determine the number of electrons for a neutral radium atom

Radium (Ra) is an alkaline earth metal, and its atomic number is 88. In a neutral atom, the number of electrons is equal to the atomic number. Thus, there are 88 electrons in a neutral radium atom.
02

Distribute electrons among atomic orbitals using the aufbau principle

By following the aufbau principle, which states that electrons fill orbitals in the order of increasing energy levels, we distribute the 88 electrons in radium as follows: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² So, the full electron configuration of radium is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s²
03

Find the noble gas before radium

Radon (Rn) is the noble gas that comes directly before radium (Ra) in the periodic table. Its atomic number is 86. So, the electron configuration of radon is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶
04

Write the noble gas abbreviated notation

We can now replace the electron configuration of radon (Rn) in the full electron configuration of radium (Ra) to obtain the noble gas abbreviated notation for radium: \[ [Rn] 7s^2 \] Thus, the ground-state electron configuration of radium (Ra) is: Full notation: \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 \) Noble gas abbreviated notation: \( [Rn] 7s^2 \)

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Most popular questions from this chapter

Order the following atoms from smallest to largest, judging from their relative positions in the periodic table: \(\mathrm{Cs}, \mathrm{Fe}, \mathrm{Ti}, \mathrm{Hf}\)

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.

Which is the correct ground-state electron configuration for antimony, Sb? (a) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 d^{3}\) (b) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{3}\) (c) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 d^{10} 5 p^{4}\) (d) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 5 p^{3}\) (e) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 f^{3}\)

Electromagnetic radiation emitted by magnesium has a wavelength of \(285.2 \mathrm{~nm}\). (a) Is this radiation visible to the eye? (b) What is the energy of this radiation?

What do \(\mathrm{F}^{-}, \mathrm{O}^{2-}, \mathrm{Na}^{+}\), and \(\mathrm{Mg}^{2+}\) all have in common?
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