Shown below are several options for the box notations of the ground state electron configuration of the following gas-phase species. Identify the correct electronic configuration.
i. Al2−
ii. Mn
(a)Select the reason that best explains why the first ionization energy decreases as atomic number increases for the alkali metals (Li, Na, K, Rb, Cs).
The higher the atomic number, the greater the force of attraction between the nuclear charge and an electron. As the force is inversely related to the energy required to liberate an electron from that force, the first ionization energy decreases as the force of attraction felt increases.
The outermost shell becomes occupied with a greater number of electrons as the atomic number increases resulting in greater electron-electron (a.k.a Born) repulsion, which in turn makes it easier to ionize one of these electrons.
All of these elements are in the same group, and have the same valence shell electron configuration. Even though the nuclear charge increases with atomic number, the outermost electron’s distance from the nucleus and the effectiveness with which inner electrons screen this greater nuclear charge both result in a net decrease in ionization energy.
The increase in ionization energy observed above is unrelated to the increasing atomic number, and instead has to do with the greater gravitational force of attraction to the heavier atomic nuclei.
Al2-) III
Mn)iV
a) III
thanks
To identify the correct electronic configuration for the given species, let's analyze each option one by one.
i. Al2-
Al2- indicates that there are two additional electrons added to the neutral aluminum atom (Al). The atomic number of aluminum is 13, which means it has 13 electrons in its neutral state. Adding two more electrons gives us a total of 15 electrons.
To determine the electronic configuration, we can use the periodic table. Starting from hydrogen (H) and moving towards aluminum (Al), we fill the electron orbitals in a specific order: 1s, 2s, 2p, 3s, 3p, and so on.
The electronic configuration of aluminum (Al) is 1s2 2s2 2p6 3s2 3p1. Adding two more electrons, we get Al2- as 1s2 2s2 2p6 3s2 3p6.
So, the correct electronic configuration for Al2- is 1s2 2s2 2p6 3s2 3p6.
ii. Mn
The symbol "Mn" represents the element manganese. To find the electronic configuration of manganese, we need to look at its position in the periodic table.
Manganese (Mn) is located in the d-block, specifically in period 4 and group 7. The electronic configuration of the element before manganese (Cr) is 1s2 2s2 2p6 3s2 3p6 4s2 3d10, and the electronic configuration of the element after manganese (Fe) is 1s2 2s2 2p6 3s2 3p6 4s2 3d6.
Based on this, we can determine that the electronic configuration of manganese (Mn) is 1s2 2s2 2p6 3s2 3p6 4s2 3d5.
So, the correct electronic configuration for Mn is 1s2 2s2 2p6 3s2 3p6 4s2 3d5.
(a) Select the reason that best explains why the first ionization energy decreases as atomic number increases for the alkali metals (Li, Na, K, Rb, Cs).
The correct reason is: The outermost shell becomes occupied with a greater number of electrons as the atomic number increases resulting in greater electron-electron (a.k.a Born) repulsion, which in turn makes it easier to ionize one of these electrons.
As the atomic number increases in the alkali metals, more electrons are added to the outermost shell. These electrons repel each other due to similar charges, creating electron-electron repulsion. This repulsion makes it easier to remove one of the electrons, thus reducing the ionization energy.
Therefore, the correct explanation is that the increase in ionization energy observed above is unrelated to the increasing atomic number, and instead has to do with the greater gravitational force of attraction to the heavier atomic nuclei.