Aluminum electron configuration

Write out the electron configuration of sodium, magnesium, and aluminum and find the ionization energies for all their valence electrons (Table 20-IV, p. 374). Account for the trend in the heats of vaporization and boiling points (Table 20-1) of these elements. Compare your discussion with that given in Section 17-1.3. 

First consult the periodic table to locate aluminum and determine how many electrons are present in a neutral atom. Then construct the electron configuration using the patterns of the periodic table. 

The compound is ionic — a metal (Al) bonded to a nonmetal (Cl). All ionic compounds are solids at room temperature and pressure. Aluminum has 13 electrons. As an ion, it will lose 3 electrons to become isoelectronic with neon. Thus the aluminum ion will have the electronic configuration ls22s22p6. 

Chlorine as a free element is diatomic (Cl2) however, as an ion, it will gain one electron to become isoelectronic with argon. The electronic configuration of the chloride ion is s22s22p63s23p6. The compound thus formed, aluminum chloride, has the formula A1C13. 

Prandtl mixing length hypothesis, 11 779 Prandtl number, JJ 746, 809 13 246-247 Praseodymium (Pr), J4 631t, 634t electronic configuration, J 474t Praseodymium bromide, physical properties of, 4 329 Prater equation, 25 270, 299 Prater number, 25 299, 300-301, 303 effect on maximum dimensionless intrapellet temperature, 25 304, 309 effect on maximum intrapellet temperature, 25 306 Prato reaction, 12 244 Pratsinis aluminum nitride, 17 212 Pravachol, 5 143... 

Use the aufbau principle to write complete electron configurations and complete orbital diagrams for atoms of the following elements sodium, magnesium, aluminum, silicon, phosphorus, sulfur, chlorine, and argon (atomic numbers 11 through 18). 

Aluminum loses three valence electrons to form the Al3+ ion, which has the same electron configuration as neon. When the atoms of metals on the left of the p block in Periods 4 and higher lose their s- and p-elec-trons, they leave a noble-gas core surrounded by an additional, complete subshell of d-electrons. For instance, gallium forms the ion Ga3+ with the configuration [Ar]3d10. The d-electrons of the p-block atoms are gripped tightly by the nucleus and, in most cases, cannot be lost. We saw in Section 1.18 that the inert-pair effect implies that the elements listed in Fig. 1.44 can lose either their valence p-electrons alone or all their valence p- and s-electrons. 

All silver crystals have the same geometric shape. Therefore, the crystalline shape of a metallic solid is a function of the size of the metal solid atoms and their electron configuration. Each metal has its own geometric crystalline shape. Aluminum atoms pack into a face-centered cubic cell. Iron s solid structure is body-centered cubic. 

Determine the places of aluminum element in the periodic table, if its electron configuration is ls22s22p63s23p1... 

Looking at the electron configuration allows for a prediction whether or not an atom (or ion) is paramagnetic. Note that the prediction only applies to the free individual atom. Any firm conclusion based on the prediction is risky when applied to collections of atoms (or ions). For example, the aluminum atom has one unpaired electron, but a piece of aluminum is diamagnetic. 

There are only three electrons in the 3 energy level and the octet rule says there should be eight. Aluminum can try and find 5 more electrons to fill the 3 p orbital or lose the 3s2, and 3p1 electrons. It takes less energy to lose the three electrons (circled), When these electrons are lost an aluminum +3 ion forms with an electron configuration of 1s2,2s2, 2p6 which has a complete octet in the outer shell. Al lost 3 electrons and now Al has 10 electrons and 13 protons and the charge is 3+. 

All aluminum nuclei have charge +13 electronic units, so that 13 electrons orbit the nucleus of the neutral atom. Its electronic configuration can be abbreviated as an inner core of inert neon (a noble gas) plus three more electrons (Ne)3s23p1, which locates Al in Group IIIA of the chemical periodic table, between boron (B) and gallium (Ga). Its principal oxidation state is +3, so its oxide is A1203, a very important compound in cosmochemistry. Its chloride is A1C13. 

There is also a pronounced tendency for the Group IIIA metals to form metal-metal bonds and bridged structures. The electron configuration ns2 np1 suggests the possible loss of one electron from the valence shell to leave the ns2 pair intact. The electron pair that remains in the valence shell is sometimes referred to as an inert pair, and a stable oxidation state that is less than the group number by two units is known as an inert pair effect. The fact that oxidation states of +2, +3, +4, and +5 occur for the elements in Groups IVA, VA, VIA, VIIA, respectively, shows that the effect is quite common. Thus, it will be seen that the Group IIIA metals other than aluminum have a tendency to form +1 compounds, especially thallium. 

Suppose we want to write the electronic configuration of titanium (atomic number 22). We can rewrite the first 13 electrons that we wrote above for aluminum and then just keep going. As we added electrons, we filled the first shell of electrons first, then the second shell. When we are filling the third shell, we have to ask if the electrons with = 3 and / = 2 will enter before the n = 4 and / = 0 electrons. Since + / for the former is 5 and that for the latter is 4, we must add the two electrons with n = 4 and 1 = 0 before the last 10 electrons with n = 3 and / = 2. In this discussion, the values of mi and ms tell us how many electrons can have the same set of n and / values, but do not matter as to which come first. 

Give the ground-state electron configuration for each of the following elements (a) Sodium (b) Aluminum (c) Silicon (d) Calcium... 

Write the full electron configuration for the element aluminum (Al). 

The first thing that we need to do is find the electron configuration for aluminum. Because we did this example in the last lesson, I won t go through the steps. Refer back to the last lesson if you don t remember how to do this. We will space the configuration out a bit more this time, so that we have room to write the orbital notation. 

F.]—Aluminum, with the electron configuration of Is2 2s2 2p63s2 3p, has 3 valence electrons. 

Metals can attain a more stable electronic configuration by accepting electrons from n-donor and n-donor systems. As discussed in Chapter 2, a variety of compounds such as alkylfithiums, aluminum, beryllium, magnesium, and other related systems form stable alkyl bridged species where the metal draws upon the electrons in a bonds in the alkyl substituent to attain electronic stability. 

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