Sodium valence electrons

The Na atom has one half-filled (3s1) and three empty orbitals (3px, 3py, 3pz). The number of valence orbitals is greater than the number of valence electrons. In the solid state, sodium atoms are surrounded by other sodium atoms. Thus, the valence electron of the sodium atom in the 3s orbital can move to the empty orbitals (3px, 3py, 3pz) of neighbouring atoms. When each sodium valence electron behaves in this way a sea of electrons is built up around the sodium atoms (now positive ions, having lost a valence electron). 

The selection rules (3.76)—(3.78) were derived without reference to the form of the radial functions hence they are valid for any one-particle central-field problem. For example, they hold well for the sodium valence electron, which moves outside a closed-shell structure. 

When alkali metal atoms are co-condensed with strong electron acceptors in matrices, electron transfer occurs spontaneously, as discussed above in the case of K + O3. If not spontaneous, electron transfer can easily be activated by photolysis. Irradiation by the Nao line (583 nm cf. ionization energy corresponding to 243 kJmol ) is sufficient to transfer the sodium valence electron to several molecular substrates e.g., CI2 CI2, B2H6 B2Hft, HI -> H + I. ... 

In the sodium atom pairs of 3/2 states result from the promotion of the 3s valence electron to any np orbital with n > 2. It is convenient to label the states with this value of n, as n P 1/2 and n f 3/2, the n label being helpful for states that arise when only one electron is promoted and the unpromoted electrons are either in filled orbitals or in an x orbital. The n label can be used, therefore, for hydrogen, the alkali metals, helium and the alkaline earths. In other atoms it is usual to precede the state symbols by the configuration of the electrons in unfilled orbitals, as in the 2p3p state of carbon. 

Because they possess an odd number of valence electrons the elements of this group can only satisfy the 18-electron rule in their carbonyls if M-M bonds are present. In accord with this, mononuclear carbonyls are not formed. Instead [M2(CO)s], [M4(CO)i2] and [M6(CO)i6] are the principal binary carbonyls of these elements. But reduction of [Co2(CO)g] with, for instance, sodium amalgam in benzene yields the monomeric and tetrahedral, 18-electron ion, [Co(CO)4] , acidification of which gives the pale yellow hydride, [HCo(CO)4]. Reductions employing Na metal in liquid NH3 yield the super-reduced [M(CO)3] (M = Co, Rh, Ir) containing these elements in their lowest formal oxidation state. 

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. 

Ionization lithium, 267 magnesium, 270 sodium, 270 Ionization energy, 267 alkaline earths, 379 and atomic number, 268 and ihe periodic table, 267 and valence electrons, 269 halogens, 353 measurement of, 268 successive, 269 table of, 268 trends, 268... 

Simons, L., Comment, phys.-math. Helsinki) 17, paper 7, "Calculation of energy levels of the valence electron in sodium." Prokofiev field instead of Hartree-Fock field. 

Sodium is in Group 1 of the periodic table and can be expected to form a +1 ion. However, the valence electron is tightly held by the effective nuclear charge—... 

Sodium has 1 valence electron, and 10 bound electrons. The first two excited states are the 3 Pi/2 and the 3 P3/2 states. Transitions to these levels give rise to the Di and D2 transitions respectively. There are two h)q)erfine levels in the 3 ground state, and four h)q)erfine levels in the 3 Pa/2 excited state (Fig. 3). There is no significant energy difference between the h)q)erfine levels in the 3 Pa/2 state. Thus, the six permitted fines appear in two groups, producing a double peaked spectral distribution, with the peaks separated by 1.772 GHz. 

Consider the number of valence electrons in each element. clO-0095. The following energies measure the strength of bonding in sodium, magnesium, and aluminum... 

Atoms go about getting eight valence electrons in the least energetic fashion. Sodium has the following electron configuration ... 

Chlorine would have to lose seven electrons to reach an electron configuration like that of neon. But if it gained one, it would have the same stable electron configuration as argon. So that is what chlorine does. If it meets an atom with a high-energy valence electron, such as sodium, the electron migrates to the chlorine atom and forms a chloride ion ... 

Note, however, that hydroxide is not on the periodic chart. That is because hydroxide is a polyatomic ion. Remember that polyatomic ions always travel together as a unit. The only way to know the charge of a polyatomic ion is to memorize it. Hydroxide has a charge of -1. So, when sodium (with a charge of +1) and a hydroxide come into contact with one another, a sodium atom gives its one valence electron to a hydroxide ion, which needs one... 

At low temperatures magnesium oxide, MgO, which adopts the sodium chloride structure, is virtually a stoichiometric phase, but at high temperatures in the MgO-A1203 system this is not so. At 1800°C the approximate composition range is from pure MgO to 5 mol % A1203 95 mol % MgO. The simplest way to account for this composition range is to assume that point defects are responsible. For this, because both Mg2+ and Al3+ cations in this system have a fixed valence, electronic compensation is unreasonable. There are then three ways to account for the composition range structurally ... 

Figure 11.1 Simple model of valency and bonding. The sodium atom (Z = 11) has electronic configuration ls22s22p63s1, drawn simply as (2, 8, 1) (i.e., showing all the n = 2 electrons as a single orbital). Chlorine (Z = 17) is s22s22p63s23p5, drawn as (2, 8, 7). In bonding to form the ionic compound NaCl, the outer (3s) electron of Na is donated to the outer orbital of Cl, giving both a full outer orbital of eight electrons, and leaving the sodium one electron short (i.e., the Na+ ion) and chlorine one extra (Cl-).

Sodium loses its valence electron and its electron configuration becomes identical to that of neon Is2 2s2 2p6. Likewise, the valence shell of chlorine becomes completely filled and its electron configuration resembles that of argon. As a result, during the reaction... 

Let us compare the metallic bonds of sodium, magnesium and aluminum, sodium (nNa) has one valence electron, magnesium (12Mg) has two and aluminum (13A1) three valence electrons. 

The number of electrons to be lost by the metal and gained by the nonmetal is determined by the number of electrons lost or gained by the atom in order to achieve a full octet. There is a rule of thumb that an atom can gain or lose one or two and, on rare occasions, three electrons, but not more than that. Sodium has one valence electron in energy level 3-... 

If it lost that one, the valence shell, now energy level 2, would be full (a more common way of showing this is with zero electrons). Chlorine, having seven valence electrons, needs to gain one more in order to complete its octet. So an electron is transferred from sodium to chlorine, completing the octet for both. 

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