Sodium second ionization energy

The second ionization energy of sodium is much larger than its first ionization energy because a core 2 p electron must be removed to create Na from Na. Removal of a core electron always requires a great deal of energy, so it is a general feature of ionic systems that ions formed by removing core electrons are not found in stable ionic compounds. 

For the element sodium, Na, the first ionization energy is 495.8 kJ/mol and the second ionization energy is about 5.5 times greater than the first ionization energy. Thus, the sodium, Na, atom gets a noble gas electron configuration after giving off one electron. 

The effect of closed shells is apparent. For example, sodium has a first ionization energy of only 495.8 kJ mol-1, whereas the second is 4562.4 kJ mol-1. The second electron removed comes from Na+, and it is removed from the filled 2p shell. For Mg, the first two ionization potentials are 737.7 and 1450.7 kJ moF1, and the difference represents the additional energy necessary to remove an electron from a +1 ion. Thus, the enormously high second ionization energy for Na is largely due to the closed shell effect. 

You can calculate the lattice enthalpy by moving around the cycle and noting that the enthalpy of formation Af/f(NaCl2) = Asd,W (Na) + A,usW (Ci2) + /i(Na) + /2(Na) - A jjW CI) -A / (NaCb)- The second ionization energy of sodium is 4562 kJ mol and is responsible for the fact that the compound does not exist as it would result in a large, positive enthalpy of formation. 

Without referring to your text, predict the trend of second ionization energies for the elements sodium through argon. Compare your answer with Table 7.5. Explain any differences. 

The second ionization energy, I2, is the energy needed to remove the second electron, and so forth, for successive removals of additional electrons. Thus, I2 for the sodium atom is the energy associated with the process... 

Construct a Born-Haber cycle for the formation of the hypothetical compound NaCl2, where the sodium ion has a 24-charge (the second ionization energy for sodium is given in Table 7.2). (a) How large would the lattice energy need to be... 

Because this process represents the removal of the first electron from a neutral sodium atom, the energy necessary to accomplish the process is called the first ionization enei. If a second electron were removed, the energy required would be called the second ionization energy, and so forth. We will focns on only the first ionization energy for representative elements. > Figure 3.16 contains values for the first ionization energy of a number of representative elements. 

Without referring to your text, predict the trend of second ionization energies for the elements sodium through argon. Compare your answer with Table 7.5. Explain any differences. Account for the fact that the line that separates the metals from the nonmetals on the periodic table is diagonal downward to the right instead of horizontal or vertical. 

The energy required to remove the second electron is the second ionization energy lEf), the energy required to remove the third electron is the third ionization energy (JEf), and so on. We represent the second ionization energy of sodium as ... 

Notice that the second ionization energy is not the energy required to remove two electrons from sodium (that quantity is the sum of lEi and IE2), but rather the energy required to remove one electron from Na. We look at trends in lEi and IE2 separately. 

For sodium, there is a huge jump between the first and second ionization energies. For magnesium, the ionization energy roughly doubles from the first to the second, but then a huge jump occurs between the second and third ionization energies. What is the reason for these jumps ... 

Is NaCl2 a stable compound Here, the answer is no. The additional lattice energy associated with NaCl2 over NaCl is not nearly enough to compensate for the very high second ionization energy of sodium (see Exercise 114). 

The quantity of alkali metal retained on the MgO surface and the concentration of the newly created ionic superbasic centres depends on the position of metal in the Periodic Table. The greater the electropositivity in the sequence sodium, potassium, caesium, the greater is the reactivity with surface acceptor centres of MgO surfaces. It is possible that metals having lower ionization energy, such as potassium or caesium (Table 1), react with these surface centres of MgO, which are not affected by sodium atoms. In consequence an oxide surface that has been heated to a particular temperature is able to bind more caesium than sodium atoms. The increase of the quantity of metal retained on MgO surfaces is not followed by a simultaneous increase in the number of newly created ionic superbasic centres. The largest quantity of such centres is formed on MgO surfaces doped with potassium. It is interesting to note that in the case of MgO-K and MgO-Cs systems two types of superbasic centres occur, one with a basic strength of 33 < H < 35, the second one with H > 35 (Table 1). ... 

The L shell can accommodate a total of 8 electrons, and as further electrons are added to form the sequence of elements beryllium, boron, carbon, etc., these electrons take their place in the second shell until finally, at the end of the second period, neon (Z = 10) is reached, with both the first and second shells fully occupied and with an electronic structure which can be symbolized as (2, 8). The addition of further electrons to form the sequence of elements sodium, magnesium, etc., of the third period requires the formation of a new shell, and in sodium (Z = 11) a single electron occupies the M shell of principal quantum number 3 again the ionization energies reflect the difference in energy between this electron and those more tightly bound in the L and K shells. 

Note Trends in successive ionization energies are also related to the size of the atom or ion under consideration. It will take considerably more energy, for example to remove the second electron from a sodium atom than it did for the fiist, because the sodium ion is considerably smaller than the sodium atom. ... 

Notice the trends in the first, second, and third ionization energies of sodium (group 1 A) and magnesium (group 2 A), as shown at left. 

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