Shielding, from nuclear charge

The overall decrease in atomic and ionic radii (Table 24.1) from La to Lu has major consequences for the chemistry of the third row of J-block metals (see Section 22.3). The contraction is similar to that observed in a period of d-block metals and is attributed to the same effect the imperfect shielding of one electron by another in the same sub-shell. However, the shielding of one 4/ electron by another is less than for one d electron by another, and as the nuclear charge increases from La to Lu, there is a fairly regular decrease in the size of the Af" sub-shell. 

Sodium, element number 11, has ten electrons in inner shells, ls 2s 2p, and one electron in an outer shell, 3r. The ten inner-shell electrons of the sodium atom screen (shield) the outer-shell electron from most of the 11 -t- nuclear charge. Recall from Chapter 4 that the third shell n = 3) is farther from the nucleus than the second shell (n = 2). Thus, we see why sodium atoms are larger than lithium atoms. Similar reasoning explains why potassium atoms are larger than sodium atoms and why the sizes of the elements in each column of the periodic table are related in a similar way. 

Across a transition series, atomic size shrinks throngh the first two or three elements because of the increasing nuclear charge. But, from then on, size remains relatively constant because shielding by the inner d electrons counteracts the increase in eff- Thus, for example, in Period 4, the third transition element, vanadium (V ... 

The atomic radius decreases from left to right across a period. Although electrons are located in the same energy level within the s and p orbitals of the elements, the nuclear charge increases from left to right within a period. These electrons are not shielded very well from the nuclear charge, and the atomic radius decreases. The radii of the common elements in organic compounds decrease in the order C > N > O. 

The decrease in atomic radius moving across the periodic table can be explained in a similar manner. Consider, for example, the third period, where electrons are being added to the third principal energy level. The added electrons should be relatively poor shields for each other because they are all at about the same distance from the nucleus. Only the ten core electrons in inner, filled levels (n = 1, n = 2) are expected to shield the outer electrons from the nucleus. This means that the charge felt by an outer electron, called the effective nuclear charge, should increase steadily with atomic number as we move across the period. As effective nuclear charge increases, the outermost electrons are pulled in more tightly, and atomic radius decreases. 

As well as being attracted to the nucleus, each electron in a many-electron atom is repelled by the other electrons present. As a result, it is less tightly bound to the nucleus than it would be if those other electrons were absent. We say that each electron is shielded from the full attraction of the nucleus by the other electrons in the atom. The shielding effectively reduces the pull of the nucleus on an electron. The effective nuclear charge, Z lle, experienced by the electron is always less than the actual nuclear charge, Ze, because the electron-electron repulsions work against the pull of the nucleus. A very approximate form of the energy of an electron in a many-electron atom is a version of Eq. 14b in which the true atomic number is replaced by the effective atomic number ... 

The electron density in d-orbitals is low near the nucleus, and so they are not very effective at shielding other electrons from the positive nuclear charge. 

The atomic radii of the second row of d-metals (Period 5) are typically greater than those in the first row (Period 4). The atomic radii in the third row (Period 6), however, are about the same as those in the second row and smaller than expected. This effect is due to the lanthanide contraction, the decrease in radius along the first row of the / block (Fig. 16.4). This decrease is due to the increasing nuclear charge along the period coupled with the poor shielding ability of /-electrons. When the d block resumes (at lutetium), the atomic radius has fallen from 217 pm for barium to 173 pm for lutetium. 

Metals in Groups 11 and 12 are easily reduced from their compounds and have low reactivity as a result of poor shielding of the nuclear charge by the d-electrons. Copper is extracted from its ores by either pyrometallurgical or bydrometallurgical processes. 

As the value of n increases, d- and /-electrons become less effective at shielding the outermost, highest-energy elec-tron(s) from the attractive charge of the nucleus. This higher effective nuclear charge makes it more difficult to oxidize the metal atom or ion. 

As the atom radii increase down group 1, the outermost electron is further from the nucleus (1) and more shielded from the nuclear charge, so is held less tightly (despite the increasing number of protons in the nucleus of each atom) (1). Always answer in terms of factors such as distance, shielding and nuclear charge. 

As we move across the period, atomic radius decreases as electrons are added to the same shell ( = 3) (1) and, therefore, the outermost electrons experience the same shielding (1) from the nuclear charge. The increase in the number of protons (1) causes the decrease in radius from Na to Ar. These factors also explain the general increase in first ionisation energy across the period. 

---