Inorganic Chemistry with Help Guides

  The energy released when an extra. electron is added to a neutrai gaseous atom is termed the electron affinity. Usually only one electron is added, forming a uninegative ion. This ~epels further electrons and energy is needed to add on a second electron: hence· the negative affinity of 0 2-. Electron affinities depend on the size and effective nuclear charge. They cannot be determined directly, but are obtained indirectly from the Born-Haber cycle.  Negative Clectron affinity values indicate that energy is.given out when the atom accepts an electron. The above values show that the halogens all evolve a large amount of energy on forming negative halide ions, and it is not surprising that these ions occur in. a large number of compounds. Energy is evolved when one electron is added to an 0 or S atom, forming the species o- and s-, but a substantial amount of energy is absorbed when two electrons are added to form 0 2- and s2- ions. Even though it requires energy to form these ...

  If a small amount of energy is supplied to an atom, then an electron may be promoted to a higher energy .level, but if the amount of. energy supplied is sufficiently large the electron may be completely removed. The energy required to remove the most loosely bound electron from an isolated gaseous atom is called the ionization energy. Ionization energies are determined from spectra and are measured in kJ mo1- 1 • It is possible to remove more than one electron from most atoms. The first ionization energy is the energy required to remove the first electron and convert M to M+; the second ionization energy is the energy required to remove the second election arid convert M+ to M2+; the third ionization energy converts M2+ to M3+, and so on. The factors that influence the ionization energy are:  1. The size of the atom.  2. The charge on the nucleus.  3. How effectively the inner electron shells screen the nuclear charge.  4. The type of electron involved (s. p, ...

 Irrespective of which set of ionic radii are used, the following trends are observed:  1. In the main groups, radii increase on descending the group, e.g. u+ = 0.76A, Na+ = l.02A, K+- = l.38A, because extra shells of electrons are added. 2. The ionic radii decrease moving from left to right across any period in the periodic table, e.g. Na+ = 1.02A, Mg2+ = 0.720Aand Al3+ = 0.535 A. This is partly due to the increased number of charges on the nucleus, and also to the increasing charge on the ions.  3. The ionic radius decreases as more electrons are ionized off, that is as the valency incteases, e.g. Cr2+ = 0.80 A (high spin), Cr3+ = 0.615 A, Cr4+ = 0.55 A, CrH = 0.49 A and Cr6+ = 0.44 A.  4. The d and f orbitals do not shield the nuclear charge very effectively. Thus there is a significant reduction in the size of ions just after 10d or 14f electrons have been filled in. The latter is called the lanthanide contraction, and results in the sizes of the second and third...

 There are several problems in obtaining an accurate set of ionic radii.  1. Though it is possible to measure the internuclear distances in a crystal very accurately by X-ray diffraction, for example the distance between Na+ and p- in NaF, there is no universally accepted formula for apportioning this to the two ions. Historically several different sets of ionic radii have been estimated. The main ones are by Goldschmidt. Pauling and Ahrens. These are all calculated from observed internuclear distances, but differ in the method used to split the distance between the ions. The most recent values, which are probably the most accurate, are by Shannon (1976) .  2. Corrections to these radii are necessary if the charge on the ion is changed.  3. Corrections must also be made for the coordination number, and the geometry.  4. The assumption that ions are spherical is probably true for ions from the s- and p-blocks with a noble gas configuration, but is probably untrue...

  Metals usually form positive ions. These are formed by removing one or more electrons from the metal atom. Metal ions are smaller than the atoms from which they were formed for two. reasons:  1. The whole of the ·outer shell of electrons is usually ionized, i.e. removed. This is one reason why catfons· are much smaller than the original metal atom.  2. A second factor is the effective nuclear charge. In an atom, the number of positive charges on the nucleus is exactly the same as the number of orbital electrons. When a positive ion is formed; the number of positive charges on the nucleus exceeds the number of orbital electrons, and the effective nuclear charge (which is the ratio of the nun:iber of charges on the nucleus to the number of electrons) is increased. This results in the remaining electrons being more strongly attracted by the nucleus. Thus the electrons are pulled in - further reducing the size.

  The size of atoms decreases fr()m left to right across a period in the periodic table. For example. on moving from lithium to beryllium one extra positive charge is added to the nucleus, and an extra orbital electron is also added. Increasing the nuclear c:harge results in all of the orbital electrons being pulled closer to the nucleus. In a given period, the alkali metal is the largest atom and the halogen the smallest. When a horizontal period contains ten transition elements the contraction in size is larger, and when in addition there are 14 inner transition eleme11ts in a horizontal period, the contraction in size is even more marked. On descending a group in the periodic table such as that containing lithium. sodium, potassiUm, rubidium and caesium, the sizes of the atoms increase due to the effect of extra shells of electrons being added: this outweighs the effect of increased nuclear charge.