ATOMIC RADIUS
According to wave mechanics there is no sharp boundary of atom however we can consider the following radius to the adopted under different form.
Half of the distance between the nuclei of two nearest packed atom is called metallic radius metallic radius > covalent radius
For example:
K(metallic) = 2.34 Å
K(covalent) = 2.03 Å
ATOMIC OR COVALENT RADIUS
If two similar atoms are linked by single covalent bond than half of the internuclear distance is referred as covalent radius
i.e. rA
If . Then
if
where Xa and Xb are electronegativity of A and B respectively.
Vander wall’s radius may be defined as one half of the distance between the nuclei of two adjacent atom belonging to two neighbouring molecules of an element in solid state.
Thus rcovalent < fmetallic < rvan der salls
Let us see the variation of atomic radius in a period. For example, consider the second period elements from Li to F. On moving from left to right, we can
see that the number of electrons in the inner shell (1s2) remains constant while the nuclear charge increases. The electrons that are added to counter balance the increasing nuclear charge are ineffective in shielding one another. Consequently, the effective nuclear charge increases steadily while the principal quantum number remains constant (n = 2). For example, the outer 2s electron in lithium is shielded from the nucleus (which has 3 protons) by the two 1s electrons. Each 1s electron shields the 2s electron by 0.85 units, so net shielding by two 1s electrons is 1.7(\(\sigma\)). Thus the effective nuclear charge of Li is 1.3. The electron configuration of Be is 1s22s2. One 2s electron shields other 2s electron by 0.35 unit while each 1s electron shields 2s electron by 0.85. So, the Zeff for Be would be 4 – [(2 X 0.85) + 0.35] = 1.95. Similarly, the Zeff for B would be 5-[(2 X 0.85) + (2 X 0.35)] = 26. Thus, it is evident that the effective nuclear charge increases, so progressively the outermost electrons are held more strongly by the nuclear and thus the atomic radius decreases steadily form Li to F.
Now, let us observe the trend of atomic radius in a group. For the alkali metals in group IA (Li, Na, K, Rb, Cs & Fr), the outermost electron resides in the ns orbital. In moving form Li to Na,the increase in Zeff is very less (Zeff for Li is 1.3 while for Na is 2.5), but orbital size increases with the increasing principal quantum number n, which supersedes the effect of increasing Zeff So, thus the size of the metal atoms increase from Li to Na and So on. Thus, atomic radius increases on moving from top to bottom in a group.
Ionic radius is the radius of a cation or an anion. When a neutral atom is converted to an ion, there would be change in size, as the Zeff will change but the number of protons in nucleus remains same.
If the atom forms an anion, its size (or radius) increases, since the nuclear charge remains the same but the repulsion resulting from the additional electron(s) enlarges the domain of the electron cloud. On the other hand, if one or more electrons are removed from an atom, it reduced the electron-electron repulsion but the nuclear charge remains the same, so the electron cloud shrinks and the cation is smaller than the atom. When a lithium atom reacts with a fluorine atom to form a LiF unit, the changes in size are very peculiar. Out of Li and F, Li is bigger in size. When lithium changes to Li+, its size decreases and when F changes to F-, its size increases. In LiF, Li+ is smaller than F- (Note that Li+ is smaller than F atom and F- is smaller than Li atom).
The variation of ionic radii in a period and a group is same as that of atomic radii. Thus, ionic radius decreases in a period while it increases in a group.
For ions derived from elements in different groups, a size comparison is meaningful only if the ions are isoelectronic.
If we examine isoelectronic ions, the cations are smaller than anions. Let us compare the radius of Na+ ion and F- ion. Both ions have same number of electrons(10), but Na (Z = 11) has more protons than F(Z = 9). Thus, Zeff of Na+ is more than that of F-, so Na+ ion is smaller in size than F- ion. Similarly, for the three isoelectronic ions of third period, Al3+, Mg2+ and Na+ they all have the same number of electrons (10) but their number of protons are 13, 12 and 11 respectively. Thus, the electron cloud in Al3+ is pulled inward more than that in Mg2+ and Mg2+ would be smaller than Na+. Thus in general in an isoelectronic cation series, the radii of tripositive ions are smaller than those of dipositive ions, which in turn are smaller than unipositive ions. Similarly, in isoelectronic anions series, the radius, increases as we go from uninegative ion to dinegatie ion and so on.
Cation is smaller than its corresponding atom
reason 1. Increase in effective nuclear charge.
2. Vanishing of outer most orbit. i.e. Na+ < Na
Anion is bigger than corresponding atom
reason 1. Decrease in effective nuclear charge.
2. Increase in interelectronic repulsion. i.e. Cl- > Cl
IONIZATION ENERGY
The stability of the outermost electrons is reflected directly in the atom’s ionization energies.
Ionization energy is the minimum energy (in kJ/mole) required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. Alternatively, ionization energy is the amount of energy in kilojoules needed to knock out one mole of electrons from one mole of isolated gaseous atoms. In this definition, gaseous atoms are specifically used because an atom in the gas phase is virtually uninfluenced by its neighbors and so there are no intermolecular forces to take into account while measuring ionization energy.
The magnitude of ionization energy is a measure of how “tightly” the electron is held in the atom. The higher the ionization energy, the more difficult it is to remove the electron. For a multi-electron atom, the amount of energy required to remove the first electron from the atom in its ground state,\(X\left( g \right)\,\,+\,\,energy\,\,\,\to \,\,\,{{X}^{+}}\left( g \right)\,\,+\,\,{{e}^{-}}\) is called the first ionization energy (I1). The second ionization energy (I2) and the third ionization have as many ionization energy values as the number of electrons present in that atom.
When an electron is removed from a neutral atom, the repulsion among the remaining electrons decreases. Since the nuclear charge remains constant, more energy is needed to remove another electron from the positively charged ion. Thus ionization energy increases in the order I1 < I2 < I3 <……
Ionization is always an endothermic process, thus ionization energies are all positive quantities.
Ionization Energy Trend in Periods
The first ionization energy of the elements in a period increases with some irregularities. This trend is due to increase in Zeff from left to right. A larger Zeff means a more tightly held outer electron and hence a higher first ionization energy.
In a period, highest ionization energy is of noble gases. The high ionization energies of the noble gases stems from the fact that they have stable ground state electron configurations (ns2np6). Among noble gases. He (1s2) has the highest ionization energy.
In a period, lowest ionization energy is of alkali metals. Alkali metals has one valence electron, which is effectively shielded by the completely filled inner shells. Consequently, it is easy to remove an electron to form a unipositive ion having stable noble gas configuration.
Ionization Energy Trend in 2nd Period
In the second period from Li to Ne, the first ionization energy increases from Li to Be, then decreases at B, then increases till N, then again decreases at O and then increases till Ne.
Explanation of Irregularity in 2nd Period
Let us look at the first ionization energies of Be & B. The electron configurations of the two elements are 1s22s2 and 1s22s22p1. Although the Zeff of B is more than Be, yet ionization energy of B is less than Be. It is because when an electron is taken out from B atom, it results in the formation of a stable configuration (1s22s2) for B+ while same process when done with Be results in configuration (1s22s1) for Be+. If stable configuration is obtained for an ion, by removing an electron from an atom, less energy is required to remove the electron. Conversely, large amount of energy is needed to remove electron from a sable configuration of 1s22s2 in Be than in B (1s22s22p1).
In case N and O, N has stable half-filled configuration (1s22s22p3) while oxygen’s electron configuration is 1s22s22p4. Thus, it is evident that knocking an electron out from N would require greater energy than required for oxygen.
Ionization Energy Trend in groups
Which atom should have a higher second ionization energy: lithium or beryllium?
The group IIA elements have higher first ionization energies than the group IA elements. The alkaline earth metals have two valence electrons. Because these two s electrons do not shield each other well the Zeff for an alkaline earth metal atom is larger than that for the preceding alkali metal. But second ionization energy is higher ion alkali metals than for alkaline earth metals because after removing one electron, alkali metal atoms have acquired stable noble gas configurations & thus the removal of second electron becomes more difficult.
Metals have relatively low ionization energies compared to non-metals. The ionization energies of the metalloids generally fall between those of metals and non-metals. This difference in ionization energies of metal & non-metals suggest why metals always form cations and non-metal form anions in ionic compound.
FACTOR AFFECTING IONIZATION ENERGY
(a) Nuclear charge I.P. \(\propto \,\,F\,\,\propto \,\,\,{{Z}_{eff}}\)
where F = Force of attraction between nucleus & valence electron.
Zeff = effective nuclear charge
(b) Atomic size: I.P. \(\propto\)F \(\propto\) \(\frac{1}{{{r}^{2}}}\)
(c) Principal Quantum Number: I.P.\(\propto\) \(\frac{1}{n}\)
where n is principal Quantum Number
(d) Screening effect: I.P. \(\propto\)\(\frac{1}{Screening\,\,effect}\)
more screening means I.P. low
\(\oplus \,\,|\,\,|\,\,|\,\,|\,\,\bullet\) (Valence electron)
less screening means I.P. high
\(\oplus \,\,|\,\,|\,\,\,\,\bullet\) (valence electron)
(e) Penetration effect:
Penetration of the orbital towards the nucleus follows the order s > P > d > f
I.P. follows the order s > p > d > f
Application of I.P.
I.P. |
Application |
Low |
More Reactive |
Low |
Metallic character |
Low |
Reducing capacity more |
Low |
Stable |
Low |
Ionic bond |
Low |
Hydroxide is stronger |