Ionization potential trend

What is ionization energy?

    Ionization energy is the minimum energy required to remove an electron completely from the gaseous atomic species. If the source energy is large enough it may possible to remove the electron completely from the pull of the positive nucleus of an atom.
    The amount of energy required to remove the most loosely bound electron, that is outermost electron from an isolated gaseous atom of an element in its lowest energy state to produce cation is known as Ionization energy.
    M(g) + Ionization Energy → M⁺(g) + e
    It is generally represented as I or IP and it is measured in electron volt (eV) or kilocalories (kcal) per gram atom. One electron volt (eV) being the energy consumption by an electron falling through a potential difference of one volt.
∴ 1 eV = (Charge of an electron) × (1 volt)
= (1.6 × 10⁻¹⁹ Coulomb) × (1 Volt)
= 1.6 × 10⁻¹⁹ Joule
∴ 1 eV = 1.6 × 10⁻¹² erg

Energy consumption for removal of an electron

    We know that the energy corresponding to the transition from n = ∞ to n = 1 gives the energy consumption or ionization potential of the hydrogen atom.
Thus the ionization potential of the hydrogen atom,
EH = (2π²me⁴/h²)[(1/n₁²) - (1/n₂²)]
∴ EH = 2.179 × 10⁻¹¹ erg
= 2.179 × 10⁻¹⁸ Joule
= (2.179 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) eV
∴ EH = 13.6 eV

Energy consumption for removal second and third electron

  1. The electrons are removed in stages one by one from an atom. The energy consumption to remove the first electron from a gaseous atom is called its first Ionization potential.
    M (g) + IP₁ → M⁺ (g) + e
  2. The energy consumption to remove the second electron from the cation is called second Ionization potential.
    M⁺ (g) + IP₂ → M⁺² (g) + e
  3. Similarly, we have third, fourth Ionization potentials.
    M⁺² (g) + IP₃ → M⁺³ (g) + e
    M⁺ (g) + IP₄ → M⁺⁴ (g) + e
Question
    Calculate the second ionization potential of helium (given ionization potential of hydrogen = 13.6 eV)
Answer
    The ground state electronic configuration of helium is 1S². The second ionization potential means the removal of the second electron from the 1S orbital against the nuclear charge of +2.
So we have: IPHe = (2π²mZ²e⁴/h²)[(1/n₁²) - (1/n₂²)]
= Z² × IPH
∴ Second Ionization potential of Helium, = 22 × 13.6
= 54.4 eV

Ionization energy trend

Ionization potential trend in periodic table
Ionization potential trend

Ionization energy trend for atomic radius

    The greater the distance of an element from the positive charge nucleus of an element, the weaker will be the attraction and hence the lower the ionization potential of that element.
    If an atom is raised to an excited state by promoting one electron to a higher energy level, then that electron is further from the nucleus and it is observed that this excited electron is more easily detached than when the electron is in its ground state.
    Thus on moving from top to bottom in a group the ionization potential of the elements decreases with the increase of their nucleus electron separation.

Ionization energy trend for the nuclear charge of an atom

    The higher the charge on the nucleus the more difficult it is to remove an electron and hence the higher the value of ionization potential.
    Thus the value of ionization potential generally increases in moving from left to right in a period, since the nuclear charge of the elements (atomic number) also increases in the same direction.
    With increasing, atomic number electrons are added to orbitals of the same principal quantum number. These electrons add little to the size of the atom, the increasing nuclear charge brings about a contraction in size. In effect, therefore, ionization potential steadily increases along a period.

Ionization energy trend for filled and half-filled orbital

    According to Hund's rule atoms having half-filled or completely filled orbital are comparatively more stable and hence more energy consumption to remove an electron from such atom.
    The ionization of such atoms is therefore relatively difficult than expected normally from their position in the periodic table. A few regulations that are seen in the increasing value of ionization energy along a period can be explained on the basis of the concept of the half-filled and completely filled orbitals.
    Be and N in the second period and Mg and P in the third period have a slightly higher value of ionization potentials than those normally expected.
    This is explained on the basis of extra stability of the completely - filled 2S - orbital in Be (2S²) and 3S - orbital in Mg (3S²) and of half-filled 2P - orbital in N (2S² 2P³) and 3P - orbital in P (3S² 3P³).

Ionization energy trend for shielding effect

    Electrons provide a shielding effect on the nucleus of an atom. The outermost electrons are shielded from the nucleus by the inner electrons.
    The radial distribution functions of the S, P, d orbitals show that for the same principal quantum number the S - orbital is the most penetrating., next is P - orbital and least penetrating is the d - orbital. shielding efficiency falls off in the order, S〉P〉d.
    As we move down a group, the number of inner - shells increases and hence the ionization potential tends to decreases.
Elements of II A Group: Be〉Mg〉Ca〉Sr〉Ba
Question
    In the first transition series electron filling up processes begins in the 3d level below a filled 4S level. During the ionization process will a 4S electron or a 3d electron be lost first? Explain with reference to chromium.
Answer
    During ionization, the 4S electron lost first.

Ionization energy trend for overall charge

    An increase in the overall charge on the ionizing species (M⁺, M⁺², M⁺³, etc) will enormously influence the ionization potential since electron withdrawal from a positively charged species is more difficult than from a neutral atom.
    The first ionization of the elements varies with their positions in the periodic table. In each of the tables, the noble gas has the highest value and the alkali metals the lowest value for the ionization potential.

Ionization energy trend for valence electrons in an atom

    If an atom has valence electronic configuration is nS² nP⁶, that is the atom attains stable noble gas configuration.
    The removal of an electron from such atoms needed greater energy thus the ionization potential of such atoms also increases.

Trend of ionization energy along a period

Ionization energy trend in periodic table
Ionization energy trend along a period
Question
    Arrange the following with the increasing order of ionization energy Li, Be, B, C, N, O, F, Ne.
Answer
    LiㄑBㄑBeㄑCㄑOㄑN〈FㄑNe
    The greater the charge on the nucleus of an atom the more energy consumed for removing an electron from the atom.
    More energy consumed means increasing the ionization potential of an atom. With the increase in nuclear charge the electrostatic attraction between the outermost electrons and the nucleus increases. Thus removal of an electron from an atom is more difficult.
    The values of ionization energy generally increase in moving left to right in a period since the nuclear charge of an element also increases in the same direction.

Ionization potential trend, energy consumption for removal of an electron and ionization energy along a period in periodic table for study chemistry

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