Factors Affecting the Lonisation Enthalpy - Anomalies in Lonisation Enthalpy, Practice Problems, FAQ

On what factors does your willingness to clear up your lunch plate depend?

Tt depends on several factors like your appetite on that day, how delicious the food is, etc. A family of five, sitting together at a table for lunch would have different levels of appetite.

Similarly, the willingness of atoms of different elements (part of the periodic table family) to lose electrons depends on several factors.

TABLE OF CONTENTS

Factors Affecting Ionisation Enthalpy
Anomalous Trends in Ionisation Enthalpy
Practice Problems
Frequently Asked Questions - FAQ

Factors Affecting Ionisation Enthalpy

Size of Atom: As the size of an atom increases, effective nuclear charge decreases and hence it is easy to remove electrons from the outermost shell. Therefore ionisation energy decreases.

Nuclear Charge: With the increase in effective nuclear charge (Z_eff), the hold of the nucleus on the outermost electrons increases. Hence, ionisation energy increases.

Shielding Effect: It is described as the effect wherein inner electrons create a shield for the electrons in outer shells which does not allow the complete nuclear charge to pull the outermost electrons.

As a result, the outermost electrons feel a lower effective nuclear charge and not the actual/complete nuclear charge. The greater the shielding effect, the lower the ionisation enthalpy.

The order of screening effect is s > p > d > f.

Effective nuclear charge can be calculated as

Z_eff = Z -

Z_eff = Effective nuclear charge

Z= Actual nuclear charge

= Screening/Shielding constant.

Penetration effect: Penetration represents how close an electron is in an orbit to the nucleus. The greater the penetration effect, the greater the proximity of an orbital (electron present in an orbital) to the nucleus. Hence, ionisation energy is higher.

As per the radial probability distribution functions, we see that the electron density of s orbitals is closer than that of p and d orbitals.

The order of penetration power of the subshells is 2s > 2p > 3s > 3p > 4s > 3d.

Ionisation enthalpy is directly proportional to penetration power.

Electronic Configuration: Orbital electronic configurations that are half-filled and fully filled are exceptionally stable. So trying to remove an electron from these orbitals will require a greater amount of energy. Hence, ionisation enthalpy for such electronic configurations is much higher.

Anomalous Trends in Ionisation Enthalpy

The first anomaly comes in the 2^ndperiod in the ionisation enthalpy values of boron and beryllium. The valence shell of Be is completely filled (2s²). So, it is more stable than the electronic configuration of B (2s²2p¹). So, more energy is required to remove the electron from Be. Hence, the first ionisation energy of Be is more than that of B. We know that the electronic configuration of beryllium is 1s² 2s² and that of boron is 1s²2s²2p¹.

The first ionisation enthalpy of nitrogen is expected to be less than that of oxygen. But it is not so. This can be explained by their electronic configuration as well.

The electronic configuration of N is [He]2s²2p³ and that of oxygen is [He]2s²2p⁴. The half-filled valence p-orbital of nitrogen gives it extra stability and thus, it has a greater first ionisation enthalpy as compared to oxygen.

Practice Problems

Q 1. Which of the following statements is incorrect?

a. The first ionisation energy of Al is less than the first ionisation energy of Mg.
b. The second ionisation energy of Mg is greater than the second ionisation energy of Na.
c. The second ionisation energy of Na is greater than the second ionisation energy of Mg.
d. The third ionisation energy of Mg is greater than the third ionisation energy of Al.

Answer: Option B)

Statement A) is correct because Mg has a filled s-orbital. So, the first ionisation energy of Al is less than that of Mg.

Statement B) is incorrect because Na after losing one electron will attain the noble gas configuration and hence has an exceptionally high second ionisation energy.

Statement C) is correct because Na after losing one electron will attain the noble gas configuration and hence has an exceptionally high second ionisation energy than Mg which incidentally has a low value due to attainment of noble gas configuration after removal of the second electron.

Statement D) is correct because the third ionisation energy of Mg is greater than the third ionisation energy of Al as a noble gas configuration is to be broken in Mg, whereas Al attains a noble gas configuration on losing three electrons.

So, option B) is the correct answer.

Q 2. Why do alkaline earth metals and pnictogens (elements belonging to the nitrogen family)

have higher values of first ionisation potential as compared to the succeeding groups (trend observed up to the 4th period)?

Answer: The alkaline earth metals have a general electronic configuration of ns². So, their outermost shell has two valence electrons, and since the s-subshell is completely filled, it requires greater energy to remove electrons. Similarly, for pnictogens, their general electronic configuration is ns²np³. Their outermost p-subshell is half-filled. Hence, they too require a greater amount of ionisation energy to remove electrons.

Q 3. The screening effect of inner electrons on the nucleus causes

a. A decrease in the ionisation energy.
b. An increase in the ionisation energy.
c. No effect on the ionisation potential.
d. An increase in attraction of the nucleus on the outermost electron.

Answer: The higher the screening effect, the lesser will be the attraction between the nucleus and the outermost electrons and hence, it is easier to remove an electron. Thus, ionisation energy decreases with the increase in screening effect. In other words, the higher the screening effect, the lesser is the ionisation energy.

So, option A) is the correct answer.

Frequently Asked Questions - FAQ

Q 1. How does ionisation enthalpy affect the metallic character of an element?
Answer: Metals have low ionisation enthalpy and readily lose electrons to form cations, while non-metals have high ionisation enthalpy and prefer forming anions by gaining electrons rather than losing.

Q 2. How can we determine the reducing nature of an element from its ionisation potential value?
Answer: Lower the value of the ionisation enthalpy, the greater the reducing power of an element. This is because a low value of ionisation potential means easier removal of electrons from the element which then can be used to reduce an oxidising agent. An oxidising agent would readily accept electrons and themselves get reduced.

Q 3. Can we determine the basic strength of an element from its ionisation enthalpy?
Answer: Lower the value of ionisation enthalpy, the easier it is to remove the electrons and the greater will be the donor properties of the element. Thus, the basic strength of the elements increases considerably with the decrease in ionisation potential values, as the base is an electron donor.

Q 4. What is the relation between ionisation energy and electron affinity in terms of determining the nature of the bond present in a compound?
Answer: Ionisation energy is the amount of energy needed to form cation from an isolated neutral gaseous atom and electron affinity is the energy needed to form anions. For ionic compounds, we need easily formable cations and anions. Hence, lower ionisation energy and higher electron affinity of two elements will facilitate the formation of ionic bonds between them.

Q 5. In which cases shielding effect is more prominent?
Answer: The shielding effect is more prominent when the inner orbitals are completely filled. For example, in Na, the 3s1electron will be shielded by its core electrons (1s2, 2s2 and 2p6).

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