Chapter 3 Classification of Elements and Periodicity in Properties Solutions
Question - 11 : - What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
(i) F–
(ii) Ar
(iii) Mg2+
(iv) Rb+
Answer - 11 : -
Atoms and ions having the same number of electrons arecalled isoelectronic species.
(i) F– ion has 9 + 1 = 10 electrons. Thus, thespecies isoelectronic with it will also have 10 electrons. Some of itsisoelectronic species are Na+ ion (11 – 1 = 10 electrons), Ne (10electrons), O2– ion (8 + 2 = 10 electrons), and Al3+ ion (13 – 3 = 10 electrons).
(ii) Ar has 18electrons. Thus, the species isoelectronic with it will also have 18 electrons.Some of its isoelectronic species are S2– ion(16 + 2 = 18 electrons), Cl– ion (17 + 1 = 18 electrons), K+ ion (19 – 1 = 18 electrons), and Ca2+ ion (20 – 2 = 18 electrons).
(iii) Mg2+ ion has 12 – 2 = 10 electrons. Thus, thespecies isoelectronic with it will also have 10 electrons. Some of itsisoelectronic species are F– ion (9 + 1 = 10 electrons), Ne (10electrons), O2– ion (8 + 2 = 10 electrons), and Al3+ ion (13 – 3 = 10 electrons).
(iv) Rb+ ion has 37 – 1 = 36 electrons. Thus, thespecies isoelectronic with it will also have 36 electrons. Some of itsisoelectronic species are Br– ion (35 + 1 = 36 electrons), Kr (36electrons), and Sr2+ ion (38 – 2 = 36 electrons).
Question - 12 : - Consider the following species:
N3–, O2–, F–, Na+, Mg2+ and Al3+
(a) What is common in them?
(b) Arrange them in the order of increasingionic radii.
Answer - 12 : -
(a) Each of the given species (ions) has the same numberof electrons (10 electrons). Hence, the given species are isoelectronic.
(b) Theionic radii of isoelectronic species increases with a decrease in themagnitudes of nuclear charge.
Thearrangement of the given species in order of their increasing nuclear charge isas follows:
N3– < O2– – < Na+ 2+ < Al3+
Nuclearcharge = +7 +8 +9 +11 +12 +13
Therefore,the arrangement of the given species in order of their increasing ionic radiiis as follows:
Al3+ < Mg2+ + < F– 2– < N3–
Question - 13 : - Explain why cation is smaller and anions larger in radii than their parent atoms?
Answer - 13 : -
Cations are formed by expelling an electron from the outermost orbit of an atom, thus cation has less electrons compared to parent atom which results in increased effective nuclear charge but the total nuclear charge remains same which results in the increased attraction of electrons towards nucleus than that of the parent atom. Thus, cations are having smaller radii than that of their parent atom.
Anions are formed by gaining an electron in the outermost orbit of an atom. Thus anion has more electrons compared to parent atom, which results in decreased effective nuclear charge but the total nuclear charge remains the same which results in the increased distance the nucleus and the valence electrons as the attraction of electrons towards nucleus decreases than that of the parent atom. Thus, anions are having larger radii than that of their parent atom.
Question - 14 : - What is the significance of the terms — ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy?
Answer - 14 : -
“Ionization enthalpy is the energy that is required to expel an electron from an isolated gaseous atom in the ground state”. Despite the fact that in the gaseous state the atoms are generally widely separated, there are a few measures of attractive forces between the atoms. To find the ionization enthalpy of any ion, it is difficult to isolate a solitary atom. This attractive force can be further diminished by bringing down the pressure. Hence, the term “isolated gaseous atom” is utilized as a part of the meaning of ionization enthalpy.
An atom’s ground state is the most stable state. Less energy is required to expel an electron if an isolated gaseous atom is present in the ground state. In this way, for the purpose of comparison, electron gain enthalpy and ionization enthalpy must be calculated for an “isolated gaseous atom” and its “ground state”.
Question - 15 : - Energy of an electron in the ground state ofthe hydrogen atom is –2.18 × 10–18 J. Calculate the ionization enthalpy ofatomic hydrogen in terms of J mol–1.
Answer - 15 : -
It is given that the energy of an electron inthe ground state of the hydrogen atom is –2.18 × 10–18 J.
Therefore, the energy required to remove that electronfrom the ground state of hydrogen atom is 2.18 × 10–18 J.
Ionizationenthalpy of atomic hydrogen = 2.18 × 10–18 J
Hence, ionization enthalpy of atomic hydrogen in terms ofJ mol–1 = 2.18 × 10–18 ×6.02 × 1023 J mol–1 =1.31 × 106 J mol–1
Question - 16 : - Among the second period elements the actual ionizationenthalpies are in the order Li< B < Be < C < O < N < F < Ne.
Explain why
(i) Be has higher ΔiH thanB
(ii) O has lower ΔiH thanN and F?
Answer - 16 : -
(i) Duringthe process of ionization, the electron to be removed from beryllium atom is a 2s-electron,whereas the electron to be removed from boron atom is a 2p-electron. Now, 2s-electronsare more strongly attached to the nucleus than 2p-electrons. Therefore, moreenergy is required to remove a 2s-electronof beryllium than that required to remove a 2p-electron of boron. Hence,beryllium has higher ΔiH thanboron.
(ii) In nitrogen, thethree 2p-electronsof nitrogen occupy three different atomic orbitals. However, in oxygen, two ofthe four 2p-electronsof oxygen occupy the same 2p-orbital.This results in increased electron-electron repulsion in oxygen atom. As aresult, the energy required to remove the fourth 2p-electron from oxygen is less ascompared to the energy required to remove one of the three 2p-electronsfrom nitrogen. Hence, oxygen has lower ΔiH than nitrogen.
Fluorine contains one electron and one proton more thanoxygen. As the electron is being added to the same shell, the increase innuclear attraction (due to the addition of a proton) is more than the increasein electronic repulsion (due to the addition of an electron). Therefore, thevalence electrons in fluorine atom experience a more effective nuclear chargethan that experienced by the electrons present in oxygen. As a result, moreenergy is required to remove an electron from fluorine atom than that requiredto remove an electron from oxygen atom. Hence, oxygen has lower ΔiH thanfluorine.
Question - 17 : - How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Answer - 17 : -
The 1st ionizationenthalpy of magnesium is higher than 1st ionization enthalpy ofsodium because,
- Magnesium is having greater atomic size than sodium.
- Magnesium is having higher effective nuclear charge than sodium.
Thus, energy requiredto expel an electron from sodium is lower than that in magnesium. Thus, the 1st ionizationenthalpy of magnesium is higher than 1st ionization enthalpy ofsodium.
The 2nd ionizationenthalpy of magnesium is lower than 2nd ionization enthalpy ofsodium is because after expelling an electron, there is still 1 electronremaining in the 3s-orbital of magnesium, whereas sodium achieves stable inertgas configuration after expelling an electron. So, magnesium still requires toexpel 1 electron to achieve stable inert gas configuration.
Thus, energy requiredto expel 2nd electron from magnesium is lower than that insodium. Thus, the 2nd ionization enthalpy of magnesium is lowerthan 2nd ionization enthalpy of sodium.
Question - 18 : - What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Answer - 18 : -
The factors because of which in elements of the main group the ionization enthalpy decreases when we move down the group are given below:
1. “Increase in the shielding effect”: Inner shells increases as we move down the group. Thus, the shielding effect of valence electrons increases by inner core electrons from the nucleus. Thus, the attractive force on electrons towards the nucleus is very strong. So, the energy required to expel a valence electron decreases as we move down the group.
2. “Increase in atomic size of elements”: Inner shells increases as we move down the group. Thus, the atomic size increases as we move down the group. Also, the distance between the valence electron and nucleus of an atom, as a result, the electrons are not strongly bounded. So, valence electrons can be expelled easily. Thus, the energy required to expel a valence electron decreases as we move down the group.
Question - 19 : - The first ionization enthalpy values (in kJ mol–1) of group 13 elements are : | B | Al | Ga | In | Tl |
| 801 | 577 | 579 | 558 | 589 |
Answer - 19 : -
Inner shells increase as we move down the group. Thus, the shielding effect of valence electrons increases by inner core electrons from the nucleus. Thus, the attractive force on electrons towards the nucleus is very strong. So, ionization enthalpy decreases as we move down the group. Hence for elements of group 13 the ionization enthalpy decreases as we move down from B to Al.
Here, Ga is having high ionization enthalpy than that of Al. This is because Al comes after the s-blocks elements, while Ga comes after the d-blocks elements. The shielding that is provided by electrons of d-block elements is not effective. So, the valence electrons are not shielded effectively. Thus, valence electrons in Ga atom experience higher effective nuclear charge compared to Al.
Further on moving down from Ga to In, the value of ionization enthalpy is decreased because of the increase in the shielding effect and an increase in atomic size.
But, Tl is having high ionization enthalpy than that of In. This is because Tl comes after the ‘4f and 5d electrons’. The shielding that is provided by these ‘4f and 5d electrons’ is not effective. So, the valence electrons are not shielded effectively. Thus, valence electrons in Tl atom experience higher effective nuclear charge compared to In.
Question - 20 : - Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F (ii) F or Cl
Answer - 20 : -
(i) O and F are the elements of the same period in the periodic table. An F-atom is having 1 electron and 1 proton more than that of O-atom as the electron is added in the same shell, thus the atomic size of O-atom is larger than F-atom. As O-atom is having 1 proton less than F-atom. So, the nucleus of O-atom cannot attract an incoming electron that strongly as that of an F-atom. Also, F-atom requires only 1 electron to achieve a stable inert gas configuration. So, the electron affinity of F(Fluorine) is more negative than that of O(oxygen).
(ii) F and Cl are the elements of the same group in the periodic table. On moving down the group the electron affinity becomes less negative. Here, the value of electron affinity of F is less negative than that of Cl. It is because the atomic size of Cl is larger than that of F. In Cl, the electron will be added to n = 3 quantum level, whereas in F, the electron will be added to n = 2 quantum level. Thus, as the electron-electron repulsion is reduced in Cl so an extra electron can easily be accommodated. So, the electron affinity of Cl is more negative compared to that of F.