Classification Of Elements And Periodicity In Properties

In the early part of the 18^th century, only a few elements were known and hence it was very easy to remember the physical and chemical properties of such elements. But with discovery of new elements, it became practically impossible to remember the properties of all the elements.

Hence the need aroused for arranging the elements in form of a table or a chart where elements with same properties can be grouped together so that no confusion arises while dealing with elements and also true knowledge can be obtained about a particular element.

The first meaningful and remarkable contribution in the field of classification of elements was done by Russian Chemist by the name Dmitri I. Mendeleev in the year of 1869 and he proposed a law called Mendeleev periodic law which says that the physical & the chemical properties of the elements are a periodic function of their atomic masses.

On the foundation of this law, he developed a periodic table which was named as Mendeleev Periodic table, where he arranged all the elements in his periodic table in terms of their atomic weight or mass. In the table he made vertical columns called groups and vertical rows called as periods for the arranging the elements. He placed the elements with similar physical and chemical properties in the same group.

Nevertheless, he did not adhere to this system for long. He observed that if the elements were arranged strictly in terms of their increasing atomic weights, then some elements could not fit within this system of categorization.

Therefore, the order of atomic weights was ignored in some cases. For instance, the atomic weight of iodine is lesser than that of tellurium. Then to tellurium [in Group VI] was placed before iodine [in Group VII] just because iodine’s properties are so similar to that of fluorine, chlorine, and bromine. Also, argon was placed before potassium.

The first meaningful and remarkable contribution in the field of classification of elements was done by Russian Chemist by the name Dmitri I. Mendeleev in the year of 1869 and he proposed a law called Mendeleev periodic law which says that the physical & the chemical properties of the elements are a periodic function of their atomic masses.

In the table he made vertical columns called groups and vertical rows called as periods for the arranging the elements. There were nine groups & seven periods. He arranged only 60 elements in periodic table as only 60 elements were known at the time of Mendeleev.

The Modern Periodic law was proposed by an English physicist by the name Moseley, in 1913 which states that the physical & the chemical properties of the elements are the periodic function of their atomic number. He proved that atomic number and not the atomic mass is the most fundamental property of an element and is also play a key role in determining its physical and chemical properties. This law becomes the base or foundation of modern periodic law. In the modern periodic table there are 18 groups & 7 periods.

The principal quantum number is a number which determines the main energy level or principal shell to which the electron of the atom belongs. It also gives the average distance of the electron from the nucleus and also specifies the energy value of the electron.

Azimuthal quantum number is the figure which determines the sub shell in a principal shell to which an electron belongs and it also signifies the number of subshell present in the principal shell.

In the periodic table, a period indicates the value of the principal quantum number [n] denotes the total number of shells present in the atom of the element. Each period starts with the filling of principal quantum number [n]. The value of principal quantum number for the sixth period is 6. For n = 6, azimuthal quantum number [L] can have values of 0, 1,2,3,4. According to Aufbau’s principle, electrons are added to different orbitals in order of their increasing energies. Thus, the 6d shell has a greater energy as compared to the 7s subshell.

In the 6^th period, electrons can be filled in only 6s, 4f, 5d, and 6 p subshells. So, the 6s subshell consists of 1 orbital, 4f subshell consists of 7 orbitals, 5d subshell has 5 orbitals and 6p subshell has three orbitals. As a result, the total number of orbitals present is 16. According to Pauli’s exclusion principle, only 2 electrons can be accommodated in each orbital. So, 32 electrons can be accommodated at a maximum level in 16 orbitals.

Therefore 32 elements would be present in the 6^th period of the periodic table.

Elements from atomic number = 87 to atomic number = 114 are present in the 7^th period of the periodic table. Therefore, the element with atomic number = 114 [Flerovium] with atomic weight 289 and a poor metal is present in 7th period and 14th group of the periodic table.

The s-block elements of the 7th period have elements with atomic number 87 and 88,

f-block elements consist of next 14 elements from atomic number 90 to 103 excluding element with atomic number = 89,

d-block element consists of next 10 elements from atomic number 104 to 112 and also elements with atomic number 89, and p-block element consists of elements from atomic number 113 to 118

Thus, the element with atomic number 114 is the second element of the p block of the 7^th period.

1^st period of the periodic table has only 2 elements and 2^nd period consists of only eight elements. The first element of the third period is sodium with atomic number 11. Now, there are nine elements in the third period. The last element of the third period is Argon which has atomic number of 18. So, the 17^th group element of third period is chlorine with atomic number = 17 and atomic mass = 35.453g. It belongs to p-block.

Lawrencium [Lr] with atomic number = 103 is a member of actinide family and hence involves the filling of 5f orbital and Berkelium [Bk] with atomic number = 97 is also a member of actinide family and hence involves the filling of 5f orbital [both of them are f block elements].

Seaborgium [Sg] with atomic number = 106 is a d block element present in 7^th period and 6^th column of periodic table.

The total number of groups in the modern periodic table is 18 and each group id distinct from every other group. The elements in the same group contains same number of valence electrons and hence the same group elements possess similar physical and chemical properties as the physical and chemical properties is dependent on the number of valence electrons in the element.

Atomic radius & ionic radius are the periodic properties which are directly or indirectly connected to the electronic configuration of the atoms & shows gradation on moving down a group or along a period.

Atomic radius is defined as the distance from the centre of the nucleus to the outer most shell containing the electrons. It evaluates the size of an atom. It is of three types:

A] Covalent radius – It is defined as the one half of the distance between the centres of the nuclei of two similar atoms bonded together by a single covalent bond.

Covalent radius = Distance between two nuclei in the bonded atoms/2

B] Metallic radium – It is defined as the one half of the distance between the centres of the nuclei of two adjacent atoms in a metallic crystal.

C] Van der Waal’s radius – It is defined as one half of the distance between the nuclei of two atoms of the same substance at their closest approach. Closest approach of two atoms means the stage when the atoms are unable to come closer without bond formation.

Ionic radii of an ion may be defined as the distance from its nucleus to the point up to which the nucleus has the influence on the electron cloud of the ion.

Ionic radius means the radius of an ion [cation or anion]. Ionic radii of an ion may be defined as the distance from its nucleus to the point up to which the nucleus has the influence on the electron cloud of the ion. The ionic radii can be determined by measuring the distances between the cations and anions in ionic crystals.

Since a cation is formed by getting a rid of electron from the atom, the cation contains lesser electrons than the parent atom, resulting in an increment in the effective nuclear charge. Thus, the size of the cation is smaller as compared to that of its parent atom. For instance, the ionic radius of sodium ion is 95 pm, whereas the atomic radius of Na atom is 186 pm.

On the other hand, in case of anion the size of the anion is larger as compared to the size of the parent atom. This is because in case of the anion the nuclear charge remains the same but more electrons get added to the atom resulting in an increased repulsion between the electrons and a reduction in the effective nuclear charge. For instance, the ionic radius of the fluoride ion is 136 pm, whereas the atomic radius of fluorine atom is 64 pm.

Atomic radius of the elements generally decreases from left to the right in a period because on moving from left to right in a period the number of shells remains the same but more and more electrons get added to the atoms and hence the effective nuclear charge increases and thereby pulling the shells closer to the nucleus and hence atomic radius decreases on moving from left to right in periodic table.

Atomic radius of the elements increases on going down the group because on going down the group there is an increase in the number of shells. As the number of shells increases, the valence shell gets farther and farther away from the nucleus resulting in reduction of overall effective nuclear charge. Therefore, the atomic size is obviously expected to increase.

Isoelectronic species are defined as those species which belong to different atoms or ions which possess same number of electrons but different magnitude of nuclear charge.

A positive charged ion denotes the loss of an electron & a negative charged ion represents the gain of an electron by a species.

The isoelectronic ions with greater nuclear charge will possess small size as compared to the ion with smaller nuclear charges.

[i] F^- ion has 9+1 = 10 electrons. The isoelectronic species of Fluoride will also have 10 electrons. Some examples of isoelectronic species are Na⁺ ion [11-1 = 10 electrons], Ne [10 electrons], O^2- ion [8+2 =10 electrons], and AL³⁺ ion [13-3 = 10 electrons].

[ii] Ar has 18 electrons. The isoelectronic species of Argon will also have 18 electrons. Some examples of isoelectronic species are S^2- ion [16+2 = 18 electrons], CL^- ion [17+1 = 18 electrons], K⁺ ion [19-1 = 18 electrons], and Ca⁺ ion [20-2 = 18 electrons].

[iii] Mg²⁺ ion has 12 – 2 = 10 electrons. The isoelectronic species of Magnesium will also have 10 electrons. Some examples of isoelectronic species are F^- ion [9+1 = 10 electrons], ne [10 electrons’, O^2- ion [8+2 = 10 electrons], and AL³⁺ ion [13-3 = 10 electrons].

[iv] Rb⁺ ion has 37 – 1 = 36 electrons. The isoelectronic species of Rubidium will also have 36 electrons. Some examples of isoelectronic species are Br^- ion [35+1 = 36 electrons], Kr [36 electrons], and Sr²⁺ ion [38-2 = 36 electrons].

A. In the given examples of isoelectronic species have a same number of electrons which is equal to 10. Hence, the given species are isoelectronic, i.e.

N^3- has 7+3 = 10 electrons

O^2- has 8+2 = 10 electrons

F^- has 9+1 = 10 electrons

Na⁺ has 11-1 = 10 electrons

Mg²⁺ has 12-2 = 10 electrons

AL³⁺ has 13-3 = 10 electrons

B. As the magnitude of the effective nuclear decreases, the ionic radii of the ions increase.

The following arrangement denotes the elements arranged in the order of increasing nuclear charge:

N^3- < O^2- < F^- < Na⁺ < Mg²⁺ < AL³⁺

Nuclear charge = +7 +8 +9 +11 +12 +13

The following arrangement denotes the elements arranged in the order of increasing ionic radii:

AL³⁺ < Mg²⁺ < Na⁺ < F^- < O^2- < N^3-

The nucleus of an atom consists of protons and hence possesses a positive charge and electrons rotate around the nucleus in fixed shells. Therefore, in the case of cations & anions are as follows:

Since a cation is formed by getting a rid of electron from the atom, the cation contains lesser electrons than the parent atom, resulting in an increment in the effective nuclear charge. Thus, the size of the cation is smaller as compared to that of its parent atom. For instance, the ionic radius of sodium ion is 95 pm, whereas the atomic radius of Na atom is 186 pm.

On the other hand, in case of anion the size of the anion is larger as compared to the size of the parent atom. This is because in case of the anion the nuclear charge remains the same but more electrons get added to the atom resulting in an increased repulsion between the electrons and a reduction in the effective nuclear charge. For instance, the ionic radius of the fluoride ion is 136 pm, whereas the atomic radius of fluorine atom is 64 pm.

Ionization enthalpy is the minimal quantity of energy which is demanded to get rid of the most loosely bound electron from a neutral isolated gaseous atom to form a cation. A cation is formed when an atom loses electrons. But for the removal electrons the atoms should be in isolated gaseous form. Though a gaseous atom is randomly separated but there exist some attractive forces between the atoms. To calculate the ionization enthalpy, it is practically impossible to isolate a single atom. By lowering the pressure, it is possible to minimize the force of attraction between the atoms. This is the reason why the term ‘isolated gaseous atom’ is used in the definition of the ionization enthalpy.

The most stable state of a gaseous atom is its ground state and in the ground state of the atom, it is very easy to remove the electron from the atom. Hence the term ground state is used in the definition.

Ionization enthalpy is the minimal quantity of energy which is demanded to get rid of the most loosely bound electron from a neutral isolated gaseous atom to form a cation.

It is given that the energy of an electron in the ground state of the hydrogen atom is -2.18 × 10^-18J.

So, Ionization enthalpy of atomic hydrogen = 2.18 × 10^-18 J.

So, converting the ionization enthalpy from Joules to J/mole

= 2.18 × 10^-18 × 6.02 × 10²³ J mol^-1

= 1.31 × 10⁶ J mol^-1

Therefore, the ionization enthalpy of hydrogen atom is 1.31 × 10⁶ J mol^-1.

(i) Beryllium has higher ionization enthalpy as compared to boron which can easily explained using the concept of symmetry factor. Beryllium has both the orbitals filled fully and boron has one half filled orbital in 2p subshell because of which beryllium has more symmetric nature as compared to boron. The electron of beryllium is places in 2s subshell and the 2s subshell is more tightly held to the nucleus as compared to 2p subshell of boron. Thus, during ionization more energy is required to remove an electron from beryllium atom as compared to boron atom. Hence, beryllium has higher ∆_i H than boron.

(ii) In nitrogen atom all three orbitals of the 2p subshell are fully filled and in case of oxygen one orbital is fully filled and the two orbitals are partially filled of 2p subshell. So, nitrogen has greater symmetricity as compared to oxygen and hence it is difficult to remove an electron from nitrogen atom as compared to oxygen atom. Thus, nitrogen has higher value of ionization enthalpy as compared to oxygen.

Fluorine has one electron and proton more as compared to oxygen atom and hence in case of fluorine atom the electron is more tightly held as compared to oxygen atom. Thus, it is difficult to remove an electron from fluorine atom as compared to oxygen atom. Thus fluorine has higher value of ionization enthalpy as compared to oxygen.

Sodium is the 2^nd member of group I [alkali metals] & magnesium is the 2^nd member of group II [alkaline earth metals]

The first ionization enthalpy of sodium is lower as compared to that of magnesium. This is because of smaller size and more symmetrical electronic configuration. This is the reason why the first ionization enthalpy of sodium is lower as compared to that of magnesium. However, the second ionization enthalpy of sodium is higher than that of magnesium. The simple reason is that after losing one electron sodium attains the noble gas configuration and thus energy required to remove the electron from the stable noble gas configuration of sodium is very much higher than that actually required to remove the second electron from magnesium atom.

Ionization enthalpy is the minimal quantity of energy which is demanded to get rid of the most loosely bound electron from a neutral isolated gaseous atom to form a cation.

The factors responsible for the decrease in ionization enthalpy in moving down the group are given as follows:

[i] Increase in the atomic size of elements: On going down the group the number of shells in the atom also increases. As a result, the size of the atom also increases gradually on going down the group. As a result, the valence electrons get farther away from the nucleus and thus weakening the effective nuclear charge of the atom and hence electrons can be easily removed from the atoms. Hence, ongoing down the group, ionization energy decreases.

[ii] Increase in the shielding effect: The number of inner shells of electrons increases on moving down a group. The inner shells or orbits of the atom acts as a shield for the outer or valence shell electrons. The inner shell shields the nuclear charge and as a result only a small amount of nuclear charge is able to attract the valence shell electrons. Hence the energy required to remove the electron reduces on moving down the group.

On going down a group the ionization enthalpies generally decrease because the size of the atom increases and screening effect of the inner shell electrons which is dominates or masks the effect of increase in nuclear charge.

Consequently, the electron becomes weakly held by the nucleus as we move down the group. The sharp decrease in ionization enthalpy from boron to aluminium is due to increase in the size of the atom. But the Gallium atom contains 10 electrons in the d subshell. The shielding tendency of d electrons is less as compared to the electrons of s and p subshell, the outer valence electrons are tightly held to the nucleus. This is the reason why the ionization enthalpy increases slightly on moving down the group from Al to Ga in spite of increase in their atomic size. The similar effect is also observed from In to Tl. In case of tellurium 14f electrons provide very low shielding effect.

(i) O and F are present in the same period of the periodic table. An F atom has one proton and one electron more than O and as an electron is being added to the same shell, the atomic size of F is smaller than that of O. As F contains one proton more than O, its nucleus can attract the incoming electron more strongly in comparison to the nucleus of O atom. Also, F needs only one more electron to attain the stable noble gas configuration. Hence, the electron gain enthalpy of F is more negative than that of O.

(ii) F and Cl belong to the same group of the periodic table. The electron gain enthalpy usually becomes less negative on moving down a group. However, in this case, the value of the electron gain enthalpy of Cl is more negative than that of F. This is because the atomic size of F is smaller than that of Cl. In F, the electron will be added to quantum level n = 2, but in Cl, the electron is added to quantum level n = 3. Therefore, there are less electron- electron repulsions in Cl and an additional electron can be accommodated easily. Hence, the electron gain enthalpy of Cl is more negative than that of F.

When an electron is added to O atom to form O^- ion, energy is released. Thus, the first electron gain enthalpy of O is negative.

On the other hand, when an electron is added to O^- ion to form O^2- ion, energy has to be given out in order to overcome the strong electronic repulsions. Thus, the second electron gain enthalpy of O is positive.

Electronegativity of an element is a variable property. It is different in different compounds. Hence, the statement which says that the electronegativity of N on Pauling scale is 3.0 in all nitrogen compounds is incorrect. The electronegativity of N is different in NH₃ and NO₂.

(A) When an atom gains an electron, its size increases. When an electron is added, the number of electrons goes up by one. This results in an increase in repulsion among the electrons. However, the number of protons remains the same. As a result, the effective nuclear charge of the atom decreases and the radius of the atom increases.

(B) When an atom loses an electron, the number of electrons decreases by one while the nuclear charge remains the same. Therefore, the interelectronic repulsions in the atom decrease. As a result, the effective nuclear charge increases. Hence, the radius of the atom decreases.

The ionization enthalpy of an atom depends on the number of electrons and protons (nuclear charge) of that atom. Now, the isotopes of an element have the same number of protons and electrons. Therefore, the first ionization enthalpy for two isotopes of the same element should be the same.

(A.) The electronic configuration of an element having 5 electrons in its outermost subshell should be ns² np³ . This is the electronic configuration of the halogen group. Thus, the element can be F, CL, Br, I, or At.

(B.) An element having two valence electrons will lose two electrons easily to attain the stable noble gas configuration. The general electronic configuration of such an element will be ns². This is the electronic configuration of group 2 elements. The elements present in group 2 are Be, Mg, Ca, Sr, Ba.

(C.) An element is likely to gain two electrons if it needs only two electrons to attain the stable noble gas configuration. Thus, the general electronic configuration of such an element should be ns² np⁴. This is the electronic configuration of the oxygen family.

(D.) Group 17 has metal, non–metal, liquid as well as gas at room temperature.

The elements present in group 1 have only 1 valence electron, which they tend to lose. Group 17 elements, on the other hand, need only one electron to attain the noble gas configuration. On moving down group 1, the ionization enthalpies decrease. This means that the energy required to lose the valence electron decreases. Thus, reactivity increases on moving down a group. Thus, the increasing order of reactivity among group 1 elements is as follows:

Li < Na < K < Rb < Cs

In group 17, as we move down the group from Cl to I, the electron gain enthalpy becomes less negative i.e., its tendency to gain electrons decreases down group 17. Thus, reactivity decreases down a group. The electron gain enthalpy of F is less negative than Cl. Still, it is the most reactive halogen. This is because of its low bond dissociation energy. Thus, the decreasing order of reactivity among group 17 elements is as follows: F > Cl > Br > I.

(i) Since n = 3, the element belongs to the 3rd period. It is a p–block element since the last electron occupies the p–orbital. There are four electrons in the p–orbital.

Thus, the corresponding group of the element is:

= Number of s–block groups + number of d–block groups + number of p–electrons

= 2 + 10 + 4 = 16

Hence, the element belongs to the 3rd period and 16th group of the periodic table. Hence, the element is Sulphur.

(ii) Since n = 4, the element belongs to the 4th period. It is a d–block element as d– orbitals are incompletely filled. There are 2 electrons in the d–orbital.

Thus, the corresponding group of the element is :

= Number of s–block groups + number of d–block groups

= 2 + 2 = 4

Hence, it is a 4th period and 4th group element. Therefore, the element is Titanium.

(iii) Since n = 6, the element is present in the 6th period. It is an f –block element as the last electron occupies the f–orbital. It belongs to group 3 of the periodic table since all f-block elements belong to group 3. Its electronic configuration is [Xe] 4f⁷ 5d¹ 6s² . Thus, its atomic number is 54 + 7 + 2 + 1 = 64. Hence, the element is Gadolinium.

(A.) Element V is likely to be the least reactive element. This is because it has the highest first ionization enthalpy (∆H₁) and a positive electron gain enthalpy (∆_egH).

(B.) Element II is likely to be the most reactive metal as it has the lowest first ionization enthalpy (∆H₁) and a low negative electron gain enthalpy (∆_egH).

(C.) Element III is likely to be the most reactive non–metal as it has a high first ionization enthalpy (∆H₁) and the highest negative electron gain enthalpy (∆_egH).

(D.) Element V is likely to be the least reactive non–metal since it has a very high first ionization enthalpy (∆H₂) and a positive electron gain enthalpy (∆_egH).

(E.) Element VI has a low negative electron gain enthalpy (∆_egH). Thus, it is a metal. Further, it has the lowest second ionization enthalpy (∆H₂). Hence, it can form a stable binary halide of the formula MX₂(X=halogen).

(F.) Element I have low first ionization energy and high second ionization energy. Therefore, it can form a predominantly stable covalent halide of the formula MX (X=halogen).

(A.) LiO₂

(B.) Mg₃N₂

(C.) AlI₃

(D.) SiO₂

(E.) PF₃ or PF₅

(F.) The element with the atomic number 71 is Lutetium (Lu). It has valency 3. Hence, the formula of the compound is LuF₃.

The value of the principal quantum number (n) for the outermost shell or the valence shell indicates a period in the Modern periodic table.

Hence, in the modern periodic table, the period indicates the value of – C. principal quantum number.

The d-block has 10 columns because a maximum of 10 electrons can occupy all the orbitals in a d subshell.

Hence, the statement related to the modern periodic table which is incorrect is - B. The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.

Nuclear mass does not affect the valence electrons.

Hence, the correct option is - (c) Nuclear mass.

The size of an isoelectronic species increases with a decrease in the nuclear charge (Z).

For example, the order of the increasing nuclear charge of F^-, Ne, and Na⁺ is as follows:

F^-(Z=9) < Ne(Z=10) < Na⁺(Z=11)

Therefore, the order of the increasing size of F^-, Ne and Na⁺ is as follows:

Na⁺ < Ne < F^-

Hence, the correct option is - (a) nuclear charge (Z).

Electrons in orbitals bearing a lower n value are more attracted to the nucleus than electrons in orbitals bearing a higher n value. Hence, the removal of electrons from orbitals bearing a higher n value is easier than the removal of electrons from orbitals having a lower n value.

Hence, the incorrect statement in relation to ionization enthalpy is: D. Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.

The metallic character of elements decreases from left to right across a period. Thus, the metallic character of Mg is more than that of Al. The metallic character of elements increases down a group. Thus, the metallic character of Al is more than that of B. Considering the above statements, we get K > Mg and the only relation out of the given relations which holds the above condition is - (d) K > Mg > Al > B.

Hence, the correct order of metallic character of the elements B, Al, Mg, and K is – (d) K > Mg > Al > B.

The non-metallic character of elements increases from left to right across a period. Thus, the decreasing order of non-metallic character is F > N > C > B. Again, the non-metallic character of elements decreases down a group. Thus, the decreasing order of non-metallic characters of C and Si are C > Si. However, Si is less non-metallic than B i.e., B > Si. Hence, the correct order of non-metallic characters of the elements B, C, N, F, and Si is F > N > C > B > Si.

Therefore, the correct option is - (c) F > N > C > B > Si.

The oxidizing character of elements increases from left to right across a period. Thus, we get the decreasing order of oxidizing property as F > O > N. Again, the oxidizing character of elements decreases down a group. Thus, we get F > Cl. However, the oxidizing character of O is more than that of Cl i.e., O > Cl. Hence, the correct order of chemical reactivity of F, Cl, O, and N in terms of their oxidizing property is F > O > Cl > N.

Therefore, the correct option is - (b) F > O > Cl > N