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Class 11th (Chemistry) Chapters
1. Some Basic Concepts Of Chemistry 2. Structure Of Atom 3. Classification Of Elements And Periodicity In Properties
4. Chemical Bonding And Molecular Structure 5. Thermodynamics 6. Equilibrium
7. Redox Reactions 8. Organic Chemistry – Some Basic Principles And Techniques 9. Hydrocarbons



Chapter 3 Classification Of Elements And Periodicity In Properties



Why Do We Need To Classify Elements ?

By the early 19th century, the number of known elements was increasing rapidly (from 31 in 1800 to 63 by 1865, and currently over 114). With such a large number, studying the individual properties of each element and their countless compounds became extremely challenging.

Scientists recognized the need for a systematic way to organize this vast amount of information. Classifying elements based on similarities in their properties would not only simplify their study but also help in understanding the relationships between different elements and potentially predict the properties of new or undiscovered ones.

The goal was to find a way to organize the elements that would rationalize the known chemical facts and provide a framework for further research and discovery.



Genesis Of Periodic Classification

The idea of classifying elements and observing trends in their properties developed gradually through the efforts of several scientists.


Early Attempts at Classification

Johann Dobereiner (early 1800s):

Element Atomic weight Element Atomic weight Element Atomic weight
Li7Ca40Cl35.5
Na23Sr88Br80
K39Ba137I127

($\text{Average wt of Li and K} = (7+39)/2 = 23$, which is the wt of Na. Similarly for other triads). Dobereiner's Law of Triads was limited, as it only applied to a few elements and was initially seen as a coincidence.

A.E.B. de Chancourtois (1862):

John Alexander Newlands (1865):

ElementLiBeBCNOF
At. wt.791112141619
ElementNaMgAlSiPSCl
At. wt.23242729313235.5
ElementKCa...............
At. wt.3940...............

Newlands' law worked well only for lighter elements up to Calcium and was not initially accepted by the scientific community. However, he was later recognized for his contribution.


Contribution Of Mendeleev And Lothar Meyer

The foundation of the modern Periodic Table is largely credited to Dmitri Mendeleev (Russian chemist) and Lothar Meyer (German chemist), working independently around 1869.

Dmitri Mendeleev is generally considered the father of the Periodic Table because he published his findings first and made bold predictions based on his classification.

Property Eka-aluminium (predicted) Gallium (found) Eka-silicon (predicted) Germanium (found)
Atomic weight68707272.6
Density/(g/cm³)5.95.945.55.36
Melting point/KLow302.93High1231
Formula of oxideE₂O₃Ga₂O₃EO₂GeO₂
Formula of chlorideECl₃GaCl₃ECl₄GeCl₄

The subsequent discovery of these elements with properties remarkably close to his predictions solidified the acceptance of Mendeleev's Periodic Table.

Mendeleev’s Periodic Table from 1905


Modern Periodic Law And The Present Form Of The Periodic Table

While Mendeleev's table was based on atomic weight, the understanding of atomic structure at the beginning of the 20th century led to a crucial refinement.


Modern Periodic Law

In 1913, Henry Moseley studied the characteristic X-ray spectra of elements. He found that the frequency of emitted X-rays was related to the element's atomic number (Z) rather than its atomic mass. Specifically, a plot of $\sqrt{\nu}$ vs Z yielded a straight line.

This experimental evidence showed that the atomic number (Z) is a more fundamental property of an element than its atomic mass.

Based on Moseley's work, the Periodic Law was modified, leading to the Modern Periodic Law:

"The physical and chemical properties of the elements are periodic functions of their atomic numbers."

Since the atomic number equals the number of protons in the nucleus (and the number of electrons in a neutral atom), the Periodic Law is essentially a consequence of the periodic recurrence of similar electronic configurations in the outermost shell of atoms, which are responsible for their chemical and physical properties.


The Long Form Of The Periodic Table

The most widely used form of the Periodic Table today is the "long form". In this table:

The Long Form of the Modern Periodic Table showing periods and groups

The number of elements in each period corresponds to the number of atomic orbitals available in the shell being filled:

The 14 elements filling the 4f orbitals (Lanthanoids) and the 14 elements filling the 5f orbitals (Actinoids) in periods 6 and 7, respectively, are placed separately at the bottom of the table to maintain its structure and highlight the similar chemistry within these series (called inner-transition elements).



Nomenclature Of Elements With Atomic Numbers > 100

Traditionally, the discoverer(s) of a new element had the privilege of naming it. However, for very heavy, unstable elements synthesized in different labs, this led to disputes over discovery claims and naming rights (e.g., element 104, claimed by both American and Soviet scientists).

To avoid controversy, IUPAC developed a systematic nomenclature for newly discovered elements with atomic numbers above 100. This temporary name is derived directly from the atomic number using numerical roots.

IUPAC Numerical Roots:

Digit Name Abbreviation
0niln
1unu
2bib
3trit
4quadq
5pentp
6hexh
7septs
8octo
9enne

Naming Rule: Combine the roots for the digits of the atomic number in order, and add the suffix "-ium". The symbol is the first letter of each root.

Example: Element with Z = 101

Once a discovery is verified and officially recognized by IUPAC, a permanent name and symbol are assigned, often honoring a scientist or place of discovery.

Atomic Number IUPAC Temporary Name Symbol IUPAC Official Name Official Symbol
101UnniluniumUnuMendeleviumMd
102UnnilbiumUnbNobeliumNo
103UnniltriumUntLawrenciumLr
104UnnilquadiumUnqRutherfordiumRf
105UnnilpentiumUnpDubniumDb
106UnnilhexiumUnhSeaborgiumSg
107UnnilseptiumUnsBohriumBh
108UnniloctiumUnoHassiumHs
109UnnilenniumUneMeitneriumMt
110UnunnilliumUunDarmstadtiumDs
111UnununiumUuuRontgeniumRg
112UnunbiumUubCoperniciumCn
113UnuntriumUutNihoniumNh
114UnunquadiumUuqFleroviumFl
115UnunpentiumUupMoscoviumMc
116UnunhexiumUuhLivermoriumLv
117UnunseptiumUusTennessineTs
118UnunoctiumUuoOganessonOg

Example 1. What would be the IUPAC name and symbol for the element with atomic number 120?

Answer:

Atomic number = 120.

Digits are 1, 2, 0.

Roots are un, bi, nil.

Combine the roots and add "-ium": un + bi + nil + ium = unbinilium.

The symbol is the first letter of each root: Ubn.

So, the IUPAC temporary name is unbinilium and the symbol is Ubn.



Electronic Configurations Of Elements And The Periodic Table

The electronic configuration of an atom, describing the arrangement of electrons in its orbitals, is the fundamental basis for its position and properties in the modern Periodic Table. The quantum numbers ($n, l, m_l, m_s$) characterize the electrons and their orbitals, and the Periodic Table visually organizes elements based on these characteristics.


Electronic Configurations In Periods

The period number ($n$) in the Periodic Table corresponds to the principal quantum number of the valence shell (outermost energy level) being filled. As you move across a period, electrons are added to orbitals within this valence shell or sometimes to inner shells that become energetically accessible (like 3d filling after 4s).

The number of elements in a period is determined by the total number of electrons that can be accommodated in all the orbitals of the energy levels being filled in that period.

The Lanthanoids (4f series) and Actinoids (5f series) are placed separately at the bottom because their filling involves inner f-orbitals, leading to similar properties within each series and keeping the main body of the table structured according to s, p, and d block filling.

Example 2. How would you justify the presence of 18 elements in the 5th period of the Periodic Table?

Answer:

The 5th period corresponds to the principal quantum number $n=5$.

According to the Aufbau principle and $(n+l)$ rule, the orbitals filled in the 5th period are:

  • 5s ($n=5, l=0$): Holds up to 2 electrons.
  • 4d ($n=4, l=2$): Holds up to 10 electrons. ($n+l = 4+2 = 6$, lower than $5+1$ for 5p, but 5s ($5+0=5$) is lower energy).
  • 5p ($n=5, l=1$): Holds up to 6 electrons. ($n+l = 5+1 = 6$, higher $n$ than 4d, so 4d fills first).

The filling order is 5s, then 4d, then 5p.

The total number of orbitals available for filling in the 5th period is 1 (in 5s) + 5 (in 4d) + 3 (in 5p) = 9 orbitals.

Since each orbital can hold a maximum of 2 electrons (Pauli exclusion principle), the maximum number of electrons that can be accommodated in these orbitals is $9 \times 2 = 18$ electrons.

Therefore, the 5th period contains 18 elements, corresponding to the filling of the 5s, 4d, and 5p subshells.


Groupwise Electronic Configurations

Elements in the same vertical column (group) of the Periodic Table have similar chemical properties because they have the same number of valence electrons and the same general outer electronic configuration.

For example, Group 1 elements (Alkali Metals) all have one electron in their outermost s subshell ($ns^1$).

Atomic number Symbol Electronic configuration Outer electronic configuration
3Li1s² 2s¹2s¹
11Na1s² 2s² 2p⁶ 3s¹3s¹
19K1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹4s¹
37Rb[Kr] 5s¹5s¹
55Cs[Xe] 6s¹6s¹
87Fr[Rn] 7s¹7s¹

This identical outermost electronic configuration leads to similar chemical behavior throughout the group.



Electronic Configurations And Types Of Elements : S-, P-, D-, F- Blocks

Based on which atomic orbital receives the last electron during the building up of electronic configuration, elements are broadly classified into four blocks: s-block, p-block, d-block, and f-block.

Periodic Table showing the classification of elements into s, p, d, and f blocks

Exceptions:


The S-Block Elements


The P-Block Elements


The D-Block Elements (Transition Elements)


The F-Block Elements (Inner-Transition Elements)

Example 3. The elements Z = 117 and 120 have not yet been discovered. In which family/group would you place these elements and also give the electronic configuration in each case.

Answer:

To determine the group and electronic configuration, we need to figure out which orbitals are being filled.

For Z = 117:

  • The period number is determined by the highest principal quantum number ($n$) of the valence shell. Element 118 (Oganesson) is the last element of Period 7, completing the 7p subshell. So, Z=117 is in Period 7.
  • The electronic configuration follows the Aufbau order: [Rn] 7s² 5f¹⁴ 6d¹⁰ 7p⁵. (Rn is Z=86, then 32 more electrons fill 7s, 5f, 6d, 7p).
  • The last electron enters the 7p subshell. Elements ending in $np^5$ belong to Group 17 (Halogens).

Element Z=117 would be in Group 17 (Halogen Family). Its electronic configuration would be [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁵.

For Z = 120:

  • Element 118 completes Period 7. So, Z=119 starts Period 8, and Z=120 is also in Period 8 ($n=8$).
  • The Aufbau filling order after 7p⁶ (element 118) goes to the 8s orbital.
  • The electronic configuration starts with the previous noble gas (element 118, Ununoctium, Uuo). So, Z=120 will have configuration [Uuo] 8s².
  • Elements ending in $ns^2$ belong to Group 2 (Alkaline Earth Metals).

Element Z=120 would be in Group 2 (Alkaline Earth Metals). Its electronic configuration would be [Uuo] 8s².


Metals, Non-Metals And Metalloids

Elements can also be broadly classified based on their general physical and chemical properties.

The metallic character of elements increases as you move down a group and decreases as you move from left to right across a period. Conversely, non-metallic character increases across a period and decreases down a group.

Example 4. Considering the atomic number and position in the periodic table, arrange the following elements in the increasing order of metallic character : Si, Be, Mg, Na, P.

Answer:

Locate the elements in the periodic table:

  • Na (Z=11): Period 3, Group 1
  • Mg (Z=12): Period 3, Group 2
  • Al (Z=13): Period 3, Group 13 (not in the list, but helpful for context)
  • Si (Z=14): Period 3, Group 14
  • P (Z=15): Period 3, Group 15
  • Be (Z=4): Period 2, Group 2

Metallic character increases down a group and decreases across a period (left to right).

Comparing elements in Period 3: Na (Group 1), Mg (Group 2), Si (Group 14), P (Group 15).

Order of decreasing metallic character in Period 3 is Na > Mg > Si > P.

Comparing Be (Period 2, Group 2) and Mg (Period 3, Group 2): Metallic character increases down Group 2, so Mg > Be.

Combining these observations:

P is the most non-metallic among the given elements in Period 3. Si is next.

In Group 2, Mg is more metallic than Be.

Comparing across periods and groups: Na is the most metallic (Group 1). P is the least metallic (Group 15). Mg and Be are in Group 2, with Mg below Be, making Mg more metallic than Be. Si is in between metals and non-metals.

Relative order: Na (most metallic) > Mg > Be > Si > P (least metallic).

Therefore, the increasing order of metallic character is: P < Si < Be < Mg < Na.



Periodic Trends In Properties Of Elements

Many physical and chemical properties of elements show gradual changes and predictable patterns as you move across periods or down groups in the Periodic Table. These systematic variations are called periodic trends and are directly linked to the electronic configuration of the atoms.


Trends In Physical Properties

We will discuss periodic trends in atomic and ionic radii, ionization enthalpy, electron gain enthalpy, and electronegativity.


Atomic Radius

Defining the exact size of an atom is difficult because the electron cloud doesn't have a sharp boundary. Atomic size is often estimated from the distance between the centers of bonded atoms.

Generally, the term "Atomic Radius" is used to refer to covalent or metallic radius.

Periodic Trends in Atomic Radius:

Noble gases' radii are typically given as van der Waals radii, which are larger than covalent radii, making direct comparison within a period tricky.


Ionic Radius

The ionic radius is the effective distance from the nucleus of an ion to its outermost electrons. It is determined from distances between ions in ionic crystals.

Isoelectronic Species: These are atoms or ions that have the same number of electrons. For isoelectronic species, ionic size is determined by the nuclear charge (number of protons).

Example of isoelectronic species with 10 electrons: O²⁻, F⁻, Ne, Na⁺, Mg²⁺, Al³⁺.

Their increasing order of nuclear charge (Z): O (8) < F (9) < Ne (10) < Na (11) < Mg (12) < Al (13).

Their decreasing order of size: O²⁻ > F⁻ > Ne > Na⁺ > Mg²⁺ > Al³⁺.

Example 5. Which of the following species will have the largest and the smallest size? Mg, Mg²⁺, Al, Al³⁺.

Answer:

We have neutral atoms Mg and Al, and their cations Mg²⁺ and Al³⁺.

Atomic radii decrease across a period (Mg is to the left of Al in Period 3). So, Atomic radius of Mg > Atomic radius of Al.

Cations are smaller than their parent atoms. So, Mg²⁺ is smaller than Mg, and Al³⁺ is smaller than Al.

Mg (Z=12) has 12 electrons. Mg²⁺ has 10 electrons.

Al (Z=13) has 13 electrons. Al³⁺ has 10 electrons.

Mg²⁺ and Al³⁺ are isoelectronic species (both have 10 electrons).

Compare Mg²⁺ and Al³⁺: Al³⁺ has a higher nuclear charge (+13) than Mg²⁺ (+12) for the same number of electrons. Therefore, Al³⁺ is smaller than Mg²⁺.

Combining these:

  • Mg is larger than Al (across period).
  • Mg is larger than Mg²⁺ (cation smaller than atom).
  • Al is larger than Al³⁺ (cation smaller than atom).
  • Mg²⁺ is larger than Al³⁺ (isoelectronic ions, higher charge = smaller size).

The largest species will be one of the neutral atoms. Since Mg is to the left of Al, Mg is larger than Al. So, Mg is the largest.

The smallest species will be one of the cations. Mg²⁺ and Al³⁺ are isoelectronic. Al³⁺ has a higher nuclear charge (+13) compared to Mg²⁺ (+12), pulling the 10 electrons more strongly. Therefore, Al³⁺ is the smallest.

Order of increasing size: Al³⁺ < Mg²⁺ < Al < Mg.


Ionization Enthalpy

Ionization Enthalpy ($\Delta_i H$) is the minimum energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. It is a measure of how difficult it is to remove an electron.

$$ \text{X(g)} \rightarrow \text{X⁺(g)} + \text{e⁻} \quad (\text{First Ionization Enthalpy, } \Delta_i H_1) $$

Subsequent ionization enthalpies ($\Delta_i H_2$, $\Delta_i H_3$, etc.) are the energies required to remove the second, third, etc., electrons. $\Delta_i H_2 > \Delta_i H_1$ because removing an electron from a positively charged ion is harder.

Ionization enthalpies are always positive (endothermic process).

Graph showing variation of first ionization enthalpies with atomic number (Z=1 to 60)

Periodic Trends in Ionization Enthalpy:

Alkali metals (Group 1) have the lowest ionization enthalpies in their periods, correlating with their high reactivity as they easily lose an electron. Noble gases (Group 18) have the highest ionization enthalpies due to their very stable closed electron shell configurations.

Example 6. The first ionization enthalpy ($\Delta_i H_1$) values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ mol⁻¹. Predict whether the first $\Delta_i H_1$ value for Al will be more close to 575 or 760 kJ mol⁻¹ ? Justify your answer.

Answer:

The elements are Na (Z=11), Mg (Z=12), Al (Z=13), Si (Z=14), all in the third period.

The trend across a period is generally increasing ionization enthalpy: Na < Mg < Al < Si (with exceptions).

The given values are: Na (496), Mg (737), Si (786).

The value for Al should be between Mg and Si, but we must consider the exception for Group 13.

The electronic configurations are:

  • Na: [Ne] 3s¹
  • Mg: [Ne] 3s² (full 3s subshell)
  • Al: [Ne] 3s² 3p¹
  • Si: [Ne] 3s² 3p²

Removing the first electron from Mg involves breaking the stable, filled 3s² subshell. Removing the first electron from Al involves removing the single electron from the 3p¹ orbital.

The 3p electron in Al is slightly higher in energy and is shielded by the inner [Ne] core and the filled 3s² subshell. It is less strongly held by the nucleus compared to a 3s electron in the same shell.

Therefore, it is easier to remove the 3p electron from Al than a 3s electron from Mg, even though Al has a higher nuclear charge.

This suggests that the ionization enthalpy of Al should be lower than that of Mg.

Given options are 575 kJ mol⁻¹ and 760 kJ mol⁻¹.

Since $\Delta_i H_1$ for Mg is 737 kJ mol⁻¹, the value for Al should be less than 737 kJ mol⁻¹.

Thus, the first ionization enthalpy for Al will be more close to 575 kJ mol⁻¹.


Electron Gain Enthalpy

Electron Gain Enthalpy ($\Delta_{eg} H$) is the enthalpy change that occurs when an electron is added to an isolated gaseous atom in its ground state to form a gaseous anion.

$$ \text{X(g)} + \text{e⁻} \rightarrow \text{X⁻(g)} \quad (\Delta_{eg} H) $$

Electron gain enthalpy can be exothermic (energy released, $\Delta_{eg} H$ is negative) or endothermic (energy absorbed, $\Delta_{eg} H$ is positive). A more negative $\Delta_{eg} H$ indicates that the atom has a greater tendency to accept an electron.

Examples:

Table showing electron gain enthalpies of some main group elements

Periodic Trends in Electron Gain Enthalpy:

Example 7. Which of the following will have the most negative electron gain enthalpy and which the least negative? P, S, Cl, F. Explain your answer.

Answer:

Locate the elements in the periodic table:

  • F (Z=9): Period 2, Group 17 (Halogen)
  • Cl (Z=17): Period 3, Group 17 (Halogen)
  • P (Z=15): Period 3, Group 15
  • S (Z=16): Period 3, Group 16

Electron gain enthalpy generally becomes more negative across a period (P $\rightarrow$ S $\rightarrow$ Cl) and less negative down a group (F $\rightarrow$ Cl, S $\rightarrow$ Se, etc.). Noble gases (like Ne) have positive $\Delta_{eg} H$, making them the least likely to gain electrons.

Comparing P, S, Cl in Period 3: Moving from left to right, $\Delta_{eg} H$ becomes more negative. So, $\Delta_{eg} H$ of P < $\Delta_{eg} H$ of S < $\Delta_{eg} H$ of Cl (increasingly negative).

Comparing F and Cl in Group 17: Cl is below F. While $\Delta_{eg} H$ generally becomes less negative down a group, there's an exception for the second period elements. F (Period 2) is smaller than Cl (Period 3). The incoming electron in F experiences more electron-electron repulsion in the compact 2p subshell than in the larger 3p subshell of Cl. Thus, Cl has a more negative electron gain enthalpy than F.

Comparing with P (Group 15): Halogens (Group 17) readily accept an electron to achieve noble gas configuration, so their $\Delta_{eg} H$ values are generally much more negative than elements in Group 15.

Therefore, among P, S, Cl, F:

  • Cl has the most negative electron gain enthalpy (highest tendency to accept an electron).
  • P, being to the left and not needing just one electron for stability, will have the least negative (or possibly even positive, though typically negative) electron gain enthalpy among the non-metals listed from Period 3. P needs 3 electrons to achieve noble gas configuration, whereas S needs 2 and Cl needs 1. Accepting the first electron is generally less favorable further from Group 17 or 16. Group 15 elements (like P) tend to have less negative $\Delta_{eg}H$ than Group 16 and 17 elements in the same period.

Thus, the element with the most negative electron gain enthalpy is Chlorine (Cl).

The element with the least negative electron gain enthalpy is Phosphorus (P).

Order of increasing (less negative to more negative) electron gain enthalpy: P < S < F < Cl.


Electronegativity

Electronegativity is a qualitative measure of the ability of an atom in a chemical compound to attract a shared pair of electrons towards itself. It is not a directly measurable quantity but is calculated based on various scales (e.g., Pauling scale, Mulliken scale).

On the widely used Pauling scale, Fluorine (F) is assigned the highest electronegativity value (4.0).

Table showing electronegativity values across periods and down groups

Electronegativity values are not constant for an element but vary depending on the atoms it is bonded to and its oxidation state.

Periodic Trends in Electronegativity:

Relationship with Metallic and Non-metallic Character:

As electronegativity increases across a period, non-metallic character increases, and metallic character decreases. As electronegativity decreases down a group, metallic character increases, and non-metallic character decreases.


Periodic Trends In Chemical Properties

The chemical properties of elements are determined by their electronic configurations, particularly the valence electrons. Periodic trends in fundamental properties like ionization enthalpy and electron gain enthalpy directly influence chemical reactivity and the nature of compounds formed.


Periodicity Of Valence Or Oxidation States

The valence of a representative element is often related to the number of electrons in its outermost shell or eight minus that number. The term oxidation state is frequently used and refers to the charge an atom appears to have in a compound based on electronegativity differences.

General Trend in Valence (for Representative Elements):

Group 12131415161718
Number of valence electrons 12345678
Valence (Common) 12343,52,61,70,8

Elements in Group 1 typically show a valence of 1 (losing one $ns^1$ electron). Group 2 shows 2 (losing two $ns^2$ electrons). Group 13 shows 3 (losing $ns^2np^1$). Group 14 shows 4 (losing $ns^2np^2$). Groups 15, 16, and 17 often show valence equal to 8 minus the number of valence electrons (gaining electrons to achieve octet), or variable valency by involving d orbitals.

Oxidation state considers the charge based on electronegativity differences. In OF₂, Oxygen is less electronegative than Fluorine, so Oxygen has a positive oxidation state (+2), while F is -1. In Na₂O, Oxygen is more electronegative than Sodium, so Oxygen is -2, and Na is +1.

Transition elements (d-block) and inner-transition elements (f-block) often exhibit variable valency/oxidation states due to the involvement of inner d or f electrons in bonding.

Example 8. Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of elements; (a) silicon and bromine (b) aluminium and sulphur.

Answer:

To predict the formula of a stable binary compound, we can use the typical valencies (combining capacity) or oxidation states of the elements.

(a) Silicon (Si) and Bromine (Br):

  • Silicon (Si) is in Group 14, Period 3. It typically has a valence of 4.
  • Bromine (Br) is in Group 17 (Halogens), Period 4. It typically has a valence of 1 (gain 1 electron) when bonding with less electronegative elements like Si.

To balance the valencies (4 for Si, 1 for Br), we need 4 Br atoms for every 1 Si atom.

Formula = Si₁Br₄ = SiBr₄.

(b) Aluminium (Al) and Sulphur (S):

  • Aluminium (Al) is in Group 13, Period 3. It typically has a valence of 3 (lose 3 electrons).
  • Sulphur (S) is in Group 16, Period 3. It typically has a valence of 2 (gain 2 electrons) when bonding with more electropositive elements like Al.

To balance the valencies (3 for Al, 2 for S), we need 2 Al atoms ($2 \times 3 = 6$) and 3 S atoms ($3 \times 2 = 6$).

Formula = Al₂S₃ = Al₂S₃.


Anomalous Properties Of Second Period Elements

The first element of each group in the s-block and p-block (Li, Be, B, C, N, O, F) exhibits some properties that are different from the other members of their respective groups. This is known as the anomalous behavior of second period elements.

Examples:

Reasons for anomalous behavior:

Example 9. Are the oxidation state and covalency of Al in [AlCl(H₂O)₅]²⁺ same ?

Answer:

In the complex ion [AlCl(H₂O)₅]²⁺:

  • The overall charge on the complex is +2.
  • Water (H₂O) is a neutral ligand (oxidation state = 0).
  • Chloride (Cl) is a ligand with a charge of -1 (oxidation state = -1).
  • Let the oxidation state of Aluminium (Al) be $x$.

The sum of oxidation states must equal the charge on the complex:

$x + (\text{oxidation state of Cl}) + 5 \times (\text{oxidation state of H₂O}) = +2$

$x + (-1) + 5 \times (0) = +2$

$x - 1 = +2$

$x = +3$

So, the oxidation state of Al is +3.

The covalency of Al in this complex is the number of bonds Al forms with the ligands. Al is bonded to one Cl atom and five H₂O molecules, totaling $1 + 5 = 6$ bonds.

So, the covalency of Al is 6.

Therefore, the oxidation state (+3) and the covalency (6) of Al in [AlCl(H₂O)₅]²⁺ are not the same.


Periodic Trends And Chemical Reactivity

The chemical reactivity of an element is closely related to its tendency to lose or gain electrons, which in turn depends on properties like ionization enthalpy and electron gain enthalpy.

Trends in Reactivity:

Metals on the left side tend to form basic oxides (e.g., Na₂O + H₂O $\rightarrow$ NaOH). Non-metals on the right side tend to form acidic oxides (e.g., Cl₂O₇ + H₂O $\rightarrow$ HClO₄). Elements in the center can form amphoteric oxides (react with both acids and bases, e.g., Al₂O₃) or neutral oxides (e.g., CO, NO).

Summary of Major Periodic Trends:

Diagram summarizing periodic trends in atomic radius, ionization enthalpy, electron gain enthalpy, and electronegativity


Exercises



Question 3.1. What is the basic theme of organisation in the periodic table?

Answer:

Question 3.2. Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?

Answer:

Question 3.3. What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?

Answer:

Question 3.4. On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.

Answer:

Question 3.5. In terms of period and group where would you locate the element with Z =114?

Answer:

Question 3.6. Write the atomic number of the element present in the third period and seventeenth group of the periodic table.

Answer:

Question 3.7. Which element do you think would have been named by

(i) Lawrence Berkeley Laboratory

(ii) Seaborg’s group?

Answer:

Question 3.8. Why do elements in the same group have similar physical and chemical properties?

Answer:

Question 3.9. What does atomic radius and ionic radius really mean to you?

Answer:

Question 3.10. How do atomic radius vary in a period and in a group? How do you explain the variation?

Answer:

Question 3.11. 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) $Mg^{2+}$

(iv) $Rb^+$

Answer:

Question 3.12. Consider the following species :

$N^{3-}, O^{2-}, F^-, Na^+, Mg^{2+}$ and $Al^{3+}$

(a) What is common in them?

(b) Arrange them in the order of increasing ionic radii.

Answer:

Question 3.13. Explain why cation are smaller and anions larger in radii than their parent atoms?

Answer:

Question 3.14. What is the significance of the terms — ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy?

Answer:

Question 3.15. Energy of an electron in the ground state of the hydrogen atom is $-2.18 \times 10^{-18}$ J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol$^{-1}$.

Hint: Apply the idea of mole concept to derive the answer.

Answer:

Question 3.16. Among the second period elements the actual ionization enthalpies are in the order

Li < B < Be < C < O < N < F < Ne.

Explain why

(i) Be has higher $\Delta_i H$ than B

(ii) O has lower $\Delta_i H$ than N and F?

Answer:

Question 3.17. 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:

Question 3.18. What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?

Answer:

Question 3.19. The first ionization enthalpy values (in kJ mol$^{-1}$) of group 13 elements are :

B Al Ga In Tl
801 577 579 558 589

How would you explain this deviation from the general trend ?

Answer:

Question 3.20. Which of the following pairs of elements would have a more negative electron gain enthalpy?

(i) O or F

(ii) F or Cl

Answer:

Question 3.21. Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.

Answer:

Question 3.22. What is the basic difference between the terms electron gain enthalpy and electronegativity?

Answer:

Question 3.23. How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?

Answer:

Question 3.24. Describe the theory associated with the radius of an atom as it

(a) gains an electron

(b) loses an electron

Answer:

Question 3.25. Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.

Answer:

Question 3.26. What are the major differences between metals and non-metals?

Answer:

Question 3.27. Use the periodic table to answer the following questions.

(a) Identify an element with five electrons in the outer subshell.

(b) Identify an element that would tend to lose two electrons.

(c) Identify an element that would tend to gain two electrons.

(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.

Answer:

Question 3.28. The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > Cl > Br > I. Explain.

Answer:

Question 3.29. Write the general outer electronic configuration of s-, p-, d- and f- block elements.

Answer:

Question 3.30. Assign the position of the element having outer electronic configuration

(i) $ns^2np^4$ for n=3

(ii) $(n-1)d^2ns^2$ for n=4, and

(iii) $(n-2)f^7(n-1)d^1ns^2$ for n=6, in the periodic table.

Answer:

Question 3.31. The first ($\Delta_iH_1$) and the second ($\Delta_iH_2$) ionization enthalpies (in kJ mol$^{-1}$) and the ($\Delta_{eg}H$) electron gain enthalpy (in kJ mol$^{-1}$) of a few elements are given below:

Elements $\Delta H_1$ $\Delta H_2$ $\Delta_{eg}H$
I 520 7300 –60
II 419 3051 –48
III 1681 3374 –328
IV 1008 1846 –295
V 2372 5251 +48
VI 738 1451 –40

Which of the above elements is likely to be :

(a) the least reactive element.

(b) the most reactive metal.

(c) the most reactive non-metal.

(d) the least reactive non-metal.

(e) the metal which can form a stable binary halide of the formula $MX_2$(X=halogen).

(f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?

Answer:

Question 3.32. Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.

(a) Lithium and oxygen

(b) Magnesium and nitrogen

(c) Aluminium and iodine

(d) Silicon and oxygen

(e) Phosphorus and fluorine

(f) Element 71 and fluorine

Answer:

Question 3.33. In the modern periodic table, the period indicates the value of :

(a) atomic number

(b) atomic mass

(c) principal quantum number

(d) azimuthal quantum number.

Answer:

Question 3.34. Which of the following statements related to the modern periodic table is incorrect?

(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.

(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.

(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.

(d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration.

Answer:

Question 3.35. Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?

(a) Valence principal quantum number (n)

(b) Nuclear charge (Z )

(c) Nuclear mass

(d) Number of core electrons.

Answer:

Question 3.36. The size of isoelectronic species — $F^-$, Ne and $Na^+$ is affected by

(a) nuclear charge (Z )

(b) valence principal quantum number (n)

(c) electron-electron interaction in the outer orbitals

(d) none of the factors because their size is the same.

Answer:

Question 3.37. Which one of the following statements is incorrect in relation to ionization enthalpy?

(a) Ionization enthalpy increases for each successive electron.

(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.

(c) End of valence electrons is marked by a big jump in ionization enthalpy.

(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.

Answer:

Question 3.38. Considering the elements B, Al, Mg, and K, the correct order of their metallic character is :

(a) B > Al > Mg > K

(b) Al > Mg > B > K

(c) Mg > Al > K > B

(d) K > Mg > Al > B

Answer:

Question 3.39. Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is :

(a) B > C > Si > N > F

(b) Si > C > B > N > F

(c) F > N > C > B > Si

(d) F > N > C > Si > B

Answer:

Question 3.40. Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is :

(a) F > Cl > O > N

(b) F > O > Cl > N

(c) Cl > F > O > N

(d) O > F > N > Cl

Answer: