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Non-Rationalised Science NCERT Notes and Solutions (Class 11th)
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Physics Chemistry Biology

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. States Of Matter 6. Thermodynamics
7. Equilibrium 8. Redox Reactions 9. Hydrogen
10. The S-Block Elements 11. The P-Block Elements 12. Organic Chemistry: Some Basic Principles And Techniques
13. Hydrocarbons 14. Environmental Chemistry



Chapter 3 Classification Of Elements And Periodicity In Properties



The Periodic Table is considered a cornerstone concept in chemistry. It provides a structured framework for organising the chemical elements, revealing patterns and trends in their properties, which is invaluable for both learning and research. It demonstrates that the elements are not a random collection but exhibit relationships and form families based on their characteristics.


Why Do We Need To Classify Elements ?

Elements are the fundamental building blocks of all matter. In the early 19th century (around 1800), only about 31 elements were known. By the mid-19th century (1865), this number had doubled to 63. Currently, over 118 elements have been identified, with many of the heavier ones being synthesised artificially. Studying the chemistry of each element and the vast number of compounds they form individually is incredibly challenging.

To manage this complexity, scientists sought a systematic method to classify elements based on their properties. Such classification would not only help organise existing knowledge about elements but also potentially predict the properties of new or undiscovered ones, guiding further research.



Genesis Of Periodic Classification

The development of the Periodic Table was a gradual process, built upon the efforts of many scientists who observed and systematised the properties of known elements.

Element Atomic weight Element Atomic weight Element Atomic weight
Li 7 Ca 40 Cl 35.5
Na 23 Sr 88 Br 80
K 39 Ba 137 I 127

However, this "Law of Triads" only worked for a limited number of elements and was not universally accepted.

Element Li Be B C N O F
At. wt. 7 9 11 12 14 16 19
Element Na Mg Al Si P S Cl
At. wt. 23 24 27 29 31 32 35.5
Element K Ca
At. wt. 39 40

Newlands' law worked reasonably well only for the lighter elements, up to calcium, and wasn't accepted by the scientific community at the time, although he was later recognised for his contribution.

The most significant steps towards the modern Periodic Table were taken by Russian chemist Dmitri Mendeleev and German chemist Lothar Meyer, working independently around 1869. Both found that when elements were arranged by increasing atomic weight, similar physical and chemical properties recurred periodically. Lothar Meyer plotted properties like atomic volume and melting point against atomic weight, showing periodic patterns and noticing variations in the length of these repeating periods. His table from 1868 closely resembled the modern one but was published after Mendeleev's work.

Mendeleev is widely credited with publishing the first clear statement of the Periodic Law:

The properties of the elements are a periodic function of their atomic weights.

Mendeleev arranged elements in rows (periods) and columns (groups) based on increasing atomic weight, placing elements with similar properties in the same vertical column. He used a broader range of properties than Meyer, including the formulas of oxides and hydrides, recognising the fundamental importance of periodicity.

Crucially, Mendeleev did not strictly adhere to the increasing order of atomic weight when it contradicted the placement based on chemical properties. He believed the atomic weight measurements might be inaccurate (which was true in some cases, like for Iodine and Tellurium). He also bravely left gaps in his table, predicting the existence and properties of undiscovered elements that would fit into those positions. For example, he predicted elements he called Eka-Aluminium and Eka-Silicon, later discovered as Gallium and Germanium, whose properties closely matched his predictions.

Property Eka-aluminium (predicted) Gallium (found) Eka-silicon (predicted) Germanium (found)
Atomic weight 68 70 72 72.6
Density / (g/cm$^3$) 5.9 5.94 5.5 5.36
Melting point /K Low 302.93 High 1231
Formula of oxide E$_2$O$_3$ Ga$_2$O$_3$ EO$_2$ GeO$_2$
Formula of chloride ECl$_3$ GaCl$_3$ ECl$_4$ GeCl$_4$

Mendeleev's successful predictions solidified the importance of his Periodic Table.

Mendeleev's Periodic Table from 1905


Modern Periodic Law And The Present Form Of The Periodic Table

At the time of Mendeleev, the internal structure of atoms was unknown. Developments in the early 20th century, particularly by English physicist Henry Moseley in 1913, provided a more fundamental basis for the periodic classification. Moseley studied the characteristic X-ray spectra of elements and found that the square root of the frequency of emitted X-rays ($\sqrt{\nu}$) plotted against the atomic number ($Z$) yielded a straight line, whereas a plot against atomic mass did not. This showed that the atomic number is a more fundamental property of an element than its atomic mass.

This led to the refinement of the Periodic Law, now known as the Modern Periodic Law:

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

The atomic number ($Z$) represents the number of protons in the nucleus and also the number of electrons in a neutral atom. Since chemical properties are determined primarily by the electronic configuration, the periodic variation in properties is a direct consequence of the periodic variation in electronic configurations as atomic number increases.

The most widely used form today is the "long form" of the Periodic Table. It organises elements in horizontal rows called periods and vertical columns called groups. Elements in the same group have similar outer electronic configurations and hence similar properties.

Long form of the Modern Periodic Table showing elements with their atomic numbers and outer electronic configurations, and group numbering from 1 to 18

According to IUPAC recommendations, groups are numbered 1 through 18, replacing older notations like IA-VIIA, VIII, IB-VIIB, and 0.

There are seven periods. The period number corresponds to the highest principal quantum number ($n$) of the valence shell being filled in that period. The number of elements in each period corresponds to the electron capacity of the orbitals being filled:

Lanthanoids and Actinoids are placed in separate panels at the bottom to keep the main body of the Periodic Table manageable and to group elements with very similar properties (within each series) together.



Nomenclature Of Elements With Atomic Numbers > 100

Traditionally, the discoverers of a new element had the privilege of naming it, which was then ratified by IUPAC. However, for elements with atomic numbers greater than 100, synthesised in different labs, competing claims for discovery led to disputes over naming (e.g., element 104 claimed by both American and Soviet scientists). To avoid this, IUPAC established a systematic temporary nomenclature for elements before their discovery is fully verified and a permanent name is agreed upon.

The temporary name is derived directly from the atomic number using numerical roots (Table 3.4) and adding the ending "-ium". The roots are combined in the order of the digits in the atomic number.

Digit Name Abbreviation
0 nil n
1 un u
2 bi b
3 tri t
4 quad q
5 pent p
6 hex h
7 sept s
8 oct o
9 enn e

The temporary symbol is derived from the abbreviations of the roots, using the first letter of each root.

Examples of IUPAC temporary names and symbols for elements with Z > 100 (Table 3.5 shows both temporary and official names):

Atomic Number Name according to IUPAC nomenclature Symbol according to IUPAC nomenclature
101 Unnilunium Unu
104 Unnilquadium Unq
118 Ununoctium Uuo

Once the discovery is verified, a permanent name and one- or two-letter symbol are assigned, often honouring a scientist or place of discovery. Official names have been announced for all elements up to Z=118 (Oganesson).


Problem 3.1. What would be the IUPAC name and symbol for the element with atomic number 120?

Answer:

The atomic number is 120.

Break down the atomic number into its digits: 1, 2, 0.

Find the corresponding numerical roots from Table 3.4:

  • Digit 1: root "un", abbreviation "u"
  • Digit 2: root "bi", abbreviation "b"
  • Digit 0: root "nil", abbreviation "n"

Combine the roots and add "-ium" at the end for the name:

un + bi + nil + ium = Unbinilium

Combine the abbreviations for the symbol:

u + b + n = Ubn

The IUPAC temporary name for the element with atomic number 120 is Unbinilium, and its symbol is Ubn.



Electronic Configurations Of Elements And The Periodic Table

The arrangement of elements in the modern Periodic Table is fundamentally based on their electronic configurations, specifically the quantum numbers of the last electron added to an atom.


(a) Electronic Configurations in Periods

Each period in the Periodic Table corresponds to the filling of electrons into the orbitals of a new principal energy level (shell), indicated by the principal quantum number ($n$) which matches the period number. The number of elements in each period is determined by the total number of electrons that can fill the subshells available in that principal energy level, following the Aufbau principle's energy ordering.

The placement of Lanthanoids and Actinoids separately is a matter of convenience to keep the main table's width manageable and group elements with very similar properties within their respective series.


Problem 3.2. How would you justify the presence of 18 elements in the 5th period of the Periodic Table?

Answer:

The period number corresponds to the principal quantum number ($n$) of the outermost shell being filled. For the 5th period, $n=5$.

According to quantum mechanics, for $n=5$, the possible values of the azimuthal quantum number ($l$) are $0, 1, 2, \dots, n-1$.

So, for $n=5$, the possible $l$ values are 0, 1, 2, and 3.

These correspond to the 5s, 5p, 5d, and 5f subshells.

However, the order of filling of orbitals is determined by the Aufbau principle and the $(n+l)$ rule. The energy ordering for $n=5$ and involved orbitals from lower shells is:

  • 4d: $n=4, l=2 \implies n+l=6$
  • 5s: $n=5, l=0 \implies n+l=5$
  • 5p: $n=5, l=1 \implies n+l=6$
  • 5d: $n=5, l=2 \implies n+l=7$
  • 5f: $n=5, l=3 \implies n+l=8$

Comparing $(n+l)$ values, the order of filling is 5s (5), then 4d (6), then 5p (6). For 4d and 5p, $(n+l)$ is 6, so we use the $n$ value; 4d has lower $n$ (4) than 5p (5), so 4d is filled before 5p.

Thus, in the 5th period, the orbitals being filled are the 5s, 4d, and 5p orbitals.

  • 5s subshell has $2l+1 = 2(0)+1 = 1$ orbital.
  • 4d subshell has $2l+1 = 2(2)+1 = 5$ orbitals.
  • 5p subshell has $2l+1 = 2(1)+1 = 3$ orbitals.

The total number of orbitals available for filling in the 5th period are $1 + 5 + 3 = 9$ orbitals (5s, five 4d, three 5p). Each orbital can accommodate a maximum of 2 electrons.

Therefore, the maximum number of electrons that can be accommodated in the 5th period is $9 \text{ orbitals} \times 2 \text{ electrons/orbital} = 18 \text{ electrons}$.

Since each element adds one electron, there are 18 elements in the 5th period of the Periodic Table.


(b) Groupwise Electronic Configurations

Elements within the same vertical column (group) in the Periodic Table share similar chemical properties. This similarity is because they have the same number and distribution of electrons in their outermost (valence) shell orbitals. The general valence shell electronic configuration is the same for all elements in a group.

For example, all Group 1 elements (Alkali Metals) have a valence shell electronic configuration of $ns^1$, where $n$ is the period number:

This consistent outer electronic configuration is the fundamental reason behind the similar chemical behavior within a group. While electronic configuration explains periodicity based on atomic number, chemical properties are the observable manifestation of these electron arrangements.



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

Based on which subshell the last electron enters, elements in the Periodic Table are classified into four blocks: s-block, p-block, d-block, and f-block. This classification is directly linked to the electronic configurations and largely dictates the chemical behavior of the elements.

Periodic Table showing the division into s, p, d, and f blocks, and also broad classification as metals, non-metals, and metalloids

There are two notable exceptions to this simple block categorisation:



Periodic Trends In Properties Of Elements

Many physical and chemical properties of elements show observable patterns or trends as we move across a period or down a group in the Periodic Table. These periodic trends are fundamentally linked to the electronic structure of atoms.


Trends In Physical Properties

We will discuss the periodic trends for several key physical properties:


(a) Atomic Radius

Defining and measuring the size of an atom is challenging because atoms are extremely small ($\approx 10^{-10}$ m) and the electron cloud surrounding the nucleus doesn't have a sharp boundary. Atomic radius is typically estimated based on the distance between the nuclei of bonded atoms:

In simplified terms, "Atomic Radius" often refers to either covalent or metallic radius, depending on the element's type. Atomic radii are determined using techniques like X-ray diffraction.

Periodic Trends in Atomic Radius:

Atom (Period II) Li Be B C N O F
Atomic radius (pm) 152 111 88 77 74 66 64
Atom (Period III) Na Mg Al Si P S Cl
Atomic radius (pm) 186 160 143 117 110 104 99
Atom (Group I) Atomic Radius (pm) Atom (Group 17) Atomic Radius (pm)
Li 152 F 64
Na 186 Cl 99
K 231 Br 114
Rb 244 I 133
Cs 262 At 140

Noble gases' atomic radii are often listed as Van der Waals radii, which are significantly larger than covalent or metallic radii of other elements in the same period. Comparing them directly requires using consistent types of radii.


(b) Ionic Radius

When an atom loses electrons, it forms a positively charged ion called a cation. When it gains electrons, it forms a negatively charged ion called an anion. Ionic radii are estimated from the distances between ions in ionic crystals.

Isoelectronic Species: These are atoms or ions that have the same total number of electrons. Examples include $\text{O}^{2-}$, $\text{F}^-$, Ne, $\text{Na}^+$, $\text{Mg}^{2+}$, and $\text{Al}^{3+}$, all having 10 electrons. In a series of isoelectronic species, the size is determined by the nuclear charge ($Z$).

Thus, for the isoelectronic series $\text{O}^{2-}, \text{F}^-, \text{Ne}, \text{Na}^+, \text{Mg}^{2+}, \text{Al}^{3+}$ (all 10 electrons), the size decreases as the nuclear charge increases: $\text{O}^{2-} > \text{F}^- > \text{Ne} > \text{Na}^+ > \text{Mg}^{2+} > \text{Al}^{3+}$.


Problem 3.5. Which of the following species will have the largest and the smallest size?

Mg, Mg$^{2+}$, Al, Al$^{3+}$.

Answer:

We have four species: neutral atoms Mg and Al, and their cations $\text{Mg}^{2+}$ and $\text{Al}^{3+}$.

First, let's compare the neutral atoms Mg and Al. They are in the same period (Period 3), with Mg ($Z=12$) to the left of Al ($Z=13$). Atomic radii decrease across a period due to increasing effective nuclear charge. So, Mg is larger than Al.

Next, let's compare each atom to its cation. A cation is smaller than its parent atom because electrons are removed, and the remaining electrons are more strongly attracted to the nucleus. So, Mg is larger than $\text{Mg}^{2+}$, and Al is larger than $\text{Al}^{3+}$.

We know Mg > Al and Mg > $\text{Mg}^{2+}$ and Al > $\text{Al}^{3+}$.

Now let's compare the cations $\text{Mg}^{2+}$ and $\text{Al}^{3+}$. Both have 10 electrons (Mg: 12-2=10, Al: 13-3=10). They are isoelectronic species.

Among isoelectronic species, size decreases as nuclear charge ($Z$) increases. Mg has $Z=12$, Al has $Z=13$. So, $\text{Al}^{3+}$ ($Z=13$) has a higher nuclear charge than $\text{Mg}^{2+}$ ($Z=12$). Therefore, $\text{Al}^{3+}$ is smaller than $\text{Mg}^{2+}$.

Putting it all together:

Mg (largest atom) > Al (smaller atom)

Mg > $\text{Mg}^{2+}$

Al > $\text{Al}^{3+}$

$\text{Mg}^{2+}$ > $\text{Al}^{3+}$ (smallest cation)

Considering the species Mg, $\text{Mg}^{2+}$, Al, $\text{Al}^{3+}$, the neutral atoms are larger than their corresponding ions. Mg is larger than Al. $\text{Mg}^{2+}$ is larger than $\text{Al}^{3+}$.

Comparing the four: Mg is larger than Al. Mg is also larger than $\text{Mg}^{2+}$ and $\text{Al}^{3+}$. So Mg is the largest.

Comparing the cations $\text{Mg}^{2+}$ and $\text{Al}^{3+}$, $\text{Al}^{3+}$ is smaller. $\text{Al}^{3+}$ is also smaller than the neutral atoms Al and Mg. So $\text{Al}^{3+}$ is the smallest.

The largest species is Mg, and the smallest species is Al$^{3+}$.


(c) Ionization Enthalpy ($\Delta_i H$)

Ionization enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. It quantifies how easily an atom loses an electron to form a positive ion (cation).

First Ionization Enthalpy ($\Delta_i H_1$): $\text{X(g)} \rightarrow \text{X}^+\text{(g)} + \text{e}^-$

Second Ionization Enthalpy ($\Delta_i H_2$): $\text{X}^+\text{(g)} \rightarrow \text{X}^{2+}\text{(g)} + \text{e}^-$

And so on for successive electrons. Ionization enthalpies are always positive values, indicating energy must be absorbed (endothermic process). Successive ionization enthalpies increase ($\Delta_i H_1 < \Delta_i H_2 < \Delta_i H_3 < \dots$) because it becomes harder to remove an electron from an increasingly positive ion.

Periodic Trends in Ionization Enthalpy:

Deviations from the general trend across a period:


Problem 3.6. The first ionization enthalpy (D$_i$ H ) values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ mol–1. Predict whether the first D$_i$ H value for Al will be more close to 575 or 760 kJ mol–1 ? Justify your answer.

Answer:

The given elements are from the third period: Na ($Z=11$), Mg ($Z=12$), Al ($Z=13$), Si ($Z=14$). The values are $\Delta_i H_1$ (Na) = 496 kJ mol$^{-1}$, $\Delta_i H_1$ (Mg) = 737 kJ mol$^{-1}$, $\Delta_i H_1$ (Si) = 786 kJ mol$^{-1}$. We need to predict $\Delta_i H_1$ (Al).

The general trend across a period is increasing ionization enthalpy: Na < Mg < Al < Si < ...

Based on the general trend, the value for Al should be greater than Mg (737 kJ mol$^{-1}$) and less than Si (786 kJ mol$^{-1}$). This would suggest the value should be closer to 760 kJ mol$^{-1}$.

However, there is an exception to this trend when moving from Group 2 to Group 13.

Let's look at the electronic configurations:

  • Na: $[Ne]3s^1$
  • Mg: $[Ne]3s^2$ (Filled s subshell, stable)
  • Al: $[Ne]3s^23p^1$ (Electron removed from p orbital)
  • Si: $[Ne]3s^23p^2$

The first electron removed from Mg is from the 3s orbital ($3s^2$). The first electron removed from Al is from the 3p orbital ($3p^1$).

Electrons in s orbitals penetrate closer to the nucleus and are shielded less effectively by inner electrons compared to electrons in p orbitals of the same principal shell. Therefore, the 3p electron in Al is better shielded by the inner core electrons (including the $3s^2$ pair) than the 3s electrons in Mg are shielded from each other. Also, removing an electron from a completely filled subshell (Mg, $3s^2$) is generally more difficult than removing an electron from a p subshell (Al, $3p^1$).

As a result, the first ionization enthalpy of Al is expected to be lower than that of Mg, despite Al having a larger nuclear charge ($Z=13$ vs $Z=12$).

Comparing the given options (575 kJ mol$^{-1}$ and 760 kJ mol$^{-1}$) with the value for Mg (737 kJ mol$^{-1}$), the value 575 kJ mol$^{-1}$ is lower than 737 kJ mol$^{-1}$, while 760 kJ mol$^{-1}$ is higher.

Therefore, the first ionization enthalpy value for Al will be more close to 575 kJ mol$^{-1}$.

Justification: The first electron is removed from the 3p orbital in Al, which is higher in energy and more shielded than the 3s orbital from which the first electron is removed in Mg. Removing the electron from the filled 3s subshell in Mg requires more energy than removing the electron from the $3p^1$ configuration in Al, despite Al having a higher nuclear charge.


(d) Electron Gain Enthalpy ($\Delta_{eg} H$)

Electron gain enthalpy is the enthalpy change that occurs when an electron is added to an isolated gaseous atom in its ground state to form a negative ion (anion).

$\text{X(g)} + \text{e}^- \rightarrow \text{X}^-\text{(g)}$

This process can be exothermic ($\Delta_{eg} H$ is negative, energy is released) or endothermic ($\Delta_{eg} H$ is positive, energy is absorbed).

Periodic Trends in Electron Gain Enthalpy:

Exceptions to the trend down a group (especially in Period 2): The electron gain enthalpies of elements in the second period (like O and F) are often less negative than those of the corresponding elements in the third period (S and Cl). This is because the second-period elements are unusually small. When an electron is added to the relatively small 2p subshell of O or F, it experiences significant repulsion from the electrons already present in the small volume. In contrast, for S or Cl, the added electron enters the larger 3p subshell, where electron-electron repulsion is less significant, leading to a more negative electron gain enthalpy.


Problem 3.7. 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:

The elements are P ($Z=15$), S ($Z=16$), Cl ($Z=17$), and F ($Z=9$).

P, S, and Cl are in the third period (Group 15, 16, and 17 respectively). F is in the second period (Group 17).

General trend across a period (P $\rightarrow$ S $\rightarrow$ Cl): Electron gain enthalpy becomes more negative. So, $\Delta_{eg}H(\text{P}) < \Delta_{eg}H(\text{S}) < \Delta_{eg}H(\text{Cl})$ (values become more negative).

General trend down a group (F $\rightarrow$ Cl): Electron gain enthalpy becomes less negative. So, $\Delta_{eg}H(\text{F}) > \Delta_{eg}H(\text{Cl})$ (values become less negative/more positive). This contradicts the general trend down a group due to the small size of F.

Comparing F and Cl: Cl is in the third period, F in the second period. Due to the small size of F, the incoming electron experiences significant repulsion from the existing electrons in the 2p orbitals. In Cl, the incoming electron enters the larger 3p orbitals, where repulsion is less. Thus, Cl has a more negative electron gain enthalpy than F. (Approximate values: F = -328 kJ/mol, Cl = -349 kJ/mol).

Comparing P, S, Cl (across period 3): $\Delta_{eg}H$ becomes more negative. P (-74 kJ/mol) < S (-200 kJ/mol) < Cl (-349 kJ/mol).

Now comparing all four values: P (-74), S (-200), Cl (-349), F (-328).

The most negative value is -349 kJ/mol (Chlorine).

The least negative value is -74 kJ/mol (Phosphorus).

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

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

Explanation:

  • Cl has the most negative value because it is a halogen and is larger than F, reducing electron repulsion for the incoming electron.
  • P has the least negative value among these elements because it is on the left side of the period (Group 15), meaning its tendency to accept an electron is lower compared to the more electronegative elements on the right. It also has a half-filled p subshell ($3p^3$), which gives it some stability and reduces the tendency to gain an electron compared to Group 16 and 17 elements.

(e) Electronegativity

Electronegativity is a qualitative measure of the ability of an atom to attract shared electrons towards itself when it is in a chemical compound (i.e., when it is bonded to another atom). Unlike ionization enthalpy or electron gain enthalpy, it is not a property of an isolated atom and cannot be directly measured experimentally. Several scales exist to assign numerical values (e.g., Pauling scale, Mulliken-Jaffe scale, Allred-Rochow scale). The most common is the Pauling scale, where Fluorine is assigned the highest value of 4.0.

Periodic Trends in Electronegativity:

Electronegativity is closely related to non-metallic character (ability to gain electrons) and inversely related to metallic character (tendency to lose electrons). High electronegativity corresponds to high non-metallic character. Low electronegativity corresponds to high metallic character.

Summary diagram showing the general trends of atomic size, ionization enthalpy, electron gain enthalpy, and electronegativity across periods and down groups

Periodic Trends In Chemical Properties

Periodic trends are also evident in the chemical properties of elements, such as valence, oxidation states, and reactivity.


(a) Periodicity of Valence or Oxidation States

The valence of an element, representing its combining capacity, is primarily determined by the number of electrons in its outermost shell. For representative (s-block and p-block) elements, the valence is often equal to the number of valence electrons (Group 1 and 2) or eight minus the number of valence electrons (Group 15-17). Group 18 elements (noble gases) have a valence of 0 (or sometimes 8).

The oxidation state of an element in a compound refers to the apparent charge on an atom based on the assumption that all bonds are ionic, with the more electronegative atom assigned the negative charge. Oxidation states can be positive or negative and often variable, especially for d and f block elements.

Periodic trends in valence (shown by common hydrides and oxides) (Table 3.9):

Group 1 2 13 14 15 16 17
Number of valence electrons 1 2 3 4 5 6 7
Valence (common) 1 2 3 4 3, 5 2, 6 1, 7
Formula of hydride (Examples) LiH, NaH, KH MgH$_2$, CaH$_2$ B$_2$H$_6$, AlH$_3$ CH$_4$, SiH$_4$, GeH$_4$, SnH$_4$ NH$_3$, PH$_3$, AsH$_3$, SbH$_3$ H$_2$O, H$_2$S, H$_2$Se, H$_2$Te HF, HCl, HBr, HI
Formula of oxide (Examples) Li$_2$O, Na$_2$O, K$_2$O MgO, CaO, SrO, BaO B$_2$O$_3$, Al$_2$O$_3$, Ga$_2$O$_3$, In$_2$O$_3$ CO$_2$, SiO$_2$, GeO$_2$, SnO$_2$, PbO$_2$ N$_2$O$_3$, N$_2$O$_5$, P$_4$O$_6$, P$_4$O$_{10}$, As$_2$O$_3$, As$_2$O$_5$, Sb$_2$O$_3$, Sb$_2$O$_5$, Bi$_2$O$_3$ SO$_3$, SeO$_3$, TeO$_3$ Cl$_2$O$_7$

Many elements, particularly transition and inner-transition metals, exhibit multiple oxidation states due to the involvement of inner d or f electrons in bonding.


Problem 3.8. 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 binary compound formed between two elements, we can often use their common valencies (or oxidation states).

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

  • Silicon (Si) is in Group 14. Common valence/oxidation state is +4.
  • Bromine (Br) is a halogen in Group 17. Common valence/oxidation state is -1.

To form a neutral compound, the total positive charge must balance the total negative charge. If Silicon has a valence of 4 and Bromine has a valence of 1, we need one Si atom and four Br atoms.

Formula: SiBr$_4$.

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

  • Aluminium (Al) is in Group 13. Common valence/oxidation state is +3.
  • Sulphur (S) is in Group 16. Common valence/oxidation state is -2.

To balance the charges, we need to find the least common multiple of the valencies (3 and 2), which is 6. We need a total positive charge of +6 and a total negative charge of -6.

Number of Al atoms = $6 / 3 = 2$. (Total positive charge = $2 \times +3 = +6$).

Number of S atoms = $6 / 2 = 3$. (Total negative charge = $3 \times -2 = -6$).

Formula: Al$_2$S$_3$.


(b) Anomalous Properties of Second Period Elements

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

Reasons for anomalous behaviour of Period 2 elements (Li to F):

  1. Small Size: They are the smallest elements in their groups.
  2. High Electronegativity: They have the highest electronegativity values in their groups.
  3. High Charge/Radius Ratio: For cations, this ratio is very high.
  4. Absence of d-orbitals in the valence shell: Elements of the second period have only 2s and 2p orbitals (a total of four orbitals) in their valence shell ($n=2$), limiting their maximum covalency to 4. Elements from the third period onwards have empty d orbitals (e.g., 3s, 3p, 3d), allowing them to expand their valence shell and exhibit covalencies greater than 4 (e.g., Al can form $\text{AlF}_6^{3-}$).
  5. Ability to form $p\pi - p\pi$ multiple bonds: Second-period elements (especially C, N, O, F) can form stable multiple bonds (double or triple bonds) by sideways overlap of their p orbitals with themselves (e.g., C=C, N$\equiv$N) and with other second-period elements (e.g., C=O, C$\equiv$N). Heavier elements in the same group have larger and more diffuse p orbitals, making $p\pi - p\pi$ overlap less effective, thus reducing their tendency to form strong multiple bonds.

Diagonal Relationship: The anomalous behaviour of second-period elements often leads to a similarity in properties between the first element of a group and the second element of the *next* group. This is called the diagonal relationship. Examples: Lithium (Group 1) shows similarities with Magnesium (Group 2), and Beryllium (Group 2) shows similarities with Aluminium (Group 13).

Property Element
Metallic radius (pm) Li (152) Be (111) B (88)
Na (186) Mg (160) Al (143)
Ionic radius M$^+$ / M$^{2+}$ / M$^{3+}$ (pm) Li$^+$ (76) Be$^{2+}$ (31)
Na$^+$ (102) Mg$^{2+}$ (72) Al$^{3+}$ (53.5)

Li ($Z/r \approx 76/76 = 1.0$) and Mg ($Z/r^2 \approx 12/72^2$? This is charge/radius, not charge/radius squared. Let's use ionic charge and ionic radius $Li^+ (1+, 76pm), Mg^{2+} (2+, 72pm), Al^{3+} (3+, 53.5pm)$ approx charge density $1/76, 2/72, 3/53.5$. The ratio is not simple) exhibit similar charge density ($Z/\text{radius}$ or $Z/\text{radius}^2$ for ions) which leads to comparable polarizing power and similarities in properties, such as forming covalent compounds more readily than other alkali/alkaline earth metals.


Problem 3.9. Are the oxidation state and covalency of Al in [AlCl(H$_2$O)$_5$]$^{2+}$ same ?

Answer:

Let's determine the oxidation state of Al in the complex ion [AlCl(H$_2$O)$_5$]$^{2+}$.

The overall charge of the complex ion is +2.

The ligands are Chloride (Cl) and Water (H$_2$O).

Chloride is a halide and usually has an oxidation state of -1 (in this case, acting as a ligand, its charge is -1).

Water is a neutral molecule, so its charge/oxidation state as a ligand is 0.

Let the oxidation state of Aluminium (Al) be $x$.

Sum of oxidation states = Overall charge of the complex ion

$x$ + (Oxidation state of Cl) + 5 $\times$ (Oxidation state of H$_2$O) = +2

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

$x - 1 + 0 = +2$

$x = +2 + 1 = +3$.

So, the oxidation state of Aluminium in [AlCl(H$_2$O)$_5$]$^{2+}$ is +3.

The covalency of the central metal atom in a complex is the number of coordinate bonds formed between the metal ion and the ligands. In [AlCl(H$_2$O)$_5$]$^{2+}$, Al is bonded to one Chloride ligand and five Water ligands.

The number of ligands directly attached to Al is 1 (from Cl) + 5 (from H$_2$O) = 6.

So, the covalency of Aluminium in [AlCl(H$_2$O)$_5$]$^{2+}$ is 6.

The oxidation state of Al is +3, and the covalency is 6. Therefore, they are not the same.


Periodic Trends And Chemical Reactivity

The chemical reactivity of elements is closely related to their tendency to lose or gain electrons, which is reflected in properties like ionization enthalpy and electron gain enthalpy.

The chemical nature of oxides provides insight into reactivity. Oxides of metals are generally basic, reacting with water to form bases. Oxides of non-metals are generally acidic, reacting with water to form acids. Oxides of elements in the middle of the table can be amphoteric (reacting with both acids and bases) or neutral.


Problem 3.10. Show by a chemical reaction with water that Na$_2$O is a basic oxide and Cl$_2$O$_7$ is an acidic oxide.

Answer:

Basic oxide: Na$_2$O

Sodium oxide (Na$_2$O) is formed from sodium, an alkali metal on the extreme left of Period 3. Metal oxides are generally basic. When a basic oxide reacts with water, it forms a base (a substance that produces hydroxide ions, OH$^-$).

Reaction of Na$_2$O with water:

$\text{Na}_2\text{O(s)} + \text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)}$

Sodium hydroxide (NaOH) is a strong base. This reaction shows that Na$_2$O is a basic oxide.

Acidic oxide: Cl$_2$O$_7$

Dichlorine heptoxide (Cl$_2$O$_7$) is formed from chlorine, a halogen on the extreme right of Period 3 (before the noble gas). Non-metal oxides are generally acidic. When an acidic oxide reacts with water, it forms an acid (a substance that produces hydronium ions, H$_3$O$^+$, or releases H$^+$).

Reaction of Cl$_2$O$_7$ with water:

$\text{Cl}_2\text{O}_7\text{(l)} + \text{H}_2\text{O(l)} \rightarrow 2\text{HClO}_4\text{(aq)}$

Perchloric acid (HClO$_4$) is a very strong acid. This reaction shows that Cl$_2$O$_7$ is an acidic oxide.



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?

Hint : Requirements for comparison purposes.

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 $\Delta_iH$ than B

(ii) O has lower $\Delta_iH$ 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 > CI > 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 $\Delta_iH_1$ $\Delta_iH_2$ $\Delta_{eg}H$
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: