CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES-Exercises

Q1. What is the basic theme of organisation in the periodic table?

Solution

The basic theme of organisation in the periodic table is the systematic arrangement of elements in order of increasing atomic number so that elements with similar chemical properties recur at regular intervals. This recurring pattern of properties is known as periodicity. The modern periodic table is founded on the modern periodic law, which states that the physical and chemical properties of elements are periodic functions of their atomic numbers.

Mathematically, the principle of periodicity can be expressed as:

\[ \text{Properties of elements} = f(Z) \]

where \( Z \) represents the atomic number. As the atomic number increases, the electronic configuration of elements changes in a systematic manner. Since chemical properties largely depend on the valence shell electronic configuration, elements having similar outer electronic configurations are placed in the same group.

The organisation of the periodic table can be understood through the following logical sequence:

\[ \begin{aligned} \text{Increasing }&\text{atomic number} \\ &\Downarrow \\ \text{Regular variation in }&\text{electronic configuration} \\ &\Downarrow \\ \text{Repetition of valen}&\text{ce shell configuration} \\ &\Downarrow \\ \text{Recurrence of simil}&\text{ar chemical properties} \end{aligned} \]

Therefore, the fundamental theme of the periodic table is periodic repetition of properties due to similar outer electronic configurations when elements are arranged in increasing order of atomic number. This systematic classification enables prediction of properties of elements and their compounds, making the periodic table a powerful tool in chemistry.

Q2. Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?

Solution

Mendeleev classified the elements primarily on the basis of their atomic masses. He arranged elements in increasing order of atomic mass and observed that elements with similar chemical properties appeared at regular intervals. This observation led him to formulate Mendeleev’s periodic law, which states that the physical and chemical properties of elements are periodic functions of their atomic masses.

This guiding idea may be represented as:

\[ \text{Properties of elements} = f(\text{atomic mass}) \]

However, Mendeleev did not strictly adhere to increasing atomic mass whenever it conflicted with chemical properties. In several cases, he deliberately reversed the order of elements to keep chemically similar elements together. For example, cobalt was placed before nickel and tellurium before iodine, even though their atomic masses suggested the opposite order.

This reasoning can be summarized as:

\[ \begin{aligned} \text{Atomic }&\text{mass order} \\ &\Downarrow \\ \text{Comparison of c}&\text{hemical properties} \\ &\Downarrow \\ \text{Priority given to simi}&\text{lar properties over mass} \end{aligned} \]

Thus, although atomic mass was the main criterion used by Mendeleev, he did not rigidly follow it. Whenever necessary, he gave greater importance to chemical properties, showing remarkable scientific insight that ultimately strengthened his periodic classification.

Q3. What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?

Solution

Mendeleev’s Periodic Law and the Modern Periodic Law differ fundamentally in the basis used to arrange elements and explain periodicity. Mendeleev proposed that the physical and chemical properties of elements are periodic functions of their atomic masses. Accordingly, he arranged elements in increasing order of atomic mass and grouped together those showing similar chemical behaviour.

This idea may be represented as:

\[ \text{Properties of elements} = f(\text{atomic mass}) \]

However, this approach led to several anomalies, such as incorrect placement of certain elements, because atomic mass does not increase uniformly with chemical properties.

The Modern Periodic Law corrected this limitation by stating that the properties of elements are periodic functions of their atomic numbers. In the modern table, elements are arranged strictly in increasing atomic number, which reflects their electronic configuration and provides a logical explanation for periodicity.

This modern concept is expressed as:

\[ \text{Properties of elements} = f(Z) \]

where \( Z \) represents the atomic number.

The contrast between the two approaches can be understood as:

\[ \begin{aligned} \text{Mendeleev: Arrangem}&\text{ent based on atomic mass} \\ &\Downarrow \\ \text{Led to inconsis}&\text{tencies in placement} \\ &\\\hline\\ \text{Modern law: Arrangem}&\text{ent based on atomic number} \\ &\Downarrow \\ \text{Explains periodicity thr}&\text{ough electronic configuration} \end{aligned} \]

Thus, the basic difference lies in the criterion of classification: Mendeleev relied on atomic mass, whereas the modern periodic table is founded on atomic number, making the modern approach scientifically more accurate and conceptually sound.

Q4. On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.

Solution

The number of elements in any period of the periodic table is governed by the availability of quantum states in the subshells being filled. Each principal quantum number \( n \) defines a shell, and within that shell electrons occupy subshells characterized by the azimuthal quantum number \( l \). For the sixth period, electrons begin filling orbitals associated with \( n = 6 \), along with inner subshells that become energetically accessible.

In the sixth period, filling occurs in the sequence \( 6s \), \( 4f \), \( 5d \), and \( 6p \). The maximum number of electrons that each subshell can accommodate is given by \( 2(2l+1) \).

Using quantum numbers, this distribution can be written as:

\[ \begin{aligned} 6s &: l=0 \Rightarrow 2(2\times0+1)=2 \text{ electrons} \\ 4f &: l=3 \Rightarrow 2(2\times3+1)=14 \text{ electrons} \\ 5d &: l=2 \Rightarrow 2(2\times2+1)=10 \text{ electrons} \\ 6p &: l=1 \Rightarrow 2(2\times1+1)=6 \text{ electrons} \end{aligned} \]

Adding the capacities of all participating subshells gives:

\[ \begin{aligned} \text{Total electrons} &= 2 + 14 + 10 + 6 \\ &= 32 \end{aligned} \]

Since each element corresponds to the addition of one electron to the ground-state configuration, the accommodation of 32 electrons implies the presence of 32 distinct elements in the sixth period. Thus, on the basis of quantum numbers and subshell capacities, it is justified that the sixth period of the periodic table contains exactly 32 elements.

Q5. In terms of period and group where would you locate the element with Z =114?

Solution

To locate the element with atomic number \( Z = 114 \) in the periodic table, we first determine its electronic configuration using the Aufbau principle and quantum rules. Starting from known noble gas cores, the configuration beyond radon \((Z=86)\) is considered.

The filling of orbitals proceeds as:

\[ \begin{aligned} Z &= 114 \\ \text{Electronic configuration} &= [Rn]\;5f^{14}\,6d^{10}\,7s^{2}\,7p^{2} \end{aligned} \]

The highest principal quantum number present is \( n = 7 \). This directly indicates that the element belongs to the seventh period.

Next, the group is identified from the valence shell configuration. The outermost shell contains:

\[ \begin{aligned} 7s^{2} + 7p^{2} &= 4 \text{ valence electrons} \end{aligned} \]

Elements having four valence electrons in the p-block are placed in Group 14.

Therefore, combining both observations:

\[ \begin{aligned} \text{Highest } n &= 7 \Rightarrow \text{Period } 7 \\ \text{Valence electrons} &= 4 \Rightarrow \text{Group } 14 \end{aligned} \]

Hence, the element with atomic number \( Z = 114 \) is located in the seventh period and Group 14 of the modern periodic table.

Q6. Write the atomic number of the element present in the third period and seventeenth group of the periodic table.

Solution

To find the atomic number of the element located in the third period and seventeenth group of the periodic table, we recall that Group 17 elements are known as halogens, and in the third period the halogen present is chlorine.

The electronic configuration of chlorine can be written as:

\[ \begin{aligned} 1s^2\,2s^2\,2p^6\,3s^2\,3p^5 \end{aligned} \]

Now, counting the total number of electrons:

\[ \begin{aligned} 2 + 2 + 6 + 2 + 5 &= 17 \end{aligned} \]

Since the atomic number of an element is equal to the number of electrons in its neutral atom, the atomic number of the element present in the third period and seventeenth group is \( 17 \).

Q7. Which element do you think would have been named by (i) Lawrence Berkeley Laboratory (ii) Seaborg’s group?

Solution

The naming of certain synthetic elements has often reflected the institutions or scientists involved in their discovery. In this context, we analyze the likely elements associated with the given sources.

Lawrence Berkeley Laboratory has been historically associated with the discovery of several transuranium elements. An element named in recognition of this institution is berkelium, symbol \( \mathrm{Bk} \), with atomic number \( 97 \). The name directly reflects the city of Berkeley, where the laboratory is located.

This association may be expressed as:

\[ \begin{aligned} \text{Lawrence Berkeley Laboratory} &\Rightarrow \text{Berkelium} \,(Z=97) \end{aligned} \]

Similarly, Seaborg’s research group played a central role in the discovery of several heavy elements. An element named in honour of the eminent nuclear chemist Glenn T. Seaborg is seaborgium, symbol \( \mathrm{Sg} \), with atomic number \( 106 \).

This relationship can be written as:

\[ \begin{aligned} \text{Seaborg’s group} &\Rightarrow \text{Seaborgium} \,(Z=106) \end{aligned} \]

Thus, the element associated with Lawrence Berkeley Laboratory is berkelium, and the element associated with Seaborg’s group is seaborgium.

Q8. Why do elements in the same group have similar physical and chemical properties?

Solution

Elements in the same group of the periodic table exhibit similar physical and chemical properties because they possess the same number of electrons in their outermost shell. Chemical behavior is primarily governed by valence electrons, as these electrons participate directly in bond formation and chemical reactions.

For elements belonging to a particular group, the general outer electronic configuration remains identical while only the principal quantum number changes. This can be expressed as:

\[ \begin{aligned} \text{Group } 1 &: ns^{1} \\ \text{Group } 2 &: ns^{2} \\ \text{Group } 17 &: ns^{2}np^{5} \end{aligned} \]

Although the value of \( n \) increases down the group, the number of valence electrons remains constant. As a result, elements show similar valency, comparable types of compounds, and analogous reactivity patterns.

This similarity may be summarized logically as:

\[ \begin{aligned} \text{Same number of }&\text{valence electrons} \\ &\Downarrow \\ \text{Similar valency }&\text{and bonding pattern} \\ &\Downarrow \\ \text{Similar chemi}&\text{cal properties} \end{aligned} \]

Physical properties such as atomic size, melting point, and density may gradually vary down the group due to increasing atomic number and additional shells, yet the fundamental chemical characteristics remain closely related because the valence shell electronic configuration is the same. Hence, elements in the same group display similar physical and chemical properties.

Q9. What does atomic radius and ionic radius really mean to you?

Solution

The atomic radius represents the size of a neutral atom and practically refers to half the distance between the nuclei of two identical atoms bonded together. Since the outer boundary of an atom is not sharply defined, atomic radius gives an approximate measure of how far the outermost electrons extend from the nucleus.

This idea may be expressed mathematically as:

\[ \begin{aligned} \text{Atomic radius} = \frac{1}{2} \times \text{internuclear distance} \end{aligned} \]

Thus, atomic radius helps us understand the overall spatial extent of an atom and plays a key role in explaining trends such as atomic size variation across periods and down groups.

Ionic radius, on the other hand, refers to the size of an ion formed when an atom gains or loses electrons. When an atom loses electrons to form a cation, its radius decreases because the remaining electrons are held more tightly by the nucleus. Conversely, when an atom gains electrons to form an anion, its radius increases due to enhanced electron–electron repulsion and reduced effective nuclear attraction per electron.

This behavior can be summarized as:

\[ \begin{aligned} \text{Cation formation} &: \text{loss of electrons} \Rightarrow \text{smaller radius} \\ \text{Anion formation} &: \text{gain of electrons} \Rightarrow \text{larger radius} \end{aligned} \]

In essence, atomic radius gives a measure of the size of a neutral atom, while ionic radius reflects how that size changes upon ion formation. Together, these concepts provide insight into bonding, reactivity, and the periodic trends observed in the elements.

Q10. How do atomic radius vary in a period and in a group? How do you explain the variation?

Solution

Atomic radius shows a definite periodic trend both across a period and down a group. Across a period, from left to right, the atomic radius gradually decreases. This happens because electrons are added to the same principal shell while the nuclear charge increases continuously. The increasing positive charge of the nucleus pulls the electron cloud closer, leading to a reduction in atomic size.

This trend may be represented as:

\[ \begin{aligned} \text{Across a period:} \quad &\text{Increase in nuclear charge} \\ &\Downarrow \\ \text{Stronger attr}&\text{action on electrons} \\ &\Downarrow \\ \text{Decrease }&\text{in atomic radius} \end{aligned} \]

On the other hand, down a group the atomic radius increases from top to bottom. This is because each successive element has an additional electron shell. Although nuclear charge also increases, the effect of inner-shell electrons shields the outer electrons from the nucleus. As a result, the outermost electrons lie farther from the nucleus, causing an increase in atomic size.

This variation can be summarized as:

\[ \begin{aligned} \text{Down a group:} \quad &\text{Addition of new shell} \\ &\Downarrow \\ \text{Increased}&\text{ shielding effect} \\ &\Downarrow \\ \text{Increase i}&\text{n atomic radius} \end{aligned} \]

Thus, atomic radius decreases across a period due to increasing effective nuclear charge within the same shell, while it increases down a group because of the addition of new shells and shielding of valence electrons. These opposing effects explain the characteristic size trends observed in the periodic table.

Q11. What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions. (i) \(\mathrm{F^–}\) (ii) Ar (iii) \(\mathrm{Mg^{2+}}\) (iv) \(\mathrm{Rb^{+}}\)

Solution

Isoelectronic species are atoms or ions that possess the same total number of electrons and hence have identical electronic configurations, even though their nuclear charges are different. Because their electron arrangements are the same, such species often show similar structural features, although their sizes may differ due to varying nuclear attraction.

To identify isoelectronic partners, we first count the number of electrons in each given atom or ion and then look for another species having the same electron count.

For the fluoride ion:

\[ \begin{aligned} \mathrm{F^-}:&\; Z=9,\; 9+1=10 \text{ electrons} \end{aligned} \]

A species with 10 electrons is neon, so Ne is isoelectronic with \( \mathrm{F^-} \).

For argon:

\[ \begin{aligned} \mathrm{Ar}:&\; Z=18,\; 18 \text{ electrons} \end{aligned} \]

A species having 18 electrons is the chloride ion, \( \mathrm{Cl^-} \), and hence it is isoelectronic with Ar.

For the magnesium ion:

\[ \begin{aligned} \mathrm{Mg^{2+}}:&\; Z=12,\; 12-2=10 \text{ electrons} \end{aligned} \]

Since neon also contains 10 electrons, Ne is isoelectronic with \( \mathrm{Mg^{2+}} \).

For the rubidium ion:

\[ \begin{aligned} \mathrm{Rb^{+}}:&\; Z=37,\; 37-1=36 \text{ electrons} \end{aligned} \]

An atom with 36 electrons is krypton, so Kr is isoelectronic with \( \mathrm{Rb^{+}} \).

Thus, the required isoelectronic pairs are Ne with \( \mathrm{F^-} \), \( \mathrm{Cl^-} \) with Ar, Ne with \( \mathrm{Mg^{2+}} \), and Kr with \( \mathrm{Rb^{+}} \). This illustrates that isoelectronic species share the same electronic configuration despite having different atomic numbers.

Q12. Consider the following species : \(N^{3–},\ O^{2–},\ F^{–},\ Na^{+},\ Mg^{2+}\) and \(\mathrm{Al^{3+}}\)
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.

Solution

Let us first examine the given species \( N^{3-},\ O^{2-},\ F^{-},\ Na^{+},\ Mg^{2+} \) and \( \mathrm{Al^{3+}} \) by calculating the number of electrons present in each. This helps in identifying any common feature among them.

\[ \begin{aligned} N^{3-} &: 7 + 3 = 10 \\ O^{2-} &: 8 + 2 = 10 \\ F^{-} &: 9 + 1 = 10 \\ Na^{+} &: 11 - 1 = 10 \\ Mg^{2+} &: 12 - 2 = 10 \\ Al^{3+} &: 13 - 3 = 10 \end{aligned} \]

From the above, it is clear that all these species contain exactly 10 electrons. Hence, they are isoelectronic and possess the same electronic configuration as neon. This common electronic structure answers part (a).

For part (b), although these ions have identical numbers of electrons, their nuclear charges are different. As nuclear charge increases, the attraction between the nucleus and the electron cloud becomes stronger, causing the ionic radius to decrease.

Thus, moving from \( N^{3-} \) to \( \mathrm{Al^{3+}} \), the nuclear charge increases progressively, leading to a continuous decrease in ionic size. This trend may be represented as:

\[ \begin{aligned} \text{Increasing nuc}&\text{lear charge} \\ \Downarrow \\ \text{decreasing io}&\text{nic radius} \end{aligned} \]

Therefore, the order of increasing ionic radii is:

\[ \begin{aligned} \mathrm{Al^{3+}} < Mg^{2+} < Na^{+} < F^{-} < O^{2-} < N^{3-} \end{aligned} \]

Hence, all the given species are isoelectronic, and their ionic radii increase in the sequence shown above due to decreasing nuclear attraction on the common electron cloud.

Q13. Explain why cation are smaller and anions larger in radii than their parent atoms?

Solution

The size of an ion differs from that of its parent atom because ion formation changes the balance between nuclear attraction and electron–electron repulsion. When a neutral atom forms a cation, it loses one or more electrons. This loss reduces the total number of electrons while the nuclear charge remains the same. As a result, the effective nuclear attraction experienced by the remaining electrons increases.

In many cases, especially for metals, the outermost shell is completely removed during cation formation. This leads to a decrease in the number of occupied shells and hence a significant reduction in size.

This change may be expressed as:

\[ \begin{aligned} \text{Atom} &\rightarrow \text{Cation} + e^- \\ \text{Fewer electrons} &\Rightarrow \text{greater effective nuclear attraction} \\ &\Rightarrow \text{smaller radius} \end{aligned} \]

On the other hand, when a neutral atom forms an anion, it gains one or more electrons. The nuclear charge remains constant, but the number of electrons increases. This addition enhances electron–electron repulsion in the valence shell and reduces the effective pull of the nucleus on each electron.

The process can be summarized as:

\[ \begin{aligned} \text{Atom} + e^- &\rightarrow \text{Anion} \\ \text{More electrons} &\Rightarrow \text{increased repulsion} \\ &\Rightarrow \text{larger radius} \end{aligned} \]

Thus, cations are smaller than their parent atoms due to increased effective nuclear attraction and possible loss of an outer shell, whereas anions are larger because of increased electron–electron repulsion and reduced effective nuclear attraction per electron.

Q14. What is the significance of the terms — ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy?

Solution

While defining ionization enthalpy and electron gain enthalpy, the terms “isolated gaseous atom” and “ground state” are used to ensure clarity and uniformity in measurement. Ionization enthalpy is defined as the enthalpy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. Similarly, electron gain enthalpy refers to the enthalpy change when an electron is added to an isolated gaseous atom in its ground state.

The term “isolated gaseous atom” signifies that the atom must be free from intermolecular forces, lattice interactions, or bonding effects. In the gaseous state, atoms are widely separated, so the measured energy change reflects only the intrinsic attraction between the nucleus and its electrons. This avoids complications arising from solid or liquid phases where additional energy would be required to overcome intermolecular or lattice forces.

This condition may be expressed as:

\[ \begin{aligned} \text{M(g)} &\rightarrow \text{M}^{+}\text{(g)} + e^- \quad (\text{Ionization enthalpy}) \\ \text{X(g)} + e^- &\rightarrow \text{X}^{-}\text{(g)} \quad (\text{Electron gain enthalpy}) \end{aligned} \]

The term “ground state” indicates that the atom must be in its lowest energy state. If the atom were in an excited state, the energy required to remove or add an electron would differ, leading to inconsistent values. Using the ground state ensures that the measured enthalpy change represents a standard and reproducible quantity.

Thus, “isolated gaseous atom” ensures absence of external interactions, and “ground state” ensures minimum internal energy. Together, these conditions provide precise and comparable definitions of ionization enthalpy and electron gain enthalpy.

Q15. Energy of an electron in the ground state of the hydrogen atom is \(\mathrm{–2.18\times 10^{–18}\ J}\). Calculate the ionization enthalpy of atomic hydrogen in terms of \(\mathrm{J\ mol^{–1}}\)

Solution

The ionization enthalpy of hydrogen corresponds to the energy required to remove the electron from a hydrogen atom in its ground state to infinity. Since the energy of the electron in the ground state is given as \( -2.18 \times 10^{-18}\,\text{J} \), the energy needed to ionize one hydrogen atom is equal in magnitude but opposite in sign.

Thus, energy required per atom is:

\[ \begin{aligned} E &= +2.18 \times 10^{-18}\,\text{J} \end{aligned} \]

To convert this value into ionization enthalpy per mole, we multiply by Avogadro’s number \( (6.022 \times 10^{23}\,\text{mol}^{-1}) \).

\[ \begin{aligned} \text{Ionization enthalpy} &= 2.18 \times 10^{-18} \times 6.022 \times 10^{23} \\ &= 13.12 \times 10^{5}\,\text{J mol}^{-1} \\ &= 1.312 \times 10^{6}\,\text{J mol}^{-1} \end{aligned} \]

Therefore, the ionization enthalpy of atomic hydrogen is approximately \( 1.31 \times 10^{6}\,\text{J mol}^{-1} \). This value represents the energy required to remove one mole of electrons from one mole of hydrogen atoms in the gaseous ground state.

Q16. 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 \(\mathrm{\Delta_i H}\) than B
(ii) O has lower \(\mathrm{\Delta_i H}\) than N and F

Solution

The general trend across the second period is a gradual increase in ionization enthalpy due to increasing nuclear charge and decreasing atomic size. However, certain deviations occur because of differences in electronic configuration and orbital stability.

For part (i), beryllium has the electronic configuration \(1s^2\,2s^2\), while boron has \(1s^2\,2s^2\,2p^1\). In boron, the electron removed during ionization belongs to the \(2p\) orbital, which is higher in energy and less penetrating than the \(2s\) orbital. Hence, it is more weakly held by the nucleus.

This may be expressed as:

\[ \begin{aligned} \text{Be}:&\;2s^2 \quad (\text{completely filled and stable}) \\ \text{B}:&\;2s^2\,2p^1 \quad (\text{electron removed from }2p) \end{aligned} \]

Since the \(2s\) electrons in Be are more strongly attracted to the nucleus than the \(2p\) electron in B, more energy is required to remove an electron from Be. Therefore, Be has a higher ionization enthalpy than B.

For part (ii), nitrogen has the configuration \(1s^2\,2s^2\,2p^3\), which corresponds to a half-filled \(2p\) subshell. This arrangement is particularly stable due to symmetrical distribution and exchange energy. Oxygen, on the other hand, has the configuration \(1s^2\,2s^2\,2p^4\), where one of the \(2p\) orbitals contains a pair of electrons.

This situation can be represented as:

\[ \begin{aligned} \text{N}:&\;2p^3 \quad (\text{half-filled, extra stability}) \\ \text{O}:&\;2p^4 \quad (\text{one paired set, increased repulsion}) \end{aligned} \]

In oxygen, the paired electrons in one orbital experience greater electron–electron repulsion, making it easier to remove one electron. Hence, oxygen has a lower ionization enthalpy than nitrogen. Fluorine, with configuration \(2p^5\), has a higher effective nuclear charge and smaller atomic size, so its electrons are held more tightly than in oxygen, giving F a higher ionization enthalpy.

Thus, Be exceeds B in ionization enthalpy because of removal from a more stable \(2s\) subshell, and O shows lower ionization enthalpy than N and F due to electron pairing in its \(2p\) orbitals and comparatively weaker nuclear attraction.

Q17. 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?

Solution

To understand this behavior, we first examine the electronic configurations of sodium and magnesium. Sodium has the configuration \(1s^2\,2s^2\,2p^6\,3s^1\), while magnesium has \(1s^2\,2s^2\,2p^6\,3s^2\). In sodium, the outermost electron lies in the \(3s\) orbital and is weakly held due to larger atomic size and lower effective nuclear charge compared to magnesium.

This situation can be represented as:

\[ \begin{aligned} \text{Na}:&\;[Ne]\,3s^1 \\ \text{Mg}:&\;[Ne]\,3s^2 \end{aligned} \]

Because sodium has only one electron in its valence shell, removal of this electron results in the stable noble gas configuration \([Ne]\). Therefore, less energy is required to remove this electron, making the first ionization enthalpy of sodium lower than that of magnesium, where an electron must be removed from a more stable \(3s^2\) configuration.

After losing one electron, sodium becomes \( \mathrm{Na^+} \) with configuration \([Ne]\), while magnesium becomes \( \mathrm{Mg^+} \) with configuration \([Ne]\,3s^1\). For the second ionization, sodium must lose an electron from the inner noble gas shell, whereas magnesium loses its remaining \(3s\) valence electron.

This may be expressed as:

\[ \begin{aligned} \mathrm{Na^+}:&\;[Ne] \Rightarrow \text{electron removed from inner shell} \\ \mathrm{Mg^+}:&\;[Ne]\,3s^1 \Rightarrow \text{electron removed from valence shell} \end{aligned} \]

Removing an electron from a noble gas configuration requires a very large amount of energy, so the second ionization enthalpy of sodium is exceptionally high. In contrast, magnesium still has a valence electron in the \(3s\) orbital after the first ionization, which can be removed comparatively easily.

Hence, sodium has a lower first ionization enthalpy than magnesium because its outer electron is more loosely held, but its second ionization enthalpy is higher because the second electron must be removed from a stable inner-shell configuration, unlike magnesium.

Q18. What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?

Solution

The ionization enthalpy of main group elements generally decreases on moving down a group because the outermost electron becomes progressively easier to remove. This trend arises from a combination of structural factors related to atomic size and electron shielding.

As we move down a group, a new principal shell is added at each successive element. This increases the distance between the nucleus and the valence electron, thereby reducing the attractive force exerted by the nucleus on that electron.

At the same time, the number of inner-shell electrons increases. These inner electrons shield the outermost electron from the full nuclear charge, further weakening the effective nuclear attraction.

This combined effect can be summarized as:

\[ \begin{aligned} \color{red}\text{Down a group} \Rightarrow \begin{cases} \color{aliceblue}\small \text{increase in atomic size} \\ \color{aliceblue}\small\text{increase in shielding effect} \\ \color{aliceblue}\small\text{decrease in effective nuclear attraction} \\ \color{aliceblue}\small\text{lower ionization enthalpy} \end{cases} \end{aligned} \]

Although nuclear charge also increases down the group, its influence is largely counterbalanced by the simultaneous increase in shielding and atomic radius. As a result, the outermost electron experiences a weaker net pull from the nucleus.

Hence, the progressive increase in atomic size along with enhanced shielding by inner electrons makes it easier to remove the valence electron, causing the ionization enthalpy of main group elements to decrease down a group.

Q19. The first ionization enthalpy values \(\mathrm{(in\ kJ\ mol^{–1})}\) of group 13 elements are : \[ \begin{array}{} \begin{aligned} B&& Al&& Ga&& In&& Tl\\ 801&& 577&& 579&& 558&& 589 \end{aligned} \end{array} \] How would you explain this deviation from the general trend ?

Solution

In general, ionization enthalpy decreases down a group because atomic size increases and the valence electron becomes farther from the nucleus. However, the given data for group 13 elements show certain irregularities that deviate from this simple trend.

The first ionization enthalpy decreases markedly from boron to aluminium due to the expected increase in atomic size:

\[ \begin{aligned} \text{B} \;(801) &> \text{Al} \;(577) \end{aligned} \]

However, the value for gallium is slightly higher than that of aluminium:

\[ \begin{aligned} \text{Al} \;(577) &< \text{Ga} \;(579) \end{aligned} \]

This anomaly arises because gallium contains completely filled \(3d^{10}\) orbitals. The \(3d\) electrons have poor shielding ability, so the effective nuclear charge experienced by the outer \(4p\) electron increases. As a result, the valence electron in gallium is held more strongly than expected, leading to a slight increase in ionization enthalpy.

A similar explanation applies to thallium. Although ionization enthalpy decreases from gallium to indium due to increasing size:

\[ \begin{aligned} \text{Ga} \;(579) &> \text{In} \;(558) \end{aligned} \]

the value increases again for thallium:

\[ \begin{aligned} \text{In} \;(558) &< \text{Tl} \;(589) \end{aligned} \]

In thallium, the presence of filled \(4f^{14}\) and \(5d^{10}\) subshells leads to poor shielding of the outer \(6p\) electron. The increased effective nuclear charge causes stronger attraction of the valence electron, raising the ionization enthalpy unexpectedly.

Thus, the deviations from the general decreasing trend in group 13 are mainly due to the poor shielding effect of intervening \(d\) and \(f\) electrons, which increases effective nuclear charge and strengthens the binding of the outermost electron.

Q20. Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F
(ii) F or Cl

Solution

Electron gain enthalpy refers to the energy change when an isolated gaseous atom accepts an electron. A more negative value indicates a stronger tendency to gain an electron and form an anion. This tendency depends mainly on atomic size, effective nuclear charge, and electron–electron repulsion in the valence shell.

For the pair oxygen and fluorine, fluorine has a higher effective nuclear charge and a smaller atomic radius. As a result, an incoming electron experiences stronger attraction toward the nucleus in fluorine than in oxygen.

This comparison may be written as:

\[ \begin{aligned} \text{F} &: \text{smaller size, higher nuclear attraction} \\ \text{O} &: \text{larger size, weaker attraction} \end{aligned} \]

Hence, fluorine has a more negative electron gain enthalpy than oxygen.

For the pair fluorine and chlorine, although fluorine is more electronegative, its very small size causes significant electron–electron repulsion in the compact \(2p\) subshell when an extra electron is added. Chlorine, being larger, can accommodate the incoming electron in its \(3p\) orbital with less repulsion.

This effect can be summarized as:

\[ \begin{aligned} \text{F} &: \text{very small size} \Rightarrow \text{greater repulsion} \\ \text{Cl} &: \text{larger size} \Rightarrow \text{less repulsion} \end{aligned} \]

Therefore, chlorine releases more energy on gaining an electron and thus has a more negative electron gain enthalpy than fluorine.

In conclusion, fluorine has a more negative electron gain enthalpy than oxygen, while chlorine has a more negative electron gain enthalpy than fluorine.

Q21. Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.

Solution

The second electron gain enthalpy of oxygen refers to the energy change when an electron is added to the already negatively charged oxide ion precursor, that is, when \( \mathrm{O^-} \) gains an electron to form \( \mathrm{O^{2-}} \).

This process can be represented as:

\[ \begin{aligned} \mathrm{O(g)} + e^- &\rightarrow \mathrm{O^- (g)} \quad (\text{first electron gain enthalpy}) \\ \mathrm{O^- (g)} + e^- &\rightarrow \mathrm{O^{2-} (g)} \quad (\text{second electron gain enthalpy}) \end{aligned} \]

The first electron gain enthalpy of oxygen is negative because energy is released when a neutral oxygen atom accepts an electron due to attraction between the nucleus and the incoming electron.

However, in the second step, the incoming electron is being added to an already negatively charged ion. Strong electrostatic repulsion exists between the added electron and the electrons present in \( \mathrm{O^-} \). Considerable external energy must be supplied to overcome this repulsion.

This situation may be summarized as:

\[ \begin{aligned} \text{Neutral atom} &\Rightarrow \text{electron addition releases energy} \\ \text{Negative ion} &\Rightarrow \text{electron addition requires energy} \end{aligned} \]

Therefore, the second electron gain enthalpy of oxygen is not negative; instead, it is positive and much higher in magnitude than the first. This is because adding an electron to an anion is energetically unfavorable due to strong electron–electron repulsion.

Q22. What is the basic difference between the terms electron gain enthalpy and electronegativity?

Solution

Electron gain enthalpy and electronegativity are both related to the tendency of an atom to attract electrons, yet they differ fundamentally in meaning, measurement, and context. Electron gain enthalpy refers to the actual energy change that occurs when an isolated gaseous atom accepts an electron to form a negative ion. It is a thermodynamic quantity with a definite numerical value and measurable units.

This process is represented as:

\[ \begin{aligned} \mathrm{X(g)} + e^- \rightarrow \mathrm{X^- (g)} \quad (\text{electron gain enthalpy}) \end{aligned} \]

Thus, electron gain enthalpy describes an energy change under well-defined conditions involving isolated atoms in the gaseous state.

Electronegativity, on the other hand, does not involve isolated atoms. It expresses the relative tendency of an atom to attract shared electrons toward itself when it is bonded to another atom in a molecule. It is not a directly measurable thermodynamic quantity and has no units; instead, it is expressed on comparative scales such as Pauling’s scale.

In simple terms, electron gain enthalpy deals with the gain of an electron by a free atom and is an absolute energy change, whereas electronegativity describes how strongly a bonded atom pulls electron density toward itself and is a relative, dimensionless property.

Therefore, the basic difference lies in the fact that electron gain enthalpy is a measurable energetic change for isolated gaseous atoms, while electronegativity is a comparative concept that applies only to atoms in chemical bonds.

Q23. How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?

Solution

The statement that the electronegativity of nitrogen on the Pauling scale is 3.0 in all nitrogen compounds is not strictly correct. While 3.0 represents the standard Pauling electronegativity value assigned to nitrogen, electronegativity itself is not a fixed atomic constant in real chemical environments.

Electronegativity is a relative property that describes the tendency of an atom to attract shared electrons in a chemical bond. This tendency depends on the chemical surroundings of the atom, such as its oxidation state, hybridization, and the nature of atoms bonded to it.

Conceptually, this may be expressed as:

\[ \begin{aligned} \text{Electronegativity} &= f(\text{bonding environment}) \end{aligned} \]

In different compounds, nitrogen may exist in varying oxidation states and bonding situations. For example, nitrogen behaves differently in ammonia, nitric acid, or nitrogen dioxide. These changes alter the effective nuclear attraction experienced by bonding electrons, leading to slight variations in its electronegativity.

Therefore, the value 3.0 should be regarded as an average reference on the Pauling scale rather than an absolute number applicable to every nitrogen compound. Hence, the statement must be interpreted cautiously, recognizing that electronegativity can vary depending on the molecular environment.

Q24. Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron

Solution

The radius of an atom changes when it gains or loses electrons because ion formation alters the balance between nuclear attraction and electron–electron repulsion within the atom.

When an atom gains an electron, it forms an anion. The added electron enters the valence shell, increasing the total number of electrons while the nuclear charge remains unchanged. As a result, electron–electron repulsion increases and the effective nuclear attraction experienced by each electron decreases slightly. The electron cloud expands due to this increased repulsion.

This change may be represented as:

\[ \begin{aligned} \text{Atom} + e^- &\rightarrow \text{Anion} \\ \text{Increase in electron–electron repulsion} &\Rightarrow \text{expansion of electron cloud} \\ &\Rightarrow \text{increase in radius} \end{aligned} \]

Therefore, the radius of an atom increases when it gains an electron.

When an atom loses an electron, it forms a cation. In this case, the total number of electrons decreases while the nuclear charge remains the same. The remaining electrons experience stronger effective nuclear attraction. In many cases, especially for metals, the entire outer shell may be removed, reducing the number of occupied shells.

This situation can be expressed as:

\[ \begin{aligned} \text{Atom} &\rightarrow \text{Cation} + e^- \\ \text{Decrease in electron–electron repulsion} &\Rightarrow \text{stronger nuclear pull} \\ &\Rightarrow \text{decrease in radius} \end{aligned} \]

Thus, the atomic radius increases when an atom gains an electron due to enhanced repulsion and decreases when it loses an electron because of increased effective nuclear attraction and possible reduction in the number of shells.

Q25. Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.

Solution

The first ionization enthalpies of two isotopes of the same element are expected to be essentially the same. This is because isotopes differ only in the number of neutrons present in the nucleus, while they have identical atomic numbers and identical electronic configurations.

Ionization enthalpy depends primarily on factors such as nuclear charge, atomic size, shielding effect, and the arrangement of electrons. Since isotopes possess the same number of protons and electrons, all these controlling factors remain unchanged.

This idea may be expressed as:

\[ \begin{aligned} \color{red}\text{Same atomic number} \Rightarrow \begin{cases} \color{aliceblue}\small\text{same electronic configuration} \\ \color{aliceblue}\small\text{same effective nuclear attraction} \\ \color{aliceblue}\small\text{same ionization enthalpy} \end{cases} \end{aligned} \]

Although isotopes have slightly different atomic masses due to different neutron numbers, neutrons do not participate in chemical interactions and have no direct influence on the attraction between the nucleus and valence electrons.

Therefore, the first ionization enthalpy of isotopes of the same element is practically identical, as the electronic environment governing electron removal remains the same in both cases.

Q26. What are the major differences between metals and non-metals?

Solution

Metals and non-metals differ fundamentally in their physical appearance, electrical behavior, and chemical nature because of differences in their electronic configurations and bonding tendencies.

Metals generally possess low ionization enthalpy and readily lose electrons to form positive ions. This tendency arises from their larger atomic size and fewer valence electrons, which makes electron removal easier. As a result, metals typically show high electrical and thermal conductivity, metallic lustre, malleability, and ductility, all of which stem from the presence of freely moving electrons in their structures.

Non-metals, in contrast, have relatively high ionization enthalpy and a strong tendency to gain electrons or share electrons during bond formation. Their smaller atomic size and higher effective nuclear charge enable them to attract electrons strongly. Consequently, non-metals are generally poor conductors of heat and electricity and often appear dull and brittle in solid form.

Chemically, metals usually form basic oxides and ionic compounds by losing electrons, whereas non-metals commonly form acidic or neutral oxides and participate mainly in covalent bonding by sharing electrons.

This contrast in electronic behavior may be summarized conceptually as:

\[ \begin{aligned} \text{Metals}:&\; \text{loss of electrons} \Rightarrow \text{formation of cations} \\ \text{Non-metals}:&\; \text{gain or sharing of electrons} \Rightarrow \text{formation of anions or covalent bonds} \end{aligned} \]

Thus, the major differences between metals and non-metals arise from their opposite tendencies toward electron loss or gain, which in turn govern their characteristic physical properties and chemical reactivity.

Q27. 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.

Solution

These questions can be answered by examining typical electronic configurations and periodic trends.

An element having five electrons in its outer subshell belongs to Group 15, where the general valence configuration is \( ns^2 np^3 \). A common example from this group is nitrogen, whose outer shell contains five electrons.

This may be expressed as:

\[ \begin{aligned} \text{N}:&\;2s^2\,2p^3 \Rightarrow 5 \text{ valence electrons} \end{aligned} \]

An element that tends to lose two electrons belongs to Group 2, since these elements have the configuration \( ns^2 \) and readily form \( +2 \) ions. Magnesium is a typical example, as it loses two electrons to attain a noble gas configuration.

\[ \begin{aligned} \text{Mg}:&\;3s^2 \Rightarrow \text{loss of two electrons} \end{aligned} \]

An element that tends to gain two electrons belongs to Group 16, where the valence configuration is \( ns^2 np^4 \). Oxygen is a representative element of this group and commonly gains two electrons to complete its octet.

\[ \begin{aligned} \text{O}:&\;2s^2\,2p^4 \Rightarrow \text{gain of two electrons} \end{aligned} \]

Finally, the group that contains metals, non-metals, a liquid, and gases at room temperature is Group 17, the halogen group. Fluorine and chlorine are gases, bromine is a liquid, iodine is a solid non-metal, and astatine shows metallic character.

Thus, nitrogen answers part (a), magnesium answers part (b), oxygen answers part (c), and Group 17 satisfies the condition in part (d).

Q28. 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.

Solution

The opposite trends in reactivity observed for Group 1 and Group 17 elements arise from their contrasting electronic tendencies. Group 1 elements, known as alkali metals, react by losing one valence electron, whereas Group 17 elements, the halogens, react by gaining one electron.

For Group 1 elements, reactivity depends on how easily the outermost electron can be removed. As we move down the group from lithium to caesium, atomic size increases and shielding by inner electrons becomes stronger. Consequently, the attraction between the nucleus and the valence electron decreases, making electron loss progressively easier.

This trend can be summarized as:

\[ \begin{aligned} \color{red}\text{Down Group 1} &\Rightarrow \begin{cases} \color{aliceblue}\small\text{increase in atomic size and shielding} \\ \color{aliceblue}\small\text{decrease in ionization enthalpy} \\ \color{aliceblue}\small\text{increase in reactivity} \end{cases} \end{aligned} \]

Hence, the reactivity increases in the order:

\[ \begin{aligned} \text{Li} < \text{Na} < \text{K} < \text{Rb} < \text{Cs} \end{aligned} \]

In contrast, Group 17 elements gain an electron during chemical reactions. Their reactivity depends on how strongly they can attract an incoming electron. Moving down the group from fluorine to iodine, atomic size increases and effective nuclear attraction on the added electron decreases. As a result, the tendency to gain an electron becomes weaker.

This behavior may be expressed as:

\[ \begin{aligned} \color{red}\text{Down Group 17} &\Rightarrow \begin{cases} \color{aliceblue}\small\text{increase in atomic size} \\ \color{aliceblue}\small\text{decrease in electron-attracting power} \\ \color{aliceblue}\small\text{decrease in reactivity} \end{cases} \end{aligned} \]

Therefore, the reactivity order among halogens is:

\[ \begin{aligned} \text{F} > \text{Cl} > \text{Br} > \text{I} \end{aligned} \]

Thus, Group 1 elements become more reactive down the group because they lose electrons more easily, while Group 17 elements become less reactive down the group because their ability to gain electrons decreases with increasing atomic size.

Q29. Write the general outer electronic configuration of s-, p-, d- and f- block elements.

Solution

The general outer electronic configuration of elements is determined by the subshell into which the differentiating electron enters. On this basis, the periodic table is divided into s-, p-, d- and f-block elements.

In s-block elements, the last electron enters the \( s \)-subshell of the outermost shell. Their general outer configuration is:

\[ \begin{aligned} ns^{1-2} \end{aligned} \]

In p-block elements, the differentiating electron enters the \( p \)-subshell of the outermost shell. Their general outer configuration is:

\[ \begin{aligned} ns^{2}\,np^{1-6} \end{aligned} \]

In d-block elements, also known as transition elements, the last electron enters the \((n-1)d\) subshell while the outermost shell contains one or two electrons in the \(ns\) orbital. Their general outer configuration is:

\[ \begin{aligned} (n-1)d^{1-10}\,ns^{1-2} \end{aligned} \]

In f-block elements, the differentiating electron enters the \((n-2)f\) subshell. These elements also contain electrons in the \((n-1)d\) and \(ns\) subshells. Their general outer configuration is:

\[ \begin{aligned} (n-2)f^{1-14}\,(n-1)d^{0-1}\,ns^{2} \end{aligned} \]

Thus, the block classification of elements is directly related to the subshell that receives the last electron, and the above expressions represent the general outer electronic configurations of the s-, p-, d- and f-block elements.

Q30. 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.

Solution

The position of an element in the periodic table is determined from its outer electronic configuration by identifying the principal quantum number (which gives the period), the type of subshell being filled (which gives the block), and the total number of valence electrons (which gives the group).

For case (i), the configuration is \( ns^2 np^4 \) with \( n=3 \). Substituting the value of \( n \), we obtain:

\[ \begin{aligned} 3s^2\,3p^4 \end{aligned} \]

The highest principal quantum number is 3, so the element belongs to the third period. Since the last electron enters a \(p\)-orbital, it lies in the p-block. The valence shell contains \(2+4=6\) electrons, placing the element in Group 16. Hence, this element is located in Period 3, Group 16 of the p-block.

For case (ii), the configuration is \( (n-1)d^2 ns^2 \) with \( n=4 \). Substituting:

\[ \begin{aligned} 3d^2\,4s^2 \end{aligned} \]

Here, the highest value of \( n \) is 4, so the element belongs to the fourth period. The differentiating electrons enter the \(3d\) subshell, indicating that it is a d-block element. The group number for d-block elements is obtained by adding the electrons in \((n-1)d\) and \(ns\):

\[ \begin{aligned} 2 + 2 = 4 \end{aligned} \]

Therefore, this element lies in Period 4, Group 4 of the d-block.

For case (iii), the configuration is \( (n-2)f^{7}(n-1)d^1ns^2 \) with \( n=6 \). Substituting:

\[ \begin{aligned} 4f^{7}\,5d^{1}\,6s^{2} \end{aligned} \]

The highest principal quantum number is 6, so the element belongs to the sixth period. Since the differentiating electrons are entering the \(4f\) subshell, the element is an f-block element, specifically a lanthanoid. Such elements are placed separately at the bottom of the periodic table and are associated with Group 3.

Thus, the positions are: (i) Period 3, Group 16, p-block; (ii) Period 4, Group 4, d-block; and (iii) Period 6, f-block (lanthanoid series, associated with Group 3).

Q31. The first \((\Delta_ i H_1 )\) and the second \((\Delta_i H_2 )\) ionization enthalpies \(\mathrm{(in\ kJ\ mol^{–1})}\) and the \((\Delta+{eg} H)\) electron gain enthalpy \(\mathrm{(in\ kJ\ mol^{–1})}\) of a few elements are given below: \[ \begin{array}{} \begin{aligned} Elements && \Delta H_1 && \Delta H_2&& \Delta_{eg}H\\ I&&520 &&7300 &&-60 \\ II&& 419&&3051 &&-48 \\ III&&1681 &&3374 &&-328 \\ IV&&1008 &&1846 &&–295 \\ V&&2372 &&5251 &&+48 \\ VI&&738 &&1451 &&–40 \\ \end{aligned} \end{array} \] 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)?

Solution

The chemical behaviour of the given elements can be inferred from their ionization enthalpies and electron gain enthalpy. Low first ionization enthalpy indicates metallic character and high reactivity as a metal, whereas highly negative electron gain enthalpy indicates strong non-metallic character and high reactivity as a non-metal. Extremely high ionization enthalpy values suggest noble gas character and very low reactivity.

Element V has exceptionally high values of \( \Delta_i H_1 = 2372 \) and \( \Delta_i H_2 = 5251 \,\text{kJ mol}^{-1} \) along with a positive electron gain enthalpy. Such a combination indicates a noble gas configuration, making it the least reactive element.

Among the metals, element II has the lowest first ionization enthalpy \( (419\,\text{kJ mol}^{-1}) \), which means it can lose an electron most easily. Hence, it is the most reactive metal.

Element III has a very high first ionization enthalpy \( (1681\,\text{kJ mol}^{-1}) \) and a highly negative electron gain enthalpy \( (-328\,\text{kJ mol}^{-1}) \), indicating a strong tendency to gain an electron. Therefore, it is the most reactive non-metal.

Element IV also has a negative electron gain enthalpy but less negative than element III. Hence, among the non-metals listed, it is comparatively less reactive and represents the least reactive non-metal.

To identify the metal forming a stable halide of formula \( MX_2 \), we look for an element with relatively low first ionization enthalpy and a reasonably low second ionization enthalpy, indicating ease of formation of a \( +2 \) oxidation state:

\[ \begin{aligned} \text{Element VI:} \quad \Delta_i H_1 = 738,\; \Delta_i H_2 = 1451 \end{aligned} \]

Since both ionization enthalpies are moderate, element VI can readily form a stable \( MX_2 \) compound.

Finally, the metal forming a predominantly covalent halide \( MX \) should have low first ionization enthalpy but extremely high second ionization enthalpy, so that only one electron is removed easily:

\[ \begin{aligned} \text{Element I:} \quad \Delta_i H_1 = 520,\; \Delta_i H_2 = 7300 \end{aligned} \]

The very high second ionization enthalpy discourages formation of a \( +2 \) state, favouring the formation of a monovalent halide \( MX \), which is often covalent in nature.

Thus, the answers are: (a) V, (b) II, (c) III, (d) IV, (e) VI, and (f) I.

Q32. 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

Solution

The formula of a stable binary compound can be predicted by considering the usual oxidation states or valencies of the combining elements and then balancing their charges to obtain electrical neutrality.

For lithium and oxygen, lithium commonly forms a \(+1\) ion while oxygen forms a \(2-\) ion. Charge balance therefore requires two lithium ions for each oxide ion:

\[ \begin{aligned} 2\,\mathrm{Li^+} + \mathrm{O^{2-}} \rightarrow \mathrm{Li_2O} \end{aligned} \]

Hence, the compound formed is \( \mathrm{Li_2O} \).

Magnesium usually forms a \(2+\) ion, whereas nitrogen forms a \(3-\) ion. Balancing these charges gives:

\[ \begin{aligned} 3\,\mathrm{Mg^{2+}} + 2\,\mathrm{N^{3-}} \rightarrow \mathrm{Mg_3N_2} \end{aligned} \]

Therefore, the formula is \( \mathrm{Mg_3N_2} \).

Aluminium generally exhibits a \(+3\) oxidation state and iodine forms a \(1-\) ion. Combining these gives:

\[ \begin{aligned} \mathrm{Al^{3+}} + 3\,\mathrm{I^-} \rightarrow \mathrm{AlI_3} \end{aligned} \]

Thus, the compound is \( \mathrm{AlI_3} \).

Silicon commonly shows a valency of 4, while oxygen has a valency of 2. The resulting neutral compound is:

\[ \begin{aligned} \mathrm{Si^{4+}} + 2\,\mathrm{O^{2-}} \rightarrow \mathrm{SiO_2} \end{aligned} \]

Hence, silicon dioxide \( \mathrm{SiO_2} \) is formed.

Phosphorus generally forms a pentahalide with fluorine because fluorine is highly electronegative. Taking phosphorus in the \(+5\) state and fluorine as \( -1 \):

\[ \begin{aligned} \mathrm{P^{5+}} + 5\,\mathrm{F^-} \rightarrow \mathrm{PF_5} \end{aligned} \]

So, the compound is \( \mathrm{PF_5} \).

Element 71 is lutetium, a lanthanoid that commonly exhibits a \(+3\) oxidation state. Fluorine forms \( -1 \) ions, giving:

\[ \begin{aligned} \mathrm{Lu^{3+}} + 3\,\mathrm{F^-} \rightarrow \mathrm{LuF_3} \end{aligned} \]

Therefore, the predicted formulas are \( \mathrm{Li_2O} \), \( \mathrm{Mg_3N_2} \), \( \mathrm{AlI_3} \), \( \mathrm{SiO_2} \), \( \mathrm{PF_5} \), and \( \mathrm{LuF_3} \).

Q33. 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.

Solution

In the modern periodic table, the period number of an element is directly related to the highest principal quantum number of electrons present in its ground-state electronic configuration. The principal quantum number \( n \) represents the main energy level or shell occupied by the outermost electrons of an atom.

As we move from one period to the next, a new principal shell begins to fill. Hence, all elements in the same period possess the same value of \( n \) for their valence shell.

This relationship can be expressed as:

\[ \begin{aligned} \text{Period number} &= \text{highest principal quantum number } (n) \end{aligned} \]

Atomic number determines the position of an element in the periodic table as a whole, atomic mass does not govern periodic classification, and the azimuthal quantum number defines subshells rather than periods.

Therefore, in the modern periodic table, the period indicates the value of the principal quantum number.

Q34. 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.

Solution

To identify the incorrect statement, we examine each option in light of subshell capacities and block structure in the modern periodic table.

Statement (a) is correct because a p-subshell consists of three orbitals, and each orbital can accommodate two electrons. Hence, a maximum of six electrons can occupy a p-subshell, which explains why the p-block contains six columns.

Statement (b) claims that the d-block has eight columns because a maximum of eight electrons can occupy a d-subshell. This is incorrect. A d-subshell actually contains five orbitals, and each orbital can hold two electrons, so it can accommodate a total of ten electrons. Accordingly, the d-block has ten columns, not eight.

This capacity is given by:

\[ \begin{aligned} \text{Maximum electrons in d-subshell} &= 5 \times 2 = 10 \end{aligned} \]

Statement (c) is correct since each block indeed contains as many columns as the maximum number of electrons that can occupy the corresponding subshell.

Statement (d) is also correct because the block designation reflects the value of the azimuthal quantum number \( l \) of the subshell that receives the last electron, with \( l=0 \) for s, \( l=1 \) for p, \( l=2 \) for d, and \( l=3 \) for f.

Therefore, the incorrect statement is option (b).

Q35. 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.

Solution

The chemistry of an element is largely determined by its valence electrons, as they are involved in chemical bonding and reactions. Various factors can influence the valence shell and, consequently, the element's chemical behavior. Let us analyze each option to determine which one does not affect the valence shell.

The valence principal quantum number \(n\) represents the energy level of the outermost electrons. A higher value of \(n\) corresponds to electrons that are farther from the nucleus, affecting the reactivity and chemical behavior of the element. Therefore, the valence principal quantum number directly affects the valence shell.

Nuclear charge \(Z\) refers to the total positive charge of the nucleus, which is directly related to the number of protons in the nucleus. An increase in nuclear charge results in a stronger attraction between the nucleus and the valence electrons, affecting the chemistry of the element. Thus, nuclear charge does affect the valence shell.

Nuclear mass, however, does not directly influence the valence shell. While it contributes to the overall mass of the atom, the mass of the nucleus does not impact the electron configuration or the behavior of the valence electrons. The mass of the nucleus primarily affects the atomic mass but not the chemical properties. Therefore, nuclear mass does not affect the valence shell.

Finally, the number of core electrons influences the shielding effect, which can affect the effective nuclear charge felt by the valence electrons. The greater the number of core electrons, the more they shield the valence electrons from the nucleus. Thus, the number of core electrons does affect the valence shell.

Therefore, the factor that does not affect the valence shell is:

\[ \boxed{\text{(c) Nuclear mass}} \]

Q36. The size of isoelectronic species — \(\mathrm{F^–}\), \(\mathrm{Ne}\) and \(\mathrm{Na^+}\) 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.

Solution

The species \(\mathrm{F^-}\), \(\mathrm{Ne}\), and \(\mathrm{Na^+}\) are called isoelectronic because each of them contains exactly 10 electrons. Their electronic configuration is therefore the same, namely \(1s^2\,2s^2\,2p^6\). Since the number of electrons and the outermost principal quantum number \((n=2)\) are identical, any difference in their sizes must arise from another factor.

Let us examine the nuclear charges:

Although all three species have the same electron cloud, the attractive force exerted by the nucleus increases as the nuclear charge increases. A higher positive charge pulls the same set of electrons closer to the nucleus, thereby reducing the size.

Hence, as nuclear charge increases from \(\mathrm{F^-}\) to \(\mathrm{Na^+}\), the radius decreases according to:

The valence principal quantum number is the same for all three, and electron–electron repulsion is comparable because the number of electrons is identical. Therefore, neither option (b) nor (c) governs the size variation. Clearly, the sizes are not equal, so option (d) is also incorrect.

Final Answer: The size of these isoelectronic species is affected by nuclear charge (Z). Hence, option (a) is correct.

Q37. 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.

Solution

Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom. Its variation follows definite periodic trends governed mainly by nuclear attraction, shielding, and the principal quantum number.

Statement (a) is correct because after each electron is removed, the remaining electrons experience a higher effective nuclear charge, so successive ionization enthalpies always increase.

Statement (b) is also correct. When removal of electrons reaches a noble gas core, the next electron must be taken from a very stable inner shell. This requires exceptionally large energy, producing the greatest jump in ionization enthalpy.

Statement (c) is correct because once all valence electrons are removed, any further ionization involves core electrons. This transition from valence shell to core shell is marked by a sharp rise in ionization enthalpy.

Statement (d) is incorrect. Electrons in orbitals with lower principal quantum number \(n\) lie closer to the nucleus and are more tightly held, so they are harder to remove. Electrons in higher \(n\) orbitals are farther from the nucleus and are removed more easily.

This can be summarized conceptually as:

Final Answer: The incorrect statement is (d).

Q38. 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

Solution

Metallic character refers to the tendency of an atom to lose electrons and form cations. This property increases down a group due to increasing atomic size and decreases across a period from left to right because effective nuclear charge increases.

Let us examine the positions of the given elements in the periodic table. Potassium lies in Group 1 and Period 4, magnesium in Group 2 and Period 3, aluminium in Group 13 and Period 3, while boron lies in Group 13 and Period 2.

Down a group, metallic character increases, so aluminium is more metallic than boron. Across Period 3 from magnesium to aluminium, metallic character decreases, hence magnesium is more metallic than aluminium. Potassium, being an alkali metal in a lower period, has the largest atomic size and weakest hold on its valence electron, making it the most metallic among all.

This trend may be summarized as:

Applying these ideas, the decreasing order of metallic character becomes:

Final Answer: The correct order is \(\mathrm{K > Mg > Al > B}\). Hence, option (d) is correct.

Q39. 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

Solution

Non-metallic character refers to the tendency of an atom to gain electrons. This property increases from left to right across a period due to increasing effective nuclear charge, and decreases down a group because atomic size increases and the attraction for incoming electrons weakens.

Among the given elements, boron, carbon, nitrogen, and fluorine all belong to Period 2, while silicon lies below carbon in Period 3. Hence, across Period 2, non-metallic character increases in the order:

Now comparing carbon and silicon, both belong to Group 14. Since non-metallic character decreases down a group, carbon (Period 2) is more non-metallic than silicon (Period 3):

Combining both trends, the overall decreasing order of non-metallic character becomes:

This arrangement reflects increasing non-metallic character across the period and its decrease on moving down the group.

Final Answer: The correct order is \(\mathrm{F > N > C > B > Si}\). Hence, option (c) is correct.

Q40. 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

Solution

Oxidizing property refers to the tendency of an element to gain electrons, that is, to oxidize other substances while itself getting reduced. A stronger oxidizing agent has a higher tendency to accept electrons. This property generally increases with increasing electronegativity and electron affinity.

Among the given elements, fluorine and chlorine belong to Group 17 (halogens), while oxygen and nitrogen belong to Groups 16 and 15 respectively. Across a period from nitrogen to fluorine, electronegativity increases sharply, making fluorine the strongest oxidizing agent. Chlorine, though less powerful than fluorine, is still a stronger oxidizing agent than oxygen because halogens readily gain one electron to achieve noble gas configuration.

Oxygen is less oxidizing than chlorine, and nitrogen shows the weakest oxidizing tendency among the four due to its relatively low electron affinity and high stability of its half-filled \(2p\) subshell.

This trend can be summarized as:

Final Answer: The correct order of oxidizing property is \(\mathrm{F > Cl > O > N}\). Hence, option (a) is correct.