Q7 (i): Lawrence Berkeley Laboratory

(i) Lawrence Berkeley Laboratory: This laboratory has been involved in the discovery of several transuranic elements. Two elements named in its honor are: Lawrencium (Lr), with atomic number 103, named after Ernest Lawrence, the inventor of the cyclotron and founder of the laboratory. Berkelium…

Q7 (ii): Seaborg's group?

(i) Lawrence Berkeley Laboratory: This laboratory has been involved in the discovery of several transuranic elements. Two elements named in its honor are: Lawrencium (Lr), with atomic number 103, named after Ernest Lawrence, the inventor of the cyclotron and founder of the laboratory. Berkelium…

Isoelectronic Species: Isoelectronic species are atoms and ions that have the same number of electrons. Examples of Isoelectronic Species: (i) F ^- Fluorine atom (F) has 9 electrons. The fluoride ion (F ^- ) has gained one electron, so it has 9 + 1 = 10 electrons. A species isoelectronic with F ^-…

Q15: Energy of an electron in the ground state of the hydrogen atom is -2.18 × 10^-18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of…

Given: Energy of an electron in the ground state of H atom ( E_1 ) = -2.18 × 10^-18 J/atom To Find: Ionization enthalpy of atomic hydrogen in J mol ^-1. Solution: Ionization is the process of removing an electron from an atom. For the hydrogen atom, this means moving the electron from its ground…

Q16 (i): Be has higher Δ_i H than B

(i) Explain why Be has higher Δ_i H than B: The electronic configuration of Beryllium (Be, Z=4) is 1s^2 2s^2. The electronic configuration of Boron (B, Z=5) is 1s^2 2s^2 2p^1. Orbital Stability: In Be, the electron is removed from a completely filled and more stable 2s orbital. In B, the electron…

Classification of Elements and Periodicity in Properties Class 11 Chemistry NCERT Solutions

Q1EXERCISES

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

Solution

The basic theme of organization in the periodic table is to classify the elements in order of their increasing atomic numbers. This arrangement places elements with similar valence shell electronic configurations, and hence similar physical and chemical properties, in the same vertical column (group). The table systematically organizes elements to show the periodic recurrence of their properties, a concept formalized by the Modern Periodic Law.

Q2EXERCISES

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

Solution

Mendeleev used atomic mass (or atomic weight) as the fundamental property to classify the elements in his periodic table. He arranged the elements in order of increasing atomic mass.

No, he did not strictly stick to this rule. In some cases, to ensure that elements with similar properties were placed in the same group, he prioritized chemical similarity over the strict order of atomic mass. For example, he placed tellurium (Te, atomic mass 127.6) before iodine (I, atomic mass 126.9) because iodine's properties were similar to those of fluorine, chlorine, and bromine.

Q3EXERCISES

What is the basic difference in approach between the Mendeleev's Periodic Law and the Modern Periodic Law?

Solution

The basic difference lies in the fundamental property used to classify the elements.

Mendeleev's Periodic Law: It is based on atomic mass. It states that the physical and chemical properties of the elements are a periodic function of their atomic masses.
Modern Periodic Law: It is based on atomic number. It states that the physical and chemical properties of the elements are a periodic function of their atomic numbers.

The modern approach recognizes that the atomic number (the number of protons) is a more fundamental property of an element than its atomic mass, as it directly determines the electronic configuration, which in turn governs the chemical properties.

Q4EXERCISES

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

Solution

To Justify: The sixth period should have 32 elements.

Justification: The sixth period corresponds to the filling of the principal energy level n=6. According to the Aufbau principle, the orbitals available for filling in this period are 6s, 4f, 5d, and 6p, in order of increasing energy.

We can calculate the maximum number of electrons that can be accommodated in each of these subshells:

6s subshell: Has 1 orbital ( $l=0, m_l=0$ ). Maximum electrons = 2.
4f subshell: Has 7 orbitals ( $l=3, m_l=-3, -2, -1, 0, +1, +2, +3$ ). Maximum electrons = $7 \times 2 = 14$ .
5d subshell: Has 5 orbitals ( $l=2, m_l=-2, -1, 0, +1, +2$ ). Maximum electrons = $5 \times 2 = 10$ .
6p subshell: Has 3 orbitals ( $l=1, m_l=-1, 0, +1$ ). Maximum electrons = $3 \times 2 = 6$ .

The total number of available orbitals is

1 + 7 + 5 + 3 = 16

. The total number of electrons that can be accommodated is the sum of the capacities of all these subshells: Total electrons =

2 + 14 + 10 + 6 = 32

.

Since each element corresponds to the filling of one electron, the sixth period can contain a maximum of 32 elements.

Q5EXERCISES

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

Solution

Given: Atomic number (Z) = 114.

To Find: The period and group of this element.

Solution:

Determine the Period: We identify the noble gas with an atomic number just less than 114. This is Radon (Rn, Z=86). Since Rn is at the end of the 6th period, the element with Z=114 must be in the next period, which is the 7th period.
Determine the Electronic Configuration and Group: After Rn (Z=86), we fill the electrons for the 7th period:
- 7s orbital: 2 electrons (up to Z=88)
- 5f orbital: 14 electrons (up to Z=102)
- 6d orbital: 10 electrons (up to Z=112)
- 7p orbital: The remaining electrons go here.
Number of electrons after Rn = $114 - 86 = 28$ . These 28 electrons are filled as: $7s^2$ , $5f^{14}$ , $6d^{10}$ . This accounts for $2+14+10 = 26$ electrons. The remaining $28 - 26 = 2$ electrons will go into the 7p orbital. So, the outer configuration is $7s^2 7p^2$ .

Since the last electron enters the p-orbital, it is a p-block element. For p-block elements, the group number is calculated as: Group Number = 10 + (number of valence s-electrons + number of valence p-electrons) Group Number = 10 + (2 + 2) = 14.

Final Answer: The element with Z=114 is located in the 7th period and Group 14.

Q6EXERCISES

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

Solution

Given:

Period = 3
Group = 17

To Find: The atomic number (Z) of the element.

Solution:

Period 3 indicates that the principal quantum number (n) of the valence shell is 3.
Group 17 is the halogen group. Elements in this group have a valence shell electronic configuration of $ns^2 np^5$ .

Combining these, the valence shell configuration for this element is

3s^2 3p^5

.

The inner shells must be completely filled. The electronic configuration of the preceding noble gas (at the end of period 2) is that of Neon (Ne, Z=10), which is

1s^2 2s^2 2p^6

.

So, the complete electronic configuration of the element is:

1s^2 2s^2 2p^6 3s^2 3p^5

The atomic number (Z) is the total number of electrons in a neutral atom. Z = 2 + 2 + 6 + 2 + 5 = 17.

The element is Chlorine (Cl).

Final Answer: The atomic number of the element is 17.

Q7EXERCISES

Which element do you think would have been named by

(i)

Lawrence Berkeley Laboratory

(ii)

Seaborg's group?

Solution

(i) Lawrence Berkeley Laboratory: This laboratory has been involved in the discovery of several transuranic elements. Two elements named in its honor are:

Lawrencium (Lr), with atomic number 103, named after Ernest Lawrence, the inventor of the cyclotron and founder of the laboratory.
Berkelium (Bk), with atomic number 97, named after the city of Berkeley, California, where the laboratory is located.

(ii) Seaborg's group: This refers to the research group led by Glenn T. Seaborg. The element named in his honor is:

Seaborgium (Sg), with atomic number 106.

Final Answer:

(i)

Lawrencium (Lr) and Berkelium (Bk).

(ii)

Seaborgium (Sg).

Q8EXERCISES

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

Solution

Elements in the same group have similar physical and chemical properties because they have the same number of valence electrons and a similar valence shell electronic configuration. Chemical properties of an element are primarily determined by the number of electrons in its outermost shell and how these electrons are arranged, as these are the electrons involved in chemical bonding. Since elements in a group have identical outer electronic structures, they tend to lose, gain, or share electrons in a similar manner, leading to similar chemical behavior, valency, and types of compounds they form.

Q9EXERCISES

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

Solution

Atomic Radius: Atomic radius refers to the size of an atom. Since an atom's electron cloud does not have a sharp, definite boundary, the atomic radius is typically defined as half the distance between the nuclei of two adjacent atoms in a bonded state. It gives an estimate of the atom's size. For a non-metal, it is the covalent radius (half the bond length in a molecule like Cl

_2

). For a metal, it is the metallic radius (half the internuclear distance in a metallic crystal).

Ionic Radius: Ionic radius refers to the size of an ion in an ionic crystal. It is the effective distance from the nucleus of the ion to the point up to which it has an influence in the ionic bond. An ionic radius is different from the atomic radius of the parent atom. A cation (positive ion) is smaller than its parent atom, while an anion (negative ion) is larger.

Q10EXERCISES

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

Solution

Variation of Atomic Radius:

1. In a Period (from left to right):

Variation: The atomic radius generally decreases across a period.
Explanation: As we move across a period, electrons are added to the same valence shell. However, the number of protons in the nucleus (the nuclear charge) increases by one for each successive element. The increased positive charge of the nucleus attracts the electrons in the same shell more strongly, pulling the electron cloud closer to the nucleus and thus decreasing the atomic size.

2. In a Group (from top to bottom):

Variation: The atomic radius increases down a group.
Explanation: As we move down a group, a new principal energy level (electron shell) is added for each element. The outermost electrons are therefore farther from the nucleus. Although the nuclear charge increases, its effect is counteracted by two factors: the increased distance of the valence shell and the increased shielding effect from the inner-shell electrons. The inner electrons shield the outer electrons from the full attractive pull of the nucleus. The increase in the number of shells is the dominant factor, leading to an increase in atomic radius.

Q11EXERCISES

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 $^{+}$

Solution

Isoelectronic Species: Isoelectronic species are atoms and ions that have the same number of electrons.

Examples of Isoelectronic Species:

(i) F $^{-}$

Fluorine atom (F) has 9 electrons. The fluoride ion (F $^{-}$ ) has gained one electron, so it has $9 + 1 = 10$ electrons.
A species isoelectronic with F $^{-}$ is Ne (Neon atom, 10 electrons) or Na $^{+}$ (Sodium ion, $11 - 1 = 10$ electrons).

(ii) Ar

Argon atom (Ar) is a noble gas with 18 electrons.
A species isoelectronic with Ar is K $^{+}$ (Potassium ion, $19 - 1 = 18$ electrons) or Cl $^{-}$ (Chloride ion, $17 + 1 = 18$ electrons).

(iii) Mg $^{2+}$

Magnesium atom (Mg) has 12 electrons. The magnesium ion (Mg $^{2+}$ ) has lost two electrons, so it has $12 - 2 = 10$ electrons.
A species isoelectronic with Mg $^{2+}$ is Ne (Neon atom, 10 electrons) or O $^{2-}$ (Oxide ion, $8 + 2 = 10$ electrons).

(iv) Rb $^{+}$

Rubidium atom (Rb) has 37 electrons. The rubidium ion (Rb $^{+}$ ) has lost one electron, so it has $37 - 1 = 36$ electrons.
A species isoelectronic with Rb $^{+}$ is Kr (Krypton atom, 36 electrons) or Sr $^{2+}$ (Strontium ion, $38 - 2 = 36$ electrons).

Q12EXERCISES

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.

Solution

Given Species: N

^{3-}

, O

^{2-}

, F

^{-}

, Na

^{+}

, Mg

^{2+}

and Al

^{3+}

(a) What is common in them? Let's determine the number of electrons in each species:

N $^{3-}$ : $7 (protons) + 3 (electrons) = 10$ electrons
O $^{2-}$ : $8 + 2 = 10$ electrons
F $^{-}$ : $9 + 1 = 10$ electrons
Na $^{+}$ : $11 - 1 = 10$ electrons
Mg $^{2+}$ : $12 - 2 = 10$ electrons
Al $^{3+}$ : $13 - 3 = 10$ electrons

What is common in all these species is that they all have 10 electrons. They are isoelectronic.

(b) Arrange them in the order of increasing ionic radii. For isoelectronic species, the ionic radius decreases as the nuclear charge (number of protons) increases. This is because a higher positive charge in the nucleus exerts a stronger pull on the same number of electrons, causing the ion to shrink.

Let's list the nuclear charge (atomic number, Z) for each species:

N $^{3-}$ : Z = 7
O $^{2-}$ : Z = 8
F $^{-}$ : Z = 9
Na $^{+}$ : Z = 11
Mg $^{2+}$ : Z = 12
Al $^{3+}$ : Z = 13

The species with the highest nuclear charge (Al

^{3+}

) will be the smallest, and the species with the lowest nuclear charge (N

^{3-}

) will be the largest.

The order of increasing ionic radii is: Al $^{3+}$ < Mg $^{2+}$ < Na $^{+}$ < F $^{-}$ < O $^{2-}$ < N $^{3-}$

Q13EXERCISES

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

Solution

Explanation for Cation Size: A cation is formed when a neutral atom loses one or more electrons. The cation is smaller than its parent atom for two main reasons:

Increased Effective Nuclear Charge: When electrons are removed, the number of protons in the nucleus remains the same. The unchanged positive nuclear charge now acts on fewer electrons. This increases the effective nuclear charge per electron, causing the nucleus to pull the remaining electrons more strongly and closer to it, resulting in a smaller radius.
Removal of Valence Shell: In many cases, the formation of a cation involves the removal of all electrons from the outermost shell. This means the new outermost shell of the ion is one level closer to the nucleus, leading to a significant decrease in size.

Explanation for Anion Size: An anion is formed when a neutral atom gains one or more electrons. The anion is larger than its parent atom because:

Increased Electron-Electron Repulsion: The addition of electrons to the valence shell increases the repulsion among the electrons. This increased repulsion causes the electron cloud to expand, making the anion larger.
Decreased Effective Nuclear Charge: The nuclear charge remains the same, but it now has to hold more electrons. The attractive force of the nucleus is distributed over a larger number of electrons, so the effective nuclear charge per electron decreases. This weaker pull allows the electron cloud to spread out, increasing the radius.

Q14EXERCISES

What is the significance of the terms - 'isolated gaseous atom' and 'ground state' while defining the ionization enthalpy and electron gain enthalpy?

Solution

The terms 'isolated gaseous atom' and 'ground state' are significant because they establish a standardized and unambiguous condition for measuring and comparing enthalpy changes.

Significance of 'Isolated Gaseous Atom':

Isolated: This specifies that the atom is completely free from the influence of neighboring atoms or molecules. In the solid or liquid state, atoms are affected by intermolecular forces of attraction, which would alter the energy required to remove or add an electron.
Gaseous: By considering the atom in the gaseous phase, we ensure that the atoms are far apart and their interatomic forces are negligible. This allows for the measurement of the energy change associated with an individual atom, making the values comparable across different elements.

Significance of 'Ground State':

The ground state is the most stable, lowest energy state of an atom. Defining the initial state as the ground state provides a consistent and reproducible reference point. If the atom were in an excited state (a higher energy state), the energy required to remove an electron (ionization enthalpy) would be lower, and the energy change on adding an electron would be different. Using the ground state ensures that the measured enthalpy values are the maximum for ionization and are consistently defined for electron gain.

Q15EXERCISES

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 $^{-1}$ .

Solution

Given:

Energy of an electron in the ground state of H atom ( $E_1$ ) = $-2.18 \times 10^{-18}$ J/atom

To Find:

Ionization enthalpy of atomic hydrogen in J mol $^{-1}$ .

Solution: Ionization is the process of removing an electron from an atom. For the hydrogen atom, this means moving the electron from its ground state (n=1) to a state of infinite distance from the nucleus (n=

\infty

), where its energy is taken as zero (

E_\infty = 0

).

The energy required to ionize a single hydrogen atom (Ionization Energy, IE) is: IE =

E_\infty - E_1

IE =

0 - (-2.18 \times 10^{-18} \text{ J})

IE =

2.18 \times 10^{-18}

J/atom

This is the energy for one atom. Ionization enthalpy is usually expressed per mole of atoms. To convert this value to J mol

^{-1}

, we must multiply it by Avogadro's constant (

N_A = 6.022 \times 10^{23}

mol

^{-1}

).

Ionization Enthalpy (

\Delta_i H

) = IE per atom

\times N_A

\Delta_i H = (2.18 \times 10^{-18} \text{ J/atom}) \times (6.022 \times 10^{23} \text{ atoms/mol})

\Delta_i H = 13.12796 \times 10^5

J mol

^{-1}

\Delta_i H \approx 1.312 \times 10^6

J mol

^{-1}

Final Answer: The ionization enthalpy of atomic hydrogen is

1.312 \times 10^6

J mol

^{-1}

.

Q16EXERCISES

Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why

(i)

Be has higher $\Delta_i H$ than B

(ii)

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

Solution

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

The electronic configuration of Beryllium (Be, Z=4) is

1s^2 2s^2

. The electronic configuration of Boron (B, Z=5) is

1s^2 2s^2 2p^1

.

Orbital Stability: In Be, the electron is removed from a completely filled and more stable 2s orbital. In B, the electron is removed from a singly occupied 2p orbital.
Penetration Effect: A 2s electron has greater penetration towards the nucleus compared to a 2p electron. This means the 2s electron is held more tightly by the nucleus.
Shielding Effect: The 2p electron in Boron is partially shielded by the inner 2s electrons, making it easier to remove.

Due to the higher stability of the filled 2s orbital and the greater penetration of 2s electrons, more energy is required to remove an electron from Be than from B. Therefore, Be has a higher first ionization enthalpy than B.

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

The electronic configuration of Nitrogen (N, Z=7) is

1s^2 2s^2 2p^3

. The electronic configuration of Oxygen (O, Z=8) is

1s^2 2s^2 2p^4

.

Comparison with Nitrogen (N): Nitrogen has a half-filled 2p subshell ( $2p_x^1 2p_y^1 2p_z^1$ ), which is an extra stable configuration according to Hund's rule. Removing an electron from N disrupts this stability, requiring a large amount of energy. In Oxygen, the 2p subshell has four electrons ( $2p_x^2 2p_y^1 2p_z^1$ ). One of the 2p orbitals is doubly occupied. The two electrons in this orbital repel each other (inter-electronic repulsion). This repulsion makes it easier to remove one of these paired electrons compared to removing an electron from the stable half-filled configuration of Nitrogen. Thus, O has a lower first ionization enthalpy than N.
Comparison with Fluorine (F): Fluorine (Z=9) is to the right of Oxygen in the same period. As we move from O to F, the nuclear charge increases while the electron is added to the same shell. The increased nuclear charge results in a stronger attraction for the valence electrons, making them harder to remove. Therefore, F has a higher ionization enthalpy than O, following the general periodic trend.

Q17EXERCISES

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

Explanation:

First Ionization Enthalpy (IE $_1$ ):

Sodium (Na): Electronic configuration is [Ne] $3s^1$ . It can easily lose its single 3s valence electron to attain the stable noble gas configuration of Neon.
Magnesium (Mg): Electronic configuration is [Ne] $3s^2$ . It has a completely filled and stable 3s orbital. Removing an electron from this stable configuration requires more energy than removing the single electron from Na. Also, Mg has a higher nuclear charge than Na.
Conclusion: Therefore, the first ionization enthalpy of Na is lower than that of Mg.
- Na(g) $\rightarrow$ Na $^{+}$ (g) + e $^{-}$ (low energy)
- Mg(g) $\rightarrow$ Mg $^{+}$ (g) + e $^{-}$ (higher energy)

Second Ionization Enthalpy (IE $_2$ ): After the first ionization, the ions formed are Na

^{+}

and Mg

^{+}

.

Sodium ion (Na $^{+}$ ): Electronic configuration is [Ne] or $1s^2 2s^2 2p^6$ . This is an extremely stable noble gas configuration. Removing a second electron from Na $^{+}$ means disrupting this stable core, which requires a very large amount of energy.
Magnesium ion (Mg $^{+}$ ): Electronic configuration is [Ne] $3s^1$ . This ion can easily lose its single 3s electron to achieve the stable noble gas configuration of Neon.
Conclusion: It is much harder to remove an electron from the stable Na $^{+}$ $^{+}$ ion than from the Mg $^{+}$ $^{+}$ ion. Therefore, the second ionization enthalpy of Na is much higher than that of Mg.
- Na $^{+}$ (g) $\rightarrow$ Na $^{2+}$ (g) + e $^{-}$ (very high energy)
- Mg $^{+}$ (g) $\rightarrow$ Mg $^{2+}$ (g) + e $^{-}$ (lower energy)

Q18EXERCISES

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 the main group elements tends to decrease down a group due to the combined effect of the following factors:

Increase in Atomic Size: As we move down a group, a new electron shell is added with each successive element. This increases the distance between the nucleus and the outermost (valence) electron. According to Coulomb's law, the force of attraction between the nucleus and the electron decreases as the distance increases, making the electron easier to remove.
Increase in Shielding (or Screening) Effect: The electrons in the inner shells shield the outermost electrons from the full attractive force of the positively charged nucleus. As we descend a group, the number of inner-shell electrons increases significantly. This enhanced shielding effect reduces the effective nuclear charge experienced by the valence electron, weakening its bond to the nucleus and lowering the energy required for its removal.

These two factors (increased atomic size and increased shielding effect) together outweigh the effect of the increasing nuclear charge as we move down a group. Consequently, the ionization enthalpy decreases.

Q19EXERCISES

The first ionization enthalpy values (in kJ mol $^{-1}$ ) of group 13 elements are : B: 801, Al: 577, Ga: 579, In: 558, Tl: 589 How would you explain this deviation from the general trend?

Solution

The general trend for ionization enthalpy (IE) is to decrease down a group. However, Group 13 shows significant deviations from this trend.

1. Decrease from B to Al:

This follows the expected trend. Aluminium is larger than Boron and has an additional inner shell of electrons, which increases the shielding effect. Both factors lead to a lower IE for Al compared to B.

2. Irregularity from Al to Ga:

The IE of Gallium (Ga) is slightly higher than that of Aluminium (Al). This is an anomaly. Gallium (Z=31) is preceded by the first series of d-block (transition) elements, which have electrons filling the 3d orbitals. The 10 electrons in the 3d orbitals provide a poor shielding effect. This poor shielding does not effectively compensate for the increase in nuclear charge from Al (13 protons) to Ga (31 protons). Consequently, the valence electrons of Ga experience a greater effective nuclear charge than those of Al, making them harder to remove. This results in a higher IE for Ga.

3. Decrease from Ga to In:

This again follows the expected trend. The increase in atomic size and shielding effect from Ga to Indium (In) outweighs the increase in nuclear charge, leading to a lower IE for In.

4. Irregularity from In to Tl:

The IE of Thallium (Tl) is higher than that of Indium (In). Thallium (Z=81) is preceded by both the d-block and f-block (lanthanoid) elements. The electrons in the 4f and 5d orbitals have a very poor shielding effect (this is related to the lanthanoid contraction). This poor shielding leads to a large increase in the effective nuclear charge experienced by the valence electrons of Tl. This effect is strong enough to overcome the increase in atomic size, resulting in a higher IE for Tl compared to In.

Q20EXERCISES

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

(i)

O or F

(ii)

F or Cl

Solution

(i) O or F

Oxygen (O) and Fluorine (F) are in the same period (Period 2). Electron gain enthalpy (EGE) generally becomes more negative as we move from left to right across a period. This is due to an increase in effective nuclear charge and a decrease in atomic size, which leads to a stronger attraction for an incoming electron.
Fluorine is to the right of Oxygen and needs only one electron to attain a stable noble gas configuration, whereas Oxygen needs two.
Therefore, F has a more negative electron gain enthalpy than O.

(ii) F or Cl

Fluorine (F) and Chlorine (Cl) are in the same group (Group 17). Generally, EGE becomes less negative down a group due to increased atomic size.
However, this is a well-known exception. The electron gain enthalpy of Cl is more negative than that of F.
Reason: The fluorine atom is very small (n=2 shell). When an electron is added, it enters this small, compact shell and experiences significant repulsion from the nine electrons already present. In chlorine, the atom is larger, and the incoming electron enters the more diffuse n=3 shell. The electron-electron repulsion is much weaker in the larger 3p orbital of chlorine. The lower repulsion in Cl makes the process of adding an electron more energetically favorable (more exothermic) than in F.
Therefore, Cl has a more negative electron gain enthalpy.

Q21EXERCISES

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

Solution

Expectation: The second electron gain enthalpy of Oxygen (O) is expected to be positive.

Justification: The first electron gain enthalpy of oxygen is the enthalpy change for the reaction: O(g) + e

^{-}

\rightarrow

O

^{-}

(g);

\Delta_{eg}H_1

= -141 kJ mol

^{-1}

(exothermic)

This process is exothermic because the neutral oxygen atom has an attraction for the incoming electron.

The second electron gain enthalpy is the enthalpy change for the reaction where an electron is added to the negatively charged oxide ion (O

^{-}

): O

^{-}

(g) + e

^{-}

\rightarrow

O

^{2-}

(g);

\Delta_{eg}H_2

= ?

In this second step, the incoming electron (which is negatively charged) is being added to an ion that is already negatively charged (O

^{-}

). There is a strong electrostatic repulsion between the like charges. To overcome this repulsion and force the electron onto the ion, energy must be supplied to the system.

Since energy is absorbed in the process, the reaction is endothermic, and the enthalpy change (

\Delta_{eg}H_2

) is positive.

Final Answer: The second electron gain enthalpy of O is expected to be positive because energy is required to overcome the strong electrostatic repulsion between the negative O

^{-}

ion and the incoming electron.

Q22EXERCISES

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

Solution

The basic differences between electron gain enthalpy and electronegativity are as follows:

Feature	Electron Gain Enthalpy	Electronegativity
Definition	It is the enthalpy change when an electron is added to an isolated gaseous atom.	It is the tendency of an atom in a chemical bond to attract a shared pair of electrons.
State of Atom	Refers to an isolated, individual atom.	Refers to an atom that is bonded to another atom in a molecule.
Nature of Property	It is a measurable, absolute quantity with units (e.g., kJ mol $^{-1}$ ).	It is a relative, dimensionless quantity calculated on various scales (e.g., Pauling scale). It is not directly measurable.
Application	Predicts the energy change for the formation of an anion from a neutral atom.	Predicts the nature of a chemical bond (ionic, polar covalent, or covalent).

Q23EXERCISES

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 is incorrect.

Reaction and Explanation: Electronegativity is not a fixed, constant property of an atom. It is a dynamic property that depends on the chemical environment of the atom, specifically:

The element it is bonded to: The ability of nitrogen to attract electrons will be different when it is bonded to a highly electronegative element like fluorine (as in NF $_3$ ) compared to when it is bonded to a less electronegative element like hydrogen (as in NH $_3$ ).
The oxidation state of the atom: An atom in a higher positive oxidation state will be more electronegative as its nucleus will have a stronger pull on the electrons.
The hybridization of the atom: The electronegativity of an atom increases with the increasing s-character of its hybrid orbitals (sp > sp $^2$ > sp $^3$ ).

The value of 3.0 for nitrogen on the Pauling scale is an average or representative value that is useful for general predictions. However, the actual electronegativity of nitrogen varies from one compound to another.

Q24EXERCISES

Describe the theory associated with the radius of an atom as it

(a)

gains an electron

(b)

loses an electron

Solution

(a) As an atom gains an electron: When a neutral atom gains one or more electrons, it forms an anion (a negatively charged ion). The radius of the anion is larger than that of its parent atom. The theory behind this is:

Increased Electron-Electron Repulsion: The addition of an extra electron into the valence shell increases the total number of electrons. This leads to increased repulsion among the electrons in the shell, causing the electron cloud to expand.
Decreased Effective Nuclear Charge: The number of protons in the nucleus remains unchanged, but this constant positive charge now has to attract a greater number of electrons. The attractive pull of the nucleus per electron decreases, allowing the electron cloud to occupy a larger volume.

(b) As an atom loses an electron: When a neutral atom loses one or more electrons, it forms a cation (a positively charged ion). The radius of the cation is smaller than that of its parent atom. The theory is:

Increased Effective Nuclear Charge: After losing an electron, the number of protons in the nucleus is greater than the number of remaining electrons. The unchanged nuclear charge pulls the smaller number of electrons more strongly, causing the electron cloud to contract.
Removal of the Outermost Shell: Often, the electron(s) lost are from the outermost valence shell. This means the entire shell is removed, and the new outermost shell of the cation is one principal quantum level closer to the nucleus, resulting in a significant decrease in radius.

Q25EXERCISES

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

Solution

Expectation: The first ionization enthalpies for two isotopes of the same element would be expected to be the same (or virtually identical).

Justification: Ionization enthalpy is the energy required to remove the most loosely bound electron from an atom. This energy primarily depends on two factors:

The nuclear charge (number of protons): This determines the strength of the electrostatic attraction on the electrons.
The electronic configuration: This determines the location of the electron (its shell and subshell) and the extent of shielding by other electrons.

Isotopes of an element have the same atomic number, which means they have the same number of protons and, in their neutral state, the same number of electrons. Therefore, they have identical nuclear charges and identical electronic configurations.

Isotopes differ only in the number of neutrons in their nucleus, which affects the atomic mass. The mass of the nucleus has a negligible effect on the electrostatic forces holding the electrons. Therefore, the energy required to remove an electron will be the same for both isotopes.

Final Answer: The first ionization enthalpies will be the same because isotopes have the same nuclear charge and electronic configuration, which are the primary factors determining ionization enthalpy.

Q26EXERCISES

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

Solution

The major differences between metals and non-metals are:

Property	Metals	Non-Metals
Physical State	Usually solids at room temperature (except Mercury, Hg).	Can be solids, liquids (e.g., Bromine, Br $_2$ ), or gases at room temperature.
Appearance	Lustrous (shiny).	Non-lustrous (dull), except for iodine and graphite.
Malleability/Ductility	Malleable (can be beaten into sheets) and ductile (can be drawn into wires).	Brittle if solid; neither malleable nor ductile.
Conductivity	Good conductors of heat and electricity.	Poor conductors of heat and electricity (except graphite).
Electronic Nature	Have low ionization enthalpies; tend to lose electrons to form cations.	Have high ionization enthalpies and electron affinities; tend to gain or share electrons.
Oxides	Form basic or amphoteric oxides (e.g., Na $_2$ O is basic, Al $_2$ O $_3$ is amphoteric).	Form acidic or neutral oxides (e.g., SO $_2$ is acidic, CO is neutral).
Location in Table	Located on the left side and in the center of the Periodic Table.	Located on the top right side of the Periodic Table.

Q27EXERCISES

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

(a) Identify an element with five electrons in the outer subshell. This refers to an element with the outer electronic configuration

np^5

. These elements belong to Group 17 (the halogens).

Example: Fluorine (F) or Chlorine (Cl).

(b) Identify an element that would tend to lose two electrons. Elements that tend to lose two electrons to form a stable +2 cation are the Group 2 elements (the alkaline earth metals).

Example: Magnesium (Mg) or Calcium (Ca).

(c) Identify an element that would tend to gain two electrons. Elements that tend to gain two electrons to form a stable -2 anion are the Group 16 elements (the chalcogens).

Example: Oxygen (O) or Sulphur (S).

(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature. This unique combination of properties is found in Group 17 (the halogens).

Gas: Fluorine (F $_2$ ) and Chlorine (Cl $_2$ ) are gases.
Liquid: Bromine (Br $_2$ ) is a liquid.
Non-metal (Solid): Iodine (I $_2$ ) is a solid non-metal.
Metal/Metalloid: Astatine (At) is a radioactive element with properties of a metalloid or metal.

Q28EXERCISES

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

Explanation:

Reactivity of Group 1 Elements (Alkali Metals):

The chemical reactivity of Group 1 elements is determined by their tendency to lose their single valence electron to form a +1 cation. This tendency is measured by their ionization enthalpy.
As we move down Group 1 from Li to Cs, the atomic size increases, and the valence electron gets farther from the nucleus. The shielding effect of the inner electrons also increases.
Both these factors lead to a decrease in ionization enthalpy down the group. It becomes progressively easier to remove the valence electron.
Since reactivity is defined by the ease of losing an electron, the reactivity increases down the group. Thus, the order is Li < Na < K < Rb < Cs.

Reactivity of Group 17 Elements (Halogens):

The chemical reactivity of Group 17 elements is determined by their tendency to gain an electron to form a -1 anion. This tendency is related to their electronegativity and electron gain enthalpy (their oxidizing power).
As we move down Group 17 from F to I, the atomic size increases. The nucleus of a larger atom has a weaker attraction for an incoming electron because the electron is added to a shell that is farther away.
Consequently, the electronegativity decreases down the group, and the electron gain enthalpy generally becomes less negative.
Since reactivity is defined by the ability to gain an electron, the reactivity decreases down the group. Thus, the order is F > Cl > Br > I.

Q29EXERCISES

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

Solution

The general outer electronic configurations for the different blocks are as follows:

s-block elements: $ns^{1-2}$ (where n = 1 to 7)
p-block elements: $ns^2 np^{1-6}$ (where n = 2 to 7)
d-block elements (Transition elements): $(n-1)d^{1-10} ns^{0-2}$ (where n = 4 to 7)
f-block elements (Inner-transition elements): $(n-2)f^{1-14} (n-1)d^{0-1} ns^2$ (where n = 6 to 7)

Q30EXERCISES

Assign the position of the element having outer electronic configuration (i) $ns^2 np^4$ for $n=3$ (ii) $(n-1)d^2 ns^2$ for $n=4$ , and (iii) $(n-2)f^7 (n-1)d^1 ns^2$ for $n=6$ , in the periodic table.

Solution

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

The outer electronic configuration is $3s^2 3p^4$ .
Period: The principal quantum number of the valence shell is n=3. Therefore, the element is in the 3rd period.
Group: The last electron enters the p-orbital, so it is a p-block element. The group number for p-block elements is 10 + (total number of valence electrons). Valence electrons = 2 (in 3s) + 4 (in 3p) = 6. Group number = 10 + 6 = 16. Therefore, the element is in Group 16.
The element is Sulphur (S).

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

The outer electronic configuration is $3d^2 4s^2$ .
Period: The highest principal quantum number is n=4. Therefore, the element is in the 4th period.
Group: The last electron enters the d-orbital, so it is a d-block element. The group number for d-block elements is the sum of electrons in the $(n-1)d$ and $ns$ subshells. Group number = 2 (in 3d) + 2 (in 4s) = 4. Therefore, the element is in Group 4.
The element is Titanium (Ti).

(iii) $(n-2)f^7 (n-1)d^1 ns^2$ for n=6

The outer electronic configuration is $4f^7 5d^1 6s^2$ .
Period: The highest principal quantum number is n=6. Therefore, the element is in the 6th period.
Group: The last electron enters the f-orbital (4f subshell), so it is an f-block element, specifically a lanthanoid. All lanthanoids and actinoids are placed in Group 3 of the periodic table.
The element is Gadolinium (Gd).

Q31EXERCISES

The first ( $\Delta_i H_1$ ) and the second ( $\Delta_i H_2$ ) ionization enthalpies (in kJ mol $^{-1}$ ) and the ( $\Delta_{eg} H$ ) electron gain enthalpy (in kJ mol $^{-1}$ ) of a few elements are given below: | Elements | $\Delta H_1$ | $\Delta H_2$ | $\Delta_{eg} H$ | |---|---|---|---| | I | 520 | 7300 | -60 | | II | 419 | 3051 | -48 | | III | 1681 | 3374 | -328 | | IV | 1008 | 1846 | -295 | | V | 2372 | 5251 | +48 | | VI | 738 | 1451 | -40 | Which of the above elements is likely to be :

(a)

the least reactive element.

(b)

the most reactive metal.

(c)

the most reactive non-metal.

(d)

the least reactive non-metal.

(e) the metal which can form a stable binary halide of the formula MX

_2

(X=halogen).

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

Solution

Analysis of the Elements:

I: Low IE $_1$ , very high IE $_2$ . This indicates a Group 1 metal (alkali metal), likely small in size (like Li).
II: Very low IE $_1$ , high IE $_2$ . This is also a Group 1 metal, more reactive than I due to lower IE $_1$ (like K).
III: High IE $_1$ , very negative EGE. This is a typical Group 17 non-metal (halogen), highly reactive (like Cl).
IV: High IE $_1$ , negative EGE (less than III). This is also a halogen, but less reactive than III (like I).
V: Very high IE $_1$ and positive EGE. This indicates a noble gas, which is very unreactive.
VI: Low IE $_1$ , and IE $_2$ is not excessively high. This pattern is typical for a Group 2 metal (alkaline earth metal) which forms a +2 ion.

Answers:

(a) the least reactive element.

The noble gas (Element V) is the least reactive due to its very high ionization enthalpy and stable electronic configuration.
Answer: V

(b) the most reactive metal.

Metal reactivity is determined by the ease of losing electrons (lowest IE $_1$ ). Element II has the lowest IE $_1$ (419 kJ mol $^{-1}$ ).
Answer: II

(c) the most reactive non-metal.

Non-metal reactivity is determined by the tendency to gain electrons (most negative EGE). Element III has the most negative EGE (-328 kJ mol $^{-1}$ ).
Answer: III

(d) the least reactive non-metal.

This would be the non-metal with the less negative EGE. Comparing III and IV, Element IV is less reactive.
Answer: IV

(e) the metal which can form a stable binary halide of the formula MX $_2$ .

An MX $_2$ halide involves a metal with a +2 oxidation state. This corresponds to the Group 2 metal, which shows a large jump after IE $_2$ . Element VI fits this profile (IE $_1$ =738, IE $_2$ =1451).
Answer: VI

(f) the metal which can form a predominantly stable covalent halide of the formula MX.

An MX halide involves a metal with a +1 oxidation state (Group 1). Covalent character is favored by small cation size and high charge, which corresponds to higher ionization enthalpy among alkali metals. Element I has a higher IE $_1$ than Element II, suggesting it's a smaller alkali metal (like Lithium), which is known to form covalent halides.
Answer: I

Q32EXERCISES

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

We predict the formulas based on the common valencies or oxidation states of the elements, which can be inferred from their group positions.

(a) Lithium and oxygen

Lithium (Li) is in Group 1, forms Li $^+$ ion (valence 1).
Oxygen (O) is in Group 16, forms O $^{2-}$ ion (valence 2).
Formula: Li $_2$ O

(b) Magnesium and nitrogen

Magnesium (Mg) is in Group 2, forms Mg $^{2+}$ ion (valence 2).
Nitrogen (N) is in Group 15, forms N $^{3-}$ ion (valence 3).
Formula: Mg $_3$ N $_2$

(c) Aluminium and iodine

Aluminium (Al) is in Group 13, forms Al $^{3+}$ ion (valence 3).
Iodine (I) is in Group 17, forms I $^{-}$ ion (valence 1).
Formula: AlI $_3$

(d) Silicon and oxygen

Silicon (Si) is in Group 14, common valence is 4.
Oxygen (O) is in Group 16, valence is 2.
Formula: SiO $_2$ (Silicon dioxide)

(e) Phosphorus and fluorine

Phosphorus (P) is in Group 15, can have valencies of 3 or 5.
Fluorine (F) is in Group 17, valence is 1. As the most electronegative element, it stabilizes the highest oxidation state of phosphorus (+5).
Formula: PF $_5$ (Phosphorus pentafluoride)

(f) Element 71 and fluorine

Element 71 is Lutetium (Lu). It is the last element of the lanthanoid series and is in Group 3. Its stable oxidation state is +3.
Fluorine (F) is in Group 17, forms F $^{-}$ ion (valence 1).
Formula: LuF $_3$

Q33EXERCISES

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

Answer: (c)

Explanation: The period number in the modern periodic table corresponds to the value of the principal quantum number (n) of the outermost or valence electron shell of the elements in that period.

Q34EXERCISES

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

Answer: (b)

Explanation:

(a) is correct. A p-subshell has 3 orbitals, which can hold a maximum of $3 \times 2 = 6$ electrons, so the p-block has 6 columns (groups 13-18).
(b) is incorrect. A d-subshell has 5 orbitals, which can hold a maximum of $5 \times 2 = 10$ electrons. Therefore, the d-block has 10 columns (groups 3-12), not 8.
(c) is correct. s-block has 2 columns, p-block has 6, d-block has 10, and f-block has 14.
(d) is correct. The block name (s, p, d, f) corresponds to the subshell receiving the last electron, which is defined by the azimuthal quantum number $l$ (s: $l=0$ , p: $l=1$ , d: $l=2$ , f: $l=3$ ).

Q35EXERCISES

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

Answer: (c)

Explanation:

(a) Valence principal quantum number (n): This determines the energy level and average distance of the valence electrons from the nucleus, directly affecting their properties.
(b) Nuclear charge (Z): This is the positive charge of the nucleus (number of protons) that attracts the valence electrons.
(d) Number of core electrons: These inner electrons shield the valence electrons from the full nuclear charge, affecting the 'effective nuclear charge' felt by the valence electrons.
(c) Nuclear mass: This is the total mass of protons and neutrons. It affects nuclear properties like stability and radioactivity but has a negligible direct effect on the chemical properties, which are governed by the electrostatic interactions of the electrons and the nucleus.

Q36EXERCISES

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.

Solution

Answer: (a)

Explanation: The species F

^{-}

, Ne, and Na

^{+}

are isoelectronic, meaning they all have the same number of electrons (10) and the same electronic configuration (

1s^2 2s^2 2p^6

).

(b) The valence principal quantum number (n=2) is the same for all of them.
(c) The electron-electron interaction is also similar since they have the same number and arrangement of electrons.
The primary factor that differs among them is the nuclear charge (Z), which is the number of protons in the nucleus:
- F $^{-}$ : Z = 9 protons
- Ne: Z = 10 protons
- Na $^{+}$ : Z = 11 protons

A higher nuclear charge exerts a stronger pull on the same number of electrons, resulting in a smaller size. Therefore, the size decreases as Z increases (Na

^{+}

< Ne < F

^{-}

). The factor affecting the size is the nuclear charge.

Q37EXERCISES

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

Answer: (d)

Explanation:

(a) is correct. It is always harder to remove an electron from a positive ion than from a neutral atom, so IE $_1$ < IE $_2$ < IE $_3$ , etc.
(b) is correct. Removing an electron from a stable, filled noble gas core requires a very large amount of energy.
(c) is correct. This is another way of stating (b). Once all valence electrons are removed, the next electron must come from the stable core, causing a large jump in ionization enthalpy.
(d) is incorrect. Orbitals with a lower principal quantum number (n) are closer to the nucleus and their electrons are held more tightly. Therefore, removal of an electron from an orbital with a lower n value is harder, not easier, than from an orbital with a higher n value.

Q38EXERCISES

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

Answer: (d)

Explanation: Metallic character increases down a group and decreases from left to right across a period. Let's locate the elements:

K: Period 4, Group 1
Mg: Period 3, Group 2
Al: Period 3, Group 13
B: Period 2, Group 13
K is in Group 1 and the lowest period (Period 4) among the choices, making it the most metallic.
In Period 3, Mg (Group 2) is to the left of Al (Group 13), so Mg is more metallic than Al.
In Group 13, Al is below B, so Al is more metallic than B.

Combining these facts, the order of decreasing metallic character is K > Mg > Al > B.

Therefore, the correct order is K > Mg > Al > B.

Q39EXERCISES

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

Answer: (c)

Explanation: Non-metallic character increases from left to right across a period and decreases down a group. Let's locate the elements:

B, C, N, F: All are in Period 2.
Si: In Period 3, below C (Group 14).
Across Period 2: Non-metallic character increases, so the order is F > N > C > B.
Down Group 14: Non-metallic character decreases, so C is more non-metallic than Si.

Combining these facts:

F is the most non-metallic element in the list.
Si is a metalloid and is the least non-metallic element in the list.
The order among the Period 2 elements is F > N > C > B. Since C > Si, the overall decreasing order is F > N > C > B > Si.

Therefore, the correct order is F > N > C > B > Si.

Q40EXERCISES

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

Answer: (b)

Explanation: Oxidizing property refers to the ability of an element to accept electrons. This property is directly related to electronegativity.

Let's compare the electronegativity values (on the Pauling scale):

F: 4.0
O: 3.5
Cl: 3.0
N: 3.0
Fluorine (F) is the most electronegative element, so it is the strongest oxidizing agent.
Oxygen (O) is the second most electronegative element, making it the second strongest oxidizing agent in this list.
Chlorine (Cl) and Nitrogen (N) have similar electronegativities. However, Chlorine has a much more negative electron gain enthalpy, making it a better electron acceptor and a stronger oxidizing agent than Nitrogen. In general, O is a stronger oxidizing agent than Cl (e.g., O can oxidize Cl in certain compounds).

The correct order of decreasing oxidizing property is F > O > Cl > N.

Therefore, the correct order is F > O > Cl > N.

Q1Problems

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

Solution

Given: Atomic number (Z) = 120

To Find: IUPAC name and symbol for the element.

Solution: According to the IUPAC nomenclature for elements with Z > 100, we use numerical roots for each digit of the atomic number.

For digit 1, the root is 'un'.
For digit 2, the root is 'bi'.
For digit 0, the root is 'nil'.

The name is formed by combining these roots and adding the suffix '-ium'. Name = un + bi + nil + ium = Unbinilium.

The symbol is formed by taking the first letter of each root. Symbol = u + b + n = Ubn.

Final Answer: The IUPAC name is Unbinilium and the symbol is Ubn.

Q2Problems

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

Solution

To Justify: The presence of 18 elements in the 5th period.

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

According to the Aufbau principle, the orbitals are filled in order of increasing energy. For n=5, the orbitals to be filled are 5s, 4d, and 5p.

5s subshell: It has 1 orbital, which can accommodate a maximum of 2 electrons.
4d subshell: It has 5 orbitals, which can accommodate a maximum of 10 electrons.
5p subshell: It has 3 orbitals, which can accommodate a maximum of 6 electrons.

The total number of electrons that can be accommodated in these orbitals is the sum of the capacities of each subshell: Total electrons = 2 (from 5s) + 10 (from 4d) + 6 (from 5p) = 18.

Since each element corresponds to the addition of one electron, the 5th period can accommodate a maximum of 18 elements.

Conclusion: The filling of 5s, 4d, and 5p orbitals, which can collectively hold 18 electrons, justifies the presence of 18 elements in the 5th period.

Q3Problems

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

Solution

Given: Two undiscovered elements with atomic numbers Z=117 and Z=120.

To Find: The group, family, and electronic configuration for each element.

Solution:

For element with Z = 117:

Position: The last discovered noble gas is Oganesson (Og) with Z=118. The element with Z=117 would be placed just before Og in the 7th period. Elements just before noble gases belong to the halogen family.
Group and Family: Group 17, Halogen family.
Electronic Configuration: The noble gas preceding the 7th period is Radon (Rn, Z=86). The electronic configuration is built up from Rn. The configuration will be: [Rn] $5f^{14} 6d^{10} 7s^2 7p^5$ .

For element with Z = 120:

Position: This element would come after the noble gas Oganesson (Og, Z=118). It would start the next period, the 8th period. The first two electrons in a new period enter the s-orbital.
Group and Family: It would have two valence electrons in the 8s orbital, placing it in Group 2, the Alkaline earth metal family.
Electronic Configuration: The preceding noble gas is Oganesson (Og, Z=118). The configuration will be: [Og] $8s^2$ .

Final Answer:

Z = 117: Group 17 (Halogen family), Electronic configuration: [Rn] $5f^{14} 6d^{10} 7s^2 7p^5$ .
Z = 120: Group 2 (Alkaline earth metal family), Electronic configuration: [Og] $8s^2$ .

Q4Problems

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

Solution

Given Elements: Si, Be, Mg, Na, P.

To Arrange: In increasing order of metallic character.

Solution: First, let's identify the position of each element in the periodic table:

Na: Period 3, Group 1
Mg: Period 3, Group 2
Be: Period 2, Group 2
Si: Period 3, Group 14
P: Period 3, Group 15

The trends for metallic character in the periodic table are:

It decreases as we move from left to right across a period.
It increases as we move down a group.

Applying the trends:

Across Period 3: Na > Mg > Si > P. (Metallic character decreases)
Down Group 2: Mg > Be. (Metallic character increases)

Combining these observations:

P is the rightmost element in period 3 from this list, making it the least metallic.
Si is to the left of P, so it is more metallic than P.
Be is in Group 2, making it a metal and more metallic than the metalloid Si.
Mg is below Be in the same group, so Mg is more metallic than Be.
Na is in Group 1, to the left of Mg, so Na is the most metallic.

The order of increasing metallic character is: P < Si < Be < Mg < Na.

Final Answer: The increasing order of metallic character is P < Si < Be < Mg < Na.

Q5Problems

Which of the following species will have the largest and the smallest size? Mg, Mg $^{2+}$ , Al, Al $^{3+}$ .

Solution

Given Species: Mg, Mg

^{2+}

, Al, Al

^{3+}

.

To Find: The species with the largest size and the species with the smallest size.

Solution: We will use the following trends for atomic and ionic radii:

Across a period: Atomic radius decreases from left to right. Mg (Group 2) and Al (Group 13) are in the same period (Period 3). Therefore, the atomic radius of Mg > Al.
Cations vs. Parent Atoms: Cations are always smaller than their parent atoms because they have fewer electrons, leading to a stronger effective nuclear charge pulling the remaining electrons closer. Thus, Mg > Mg $^{2+}$ and Al > Al $^{3+}$ .
Isoelectronic Species: Species with the same number of electrons are called isoelectronic. Mg $^{2+}$ (12 protons, 10 electrons) and Al $^{3+}$ (13 protons, 10 electrons) are isoelectronic. For isoelectronic species, the one with the higher nuclear charge (more protons) will be smaller because of the stronger attraction on the electrons. Therefore, Mg $^{2+}$ > Al $^{3+}$ .

Combining the trends:

From (1) and (2), Mg is the largest species as it is the largest parent atom.
From (3), Al $^{3+}$ is the smallest of the two ions. Since ions are smaller than their parent atoms, Al $^{3+}$ is the smallest species overall.

Final Answer:

Largest size: Mg
Smallest size: Al $^{3+}$

Q6Problems

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

Solution

Given:

First IE of Na = 496 kJ mol $^{-1}$
First IE of Mg = 737 kJ mol $^{-1}$
First IE of Si = 786 kJ mol $^{-1}$

To Predict: The first ionization enthalpy of Al, choosing between 575 kJ mol

^{-1}

and 760 kJ mol

^{-1}

.

Justification: The general trend for first ionization enthalpy (IE) is that it increases across a period from left to right due to increasing nuclear charge. Based on this trend alone, one would expect the IE of Al to be between that of Mg (737) and Si (786).

However, there is an exception to this trend when comparing Group 2 and Group 13 elements.

Electronic configuration of Mg (Z=12): [Ne] $3s^2$
Electronic configuration of Al (Z=13): [Ne] $3s^2 3p^1$

To ionize Mg, an electron must be removed from the stable, completely filled 3s orbital. To ionize Al, an electron is removed from the single-occupied 3p orbital.

The 3p electron in Al is at a slightly higher energy level than the 3s electrons. Additionally, the 3p electron is shielded from the nucleus by the inner 3s electrons. This shielding makes the 3p electron less tightly bound and easier to remove compared to a 3s electron from Mg, despite Al having a higher nuclear charge.

Therefore, the first ionization enthalpy of Al is expected to be lower than that of Mg.

Given the value for Mg is 737 kJ mol

^{-1}

, the value for Al must be less than 737 kJ mol

^{-1}

. Comparing the two options:

575 kJ mol $^{-1}$ is less than 737 kJ mol $^{-1}$ .
760 kJ mol $^{-1}$ is greater than 737 kJ mol $^{-1}$ .

Thus, the value for Al will be closer to 575 kJ mol

^{-1}

.

Final Answer: The first

\Delta_i H

value for Al will be more close to 575 kJ mol $^{-1}$ . This is because the electron to be removed from Al is a 3p electron, which is more effectively shielded by the 3s electrons and is easier to remove than a 3s electron from Mg.

Q7Problems

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

Solution

Given Elements: P, S, Cl, F.

To Find: The element with the most negative electron gain enthalpy and the one with the least negative.

Explanation: Electron gain enthalpy (EGE) is the enthalpy change when an electron is added to an isolated gaseous atom. A more negative value indicates a greater tendency to accept an electron.

Trends in Electron Gain Enthalpy:

Across a period (left to right): EGE generally becomes more negative due to an increase in effective nuclear charge and smaller atomic size. Following this, for the third-period elements P, S, and Cl, the order of increasingly negative EGE is P < S < Cl.
Down a group: EGE generally becomes less negative due to an increase in atomic size, which means the added electron is farther from the nucleus and experiences less attraction.

Analysis of the given elements:

P, S, Cl are in the 3rd period. Based on the trend across a period, Cl will have a more negative EGE than S, which in turn is more negative than P. Thus, P has the least negative EGE among these three.
F and Cl are in Group 17. Based on the general trend down a group, F should have a more negative EGE than Cl. However, this is a known exception. The EGE of F is less negative than that of Cl. This is because the fluorine atom is very small (n=2 shell). The incoming electron experiences significant electron-electron repulsion from the already present nine electrons in a small volume. In chlorine (n=3 shell), the atom is larger, and the incoming electron is added to a larger orbital, resulting in much less electron-electron repulsion. This makes the electron addition process more favorable (more exothermic) for Cl.

Conclusion:

Most Negative EGE: Comparing all four, Cl has the most negative electron gain enthalpy.
Least Negative EGE: Phosphorus (P) is in Group 15 and has a stable half-filled $p^3$ configuration, making it less inclined to accept an electron compared to S and Cl. Thus, P will have the least negative EGE.

Final Answer:

Most negative electron gain enthalpy: Chlorine (Cl)
Least negative electron gain enthalpy: Phosphorus (P)

Q8Problems

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.

Solution

To Predict: Formulas of compounds from given pairs of elements.

Solution: We determine the common valence (or oxidation state) of each element based on its group number and then use the criss-cross method to find the formula.

(a) silicon and bromine

Silicon (Si): It is in Group 14. Its common valence is 4.
Bromine (Br): It is in Group 17 (a halogen). Its common valence is 1.

To form a neutral compound, one silicon atom (valence 4) will combine with four bromine atoms (valence 1). Si

^{4+}

Br

^{1-}

Criss-crossing the valencies gives Si

_1

Br

_4

.

Formula: SiBr $_4$

(b) aluminium and sulphur

Aluminium (Al): It is in Group 13. Its common valence is 3.
Sulphur (S): It is in Group 16. Its common valence is 2.

To form a neutral compound, two aluminium atoms (total charge 2 x 3 = +6) will combine with three sulphur atoms (total charge 3 x 2 = -6). Al

^{3+}

S

^{2-}

Criss-crossing the valencies gives Al

_2

S

_3

.

Formula: Al $_2$ S $_3$

Final Answer: (a) The formula for the compound of silicon and bromine is SiBr $_4$ . (b) The formula for the compound of aluminium and sulphur is Al $_2$ S $_3$ .

Q9Problems

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

Solution

Given Complex Ion: [AlCl(H

_2

O)

_5

]

^{2+}

To Determine: If the oxidation state and covalency of Aluminium (Al) are the same.

Solution:

1. Calculation of Oxidation State: Let the oxidation state of Al be 'x'. The oxidation state of a chloride ion (Cl) is -1. Water (H

_2

O) is a neutral molecule, so its oxidation state is 0. The overall charge of the complex ion is +2.

Setting up the equation: x + (-1) + 5(0) = +2 x - 1 = +2 x = +3

So, the oxidation state of Al is +3.

2. Determination of Covalency: Covalency refers to the number of bonds formed by the central atom. In this complex, the central Aluminium atom is bonded to one chloride ligand and five water ligands.

Number of bonds = 1 (from Cl) + 5 (from H

_2

O) = 6

So, the covalency of Al is 6.

Conclusion: The oxidation state of Al is +3, while its covalency is 6. These two values are not the same.

Final Answer: No, the oxidation state (+3) and the covalency (6) of Al in [AlCl(H

_2

O)

_5

]

^{2+}

are not the same.

Q10Problems

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

Solution

To Show: Na

_2

O is a basic oxide and Cl

_2

O

_7

is an acidic oxide using reactions with water.

Solution:

1. Reaction of Na $_2$ O (Sodium Oxide) with water: Metal oxides, especially those of alkali and alkaline earth metals, are typically basic. When a basic oxide reacts with water, it forms a base (a metal hydroxide).

Sodium oxide (Na

_2

O) reacts with water (H

_2

O) to form sodium hydroxide (NaOH), which is a strong base.

Chemical Equation:

Na_2O(s) + H_2O(l) \rightarrow 2NaOH(aq)

Since the product, NaOH, is a base, Na

_2

O is a basic oxide.

2. Reaction of Cl $_2$ O $_7$ (Dichlorine Heptoxide) with water: Non-metal oxides are typically acidic. When an acidic oxide reacts with water, it forms an acid.

Dichlorine heptoxide (Cl

_2

O

_7

) reacts with water (H

_2

O) to form perchloric acid (HClO

_4

), which is a strong acid.

Chemical Equation:

Cl_2O_7(l) + H_2O(l) \rightarrow 2HClO_4(aq)

Since the product, HClO

_4

, is an acid, Cl

_2

O

_7

is an acidic oxide.

Final Answer: The reactions Na

_2

O + H

_2

O

\rightarrow

2NaOH (a base) and Cl

_2

O

_7

+ H

_2

O

\rightarrow

2HClO

_4

(an acid) demonstrate that Na

_2

O is a basic oxide and Cl

_2

O

_7

is an acidic oxide.

Classification of Elements and Periodicity in PropertiesClass 11 Chemistry NCERT Solutions

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

Solution

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

Solution

What is the basic difference in approach between the Mendeleev's Periodic Law and the Modern Periodic Law?

Solution

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

Solution

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

Solution

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

Solution

Which element do you think would have been named by

Lawrence Berkeley Laboratory

Seaborg's group?

Solution

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

Solution

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

Solution

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

Solution

What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.

F−^{-}−

Ar

Mg2+^{2+}2+

Rb+^{+}+

Solution

Consider the following species : N3−^{3-}3−, O2−^{2-}2−, F−^{-}−, Na+^{+}+, Mg2+^{2+}2+ and Al3+^{3+}3+

Solution

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

Solution

What is the significance of the terms - 'isolated gaseous atom' and 'ground state' while defining the ionization enthalpy and electron gain enthalpy?

Solution

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

Solution

Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why

Be has higher ΔiH\Delta_i HΔi​H than B

O has lower ΔiH\Delta_i HΔi​H than N and F ?

Solution

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

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

Solution

The first ionization enthalpy values (in kJ mol−1^{-1}−1) of group 13 elements are : B: 801, Al: 577, Ga: 579, In: 558, Tl: 589 How would you explain this deviation from the general trend?

Solution

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

O or F

F or Cl

Solution

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

Solution

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

Solution

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

Solution

Describe the theory associated with the radius of an atom as it

Solution

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

Solution

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

Solution

Use the periodic table to answer the following questions.

Solution

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

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

Solution

Assign the position of the element having outer electronic configuration (i) ns2np4ns^2 np^4ns2np4 for n=3n=3n=3 (ii) (n−1)d2ns2(n-1)d^2 ns^2(n−1)d2ns2 for n=4n=4n=4, and (iii) (n−2)f7(n−1)d1ns2(n-2)f^7 (n-1)d^1 ns^2(n−2)f7(n−1)d1ns2 for n=6n=6n=6, in the periodic table.

Solution

Solution

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

Solution

In the modern periodic table, the period indicates the value of :

Solution

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

Solution

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?

Solution

F $^{-}$

Mg $^{2+}$

Rb $^{+}$

Consider the following species : N $^{3-}$ , O $^{2-}$ , F $^{-}$ , Na $^{+}$ , Mg $^{2+}$ and Al $^{3+}$

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 $^{-1}$ .

Be has higher $\Delta_i H$ than B

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

The first ionization enthalpy values (in kJ mol $^{-1}$ ) of group 13 elements are : B: 801, Al: 577, Ga: 579, In: 558, Tl: 589 How would you explain this deviation from the general trend?

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

Assign the position of the element having outer electronic configuration (i) $ns^2 np^4$ for $n=3$ (ii) $(n-1)d^2 ns^2$ for $n=4$ , and (iii) $(n-2)f^7 (n-1)d^1 ns^2$ for $n=6$ , in the periodic table.

The size of isoelectronic species - F $^{-}$ , Ne and Na $^{+}$ is affected by

Which of the following species will have the largest and the smallest size? Mg, Mg $^{2+}$ , Al, Al $^{3+}$ .

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

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

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