Classification of Elements and Periodicity in Properties
This chapter explores the periodic classification of elements, which is based on their recurring properties. It delves into the historical context of the periodic table and the contributions of scientists like Döbereiner, Newlands, Mendeleev, and Moseley.
Key Points:
Döbereiner’s Triads: Observed that certain groups of three elements had similar properties.
Newlands’ Octaves: Arranged elements in order of increasing atomic weight and proposed a pattern of repeating properties every eighth element.
Mendeleev discovered a cyclical recurrence of elemental properties when arranged according to their atomic weights.
Emphasizing the significance:
Modern Periodic Table: Arranged elements based on their atomic number, which is the number of protons in the nucleus.
Periodic Trends: Properties of elements show regular patterns when arranged in the periodic table. These include:
Atomic radius increases as you descend a group, but it decreases as you move from left to right across a period.”
Ionic radius generally increases down a group and decreases across a period, with exceptions for anions.
The ionization energy of elements decreases as you move down a group in the periodic table and increases as you move across a period.
Electron Affinity: Increases down a group and decreases across a period (with some exceptions).
Electronegativity follows a periodic trend, decreasing down a group and increasing across a period.
Periodic Table Organization:
Periods are horizontal rows that group elements with the same number of electron shells.
Groups: Vertical columns, representing elements with similar chemical properties due to having the same number of valence electrons.
Representative Elements: Groups 1A, 2A, 3A, 4A, 5A, 6A, 7A, and 8A.
Transition Elements: Groups 3B to 8B.
Inner Transition Elements: Lanthanides (4f) and Actinides (5f).
Understanding the periodic table is essential for predicting the properties of elements and understanding chemical reactions.
1 What is the basic theme of organisation in the periodic table?
Ans : The basic theme of organization in the periodic table is the periodic recurrence of similar properties of elements. Elements are arranged in a systematic way based on their atomic number, which is the number of protons in their nucleus. This arrangement reveals patterns in the properties of elements, such as their reactivity, electronegativity, and atomic size.
In periods (horizontal rows) and groups (vertical columns), the periodic table is divided. Elements within the same group share similar chemical properties due to having the same number of valence electrons (electrons in the outermost shell). Elements within the same period show a gradual change in properties as their atomic number increases.
This organization allows for the prediction of the properties of elements based on their position in the table, making it a valuable tool for chemists and scientists.
2 Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?
Ans : Mendeleev primarily used the atomic mass of elements to classify them in his periodic table. He arranged the elements in order of increasing atomic mass, which led to the discovery of periodic trends in their properties.
However, it’s important to note that Mendeleev did make some exceptions to this strict ordering. In a few cases, he reversed the order of elements based on their observed properties, such as tellurium and iodine. This was done to maintain the periodic pattern of properties, even though it went against the strict order of atomic masses.
Later, with the discovery of atomic number, it was found that the periodic table is better organized when elements are arranged in order of increasing atomic number rather than atomic mass. This led to the modern periodic table.
3 What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?
Ans : Mendeleev’s Periodic Law used atomic weight as the basis for classification, while the Modern Periodic Law is based on atomic number. This shift was crucial because atomic number is a more fundamental property of an element than atomic mass, as it reflects the number of protons in the nucleus, which determines the element’s identity.
4 On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Ans : The maximum number of electrons that can be accommodated in a given energy level (n) is given by the formula:
2n²
For the sixth period, n = 6. Therefore, the maximum number of electrons in the sixth period is:
2 * 6² = 72
Now, let’s break down the electron configuration for the sixth period:
6s: 2 electrons
4f: 14 electrons
5d: 10 electrons
6p: 6 electrons
Adding up the electrons in each subshell:
2 + 14 + 10 + 6 = 32
Therefore, based on the quantum numbers and the maximum number of electrons that can be accommodated in the sixth energy level, the sixth period of the periodic table should have 32 elements.
5. In terms of period and group where would you locate the element with Z =114?
Ans : The element with atomic number Z = 114 is flerovium (Fl).
It is located in the 7th period and the p-block of the periodic table.
In groups 13-18 of the periodic table, the p-block elements are found.
Flerovium belongs to group 14.
6. Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Ans : The third period and seventeenth group of the periodic table corresponds to the element chlorine (Cl).
Hence, the atomic number of the element = 17.
7 Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group?
Ans : (i) Lawrence Berkeley Laboratory
Given its name, Lawrence Berkeley Laboratory would likely have been involved in the discovery and naming of an element related to Lawrence. Indeed, the element Lawrencium (Lr), with the atomic number 103, was discovered and named after Ernest Lawrence, the director of the laboratory.
(ii) Seaborg’s Group
Glenn Seaborg was a prominent nuclear chemist who discovered several elements. One element named after him is Seaborgium (Sg), which has the atomic number 106. Seaborg’s group was also involved in the discovery of other elements like americium, curium, berkelium, californium, einsteinium, fermium, mendelevium, nobelium, and lawrencium.
8 Why do elements in the same group have similar physical and chemical properties?
Ans : Elements in the same group of the periodic table have similar physical and chemical properties primarily due to their similar electron configurations.
The electron configuration of an element refers to the arrangement of its electrons in orbitals. Elements in the same group have the same number of valence electrons, which are the outermost electrons that participate in chemical bonding.
Key reasons for similarity in properties:
Similar chemical reactivity: Elements in the same group tend to react in similar ways because they have the same number of valence electrons. For example, elements in group 1 (alkali metals) are highly reactive and tend to lose an electron to form positive ions.
Similar physical properties: Elements in the same group often have similar physical properties, such as melting points, boiling points, and densities. This is because the valence electrons influence the interatomic forces between atoms, which in turn affect these physical properties.
Similar trends in atomic size and ionization energy: As you move down a group, the atomic size generally increases, and the ionization energy (the energy required to remove an electron) decreases. These trends are due to the increasing number of electron shells and the increasing distance between the nucleus and the outermost electrons.
In conclusion, the similarity in electron configurations of elements within a group leads to similar physical and chemical properties. This periodic pattern is a fundamental principle of the modern periodic table.
9. What does atomic radius and ionic radius really mean to you?
Ans : Atomic radius and ionic radius are measures of the size of an atom or ion, respectively. They are important concepts in understanding the properties of elements and compounds.
Atomic Radius
Definition: The atomic radius is the average distance between the nucleus and the outermost electrons of an atom.
Trends in the Periodic Table:
Increases down a group: Atoms get larger as you move down a group due to the addition of electron shells.
Decreases across a period: Atoms get smaller as you move across a period due to the increasing nuclear charge pulling the electrons closer to the nucleus.
Ionic Radius
Definition: The ionic radius is the average distance between the nucleus and the outermost electrons of an ion.
Trends:
Cations (positive ions) are smaller than their parent atoms: Losing electrons reduces the repulsion between electrons, allowing the nucleus to pull them closer.
Anions (negative ions) are larger than their parent atoms: Gaining electrons increases the repulsion between electrons, pushing them farther apart.
In essence, atomic radius and ionic radius help us understand:
The size of atoms and ions
How the size of atoms and ions changes within the periodic table
The factors that influence the size of atoms and ions
These concepts are crucial for understanding chemical bonding, properties of elements and compounds, and other areas of chemistry.
10 How do atomic radius vary in a period and in a group? How do you explain the variation?
Ans : In the Periodic table, atomic radius have variations
Variation Across a Period
Decreases: As you move from left to right across a period, the atomic radius generally decreases.
Reason: The number of protons in the nucleus increases, which increases the nuclear charge. This stronger nuclear charge pulls the electrons closer to the nucleus, making the atoms smaller.
Variation Down a Group
Increases: As you move down a group, the atomic radius generally increases.
Reason: Electrons are added to new energy levels (shells) as you move down a group. These outer electrons are further away from the nucleus, making the atoms larger.
In summary, atomic radius decreases across a period due to increased nuclear charge and increases down a group due to the addition of electron shells.
11 What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions. (i) F– (ii) Ar (iii) Mg2+ (iv) Rb+
Ans : Isoelectronic species are atoms or ions that consists the same number of electrons.
Here are some examples of isoelectronic species for the given atoms or ions:
(i) F⁻: This ion has 10 electrons (9 from the neutral fluorine atom + 1 from the gained electron). Some isoelectronic species of F⁻ include:
* Ne (neutral neon atom)
* Na⁺ (sodium ion)
(ii) Ar: Argon has 18 electrons. Some isoelectronic species of Ar include:
* K⁺ (potassium ion)
* Cl⁻ (chloride ion)
* Ca²⁺ (calcium ion)
(iii) Mg²⁺: This ion has 10 electrons (12 from the neutral magnesium atom – 2 lost electrons). Some isoelectronic species of Mg²⁺ include:
* Na⁺ (sodium ion)
* Ne (neutral neon atom)
(iv) Rb⁺: This ion has 36 electrons (37 from the neutral rubidium atom – 1 lost electron). Some isoelectronic species of Rb⁺ include:
* Kr (krypton atom)
* Sr²⁺ (strontium ion)
* Br⁻ (bromide ion)
In general, isoelectronic species often have similar chemical properties due to their identical electron configurations.
12Consider the following species : N3–, O2–, F– , Na+, Mg2+ and Al3+
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
Ans : (a) Common Feature:
All of these species have the same number of electrons, which is 10 electrons. This makes them isoelectronic species.
(b) Increasing Ionic Radii:
Ionic radius generally increases as the number of electron shells increases and decreases as the nuclear charge increases.
Hence, Increasing order of ionic radii is:
Al³⁺ (smallest ionic radius due to the highest nuclear charge)
Mg²⁺
Na⁺
O²⁻
F⁻
N³⁻ (largest ionic radius due to the highest negative charge, which increases electron-electron repulsion)
In summary, the given species are isoelectronic, and their ionic radii increase with increasing negative charge and decreasing positive charge.
13 Explain why cation are smaller and anions larger in radii than their parent atoms?
Ans : Due to lost of electrons the Cations are smaller than their parent atoms. When an atom loses an electron to form a cation, the remaining electrons are pulled closer to the nucleus by the stronger electrostatic attraction. This results in a smaller ionic radius compared to the neutral atom.
In the case of regained electrons the Anions are larger than their parent atoms. When an atom gains an electron to form an anion, the increased electron-electron repulsion pushes the electrons farther apart, leading to a larger ionic radius compared to the neutral atom.
In summary, the change in ionic radius compared to the parent atom is primarily due to the change in the balance between the attractive force of the nucleus and the repulsive force between electrons.
14. What is the significance of the terms — ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy?
(Hint : Requirements for comparison purposes. )
Ans : The terms “isolated gaseous atom” and “ground state” are crucial when defining ionization enthalpy and electron gain enthalpy because they establish a standard reference point for comparison.
Isolated gaseous atom: This means that the atom is considered to be in a vacuum, far away from any other atoms or molecules. This ensures that the energy measured is solely due to the removal or addition of an electron to the isolated atom, without any interference from external factors.
Ground state: The ground state refers to the lowest energy state of an atom. When defining ionization enthalpy and electron gain enthalpy, it is important to specify that the atom is in its ground state to ensure that the measurements are consistent and comparable.
By using these terms, we can establish a standard set of conditions under which the ionization enthalpy and electron gain enthalpy can be measured and compared. This allows us to make meaningful comparisons between different elements and understand the trends in these properties across the periodic table.
15. Energy of an electron in the ground state of the hydrogen atom is –2.18×10–18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol–1.( Hint: Apply the idea of mole concept to derive the answer.)
Ans : The ionization enthalpy is the energy required to remove one electron from one mole of atoms in their ground state.
To find ground state energy of a hydrogen atom, we use formula:
E (ground state)
= -2.18 x 10^-18 eV/atom x 6.022 x 10^23 atoms/mol
= -1.312 x 10^6 J/mol
The ionization enthalpy is equal to the difference between the energy of the ionized state (E∞) and the ground state energy. Since the ionized state has an electron at infinity, its energy is zero. Therefore, the ionization enthalpy is:
Ionization enthalpy = E∞ – E (ground state) = 0 – (-1.312 x 10^6 J/mol)
= 1.312 x 10^6 J/mol
16. Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why ?
(i) Be has higher ∆i H than B
(ii) O has lower ∆i H than N and F? 2024-25 Classification of Elements and Periodicity in Properties 97
Ans : (i) Be has higher ionization enthalpy than B
Electron Configuration: Be (1s² 2s²) has a completely filled 2s orbital, while B (1s² 2s² 2p¹) has one electron in the 2p orbital.
Stability: Half-filled and completely filled orbitals are relatively stable. Removing an electron from Be disrupts this stable configuration, requiring more energy.
Therefore, Be has a higher ionization enthalpy than B.
(ii) Among N, F & O, i.e O has the lowest ionization enthalpy.
Electron Configuration: O (1s² 2s² 2p⁴) has two unpaired electrons in the 2p orbital, while N (1s² 2s² 2p³) has three unpaired electrons. F (1s² 2s² 2p⁵) has four unpaired electrons.
Stability: Half-filled orbitals are slightly more stable than partially filled orbitals. Removing an electron from O disrupts a half-filled orbital, requiring less energy compared to removing an electron from N or F, which would disrupt a more stable configuration.
Therefore, O < N and F.
17 How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Ans : The electron configurations of sodium (Na) and magnesium (Mg) are:
Na: 1s² 2s² 2p⁶ 3s¹
Mg: 1s² 2s² 2p⁶ 3s²
In both cases, the first electron to be removed is from the 3s orbital. However, the nuclear charge of sodium (+11) is lower than that of magnesium (+12). This means that the nucleus of sodium has a weaker attraction for its electrons, making it easier to remove the first electron.
Consequently, sodium exhibits a lower first ionization energy than magnesium.
After the loss of the first electron, the electron configurations become:
Na⁺: 1s² 2s² 2p⁶
Mg⁺: 1s² 2s² 2p⁶ 3s¹
In the case of sodium, the second electron must be removed from a stable noble gas configuration (neon), which requires more energy. In magnesium, the second electron can be removed from the 3s orbital, which is less stable.
As a result, sodium exhibits a higher second ionization enthalpy compared to magnesium
18. What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Ans : The ionization enthalpy of elements generally decreases down a group in the periodic table due to several factors:
Atomic radius increases : The atomic radius increases, when you move down a group. This means that the outermost electron is further away from the nucleus, which reduces the electrostatic attraction between the nucleus and the electron.
Shielding effect: The inner electrons shield the outermost electron from the full attractive force of the nucleus. This shielding effect increases as you move down a group, further reducing the attraction between the nucleus and the outermost electron.
Penetration effect: The ability of an electron to penetrate the inner electron shells and get closer to the nucleus varies among different orbitals. Electrons in s orbitals penetrate more deeply than those in p, d, or f orbitals. As you move down a group, the outermost electrons are in higher energy levels with more penetrating orbitals, which can slightly increase the attraction to the nucleus. However, the overall trend is still a decrease in ionization enthalpy due to the first two factors.
In summary, the decrease in ionization enthalpy down a group is primarily due to the increase in atomic radius and the shielding effect of inner electrons, which outweigh the slight increase in penetration effect.
19. The first ionization enthalpy values (in kJ mol–1) of group 13 elements are :
B Al Ga In Tl
801 577 579 558 589
How would you explain this deviation from the general trend ?
Ans : The decrease in the first ionization enthalpy (∆iH₁) from boron (B) to aluminum (Al) is primarily due to the increase in atomic size of aluminum. As the atomic size increases, the outermost electron is further away from the nucleus, and the electrostatic attraction between the nucleus and the electron decreases, making it easier to remove.
In gallium (Ga), there are 10 3d electrons, which are not as effective at shielding the outer electrons from the nuclear charge as s and p electrons. This results in an unexpected increase in the effective nuclear charge experienced by the outermost electron, leading to a higher first ionization enthalpy.
Similarly, in thallium (Tl), the presence of 14 4f electrons, which have a very poor shielding effect, also increases the effective nuclear charge on the outermost electron, resulting in a higher first ionization enthalpy.
20. Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F
(ii) F or Cl
Ans : (i) F has a more negative electron gain enthalpy than O.
Reason:
Both oxygen and fluorine belong to group 17 (halogens) and have a strong tendency to gain electrons.
However, fluorine has a smaller atomic size than oxygen, which means that the incoming electron experiences a stronger electrostatic attraction from the nucleus in fluorine. This stronger attraction makes it more favorable for fluorine to gain an electron, resulting in a more negative electron gain enthalpy.
(ii) Cl has a more negative electron gain enthalpy than F.
Reason:
While fluorine has a smaller atomic size, chlorine has a larger atomic size.
Despite the larger size, chlorine has a more negative electron gain enthalpy than fluorine due to a phenomenon known as anomalous behavior. In chlorine, the incoming electron is added to the 3p subshell, which is larger than the 2p subshell in fluorine. This larger size means that the incoming electron experiences less electron-electron repulsion in chlorine, making it more favorable for chlorine to gain an electron.
21 Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
Ans : The second electron gain enthalpy of oxygen (O) is significantly more positive than the first.
Here’s why:
Electron Configuration: Oxygen has the electron configuration [He]2s²2p⁴. When it gains one electron to form O⁻, it becomes [He]2s²2p⁵, which is a relatively stable configuration with a half-filled p subshell.
Repulsion: Adding a second electron to O⁻ to form O²⁻ would mean placing two electrons in the same 2p orbital. This leads to increased electron-electron repulsion, making it significantly more difficult to add the second electron.
Energy Requirement: The energy required to overcome this repulsion and add the second electron is significantly higher than the energy required to add the first electron. This results in a positive second electron gain enthalpy for oxygen.
In conclusion, the second electron gain enthalpy of oxygen is positive and significantly higher than the first electron gain enthalpy due to the increased electron-electron repulsion in the O²⁻ ion.
22 What is the basic difference between the terms electron gain enthalpy and electronegativity?
Ans : Electron gain enthalpy and electronegativity are related concepts, but they have distinct meanings:
Electron gain enthalpy (EGE):
Definition: The energy change that occurs when a neutral gaseous atom accepts an additional electron to form a negative ion.
Focus: The tendency of an isolated atom to gain an electron.
Measurement: A quantitative property that can be measured experimentally.
Electronegativity:
Definition: The relative ability of an atom in a molecule to attract shared electrons towards itself.
Focus: The tendency of an atom to attract electrons in a chemical bond.
Measurement: A qualitative property that is usually estimated using empirical scales like the Pauling scale.
Key differences:
Context: Electron gain enthalpy refers to an isolated atom, while electronegativity is a property of an atom in a molecule.
Focus: Electron gain enthalpy focuses on the energy change during the addition of an electron, while electronegativity focuses on the attraction of shared electrons in a bond.
Measurement: Electron gain enthalpy is a measurable quantity, while electronegativity is often estimated using empirical scales.
In summary, electron gain enthalpy is a measure of an atom’s tendency to gain an electron, while electronegativity is a measure of an atom’s ability to attract shared electrons in a molecule.
23 How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
Ans : The statement that the electronegativity of N on the Pauling scale is 3.0 in all nitrogen compounds is not entirely accurate.
While electronegativity is generally a characteristic property of an element, it can vary slightly depending on the specific chemical environment in a molecule.
Here are some factors that can influence the electronegativity of nitrogen:
Bonding partners: The electronegativity of nitrogen can vary slightly depending on the elements it is bonded to. For example, nitrogen may have a slightly higher electronegativity when bonded to a more electronegative element like oxygen compared to when it is bonded to a less electronegative element like hydrogen.
Hybridization: The hybridization state of nitrogen can also affect its electronegativity. Different hybridization states can lead to variations in the electron density around the nitrogen atom.
Therefore, while the electronegativity of nitrogen is generally around 3.0 on the Pauling scale, it can exhibit slight variations depending on the specific chemical context.
24 Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron
Ans : Atomic Radius and Electron Gain/Loss
(a) Gaining an Electron (Anion Formation)
Increase in Radius: When an atom gains an electron to form an anion, the ionic radius is generally larger than the atomic radius of the neutral atom.
Reason: The addition of an electron increases the electron-electron repulsion, causing the electron cloud to expand. This results in a larger distance between the nucleus and the outermost electrons.
(b) Losing an Electron (Cation Formation)
Decrease in Radius: When an atom loses an electron to form a cation, the ionic radius is generally smaller than the atomic radius of the neutral atom.
Reason: The loss of an electron reduces the electron-electron repulsion, allowing the remaining electrons to be pulled closer to the nucleus. This leads to a decrease in the distance between the nucleus and the outermost electrons.
In summary, the ionic radius of an atom changes when it gains or loses electrons due to the alteration in electron-electron repulsion and the resulting change in the size of the electron cloud.
25. Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Ans : There is no difference in the first ionization enthalpy between isotopes of the same element.
Here’s why:
Isotopes have the same number of protons: Isotopes are atoms of the same element with different numbers of neutrons. However, they have the same number of protons, which determines the element’s identity and its chemical properties.
Ionization enthalpy depends on nuclear charge: The first ionization enthalpy is the energy required to remove the outermost electron from an atom. This energy depends primarily on the attractive force between the nucleus (protons) and the outermost electron.
Isotopes have the same nuclear charge: Since isotopes have the same number of protons, they have the same nuclear charge.
Therefore, the first ionization enthalpies for two isotopes of the same element should be the same, as they depend on the same nuclear charge and electron configuration.
26 What are the major differences between metals and non-metals?
Ans : Key Differences Between Metals and Non-Metals
Physical Properties
Appearance: Metals are generally shiny, malleable, ductile, and good conductors of heat and electricity. Non-metals are often dull, brittle, and poor conductors.
Hardness: Metals tend to be harder than non-metals, with exceptions like lead and sodium.
Density: Metals are generally denser than non-metals.
Melting and Boiling Points: Metals typically have higher melting and boiling points than non-metals.
Chemical Properties
Reactivity: Metals are generally more reactive than non-metals. Metals typically lose electrons to become positively charged ions, while non-metals gain electrons to become negatively charged ions.
Oxidation: Metals are oxidized (lose electrons) in chemical reactions. Non-metals are reduced (gain electrons).
Formation of Compounds: Metals often form ionic compounds with non-metals, while non-metals can form covalent compounds with each other or with metals.
In summary, metals and non-metals exhibit distinct physical and chemical properties. Metals are generally shiny, malleable, ductile, and conductive, while non-metals are often dull, brittle, and non-conductive. Metals tend to lose electrons, while non-metals tend to gain electrons.
27 Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.
Ans : (a) Elements with five electrons in the outer subshell belong to Group 15 of the periodic table, also known as the pnictogens. Examples include:
Nitrogen (N)
Phosphorus (P)
Arsenic (As)
Antimony (Sb)
Bismuth (Bi)
(b) Elements that tend to lose two electrons are those in Group 2 of the periodic table, also known as the alkaline earth metals. Examples include:
Beryllium (Be)
Magnesium (Mg)
Calcium (Ca)
Strontium (Sr)
Barium (Ba)
(c) Elements that tend to gain two electrons are those in Group 16 of the periodic table, also known as the chalcogens. Examples include:
Oxygen (O)
Sulfur (S)
Selenium (Se)
Tellurium (Te)
(d) Group 17 (halogens) contains elements in all of these states at room temperature:
Fluorine (F): Gas
Chlorine (Cl): Gas
Bromine (Br): Liquid
Iodine (I): Solid (but can sublime)
Astatine (At): Solid (radioactive)
28 The increasing order of reactivity among group 1 elements is Li < Na < K < Rb CI > Br > I. Explain.
Ans : The increasing order of reactivity among group 1 elements (Li, Na, K, Rb, Cs) is Li < Na < K < Rb < Cs. Explained this by the following factors:
Atomic Radius: As we move down group 1, the atomic radius increases. This means that the outermost electron is further away from the nucleus, making it easier to remove.
Shielding Effect: The inner electrons shield the outermost electron from the full attractive force of the nucleus. This shielding effect increases as we move down the group, making it easier to remove the outermost electron.
Effective Nuclear Charge: The effective nuclear charge (Zeff) is the net positive charge experienced by an outer electron. As we move down group 1, the increase in atomic radius and shielding effect outweigh the increase in nuclear charge. This leads to a decrease in Zeff, making it easier to remove the outermost electron.
In summary, the increasing reactivity of group 1 elements from Li to Cs is primarily due to the decrease in effective nuclear charge and the increasing ease of removing the outermost electron.
29.Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Ans : The electron configurations of the elements in the periodic table can be classified into four blocks based on the orbital that fills with electrons:
(i) s-Block elements: The outermost electrons occupy the ns orbital, where n is 2 to 7.
Hydrogen, lithium, and sodium are representative examples
(ii) p-Block elements: The outermost electrons occupy the np orbital, where n is 2 to 6.
Examples include boron (B), carbon (C), and nitrogen (N).
(iii) d-Block elements: The outermost electrons occupy the (n-1)d orbital, where n is 4 to 7.
Examples include scandium (Sc), titanium (Ti), and vanadium (V).
(iv) f-Block elements: The outermost electrons occupy the (n-2)f orbital, where n is 6 or 7.
Examples include lanthanum (La) and cerium (Ce).
30. Assign the position of the element having outer electronic configuration
(i) ns2np4 for n=3
(ii) (n-1)d2ns2 for n=4, and
(iii) (n-2) f 7 (n-1)d1ns2 for n=6, in the periodic table.
Ans : (i) ns²np⁴ for n=3
This represents the electronic configuration of an element in the third period and the fourth group of the periodic table. The specific element is silicon (Si).
(ii) (n-1)d²ns² for n=4
For n=4, (n-1) becomes 3. So, the configuration is 3d²4s². This represents an element in the fourth period and the fourth group of the periodic table. The specific element is titanium (Ti).
(iii) (n-2)f⁷(n-1)d¹ns² for n=6
For n=6, (n-2) becomes 4 and (n-1) becomes 5. So, the configuration is 4f⁷5d¹6s². This represents an element in the sixth period and the lanthanide series (specifically, the seventh element in the series). The specific element is gadolinium (Gd).
31 .
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 MX2 (X=halogen)
(f) the metal which can form a predominantly stable covalent halide of the formula MX(X=halogen
Ans : (a)The element (V) has the highest first ionization enthalpy (∆iH₁) and a positive electron gain enthalpy (∆egH). This indicates that it is the least reactive element among the given elements.
(b)The least reactive metal is the one with the lowest first ionization enthalpy (∆iH₁) and a low negative electron gain enthalpy (∆egH). Among the given elements, element II has the lowest ∆iH₁ and a low negative ∆egH, indicating that it is the most reactive metal. The values of ∆iH₁, ∆iH₂, and ∆egH for element II match those of potassium (K), suggesting that element II is likely potassium.
(c) The element with the highest first ionization enthalpy (∆iHi) and a very high negative electron gain enthalpy (∆egH) is the most reactive non-metal. This element is fluorine (F). The values of ∆iHi, ∆iH₂, and ∆egH for this element match the values for fluorine.
(d)The element IV has a high negative electron gain enthalpy (ΔegH), indicating that it readily accepts an electron. However, its first ionization enthalpy (ΔiH₁) is not exceptionally high, suggesting that it is not difficult to remove an electron from it. This combination of properties suggests that it is the least reactive non-metal.
The ionization enthalpies (ΔiH₁, ΔiH₂) and electron gain enthalpy (ΔegH) of element IV closely match those of iodine (I), suggesting that it is a halogen element.
(e)The element in group VI has a low first ionization enthalpy (∆iHi) compared to alkali metals, indicating that it is an alkaline earth metal. This means it will form binary halides with the formula MX2, where X is a halogen.
The values of the first ionization enthalpy (∆iHi), second ionization enthalpy (∆iH2), and electron gain enthalpy (∆egH) for this element match those of magnesium (Mg).
(f)The element I has a low first ionization enthalpy (∆iH₁) but a very high second ionization enthalpy (∆iH₂). This indicates that it is an alkali metal. Alkali metals have a single valence electron that is relatively easy to remove, resulting in a low first ionization enthalpy. However, once the first electron is removed, the remaining electrons are held more tightly by the nucleus, making it difficult to remove a second electron.
Since the element I forms predominantly stable covalent halides of the formula MX (X=halogen), it must be the least reactive alkali metal. This is because the electronegativity difference between the alkali metal and the halogen is relatively small, leading to the formation of covalent bonds rather than ionic bonds.
The values of ∆iH₁, ∆iH₂, and electron gain enthalpy (∆egH) for I match those of lithium (Li), further confirming the identity of the element
32 Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.
(a) Lithium and oxygen
(b) Magnesium and nitrogen
(c) Aluminium and iodine
(d) Silicon and oxygen
(e)Phosphorus and fluorine
(f) Element 71 and fluorine
Ans : To predict the formulas of the stable binary compounds, we need to consider the valencies of the elements involved. Valency is the combining capacity of an element, and it is often equal to the number of electrons an atom needs to gain or lose to achieve a stable noble gas configuration.
(a) Lithium and oxygen
Lithium belongs to group 1 of the periodic table and has a valency of 1.
Oxygen belongs to group 16 of the periodic table and has a valency of 2.
To achieve a stable configuration, lithium will lose 1 electron, and oxygen will gain 2 electrons. The compound will have the formula Li₂O.
(b) Magnesium and nitrogen
Magnesium belongs to group 2 of the periodic table and has a valence of 2.
Nitrogen is a member of group 15 with a valency of 3.
To achieve a stable configuration, magnesium will lose 2 electrons, and nitrogen will gain 3 electrons. The compound formula is Mg₃N₂.
(c) Aluminum and iodine
Aluminum is a group 13 element with a valency of 3.
Iodine is a member of group 17 and has a valency of 1.
To achieve a stable configuration, aluminum will lose 3 electrons, and iodine will gain 1 electron. The compound will have the formula AlI₃.
(d) Silicon and oxygen
Silicon, a member of group 14 in the periodic table, exhibits a valency of 4.
Oxygen belongs to group 16 of the periodic table and has a valency of 2.
To achieve a stable configuration, silicon will share 4 electrons with 2 oxygen atoms. The compound will have the formula SiO₂.
(e) Phosphorus and fluorine
Phosphorus belongs to group 15 of the periodic table and has a valency of 5.
Fluorine belongs to group 17 of the periodic table and has a valency of 1.
To achieve a stable configuration, phosphorus will share 5 electrons with 5 fluorine atoms. The compound will have the formula PF₅.
(f) Element 71 and fluorine
Element 71 is lutetium (Lu), which is in group 3 and has a valency of 3. Fluorine has a valency of 1.
To achieve a stable configuration, lutetium will lose 3 electrons, and fluorine will gain 1 electron. The compound will have the formula LuF₃.
33. In the modern periodic table, the period indicates the value of :
(a) atomic number
(b) atomic mass
(c) principal quantum number
(d) azimuthal quantum number.
Ans : The period number in the periodic table corresponds to the principal quantum number of the elements within that period.
Here’s a breakdown of the other options:
Atomic number: The atomic number is the number of protons in an atom, which determines the element’s identity. While there’s a general trend of increasing atomic number across a period, it’s not the defining characteristic.
Atomic mass: Atomic mass increases across a period, but it’s not a direct indicator of the period number.
Azimuthal quantum number: The azimuthal quantum number (l) determines the shape of the orbital. While there is a relationship between l and the period, the period number primarily reflects the principal quantum number.
Therefore, the correct answer is (c) principal quantum number.
34 Which of the following statements related to the modern periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration.
Ans : The incorrect statement is (b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
The correct number of columns in the d-block is 10. This is because a d-subshell can hold a maximum of 10 electrons, distributed across 5 orbitals (dxy, dxz, dyz, dx²-y², and dz²).
35.Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell? (a) Valence principal quantum number (n)
(b) Nuclear charge (Z )
(c) Nuclear mass
(d) Number of core electrons.
Ans : The answer is (c) Nuclear mass.
Here’s a breakdown of why the other factors do affect the valence shell:
Valence principal quantum number (n): This directly determines the energy level of the valence shell.
Nuclear charge (Z): The positive charge of the nucleus attracts the valence electrons, affecting their energy and behavior.
Number of core electrons: Core electrons shield the valence electrons from the full attractive force of the nucleus, influencing their effective nuclear charge.
Nuclear mass, however, does not directly affect the valence electrons.
The nucleus is primarily composed of protons and neutrons, and the number of neutrons does not significantly impact the electrostatic interaction between the nucleus and the valence electrons. Therefore, nuclear mass does not influence the valence shell and the chemistry of an element.
36.The size of isoelectronic species — F– , Ne and Na+ is affected by
(a) nuclear charge (Z )
(b) valence principal quantum number (n)
(c) electron-electron interaction in the outer orbitals
(d) none of the factors because their size is the same.
Ans : The size of isoelectronic species is primarily affected by nuclear charge (Z).
Here’s a breakdown of why:
Isoelectronic species share the same electron count.
Smaller Z (lower nuclear charge): A smaller nuclear charge means a weaker attraction between the nucleus and the electrons. This allows the electrons to spread out more, resulting in a larger atomic/ionic radius.
Larger Z (higher nuclear charge): A larger nuclear charge means a stronger attraction between the nucleus and the electrons. This pulls the electrons closer together, resulting in a smaller atomic/ionic radius.
While valence principal quantum number (n) and electron-electron interaction can also influence atomic/ionic size, they are not as significant as nuclear charge in the case of isoelectronic species.
37 Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Ans : Incorrect = (d)
Because, Ionization enthalpy generally increases with increasing principal quantum number (n): This is because electrons in higher energy levels (higher n values) are further away from the nucleus and experience less attraction, making them easier to remove.
38 Considering the elements B, Al, Mg, and K, the correct order of their metallic character is :
(a) B > Al > Mg > K
(b) Al > Mg > B > K
(c) Mg > Al > K > B
(d) K > Mg > Al > B
Ans : The correct order of metallic character among the elements B, Al, Mg, and K is:
K > Mg > Al > B
Explanation:
Metallic character increases down a group: As we move down a group in the periodic table, the metallic character of elements generally increases. This means that elements at the bottom of a group are more likely to exhibit metallic properties like conductivity and malleability.
Metallic character decreases across a period: As we move from left to right across a period, the metallic character generally decreases. Elements on the left side of the periodic table are more metallic, while elements on the right side are more non-metallic.
In this case, K is in group 1, Mg is in group 2, Al is in group 13, and B is in group 13. Therefore, the order of metallic character is K > Mg > Al > B.
39.Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is :
(a) B > C > Si > N > F
(b) Si > C > B > N > F
(c) F > N > C > B > Si
(d) F > N > C > Si > B
Ans : The non-metallic character of elements increases across a period and decreases down a group in the periodic table.
Among the elements B, C, N, and F, fluorine (F) has the highest non-metallic character, followed by nitrogen (N), carbon (C), and boron (B). This is because the atomic size decreases across a period, leading to a stronger attraction between the nucleus and the outermost electrons, which increases the non-metallic character.
Within a group, the non-metallic character decreases from top to bottom. Therefore, carbon (C) is more non-metallic than silicon (Si).
Combining these trends, the correct sequence of decreasing non-metallic character is F > N > C > B.
Therefore, option ‘C’ is correct.
40 Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is :
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) Cl > F > O > N
(d) O > F > N > Cl
Ans : The oxidizing character of elements increases across a period and decreases down a group in the periodic table.
Among the elements F, O, and N, fluorine (F) has the highest oxidizing power, followed by oxygen (O) and nitrogen (N). This is because the atomic size decreases across a period, leading to a stronger attraction between the nucleus and the outermost electrons, which increases the oxidizing power.
Within a group, the oxidizing power decreases from top to bottom. Therefore, fluorine (F) is a stronger oxidizing agent than chlorine (Cl). Additionally, oxygen (O) is more electronegative than chlorine (Cl), which also contributes to its stronger oxidizing power.
Therefore, the overall decreasing order of oxidizing power is F > O > N.
Option ‘b’ is correct.
