GSEB Class 11 Chemistry Solutions Chapter 3 Classification of Elements and Periodicity in Properties

 

Question 1. What is the basic theme of organization in the periodic table?
Answer: The primary goal is to make simpler and more organized the examination of all elements' characteristics. Based on similar chemical features, the different elements are currently separated into various groups. Therefore, the main idea behind arranging elements in the periodic table is how electrons are set up in their atoms and their chemical reactions, which comes from having similar electron setups.
In simple words: The main idea of the periodic table is to organize elements based on their electron setup and chemical behavior.

Exam Tip: Remember to highlight both the goal (simplification/systematization) and the fundamental basis (electronic configuration/chemical properties) when explaining the periodic table's organization.

 

Question 2. Which important property did Mendeleev use to classify the elements in his periodic table?
Answer: Mendeleev classified elements mainly by their atomic masses. He put elements into horizontal rows and vertical columns, arranging them by increasing atomic mass so that elements with like characteristics appeared in the same vertical column or group. He noticed that some elements did not fit his classification system when strictly following the atomic mass order. Sometimes, he disregarded the atomic mass sequence. For instance, iodine, with a lower atomic mass (\( = 126.9 \)) than tellurium (\( = 127.6 \)), was positioned in Group VII with other halogens due to shared properties.
In simple words: Mendeleev primarily used atomic mass to classify elements, arranging them so similar properties aligned, even if it meant sometimes ignoring strict atomic mass order.

Exam Tip: Always specify "atomic weights" or "atomic masses" as Mendeleev's primary criterion, and mention his flexibility in prioritizing chemical properties over strict atomic weight order in some cases.

 

Question 3. What is the basic difference in approach between Mendeleev's Periodic Law and the Modern Periodic Law?
Answer: Mendeleev's Periodic Law relied on atomic masses of elements, while the Modern Periodic Law uses atomic numbers of elements. Mendeleev declared: “Element properties are periodic, depending on their atomic weights”. The Modern Periodic Law says: “The physical and chemical characteristics of elements are periodic functions of their atomic numbers”.
In simple words: Mendeleev's law used atomic mass, but the Modern Periodic Law uses atomic number to explain element properties.

Exam Tip: Clearly state the defining property for each law – atomic weight for Mendeleev and atomic number for the Modern Periodic Law – and include the precise wording of both laws for full marks.

 

Question 4. On the basis of quantum numbers justify that the sixth period of the periodic table should have 32 elements.
Answer: For \( n = 6 \), the values of \( l \) are \( 0, 1, 2, 3, 4, 5 \). Orbital energy increases in this sequence: \( 6s < 4f < 5d < 6p \). There are 16 orbitals in total: \( s=1, p=3, d=5, f=7 \). Each orbital can hold a maximum of 2 electrons. Consequently, the sixth period can contain 32 elements.
In simple words: Because the sixth period includes \(6s\), \(4f\), \(5d\), and \(6p\) orbitals, making a total of 16 orbitals, and each orbital holds 2 electrons, it can fit 32 elements.

Exam Tip: When justifying the number of elements in a period, list the orbitals filled (e.g., \(6s, 4f, 5d, 6p\)), calculate the total number of orbitals, and then multiply by two for the maximum electron capacity.

 

Question 5. In terms of period and group, where would you locate the element with \( Z = 114 \)?
Answer: For an element with atomic number \( Z = 114 \), its likely symbol is Uuq. This element would be found in the 7th period and 14th group. Its electron configuration is \( 7s^2 7p^2 \).
In simple words: An element with atomic number 114, called Uuq, is in the 7th period and 14th group, with an electron setup of \( 7s^2 7p^2 \).

Exam Tip: To locate an element, identify its period from the highest principal quantum number (\(n\)) and its group from the number of valence electrons (or \(p\)-block position calculation).

 

Question 6. Write the atomic number of the element present in the third period and a seventeenth group of the periodic table.
Answer: The element in the third period and seventeenth group has an atomic number of 17, with an electron configuration of \( 2, 8, 7 \). This element is Chlorine.
In simple words: The element in the third period, group 17, is Chlorine, which has an atomic number of 17.

Exam Tip: Remember that the period number corresponds to the principal quantum number of the outermost shell, and for \(p\)-block elements, the group number is 10 + valence electrons (or 10 + group number in the \(p\)-block for simpler systems).

 

Question 7. Which element do you think would have been named by
1. Lawrence Berkeley Laboratory
2. Seaborg's group?
Answer:
1. Berkelium (\(Bk\)), with an atomic number of 97, and Lawrencium (\(Lr\)), with \( Z = 103 \).
2. Seaborgium (\(Sg\)), with an atomic number of 106.
In simple words: The elements named by Lawrence Berkeley Lab are Berkelium and Lawrencium, while Seaborg's group named Seaborgium.

Exam Tip: Associate element names with the laboratories or scientists who discovered or made significant contributions to their study, especially for transuranic elements named after places or people.

 

Question 8. Why do elements in the same group have similar physical and chemical properties?
Answer: Elements within the same vertical column or group share similar properties because they possess identical valence shell electron configurations, meaning they have the same number of electrons in their outermost orbitals. Therefore, these shared properties arise from the similar electron arrangements in the outermost shells of their atoms. For instance, all Group I elements (alkali metals) have an \(ns^1\) valence shell electron configuration.
In simple words: Elements in the same group act alike because they have the same number of electrons in their outermost shell.

Exam Tip: The key to explaining similar properties within a group is always the "same number of valence electrons" or "similar outermost electronic configuration."

 

Question 9. What do atomic radius and ionic radius really mean? to you?
Answer: For non-metals such as chlorine, the atomic radius is actually the covalent radius, which is half the distance between two bonded chlorine atoms in a \(Cl_2\) molecule. The bond distance in a \(Cl_2\) molecule measures \(198\text{ pm}\). So, the atomic or covalent radius is \( \frac{198}{2} = 99\text{ pm} \). For instance, the space between two neighboring copper atoms in solid copper is \(256\text{ pm}\), so the metallic radius (or atomic radius) of copper is \(128\text{ pm}\). However, the ionic radius refers to the radius of an ion, whether it's a cation or an anion. We estimate it by measuring the distances between positive and negative ions in ionic crystals.
In simple words: Atomic radius is how big an atom is, often measured as half the distance between two similar atoms. Ionic radius is the size of an ion, which is measured from the distances between ions in crystals.

Exam Tip: Differentiate atomic radius (covalent for non-metals, metallic for metals) from ionic radius. For each, describe how it is measured and what it represents (half bond distance, ion size).

 

Question 10. How does atomic radius vary in a period and in a group? How do you explain the variation?
Answer: Across a period, atomic radii of elements typically decrease from left to right. This occurs because, within the same period, outer electrons stay in the same valence shell, but the effective nuclear charge grows as the atomic number rises, leading to stronger attraction of electrons to the nucleus. Within a group, atomic radii increase from top to bottom. As you move down a group, the nuclear charge grows with atomic number, but simultaneously, the principal quantum level increases steadily, while the number of outermost electrons stays constant. As a result, the distance of the outermost electrons from the nucleus slowly expands down the group, because the impact of the added energy level is more noticeable than the effect of the increased nuclear charge.
In simple words: Atomic radius shrinks across a period because nuclear pull increases, but it grows down a group because new shells are added, making atoms bigger.

Exam Tip: When explaining periodic trends, always cite the two main opposing factors: increasing nuclear charge and increasing shielding effect/number of shells, and specify which one dominates in each direction (period vs. group).

 

Question 11. Identify an isoelectronic species for each of the following atoms or ions:
1. \(F^-\)
2. \(Ar\)
3. \(Mg^{2+}\)
4. \(Rb^+\)
Answer: Isoelectronic species, whether atoms or ions, are ones that have an equal quantity of electrons. For instance, \(O^{2-}\), \(F^-\), \(Na^+\), and \(Mg^{2+}\) all possess 10 electrons. These form an isoelectronic series.
1. \(F^-\) is isoelectronic with \(Na^+\) (\(10e^-\))
2. \(Ar\) is isoelectronic with \(K^+\) (\(18e^-\))
3. \(Mg^{2+}\) is isoelectronic with \(O^{2-}\) (\(10e^-\))
4. \(Rb^+\) is isoelectronic with \(Br^-\) (\(36e^-\))
In simple words: Isoelectronic means having the same number of electrons. For example, \(F^-\) has the same number of electrons as \(Na^+\).

Exam Tip: To identify an isoelectronic species, first count the total number of electrons in the given atom or ion. Then, find another atom or ion that has the same electron count.

 

Question 12. Consider the following species : \(N^{3-}, O^{2-}, F^-, Na^+, Mg^{2+}\) and \(Al^{3+}\)
1. What is common in them?
2. Arrange them in the order of increasing ionic radii.
Answer:
1. The species \(N^{3-}, O^{2-}, F^-, Na^+\), and \(Al^{3+}\) are all isoelectronic, each having 10 electrons.
2. The increasing order of ionic radii for these species is: \(Al^{3+} < Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}\).
In simple words: All these particles have 10 electrons. Their sizes go from smallest to largest in this order: \(Al^{3+} < Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}\).

Exam Tip: For isoelectronic species, ionic radii decrease as nuclear charge increases. More protons pulling the same number of electrons means a smaller radius.

 

Question 13. Explain why cations are smaller and anions larger in radii than their parent atoms?
Answer: Cations are consistently smaller than their original atoms because they contain fewer electrons, yet the nuclear charge stays the same. The remaining electrons are pulled more strongly by the protons in the nucleus, making their radii smaller than the original atoms. Anions are always bigger than their parent atoms because adding one or more electrons leads to greater repulsion among electrons and a reduction in the effective nuclear charge.
In simple words: Cations are smaller than their parent atoms because they lose electrons, increasing nuclear attraction. Anions are larger because they gain electrons, increasing electron repulsion.

Exam Tip: For cations, remember that losing electrons reduces electron-electron repulsion and increases the effective nuclear charge. For anions, gaining electrons increases electron-electron repulsion and decreases the effective nuclear charge, causing expansion.

 

Question 14. What is the significance of the terms 'isolated gaseous atom' and 'ground state' while defining the ionization enthalpy and electron gain enthalpy?
Answer: To define ionization enthalpy, the smallest amount of energy needed to detach the most loosely held electron from a lone gaseous atom is required, forming a positive ion \( (X (g) + \text{energy} \rightarrow X^+ (g) + e^-) \). When defining electron gain enthalpy, an energy change accompanies the process \( (X (g) + e^- \rightarrow X^- (g) \Delta H = \Delta H_{eg}) \). Energy is typically released when a neutral gaseous atom \(X\) gains an electron, turning it into a negative ion. The atom must be in its lowest energy, or ground state. This condition is essential for making proper comparisons.
In simple words: "Isolated gaseous atom" means no other atoms interfere, and "ground state" means the atom is in its most stable energy level. These conditions ensure consistent and accurate measurements of ionization and electron gain enthalpies.

Exam Tip: Always emphasize that these terms ensure that the measured energy changes are intrinsic properties of the atom, free from external influences (isolated gaseous) and in its most stable state (ground state), allowing for valid comparisons.

 

Question 15. The 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 \( J \text{ mol}^{-1} \).
Answer: An electron in the ground state possesses energy \( E = -2.18 \times 10^{-18} \text{ J} \). To remove this electron, \( +2.18 \times 10^{-18} \text{ J} \) of energy is needed. The reaction is \( H (g) \rightarrow H^+ (g) + e^- \), where \( \Delta E \) (Ionization Energy) equals \( +2.18 \times 10^{-18} \text{ J/electron} \). To find the molar value, we use the mole concept: Expressed per mole, the value is \( (2.18 \times 10^{-18} \text{ J}) \times (6.022 \times 10^{23}) = 13.12 \times 10^5 \text{ J mol}^{-1} \) (since \( 1 \) mole of electrons is \( 6.022 \times 10^{23} \) electrons).
In simple words: The energy to remove one electron from hydrogen is \( +2.18 \times 10^{-18} \text{ J} \). To find this for a whole mole of hydrogen, multiply this by Avogadro's number, giving \( 13.12 \times 10^5 \text{ J mol}^{-1} \).

Exam Tip: Remember that ionization enthalpy is the *energy required* to remove an electron, so the sign will be positive. Always convert per atom/electron values to molar values using Avogadro's number if asked for \( \text{J mol}^{-1} \).

 

Question 16. Among the second period elements the actual ionization enthalpies are in the order Li < B < Be. < C < O < N < F < Ne. Explain why:
1. Be has higher \( \Delta_i H \) than B
2. O have lower \( \Delta_i H \) than N and F?
Answer:
1. The ionization enthalpy (\( \Delta_i H \)) of Beryllium (\(Be\)) is greater than that of Boron (\(B\)). Beryllium's electron arrangement is \( 1s^2 2s^2 \), while Boron's is \( 1s^2 2s^2 2p^1 \). For Beryllium, an electron must be taken from an \(s\)-orbital, but for Boron, it's removed from a \(p\)-orbital. Removing an \(s\)-electron is challenging because it sits closer to the nucleus than a \(p\)-electron. Thus, more energy is needed to extract an electron from \( 2s \) in Beryllium than from \( 2p \) in Boron, making Beryllium's ionization energy higher.
2. Oxygen (\(O\)) has an electron configuration of \( 1s^2 2s^2 2p_x^2 2p_y^1 2p_z^1 \), which is neither half-filled nor completely filled. Nitrogen (\(N\)), however, has \( 1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1 \), which is exactly half-filled. It is hard to detach an electron from Nitrogen's valence shell because its \(p\)-subshell is precisely half-filled, giving it greater stability. Oxygen's electron configuration is neither fully filled nor half-filled. Consequently, it is simpler to remove one electron from an Oxygen atom. Fluorine (\(F\)), with its higher nuclear charge, has a greater ionization energy than both Oxygen (\(O\)) and Nitrogen (\(N\)).
In simple words: 1. Beryllium's ionization energy is higher than Boron's because it's harder to remove an electron from a stable \(s\)-orbital closer to the nucleus. 2. Oxygen has lower ionization energy than Nitrogen because Nitrogen's half-filled \(p\)-subshell is very stable, making it harder to remove an electron. Fluorine has higher ionization energy due to stronger nuclear attraction.

Exam Tip: When explaining exceptions to ionization enthalpy trends, focus on electron configuration stability (e.g., half-filled or full-filled subshells) and the penetration effect of orbitals (s-electrons are harder to remove than p-electrons due to closer proximity to the nucleus).

 

Question 17. How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Answer: For both elements, the initial electron must be taken from the \( 3s \)-orbital, but Sodium (\(Na\)) has a lower nuclear charge than Magnesium (\(Mg\)). Therefore, Sodium's ionization energy is less than Magnesium's.
\( Na = 1s^2 2s^2 2p^6 3s^1 \quad \xrightarrow{IE_1} \quad Na^+ = 1s^2 2s^2 2p^6 \)
\( Mg = 1s^2 2s^2 2p^6 3s^2 \quad \xrightarrow{IE_1} \quad Mg^+ = 1s^2 2s^2 2p^6 3s^1 \)
After losing the first electron, \( Na^+ \) has the electron configuration of \( 1s^2 2s^2 2p^6 \), which is a very stable noble gas configuration, making it very hard to remove a second electron from \( Na^+ \). For \( Mg^+ \), after the first electron is lost, its electron configuration becomes \( 1s^2 2s^2 2p^6 3s^1 \). The second electron is removed from the \( 3s \) orbital, which is simpler to do. This also explains why the second ionization enthalpy of Sodium is much greater than that of Magnesium.
In simple words: Sodium has a lower first ionization energy than Magnesium because it has fewer protons pulling the outer electron. However, Sodium's second ionization energy is much higher because removing another electron means breaking a very stable noble gas configuration, which requires a lot of energy. For Magnesium, the second electron is easier to remove as it still comes from an outer shell.

Exam Tip: When comparing successive ionization enthalpies, always consider the stability of the electron configuration after each electron removal. Noble gas configurations are highly stable and require significantly more energy to disrupt.

 

Question 18. What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Answer:
1. As one moves down a group, the atomic size steadily grows because a new principal energy shell is added with each subsequent element. Consequently, the outermost electrons are further from the nucleus. This leads to a weaker attraction between the nucleus and valence electrons, so ionization enthalpy should drop.
2. When new shells are added, the quantity of inner shell electrons that shield the valence electrons from the nucleus increases. This means the shielding or screening effect becomes stronger. As a consequence, the nucleus's attraction for the valence electrons diminishes even more, causing the ionization enthalpy to decrease.
3. Nuclear charge rises down the group with a growing atomic number, which would typically cause the force of attraction between the nucleus and valence electrons to strengthen, thereby increasing ionization enthalpy. However, the combined impact of expanding atomic size and stronger screening effect more than offsets the influence of the rising nuclear charge. As a result, valence electrons are less and less strongly held by the nucleus, causing ionization enthalpies to progressively decrease down the group.
In simple words: Ionization enthalpy decreases down a group mainly because atoms get bigger and inner electrons shield the outer ones better. These effects overpower the increased nuclear charge, making it easier to remove outer electrons.

Exam Tip: Always discuss the interplay of three factors: increasing atomic size, increasing shielding effect (which dominates), and increasing nuclear charge (which is compensated for), to explain the decrease in ionization enthalpy down a group.

 

Question 19. The first ionization enthalpy values (in \( \text{kJ mol}^{-1} \)) of group 13 elements are :
B Al Ga In Tl
801 577 579 558 589
How would you explain this deviation from the general trend?
Answer: When descending Group 13 from Boron (\(B\)) to Aluminum (\(Al\)), the first ionization enthalpy (\(IE_1\)) drops due to an increase in atomic size and a stronger shielding effect, which overrides the impact of the rising nuclear charge. Nevertheless, Gallium's (\(Ga\)) \(IE_1\) is only marginally greater than Aluminum's (\(Al\)) (by \( 2 \text{ kJ mol}^{-1} \)), and Thallium's (\(Tl\)) \(IE_1\) is considerably higher than Aluminum, Gallium, and Indium (\(In\)). This happens because Aluminum (\(Al\)) comes right after the s-block elements, whereas Gallium (\(Ga\)) and Indium (\(In\)) follow the d-block elements, and Thallium (\(Tl\)) follows both d- and f-block elements. These additional \(d\) and \(f\) electrons do not effectively shield or screen the outer shell electrons from the nucleus.
In simple words: The ionization energy trend in Group 13 is unusual because of poor shielding by \(d\) and \(f\) electrons. While it generally decreases from Boron to Aluminum, Gallium's is slightly higher than Aluminum's, and Thallium's is significantly higher than Aluminum, Gallium, and Indium, due to these \(d\) and \(f\) electrons not blocking the nuclear charge well.

Exam Tip: For irregular ionization enthalpy trends in transition metal or inner transition metal series (like Group 13), remember to attribute the deviations to the poor shielding effect of \(d\) and \(f\) electrons, which causes the effective nuclear charge to increase more than expected.

 

Question 20. Which of the following pairs of elements would have a more negative electron gain enthalpy?
1. O or F
2. F or Cl
Answer:
1. Fluorine (\(F\)), with an electron gain enthalpy of \( -328\text{ kJ mol}^{-1} \), has a more negative electron gain enthalpy compared to Oxygen (\(O\)), which is \( -141\text{ kJ mol}^{-1} \).
2. Chlorine (\(Cl\)), with an electron gain enthalpy of \( -349\text{ kJ mol}^{-1} \), possesses a more negative electron gain enthalpy than Fluorine (\(F\)), which is \( -328\text{ kJ mol}^{-1} \).
In simple words: 1. Fluorine has a more negative electron gain enthalpy than Oxygen. 2. Chlorine has a more negative electron gain enthalpy than Fluorine.

Exam Tip: Generally, electron gain enthalpy becomes more negative across a period. Group 17 elements (halogens) have very negative electron gain enthalpies. Chlorine often has a more negative electron gain enthalpy than fluorine due to less electron-electron repulsion in its larger atomic size.

 

Question 21. Would you expect the second electron gain enthalpy of O as positive, more negative, or less negative than the first? Justify your answer.
Answer: When an electron is added to an Oxygen atom, the \( O^- \) ion is formed, and energy is released. However, when a second electron is added to the \( O^- \) ion to create an \( O^{2-} \) ion, energy must be absorbed to overcome the strong electrostatic repulsion between the negatively charged \( O^- \) ion and the incoming electron. Therefore, the second electron gain enthalpy for Oxygen is positive.
Here is the first electron gain enthalpy:
\( O (g) + e^- (g) \rightarrow O^- (g); \Delta H_{eg} = -141 \text{ kJ mol}^{-1} \) (energy is given off)
Here is the second electron gain enthalpy:
\( O^- (g) + e^- (g) \rightarrow O^{2-} (g); \Delta H_{eg} = +780 \text{ kJ mol}^{-1} \) (energy is taken in)
In simple words: The second electron gain enthalpy for Oxygen is positive. This is because adding a second electron to an already negative \( O^- \) ion requires energy to overcome the repulsion between the negative ion and the new electron.

Exam Tip: Remember that adding a second electron to an anion (which is already negatively charged) always requires energy input to overcome electron-electron repulsion, making the second electron gain enthalpy (and subsequent ones) positive.

 

Question 22. What is the basic difference between the terms electron gain enthalpy and electronegativity?
Answer: Electronegativity is distinct from electron gain affinity because the latter concerns atoms in their separate states, while the former describes an atom's property within a bonded state, specifically within a molecule. Electronegativity is purely a qualitative concept and cannot be quantitatively measured.
In simple words: Electron gain enthalpy is about an atom gaining an electron in isolation, while electronegativity is about an atom's pull on shared electrons in a bond. Electronegativity is a qualitative measure, not quantitative.

Exam Tip: Clearly distinguish between electron gain enthalpy (energy change for an isolated atom gaining an electron) and electronegativity (tendency of an atom in a molecule to attract shared electrons in a bond). Note that enthalpy is quantitative, while electronegativity is a relative, qualitative measure.

 

Question 23. How would you react to the statement that the electronegativity of N on the Pauling scale is 3.0 in all the nitrogen compounds?
Answer: Although Nitrogen's electronegativity on the Pauling scale is 3.0, an element's electronegativity is not always fixed. It changes based on the element it's bonded to, its hybridization state, and its oxidation state. Electronegativity rises as the percentage of \(s\)-character in a hybrid orbital increases, or as the element's oxidation state increases. For example, the electronegativity of Nitrogen in \(NO_2\), where its oxidation state is +4, is greater than in \(NO\), where its oxidation state is +2. The 3.0 electronegativity value given to Nitrogen is just an arbitrary assignment, as electronegativity itself is not a quantity that can be measured precisely.
In simple words: The statement is incorrect. An element's electronegativity, even for Nitrogen, isn't constant; it changes with what it's bonded to, its hybridization, and its oxidation state. So, a single value like 3.0 can't apply to all its compounds.

Exam Tip: When discussing electronegativity, highlight that it's not a fixed value for an element but varies with its chemical environment, specifically its bonding partner, hybridization, and oxidation state. Provide examples to illustrate these variations.

 

Question 24. Describe the theory associated with the radius of an atom as it
1. gains an electron
2. loses an electron
Answer:
1. When an atom acquires one or more electrons, it transforms into a negatively charged ion known as an anion. An anion's radius will be larger than its original atom's because adding electrons causes more repulsion between them and a reduction in the effective nuclear charge. Consequently, the electron cloud grows, meaning the ionic radius expands.
2. When an atom loses one or more electrons, it becomes a positively charged ion, which is called a cation. A cation's radius is consistently smaller than that of its parent atom. This can be attributed to the following factors.
• The valence shell might disappear. Sometimes, losing electrons causes the entire valence shell to vanish. Removing the outermost shell reduces the size.
• The effective nuclear charge increases. A cation has fewer electrons than its corresponding atom, while the total nuclear charge remains constant. This boosts the effective nuclear charge. As a consequence, electrons are drawn closer to the nucleus, resulting in a smaller ionic radius.
In simple words: 1. When an atom gains electrons, it becomes a larger anion due to increased electron repulsion. 2. When an atom loses electrons, it becomes a smaller cation because electron repulsion decreases, and the remaining electrons are pulled closer by the nucleus.

Exam Tip: Explain the size change for both cations and anions by focusing on the change in electron-electron repulsion and the effective nuclear charge. For cations, also mention the possible loss of an entire valence shell.

 

Question 25. Would you expect the first ionization enthalpies of two isotopes of the same element to be the same or different? Justify your answer.
Answer: Isotopes of an element only vary in their neutron count. They possess the same number of protons and electrons. Ionization energy values are mainly influenced by: 1. nuclear charge and 2. the atom's size. Ionization energy is not affected by the number of neutrons. Therefore, the ionization energy value stays constant for different isotopes of the same element.
In simple words: The first ionization enthalpies of isotopes are the same. This is because isotopes only differ in neutron count, which doesn't affect the number of protons or electrons, or the atom's size—the main factors for ionization energy.

Exam Tip: Remember that ionization enthalpy depends primarily on electronic configuration and nuclear charge, which are identical for isotopes. The mass difference (due to neutrons) does not significantly impact these electronic properties.

 

Question 26. What are the major differences between metals and non-metals?
Answer: Elements that readily give up electrons to form positive ions (cations) and display electropositive behavior are known as metals. Conversely, elements that tend to gain electrons and display electronegative behavior to form negative ions (anions) are called non-metals. As a result, metals serve as potent reducing agents, characterized by low ionization enthalpies, less negative electron gain enthalpies, and small electronegativity values. They typically create basic oxides and ionic compounds. Non-metals, conversely, are powerful oxidizing agents, possessing high ionization enthalpies, notably high electron gain enthalpies, and significant electronegativity values. They typically form acidic oxides and primarily covalent compounds. The chemical distinctions between metals and non-metals are presented in detail below:

Students Note: Please pick-up only a few properties from the above detailed properties to distinguish between metals and non-metals.
In simple words: Metals usually give away electrons, forming positive ions and acting as reducing agents. Non-metals usually gain electrons, forming negative ions and acting as oxidizing agents. The table above gives more detailed differences in their electron arrangement, how they form ions, and their reactions.

Exam Tip: When listing differences between metals and non-metals, ensure you cover key characteristics like electron tendency, ion formation, oxide nature (basic/acidic), and reactivity (reducing/oxidizing agents). Using a table format helps organize your points clearly.

 

Question 27. Use the periodic table to answer the following questions.
1. Identify an element with five electrons in the outer subshell.
2. Identify an element that would tend to lose two electrons.
3. Identify an element that would tend to gain two electrons.
4. Identify the group having metal, non-metal, liquid as well as gas at room temperature.
Answer:
1. Fluorine \( F = 1s^2, 2s^2, 2p^5 \) has five electrons in its 2p subshell.
2. Calcium \( Ca = 1s^2, 2s^2 2p^6 3s^2 3p^6 4s^2 \) tends to give away two electrons.
3. Oxygen \( O = 1s^2, 2s^22p^4 \) tends to acquire two electrons.
4. Mercury (Hg) is a metal that is liquid at room temperature; it is a transition metal from group 12. Nitrogen \( N_2 \) (group 15) or Oxygen \( O_2 \) (group 16) or Fluorine \( F_2 \)/Chlorine \( Cl_2 \) (group 17) and inert gases (group 18) are non-metals that are gases at room temperature. Bromine (group 17) is a non-metal that is liquid at room temperature. The group that includes metals, non-metals, and elements that are liquid and gas at room temperature is group no. 1. It contains \( H_2 \) (a gas and a non-metal at room temperature), and Cesium (Cs) is a liquid metal.
In simple words: The elements are identified based on their electron configurations and typical chemical behaviors within the periodic table. This includes elements that lose electrons, gain electrons, and those with specific physical states at room temperature.

Exam Tip: Remember key properties like electron configurations for specific elements, trends in electron loss/gain, and common states of matter for elements at room temperature.

 

Question 28. The increasing order of reactivity among group 1 elements is \( Li < Na < K < Rb < Cs \) whereas that among group 17 elements is \( F > Cl > Br > I \). Explain.
Answer: Group I elements are known as alkali metals. They produce unipositive ions \( M^+ \) by losing one electron from their outer electron shells. As ionization enthalpy decreases from top to bottom in group I, their ability to produce unipositive ions grows from top to bottom. Therefore, reactivity grows down the group from Li to Cs. In other words, the order of reactivity in group I elements is \( Li < Na < K < Rb < Cs \). Group 17 elements, also known as halogens, have an electronic configuration of \( ns^2 np^5 \) in their outer electron shells, aiming to achieve \( ns^2 np^6 \) by gaining one electron. All halogens are highly reactive elements due to two main reasons: (1) Low bond dissociation energy: Halogens possess low bond dissociation energy \( (X_2 \rightarrow 2X) \). \( F_2 \) has the lowest bond dissociation energy, which makes it the most reactive. (2) High Electron Affinity: Because of high electron affinity, these elements show a strong tendency to acquire an electron, making them highly reactive. (3) High reduction potentials:
\( \frac{1}{2} F_2 + e^- \rightarrow F^- \) \( E_{Red}^0 = +2.87V \)
\( \frac{1}{2} I_2 + e^- \rightarrow I^- \) \( E_{Red}^0 = +0.53V \)
\( F_2 \) possesses maximum reduction potential. Hence, the order of reactivity is \( F > Cl > Br > I \).
In simple words: Alkali metals become more reactive as you go down the group because it's easier to lose an electron. Halogens become more reactive as you go up the group because it's easier to gain an electron.

Exam Tip: Understand the periodic trends for ionization enthalpy and electron affinity to explain reactivity variations within groups.

 

Question 29. Write the general outer electronic configuration of s, p, d, and f-block elements.
Answer: The elements of s-block have an outer electronic configuration of \( ns^{1-2} \). The elements of p-block have an outer electronic configuration of \( ns^2 np^{1-6} \), except for Helium, which is \( 1s^2 \). d-block elements have an outer electronic configuration of \( (n - 1)d^{1-10} ns^{0-2} \). f-block elements have a general outer electronic configuration of \( (n - 2)f^{0-14} (n - 1)d^{0-1} ns^2 \).
In simple words: This lists the general electron arrangements for the outermost shell of elements in the s, p, d, and f blocks of the periodic table, showing how many electrons can be in each type of orbital.

Exam Tip: Memorize these general electronic configurations as they are fundamental for understanding element placement and properties in the periodic table.

 

Question 30. Assign the position of the element having outer electronic configuration
1. \( ns^2 np^4 \) for \( n = 3 \)
2. \( (n - 1)d^2 ns^2 \) for \( n = 4 \), and
3. \( (n - 2)f^7 (n - 1)d^1 ns^2 \) for \( n = 6 \), in the periodic table.
Answer:
(1) \( ns^2 np^4 \) for \( n = 3 \) means \( 3s^2 3p^4 \). The complete electronic configuration is \( 1s^2, 2s^2 2p^6, 3s^2 3p^4 \). The atomic number is \( 2 + 2 + 6 + 2 + 4 = 16 \). This element is Sulfur, found in the 3rd period and Group 16 (p-block).
(2) \( (n - 1)d^2 ns^2 \) for \( n = 4 \) means \( 3d^2 4s^2 \). The complete electronic configuration is \( 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^2 4s^2 \). The atomic number is 22, and the element is Titanium. This is a transition element, located in the 4th period and Group 4.
(3) \( (n - 2)f^7 (n - 1)d^1 ns^2 \) for \( n = 6 \) means \( 4f^7 5d^1 6s^2 \). Its complete electronic configuration is \( 1s^2, 2s^2 2p^6, 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 4f^7 5s^2 5p^6 5d^1 6s^2 \). This element is Gadolinium (Gd). It is an inner transition element, belonging to the Lanthanoid series or 4f series. It is an f-block element.
In simple words: By looking at the outermost electrons and counting all electrons, we can figure out which element it is and where it belongs on the periodic table—its period and group.

Exam Tip: To identify an element's position, first determine the atomic number from the complete electronic configuration. The highest principal quantum number (n) indicates the period, and the subshell filling pattern determines the block and group.

 

Question 31. 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 :

Which of the above elements is likely to be :
1. the least reactive element,
2. the most reactive metal.
3. the most reactive non-metal
4. the least reactive non-metal.
5. the metal which can form a stable binary halide of the formula \( MX_2 \) (X = hydrogen).
6. the metal which can form a predominantly stable covalent halide of the formula MX (X = halogen)?
Answer:
(1) Element V is the least reactive element. It is Helium. It shows no tendency to lose electrons (as indicated by its high \( \Delta_i H_1 \) and higher \( \Delta_i H_2 \)), nor can it gain electrons (positive electron gain enthalpy value means energy must be absorbed to add an electron).
(2) Element II is the most reactive metal, which is Potassium (K). Its \( \Delta_i H_1 \) is the lowest, and \( \Delta_i H_2 \) is very high, meaning it can readily lose one electron from its valence shell (4th) to become \( K^+ \). It cannot easily lose the second electron due to the stable Argon core.
(3) Element III is the most reactive non-metal, which is Fluorine. Its \( \Delta_i H_1 \) and \( \Delta_i H_2 \) are very high, meaning it cannot readily lose electrons. However, its electron gain enthalpy has a large negative value, indicating it can readily accept an electron.
(4) Element IV is the least reactive non-metal, which is Iodine. Its \( \Delta_i H_1 \) and \( \Delta_i H_2 \) are very high, meaning it cannot readily lose electrons. On the other hand, its electron gain enthalpy has a large negative value (but not as negative as Fluorine), indicating that it can still accept an electron.
(5) Element VI is the metal that can form a stable binary halide of the formula \( MX_2 \) (X = halogen). It is Magnesium (Mg). \( Mg + X_2 \rightarrow MgX_2 \). Its \( \Delta_i H_1 \) and \( \Delta_i H_2 \) are not very high, but its electron gain enthalpy is very small; \( \Delta_i H_2 \) is not significantly larger compared to \( \Delta_i H_1 \). Because of its electropositive nature and tendency to lose two electrons, it can readily form a stable binary halide \( MX_2 \).
(6) Element I is the metal that can form a stable covalent halide. It is Lithium (Li). Its \( \Delta_i H_2 > > \Delta_i H_1 \), which means it can easily lose one electron but not the second. Because of its small size, it forms a stable covalent halide of the formula MX.
In simple words: We identified elements based on their energy values for losing or gaining electrons. Elements with low first ionization energy are reactive metals, while those with very negative electron gain enthalpy are reactive non-metals. Stable halides depend on how many electrons an element typically loses or gains.

Exam Tip: Analyze ionization enthalpies and electron gain enthalpies to determine an element's metallic/non-metallic character and reactivity. A large jump between successive ionization enthalpies indicates a stable electron configuration, like a noble gas, after removing a certain number of electrons.

 

Question 32. Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.
1. Lithium and oxygen
2. Magnesium and nitrogen
3. Aluminum and iodine
4. Silicon and oxygen
5. Phosphorus and fluorine
6. Element Lu and fluorine
Answer:
1. Lithium oxide — \( Li_2O \)
2. Magnesium nitride — \( Mg_3N_2 \)
3. Aluminum iodide — \( AlI_3 \)
4. Silicon oxide or silica — \( SiO_2 \)
5. Phosphorus fluoride — \( PF_5 \)
6. Element (Lu) fluoride — \( EF_3 \) [\( LuF_3 \)]
In simple words: We combined each pair of elements to form stable compounds, making sure the number of atoms balances out based on how many electrons each element typically gains or loses.

Exam Tip: To predict compound formulas, determine the common valencies (combining power) or oxidation states of each element. Then, cross-multiply the valencies to balance the charges and form a neutral compound.

 

Question 33. In the modern periodic table, the period indicates the value of:
1. atomic number
2. atomic mass
3. principal quantum number
4. azimuthal quantum number
Answer: 3. Principal quantum number.
In simple words: In the periodic table we use today, the row number (period) tells us the main energy level where an element's outermost electrons are found.

Exam Tip: Remember that the period number directly corresponds to the principal quantum number (n) of the outermost electron shell being filled.

 

Question 34. Which of the following statements related to the modern periodic table is incorrect?
1. The p-block has 6 columns because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
2. The d-block has 8 columns because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
3. Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
4. The block indicates the value of an azimuthal quantum number (0 for the last subshell that received electrons in building up the electronic configuration.
Answer: 2. the d-block has 10 columns because a maximum of 10 electrons can occupy all the five d-orbitals of the d-subshell.
In simple words: The statement that the d-block has 8 columns is wrong. It actually has 10 columns because each d-subshell can hold a total of 10 electrons across its five orbitals.

Exam Tip: Recall the maximum electron capacity of each subshell (s-2, p-6, d-10, f-14) to determine the number of columns in each block of the periodic table.

 

Question 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?
1. Valence principal quantum number (M)
2. Nuclear charge (Z) (c) Nuclear mass
3. A number of core electrons.
Answer: 3. Nuclear mass does not affect the valence shell because the nucleus consists of protons and neutrons whereas protons, i.e., nuclear charge affects the valence shell, but neutrons do not. Hence option (c) is wrong.
In simple words: Nuclear mass doesn't change how an element's outer electrons behave. Only the nuclear charge (number of protons) and the number of other electrons impact the valence shell, not the total weight of the nucleus.

Exam Tip: Understand that chemical properties primarily depend on electron configuration and the effective nuclear charge, which is influenced by protons and core electrons, not neutron count or nuclear mass.

 

Question 36. The size of isoelectronic species - \( F^-, Ne \) and \( Na^+ \) is affected by
1. nuclear charge' (Z)
2. valence principal quantum number (n)
3. electron-electron interaction in the outer orbitals
4. none of the factors because their size is the same.
Answer: 1. Nuclear charge (Z).
In simple words: For atoms and ions that have the same number of electrons, their size is mainly determined by the nuclear charge, which is the number of protons in their nucleus.

Exam Tip: For isoelectronic species, the greater the nuclear charge (Z), the stronger the attraction for electrons, and thus, the smaller the ionic/atomic radius. Conversely, lower nuclear charge leads to a larger size.

 

Question 37. Which one of the following statements is incorrect in relation to ionization enthalpy?
1. Ionization enthalpy increases for each successive electron.
2. The greatest increase in ionization enthalpy is experienced in the removal of electrons from the core noble gas configuration.
3. The end of valence electrons is marked by a big jump in ionization enthalpy.
4. Removal of electrons from orbitals bearing a lower n value is easier than from orbital having a higher n value.
Answer: 4. the statement is incorrect. It is easier to remove an electron from orbitals bearing a higher n value than from orbital having a lower n value.
In simple words: The incorrect statement is that it's easier to remove electrons from lower energy shells. It's actually easier to remove electrons from higher energy shells (higher 'n' value) because they are further from the nucleus and less strongly held.

Exam Tip: Ionization enthalpy is the energy required to remove an electron. Electrons in higher energy shells (larger 'n') are further from the nucleus and experience less attraction, making them easier to remove, hence requiring less ionization energy.

 

Question 38. Considering the elements B, Al, Mg and K, the correct order of their metallic character is :
1. B > Al > Mg > K
2. Al > Mg > B > K
3. Mg > Al > K > B
4. K > Mg > Al > B
Answer: 4. K > Mg > Al > B
In simple words: The correct order for metallic character, which means how easily an element loses electrons, is from Potassium to Magnesium, then Aluminum, and finally Boron. This order shows that metallic character increases as you go down a group and decreases as you go across a period.

Exam Tip: Metallic character increases down a group (due to larger atomic size and lower ionization energy) and decreases across a period (due to increasing nuclear charge and higher ionization energy).

 

Question 39. Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is:
1. B > C > Si > N > F
2. Si > C > B > N > F
3. F > N > C > B > Si
4. F > N > C > Si > B
Answer: 3. F > N > C > B > Si is correct.
In simple words: Non-metallic character increases from left to right across a period and decreases down a group. So, Fluorine is the most non-metallic, followed by Nitrogen, Carbon, Boron, and finally Silicon.

Exam Tip: Non-metallic character generally increases from left to right across a period and decreases down a group, due to increasing electronegativity and electron gain enthalpy. Always consider both horizontal and vertical trends.

 

Question 40. Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is :
1. F > Cl > O > N
2. F > O > Cl > N
3. Cl > F > O > N
4. O > F > N > Cl
Answer: 2. F > O > Cl > N is correct.
In simple words: The ability to act as an oxidizing agent, meaning to gain electrons, is strongest in Fluorine, then Oxygen, followed by Chlorine, and weakest in Nitrogen among these elements. This order reflects how easily these elements attract and accept electrons.

Exam Tip: Oxidizing power correlates with electronegativity and electron gain enthalpy. Elements with high electronegativity and very negative electron gain enthalpy are strong oxidizing agents. Fluorine is the most electronegative element.