Samacheer Kalvi Class 11 Chemistry Solutions Chapter 3 Periodic Classification of Elements

Textual Questions:

I. Choose the Best Answer:

Question 1. What would be the IUPAC name for an element with atomic number 222?
(a) bibibiium
(b) bididium
(c) didibium
(d) bibibium
Answer: (d) bibibium
In simple words: The IUPAC name is formed by combining the roots for each digit of the atomic number. For 222, 'bi' stands for 2, so putting them together with the '-ium' suffix gives bibibium.

🎯 Exam Tip: Remember the IUPAC numerical roots for digits 0-9 (nil, un, bi, tri, quad, pent, hex, sept, oct, enn) and apply them in sequence, adding '-ium' at the end.

Question 2. The electronic configuration of the elements A and B are \( 1s^2, 2s^2, 2p^6, 3s^2 \) and \( 1s^2, 2s^2, 2p^5 \) respectively. The formula of the ionic compound that can be formed between these elements is
(a) AB
(b) AB2
(c) A2B
(d) none of the above
Answer: (b) AB2
In simple words: Element A has 2 outer electrons to lose (\(A^{2+}\)), while element B needs 1 electron to gain (\(B^{-}\)). To balance the charges, one A atom combines with two B atoms, forming \(AB_2\).

🎯 Exam Tip: Determine the probable charge of each ion by looking at its valence electron count, then use these charges to balance them in the compound formula.

Question 3. The group of elements in which the differentiating electron enters the anti penultimate shell of atoms are called
(a) p-block elements
(b) d-block elements
(c) s-block elements
(d) f-block elements
Answer: (d) f-block elements
In simple words: The f-block elements are characterized by their last electron entering a shell that is two layers inside the outermost electron shell.

🎯 Exam Tip: Recall that s-block elements fill the outermost s-orbital, p-block the outermost p-orbital, d-block the penultimate d-orbital, and f-block the anti-penultimate f-orbital.

Question 4. Which of the following options the order of arrangement does not agree with the variation of property indicated against it?
(a) I < Br < Cl < F (increasing electron gain enthalpy)
(b) Li < Na < K < Rb (increasing metallic radius)
(c) Al3+ < Mg2+ < Na+ < F¯ (increasing ionic size)
(d) B < C < O < N (increasing first ionisation enthalpy)
Answer: (a) I < Br < Cl < F (increasing electron gain enthalpy)
In simple words: The statement about electron gain enthalpy is incorrect because chlorine actually releases more energy than fluorine when gaining an electron, due to fluorine's very small size causing electron repulsion.

🎯 Exam Tip: Remember the anomaly in electron gain enthalpy where Chlorine has a more negative (releases more energy) value than Fluorine due to less electron-electron repulsion in its larger 3p orbital.

Question 5. Which of the following elements will have the highest electronegativity?
(a) Chlorine
(b) Nitrogen
(c) Cesium
(d) Fluorine
Answer: (d) Fluorine
In simple words: Fluorine is the 'greediest' element for electrons when it forms a bond, making it the most electronegative element in the entire periodic table.

🎯 Exam Tip: Fluorine is the most electronegative element, sitting at the top right of the periodic table (excluding noble gases). Electronegativity increases up a group and across a period.

Question 6. Various successive ionisation enthalpies (in kJ mol\(^{-1}\)) of an element are given below.

The element is
(a) phosphorus
(b) Sodium
(c) Aluminium
Answer: (c) Aluminium
In simple words: The sudden large jump in energy from IE3 to IE4 means that after losing three electrons, the atom reaches a very stable state, making it hard to remove a fourth. This pattern matches aluminium, which has three valence electrons.

🎯 Exam Tip: A significant increase in ionization energy between successive steps indicates that an electron is being removed from a more stable, inner electron shell, typically a noble gas configuration.

Question 7. In the third period the first ionization potential is of the order.
(a) Na > Al > Mg > Si > P
(b) Na < Al < Mg < Si < P
(c) Mg > Na > Si > P > Al
(d) Na < Al < Mg < P < Si
Answer: (b) Na < Al < Mg < Si < P
In simple words: While ionization potential generally increases across a period, aluminium's is lower than magnesium's because it's easier to remove an electron from its p-orbital than from magnesium's stable s-orbital.

🎯 Exam Tip: Remember the exceptions to the general trend of increasing ionization energy across a period: Group 13 elements (like Al) often have lower IE1 than Group 2 elements (like Mg), and Group 16 (like O) can be lower than Group 15 (like N).

Question 8. Identify the wrong statement.
(a) Amongst the isoelectronic species, smaller the positive charge on cation, smaller is the ionic radius
(b) Amongst isoelectric species greater the negative charge on the anion, larger is the ionic radius
(c) Atomic radius of the elements increases as one moves down the first group of the periodic table
(d) Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table.
Answer: (a) Amongst the isoelectronic species, smaller the positive charge on cation, smaller is the ionic radius
In simple words: This statement is wrong because for ions with the same number of electrons, a smaller positive charge means there are fewer protons pulling the electrons, which actually makes the ion larger.

🎯 Exam Tip: For isoelectronic species, remember that the ionic radius decreases as the positive nuclear charge increases. Conversely, for anions, increasing negative charge means increasing ionic radius.

Question 9. Which one of the following arrangements represent the correct order of least negative to most negative electron gain enthalpy
(b) Al < Ca Answer: (d) Ca < Al < C
In simple words: This order correctly shows elements from those that resist gaining electrons (like calcium) to those that strongly attract them (like fluorine), based on how much energy is released or absorbed.

🎯 Exam Tip: Elements in Group 2 often have positive electron gain enthalpy because adding an electron disrupts their stable \(s^2\) configuration. Halogens, on the other hand, have highly negative electron gain enthalpy.

Question 10. The correct order of electron gain enthalpy with negative sign of F, Cl, Br and I having atomic number 9, 17, 35 and 53 respectively is
(a) I > Br > Cl > F
(b) F > Cl > Br > I
(c) Cl > F > Br > I
(d) Br > I > Cl > F
Answer: (c) Cl > F > Br > I
In simple words: Chlorine releases the most energy when it gains an electron. Fluorine is next, then bromine, and iodine releases the least. This is because fluorine's small size creates electron-electron repulsion.

🎯 Exam Tip: Remember that Chlorine (Cl) has a more negative electron gain enthalpy than Fluorine (F) due to less electron-electron repulsion in its larger atomic size, even though F is more electronegative.

Question 11. Which one of the following is the least electronegative element?
(a) Bromine
(b) Chlorine
(c) Iodine
(d) Hydrogen
Answer: (d) Hydrogen
In simple words: Hydrogen attracts electrons less strongly than any of the halogens listed, making it the least electronegative among these options.

🎯 Exam Tip: Electronegativity decreases down a group. While iodine is the least electronegative halogen here, compare its value to other elements outside the group if listed as options.

Question 12. The element with positive electron gain enthalpy is
(a) Hydrogen
(c) Argon
Answer: (c) Argon
In simple words: Argon, being a noble gas with a full outer electron shell, is very stable and resists gaining an extra electron, so energy is needed to add one, resulting in a positive electron gain enthalpy.

🎯 Exam Tip: Noble gases and elements with stable half-filled or completely filled subshells often have positive electron gain enthalpies, meaning they absorb energy to accept an electron.

Question 13. The correct order of decreasing electronegativity values among the elements X, Y, Z and A with atomic numbers 4, 8, 7 and 12 respectively
(a) Y > Z > X > A
(b) Z > A > Y > X
(c) X > Y > Z > A
(d) X > Y > A>Z
Answer: (a) Y > Z > X > A
In simple words: Electronegativity increases across a period and decreases down a group. So, Oxygen (Y, Z=8) is highest, followed by Nitrogen (Z, Z=7), then Beryllium (X, Z=4), and finally Magnesium (A, Z=12) is the lowest.

🎯 Exam Tip: When comparing electronegativity, first locate elements by period and group. Elements higher up and further to the right (excluding noble gases) are generally more electronegative.

Question 14. Assertion: Helium has the highest value of ionisation energy among all the elements known
Reason: Helium has the highest value of electron affinity among all the elements known
(a) Both assertion and reason are true and reason is correct explanation for the assertion
(b) Both assertion and reason are true but the reason is not the correct explanation for the assertion
(c) Assertion is true and the reason is false
(d) Both assertion and the reason are false
Answer: (c) Assertion is true and the reason is false
In simple words: It is true that helium needs the most energy to lose an electron because of its small size and stable configuration. However, it's false that helium readily accepts an electron; as a noble gas, it strongly resists gaining electrons.

🎯 Exam Tip: Ionization energy and electron affinity trends are often inverse. Helium is a prime example: extremely high ionization energy but very low (or positive) electron affinity.

Question 15. The configuration of the atom having maximum difference in first and second ionisation energies is
(a) \( 1s^2, 2s^2, 2p^6, 3s^1 \)
(b) \( 1s^2, 2s^2, 2p^6, 3s^2 \)
(c) \( 1s^2, 2s^2, 2p^6, 3s^2, 3s^2, 3p^6, 4s^1 \)
(d) \( 1s^2, 2s^2, 2p^6, 3s^2, 3p^1 \)
Answer: (a) \( 1s^2, 2s^2, 2p^6, 3s^1 \)
In simple words: This configuration represents sodium, which has only one electron in its outermost shell. Removing this first electron is easy, but removing the second electron means breaking into a very stable, full electron shell, causing a huge jump in energy.

🎯 Exam Tip: A significant increase between successive ionization energies (e.g., IE1 vs IE2) indicates that a stable noble gas core electron configuration is being broken.

Question 16. Which of the following is second most electronegative element?
(a) Chlorine
(b) Fluorine
(c) Oxygen
(d) Sulphur
Answer: (a) Chlorine
In simple words: Fluorine is known to be the most electronegative element. Chlorine, being a halogen and right below fluorine, also has a very strong pull on shared electrons in a bond, making it the second most electronegative among some important elements.

🎯 Exam Tip: Fluorine is universally recognized as the most electronegative element. While Oxygen is the second globally, within a specific comparative context, Chlorine can be considered as having the second highest pulling power due to its position in the halogen group.

Question 17. IE\(_{1}\) and IE\(_{2}\) of Mg are 179 and 348 kcal mol\(^{-1}\) respectively. The energy required for the reaction \( Mg \rightarrow Mg^{2+} + 2e^{-} \) is
(a) + 169 kcal mol\(^{-1}\)
(b) -169 kcal mol\(^{-1}\)
(c) +527 kcalmol\(^{-1}\)
(d) -527 kcal mol\(^{-1}\)
Answer: (c) +527 kcalmol\(^{-1}\)
In simple words: To remove two electrons from a magnesium atom, you add up the energy needed for the first electron and the energy needed for the second electron. This gives a total of 527 kcal mol\(^{-1}\).

🎯 Exam Tip: The total energy required to remove multiple electrons is the sum of the individual successive ionization energies. Since energy is absorbed, the value is always positive.

Question 18. The order of screening effect is
(a) s > p > d >f
(b) s > p > f > d
(c) f > d > p > S
(d) f > p > s > d
Answer: (a) s > p > d >f
In simple words: The s-orbitals are best at blocking the nucleus's positive charge from outer electrons, followed by p, then d, and f-orbitals are the least effective due to their shapes.

🎯 Exam Tip: Remember that s-orbitals are spherical and penetrate closest to the nucleus, providing the most effective shielding, while f-orbitals are very diffuse and provide the least.

Question 19. Which of the following orders of ionic radii is correct?
(a) H > H+ > H
(b) Na+ > F¯ > O2-
(c) F > O²- > Na+
(d) None of these
Answer: (d) None of these
In simple words: All the given orders are incorrect. For ions with the same number of electrons (isoelectronic species), the one with fewer positive charges in its center will be larger. So, the correct order for \(Na^{+}\), \(F^{-}\), and \(O^{2-}\) is \(Na^{+} < F^{-} < O^{2-}\).

🎯 Exam Tip: When comparing ionic radii for isoelectronic species, remember that increased nuclear charge (more protons) leads to a smaller ion because the electrons are pulled closer to the nucleus.

Question 20. The First ionisation potential of Na, Mg and Si are 496, 737 and 786 kJ mol\(^{-1}\) respectively. The ionisation potential of Al will be closer to
(a) 760 kJ mol\(^{-1}\)
(b) 575 kJ mol\(^{-1}\)
(c) 801 kJ mol\(^{-1}\)
(d) 419 kJ mol\(^{-1}\)
Answer: (b) 575 kJ mol\(^{-1}\)
In simple words: Even though aluminium is to the right of magnesium, its first ionization energy is actually lower. This is because it's easier to remove the single electron from aluminium's p-orbital than from magnesium's stable, full s-orbital.

🎯 Exam Tip: Always be aware of exceptions to general periodic trends. For example, Group 13 elements often have a lower first ionization energy than Group 2 elements due to the easier removal of an electron from a p-orbital.

Question 21. Which one of the following is true about metallic character when we move from period and top to bottom in a group?
(a) Decreases in a period and increases along the group
(b) Increases in a period and decreases in a group
(c) Increases both in the period and the group
(d) Decreases both in the period and in the group
Answer: (a) Decreases in a period and increases along the group
In simple words: Metallic character means how easily an element loses electrons. As you move across a row, it becomes harder to lose electrons, but as you move down a column, it becomes easier.

🎯 Exam Tip: Metallic character increases as you move down a group (due to increasing atomic size and shielding) and decreases as you move across a period (due to increasing effective nuclear charge).

Question 22. How does electron affinity change when we move from left to right in a period in the periodic table?
(a) Generally increases
(b) Generally decreases
(c) Remains unchanged
(d) First increases and then decreases
Answer: (a) Generally increases
In simple words: As you move from left to right across a row, atoms become more willing to accept an extra electron, because their nucleus can pull it in more strongly.

🎯 Exam Tip: Electron affinity (or electron gain enthalpy) generally increases in magnitude (becomes more negative) across a period as effective nuclear charge rises and atomic size decreases, making it easier to attract an additional electron.

Question 23. Which of the following pairs of elements exhibit diagonal relationship?
(a) Be and Mg
(b) Li and Mg
(c) Be and B
(d) Be and Al
Answer: (d) Be and Al
In simple words: A diagonal relationship is when an element from the second row behaves similarly to an element diagonally below it in the third row. Beryllium and aluminium are one such example.

🎯 Exam Tip: Common diagonal relationships to remember are Li-Mg, Be-Al, and B-Si. These similarities arise from comparable charge density and polarizing power.

II. Write Brief Answer to the Following Questions:

Question 24. Define modern periodic law.
Answer: The modern periodic law states that the physical and chemical properties of elements are a repeating function of their atomic numbers. This means if you arrange elements by their atomic number, their properties will show a regular pattern. This law is fundamental to understanding the organization of the periodic table.
In simple words: The modern periodic law says that if we put elements in order by their atomic number, their properties will repeat in a regular way.

🎯 Exam Tip: Clearly distinguish the modern periodic law (based on atomic number) from Mendeleev's periodic law (based on atomic weight).

Question 25. What are isoelectronic ions? Give examples.
Answer: Isoelectronic ions are ions that have the exact same number of electrons in their electron shells. Even though they come from different atoms, their electron count is identical. For example, \(Na^{+}\) (sodium ion) has an electron configuration of \(1s^2 2s^2 2p^6\), and \(F^{-}\) (fluoride ion) also has \(1s^2 2s^2 2p^6\). Both of these ions contain 10 electrons, making them isoelectronic.
In simple words: Isoelectronic ions are like cousins who have the same number of toy cars, even if they come from different families. For example, the sodium ion and the fluoride ion both have 10 electrons.

🎯 Exam Tip: When providing examples for isoelectronic species, choose ions that clearly show a different atomic number but the same electron count, often corresponding to a noble gas configuration.

Question 26. What is an effective nuclear charge?
Answer: Effective nuclear charge is the net positive charge from the nucleus that an electron actually experiences. Because inner electrons "shield" or block some of the nucleus's full positive charge, the outer valence electrons feel a slightly weaker pull. It is calculated using the formula: \( Z_{eff} = Z - S \), where \( Z \) is the actual atomic number (number of protons) and \( S \) is the shielding constant, determined by factors like Slater's rules.
In simple words: Effective nuclear charge is how much positive pull an outer electron really feels from the nucleus, after other electrons block some of that pull.

🎯 Exam Tip: Emphasize that effective nuclear charge is always less than the actual nuclear charge due to the screening effect of inner electrons.

Question 27. Is the definition given below for ionization enthalpy is correct? "lonisation enthalpy is defined as the energy required to remove the most loosely bound electron from the valence shell of an atom"
Answer: The given statement, "Ionisation enthalpy is defined as the energy required to remove the most loosely bound electron from the valence shell of an atom," is not fully correct. The complete and precise definition of ionization energy (or enthalpy) is the *minimum* amount of energy required to remove the most loosely bound electron from the valence shell of an *isolated neutral gaseous atom in its ground state*. The conditions "minimum," "isolated," "gaseous," and "ground state" are essential for an accurate definition.
In simple words: No, the first definition is missing some important parts. The correct definition says it's the *smallest* energy needed to take away an electron from a *single, neutral gas atom* that is in its most basic, calm state.

🎯 Exam Tip: For full marks on definitions, always include the crucial conditions: "minimum energy," "isolated," "gaseous," and "ground state."

Question 28. Magnesium loses electrons successively to form \( Mg^{+} \), \( Mg^{2+} \), and \( Mg^{3+} \) ions. Which step will have the highest ionization energy and why?
Answer: The third ionization energy (IE3) will be the highest. Magnesium loses electrons in three steps: \( Mg_{(g)} + IE1 \rightarrow Mg^{+}_{(g)} + e^{-} \), \( Mg^{+}_{(g)} + IE2 \rightarrow Mg^{2+}_{(g)} + e^{-} \), and \( Mg^{2+}_{(g)} + IE3 \rightarrow Mg^{3+}_{(g)} + e^{-} \). With each electron removed, the remaining electrons experience a stronger pull from the nucleus because the positive nuclear charge stays the same while the number of electrons decreases. This increased effective nuclear charge makes it progressively harder to remove each subsequent electron. Therefore, \( IE1 < IE2 < IE3 \), and removing the third electron will require the most energy, as it means breaking into a very stable, filled electron shell (like that of a noble gas).
In simple words: Taking the third electron from magnesium will need the most energy. This is because after losing the first two, the remaining electrons are held much more tightly by the nucleus.

🎯 Exam Tip: Successive ionization energies always increase. The most significant jump occurs when an electron is removed from a stable, noble gas electron configuration.

Question 29. Define electronegativity.
Answer: Electronegativity is defined as an atom's relative ability to attract shared electrons towards itself when it is part of a covalent bond in a molecule. This property helps predict how electrons will be distributed between two bonded atoms.
In simple words: Electronegativity is how strongly an atom pulls on the shared electrons in a chemical bond.

🎯 Exam Tip: Distinguish electronegativity (attraction in a bond) from electron affinity (energy change when an electron is added to an isolated atom).

Question 30. How would you explain the fact that the second ionisation potential is always higher than the first ionisation potential?
Answer: The second ionization potential is always higher than the first ionization potential because after losing one electron, the atom becomes a positively charged ion. This positive ion has the same number of protons pulling on fewer electrons, which increases the effective nuclear charge experienced by the remaining electrons. As a result, the remaining electrons are held much more tightly by the nucleus, making it significantly harder to remove a second electron compared to the first.
In simple words: Taking the second electron is always harder than taking the first. This is because once you remove one electron, the atom becomes positive and holds onto its remaining electrons much more tightly.

🎯 Exam Tip: The increase in effective nuclear charge on remaining electrons is the key explanation for why successive ionization energies are always higher.

Question 31. The energy of an electron in the ground state of the hydrogen atom is \( -2.8 \times 10^{-18} \text{ J} \). Calculate the ionization enthalpy of atomic hydrogen in terms of kJ mol\(^{-1}\).
Answer: To calculate the ionization enthalpy of atomic hydrogen in kJ mol\(^{-1}\), we first recognize that ionization energy is the positive value of the electron's ground state energy. Given the energy of an electron in the ground state of a hydrogen atom is \( -2.8 \times 10^{-18} \text{ J} \).
The ionization energy for one hydrogen atom is \( +2.8 \times 10^{-18} \text{ J} \).
To find the energy per mole, we multiply by Avogadro's number (\( 6.022 \times 10^{23} \text{ mol}^{-1} \)):
\( \text{Ionization energy} = (2.8 \times 10^{-18} \text{ J/atom}) \times (6.022 \times 10^{23} \text{ atoms/mol}) \)
\( = 16.8616 \times 10^{5} \text{ J/mol} \)
\( = 1,686,160 \text{ J/mol} \)
To convert to kJ/mol, divide by 1000:
\( = 1686.16 \text{ kJ/mol} \)
Therefore, the ionization enthalpy is approximately \( 1686 \text{ kJ/mol} \). This energy represents the strength with which the electron is bound to the hydrogen atom.
In simple words: To find the energy needed to take an electron from one mole of hydrogen atoms, we take the electron's energy in a single atom, make it positive, and then multiply it by how many atoms are in a mole. Then we change Joules to kilojoules.

🎯 Exam Tip: Ionization enthalpy is numerically equal to the negative of the electron's energy in its ground state. Always remember to convert from Joules per atom to kJ per mole using Avogadro's number.

Question 32. The electronic configuration of an atom is one of the important factors which affects the value of ionization potential and electron gain enthalpy. Explain.
Answer: The electronic configuration of an atom significantly impacts both its ionization potential and electron gain enthalpy. Atoms with half-filled or completely filled valence electron shells are particularly stable. This stability means that it takes a lot of energy to remove an electron from them (high ionization potential), and it is also difficult for them to accept an additional electron (low or even positive electron gain enthalpy), as adding an electron would disrupt their stable state. For instance, Beryllium (\(1s^2 2s^2\)) has a completely filled s-subshell, and Nitrogen (\(1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1\)) has a half-filled p-subshell. Both of these elements, as well as noble gases like Argon with their full outer shells, are very stable, leading to high ionization energies and nearly zero or positive electron affinities.
In simple words: How electrons are arranged in an atom greatly changes how much energy it takes to pull one off or add one. Atoms with shells that are perfectly full or exactly half-full are very stable and don't like to lose or gain electrons easily.

🎯 Exam Tip: Focus on the stability associated with half-filled and completely filled subshells when explaining the influence of electronic configuration on ionization potential and electron gain enthalpy.

Question 33. In what period and group will an element with Z = 118 will be present?
Answer: The element with atomic number \(Z = 118\) is Oganesson (formerly Ununoctium). It is located in the 7th period and the 18th group of the periodic table. This element is a synthetic transactinide and the second element that has its name after a person (Yuri Oganessian).
In simple words: The element with 118 protons, called Oganesson, can be found in the 7th row and the 18th column of the periodic table.

🎯 Exam Tip: For higher atomic number elements, remember their temporary IUPAC names (e.g., Ununoctium) and their final official names (e.g., Oganesson) along with their positions.

Question 34. Justify that the fifth period of the periodic table should have 18 elements on the basis of quantum numbers.
Answer: The fifth period of the periodic table indeed has 18 elements. This can be justified by considering the quantum numbers and the orbitals available for filling electrons in this period. For the fifth period, the principal quantum number is \(n = 5\). Electrons fill the \(5s\), \(4d\), and \(5p\) orbitals. There is 1 \(5s\) orbital, 5 \(4d\) orbitals, and 3 \(5p\) orbitals, totaling \(1 + 5 + 3 = 9\) orbitals. Since each orbital can hold a maximum of two electrons, the total number of electrons that can be accommodated in these orbitals is \(9 \times 2 = 18\). Thus, the fifth period contains 18 elements, starting from Rubidium (Rb) and ending with Xenon (Xe).
In simple words: The fifth row of the periodic table holds 18 elements. This is because in this row, electrons can fill 9 different "spaces" (orbitals: one 5s, five 4d, and three 5p orbitals). Since each space can hold two electrons, a total of 18 electrons, and thus 18 elements, fit in this period.

🎯 Exam Tip: When justifying the number of elements in a period, identify all the subshells that are filled in that period (e.g., \(ns\), \((n-1)d\), \(np\)) and multiply the total number of orbitals by two.

Question 35. Elements a, b, c and d have the following electronic configurations:
a: \( 1s^2, 2s^2, 2p^6 \)
b: \( 1s^2, 2s^2, 2p^6, 3s^2, 3p^1 \)
c: \( 1s^2, 2s^2, 2p^6, 3s^2, 3p^6 \)
d: \( 1s^2, 2s^2, 2p^1 \)
Which elements among these will belong to the same group of the periodic table?
Answer: Elements that belong to the same group in the periodic table share a similar valence electronic configuration, meaning they have the same number of electrons in their outermost shell.
- Elements 'a' (\(1s^2 2s^2 2p^6\), Neon, \(Z=10\)) and 'c' (\(1s^2 2s^2 2p^6 3s^2 3p^6\), Argon, \(Z=18\)) both have 8 valence electrons. Their outermost shells are completely filled, classifying them as noble gases and placing them in Group 18 of the periodic table.
- Elements 'b' (\(1s^2 2s^2 2p^6 3s^2 3p^1\), Aluminium, \(Z=13\)) and 'd' (\(1s^2 2s^2 2p^1\), Boron, \(Z=5\)) both have 3 valence electrons in their outermost shell. This makes them members of Group 13 of the periodic table.
In simple words: Elements 'a' and 'c' are noble gases because their outer shells are full. Elements 'b' and 'd' are in the boron group, as they both have three electrons in their outermost shell.

🎯 Exam Tip: To identify elements in the same group, compare their valence electronic configurations; identical or analogous configurations in the outermost shell indicate they belong to the same group.

Question 36. Give the general electronic configuration of lanthanides and actinides?
Answer: The general electronic configurations for lanthanides and actinides are specific to their f-block nature:
- For **lanthanides**: The general electronic configuration is \( 4f^{1-14} 5d^{0-1} 6s^2 \). This means they typically have two electrons in the 6s orbital, and the inner 4f subshell is being filled with 1 to 14 electrons, sometimes with one electron in the 5d orbital.
- For **actinides**: The general electronic configuration is \( 5f^{1-14} 6d^{0-1} 7s^2 \). Similar to lanthanides, they have two electrons in the 7s orbital, with the inner 5f subshell filling up from 1 to 14 electrons, and sometimes one electron in the 6d orbital. These elements are known for their unique magnetic and optical properties.
In simple words: Lanthanides usually have electrons in the 4f, 5d, and 6s orbitals. Actinides usually have electrons in the 5f, 6d, and 7s orbitals.

🎯 Exam Tip: Remember the specific principal quantum numbers for the f-orbitals (4f for lanthanides, 5f for actinides) and the \(nd^{0-1} (n+1)s^2\) pattern.

Question 37. Why do halogens act as oxidizing agents?
Answer: Halogens act as strong oxidizing agents because they have a valence electronic configuration of \( ns^2 np^5 \). This means they are only one electron short of achieving a very stable, completely filled noble gas electron configuration (\( ns^2 np^6 \)). Due to this strong desire for one more electron, halogens readily accept an electron from other elements. When an element gains electrons, it is reduced, and in this process, it causes the other element to lose electrons (oxidize), thus acting as an oxidizing agent. They possess a high electron affinity, reflecting this strong tendency to gain electrons.
In simple words: Halogens are good at taking electrons from other substances. They do this because they need just one more electron to fill their outer shell and become very stable, like a noble gas.

🎯 Exam Tip: The desire of halogens to complete their octet by gaining one electron is the primary reason for their strong oxidizing nature and high electron affinity.

Question 38. Mention any two anomalous properties of second-period elements.
Answer: Second-period elements exhibit some unique or "anomalous" properties compared to the other elements in their respective groups, primarily due to their very small size, high electronegativity, and the absence of d-orbitals. Two such properties are:
1. **Lithium's Covalent Character:** In Group 1, lithium (a second-period element) forms compounds that tend to have more covalent characteristics, unlike the other alkali metals in its group, which primarily form ionic compounds.
2. **Beryllium's Covalent Character:** Similarly, in Group 2, beryllium (another second-period element) forms compounds with a greater degree of covalent character, while the rest of the alkaline earth metals in its family mostly form ionic compounds. These differences make them stand out within their groups.
In simple words: Elements in the second row of the periodic table act a bit differently from their family members below them. For example, lithium and beryllium make compounds that share electrons more (covalent) than the other elements in their groups, which usually just give or take electrons (ionic).

🎯 Exam Tip: Anomalous behavior of second-period elements is often attributed to their small size, high electronegativity, high polarizing power, and lack of d-orbitals in their valence shell.

Question 39. Explain the Pauling method for the determination of ionic radius.
Answer: The ionic radius is the distance from the center of an ion's nucleus to the edge of its electron cloud. Pauling's method helps calculate the ionic radius of ions in a crystal where ions have single charges. He assumed that ions in a crystal are perfectly round and touch each other.
So, \( d = r_{c+} + r_{a-} \) (1)
Here, 'd' is the total distance between the centers of the cation \( C^+ \) and anion \( A^- \). \( r_{c+} \) and \( r_{a-} \) are their respective radii.
Pauling also thought that an ion's radius (for noble gas electron arrangement like \( Na^+ \) and \( Cl^- \)) gets smaller when the effective nuclear charge pulling on its outer electrons is stronger. This means:
\( r_{c+} \propto \frac{1}{\left(Z_{e f f}\right)^{C+}} \) (2)
\( r_{a-} \propto \frac{1}{\left(Z_{e f f}\right)^{A-}} \) (3)
\( Z_{eff} \) is the effective nuclear charge, calculated as \( Z_{eff} = Z - S \).
By dividing equation (2) by (3), we get:
\[ \frac{r_{C^+}}{r_{A^-}}=\frac{\left(Z_{e f f}\right)^{A-}}{\left(Z_{e f f}\right)^{C+}} \] Using these equations, we can calculate the ionic radii of cations and anions. Pauling's method is useful for estimating ion sizes even when direct measurement is difficult.
In simple words: Pauling's method finds an ion's size by looking at the distance between ions in a crystal. It assumes ions are spheres and relates their size to how strongly their nucleus pulls on electrons.

🎯 Exam Tip: Remember to clearly state Pauling's assumptions (spherical ions, contact in crystal lattice, inverse proportionality to effective nuclear charge) for full marks.

Question 40. Explain the periodic trend of ionisation potential.
Answer:(a) Ionization energy is the smallest amount of energy needed to take away the electron that is easiest to remove from the outermost electron shell of an atom that is alone and in its most stable energy state.
(b) **Changes across a period:**
As you move from left to right across a period, the ionization energy generally increases. This happens for two main reasons:
• The nuclear charge goes up across a period.
• The atomic size gets smaller across a period.
Because of these, the outer electrons are pulled more strongly by the nucleus, so it takes more energy to remove them. Ionization energy helps us understand how easily an atom can form positive ions.
(c) **Changes down a group:**
As you move from top to bottom down a group, the ionization energy generally decreases. This is due to:
• A steady increase in the atom's size.
• An increase in the shielding effect, where inner electrons block the pull of the nucleus on the outer electrons more effectively.
These changes show that ionization enthalpy is a periodic property.

🎯 Exam Tip: When explaining trends, always mention both the change in nuclear charge and atomic size (and shielding effect for groups) as these are key factors.

Question 41. Explain the diagonal relationship.
Answer: When you move diagonally across the periodic table, elements in the second and third periods show some similarities. Even though this similarity is not as strong as elements in the same group, it is clearly seen in specific pairs of elements, as shown below:
The similarities in properties between these diagonally placed elements are called the 'diagonal relationship'. This relationship helps predict properties of elements in these rows.
In simple words: Diagonal relationship means elements in the second and third rows of the periodic table, when looked at diagonally, have similar properties. Like lithium acts a bit like magnesium.

🎯 Exam Tip: When describing diagonal relationships, always provide examples of the pairs of elements that exhibit this unique similarity.

Question 42. Why the first Ionisation enthalpy of sodium is lower than that of magnesium while its second ionisation enthalpy Is higher than that of magnesium? Explain.
Answer: The first ionization energy of magnesium is higher than that of sodium. This is because magnesium has a stronger pull from its nucleus and is a bit smaller in size compared to sodium.
However, the second ionization energy of sodium is much higher than that of magnesium. After sodium loses its first electron, \( Na^+ \) is formed. This \( Na^+ \) ion has an electron arrangement like a noble gas (neon), which is very stable because its electron shells are completely full. Because of this high stability, it takes a lot of energy to remove a second electron from \( Na^+ \).
On the other hand, after magnesium loses its first electron, \( Mg^+ \) is formed. This \( Mg^+ \) ion still has one electron in its outermost (3s) orbital, so it is not as stable as \( Na^+ \). As a result, it takes less energy to remove a second electron from magnesium than from sodium. This difference shows how important electron configuration is for an atom's energy levels.
In simple words: Sodium loses its first electron easily because it helps it become stable. But once it's stable, removing a second electron is very hard. Magnesium needs more energy to lose its first electron, but less energy to lose its second, as it becomes more stable after losing two electrons.

🎯 Exam Tip: Always relate ionization energy trends to electron configurations (especially stable full or half-full shells) and effective nuclear charge for a complete explanation.

Question 43. By using Pauling's method calculate the ionic radii of \( K^+ \) and \( Cl^- \) ions in the potassium chloride crystal. Given that \( d_{K^+ - Cl^-} = 3.14 \) Å.
Answer: We know that:
\( d = r_{K^+} + r_{Cl^-} = 3.14 \) Å
According to Pauling's method:
\[ \frac{r_{K^+}}{r_{Cl^-}}=\frac{\left(Z_{e f f}\right)^{Cl^-}}{\left(Z_{e f f}\right)^{K^+}} \] First, calculate the effective nuclear charge for \( Cl^- \) and \( K^+ \):
\( (Z_{eff})^{Cl^-} = Z - S = 17 - 10.9 = 6.1 \)
\( (Z_{eff})^{K^+} = Z - S = 19 - 16.8 = 2.2 \)
Now, substitute these values into the ratio:
\[ \frac{r_{K^+}}{r_{Cl^-}}=\frac{6.1}{2.2} = 2.77 \] So, \( r_{K^+} = 2.77 \ r_{Cl^-} \)
Substitute this into the first equation:
\( 2.77 \ r_{Cl^-} + r_{Cl^-} = 3.14 \) Å
\( 3.77 \ r_{Cl^-} = 3.14 \) Å
\( r_{Cl^-} = \frac{3.14}{3.77} = 0.83 \) Å
Now find \( r_{K^+} \):
\( r_{K^+} = (3.14 - 0.83) \) Å \( = 2.31 \) Å
This means the \( K^+ \) ion has an ionic radius of 2.31 Å, and the \( Cl^- \) ion has an ionic radius of 0.83 Å. Pauling's method is useful for estimating ionic radii, which helps us understand crystal structures.
In simple words: Using Pauling's method and the given distance between potassium and chloride ions, we calculate the effective pull from their nuclei. Then, we use these values to find that \( K^+ \) has a radius of 2.31 Å and \( Cl^- \) has a radius of 0.83 Å.

🎯 Exam Tip: Remember to show all steps clearly, including the calculation of effective nuclear charge \( (Z_{eff}) \) for both ions, to get full credit for Pauling's method problems.

Question 44. Explain the following, give appropriate reasons.
(i) Ionisation potential of N is greater than that of O:
Answer: Nitrogen has an electron arrangement of \( 1s^2, 2s^2, 2p^3 \), which means it needs more energy to lose an electron than oxygen does. Nitrogen's half-filled 2p electron shell is very stable. Because of this, it needs higher energy to pull an electron away from its 2p orbital.
On the other hand, taking one 2p electron from oxygen (\( 1s^2, 2s^2, 2p^4 \)) makes its 2p shell half-filled (\( 1s^2, 2s^2, 2p^3 \)), which is a more stable arrangement. This makes it comparatively easier to remove a 2p electron from oxygen. Atoms tend to prefer stable electron arrangements, which affects how much energy they need to lose or gain electrons.
In simple words: Nitrogen is more stable with its half-filled outer shell, so it's harder to remove an electron from it than from oxygen. Oxygen becomes more stable by losing one electron, making it easier.

🎯 Exam Tip: Always refer to the stability of half-filled or completely filled orbitals when comparing ionization energies of adjacent elements in a period.

(ii) First ionisation potential of the C-atom is greater than that of the B atom, whereas the reverse is true is for the second ionisation potential.
Answer: The first ionization energy of carbon is higher than boron because ionization energy generally goes up as you move across a period.
For the first ionization potential:
\( C (1s^2, 2s^2, 2p^2) + IE_1 \rightarrow C^+ (1s^2, 2s^2, 2p^1) \)
\( B (1s^2, 2s^2, 2p^1) + IE_1 \rightarrow B^+ (1s^2, 2s^2) \)
For the second ionization potential:
\( C^+ (1s^2, 2s^2, 2p^1) + IE_2 \rightarrow C^{2+}(1s^2, 2s^2) \)
\( B^+ (1s^2, 2s^2) + IE_2 \rightarrow B^{2+}(1s^2, 2s^1) \)
The \( B^+ \) ion has a full 2s electron shell, which is more stable than the partially filled outer shell of \( C^+ \). Because of this, it takes more energy to remove a second electron from Boron than from carbon. Understanding these energy changes helps us predict how elements will react and form bonds.
In simple words: It's harder to remove the first electron from carbon than from boron. But it's harder to remove the second electron from boron than from carbon, because boron becomes very stable after losing its first electron.

🎯 Exam Tip: Compare electron configurations carefully when explaining exceptions or reversals in ionization energy trends for successive removals of electrons.

(iii) The electron affinity values of Be, Mg, and noble gases are zero, and those of N (0.02 eV) and P (0.80 eV) are very low.
Answer: Beryllium (Be), Magnesium (Mg), and noble gases have very stable electron shells that are completely full. Because of this, adding another electron is not easy and would actually need energy to happen. Adding extra electrons would upset their stable electron arrangement, which is why they have almost no desire to gain electrons (electron affinity is nearly zero).
Similarly, Nitrogen (N) and Phosphorus (P) have stable half-filled electron shells. Adding more electrons would upset their stable electron arrangement, which is why they also have a very low desire to gain electrons. This shows that atoms prefer to keep their stable electron setups, whether they are completely full or half-filled.
In simple words: Elements like Be, Mg, and noble gases don't want to gain electrons because their outer shells are already full and stable. Nitrogen and Phosphorus also don't want to gain electrons because their outer shells are already half-filled and stable.

🎯 Exam Tip: When explaining electron affinity values, always link them to the stability of electron configurations, especially completely filled or half-filled orbitals.

(iv) The formation of \( F^- (g) \) from \( F(g) \) is exothermic while that of \( O^{2-}(g) \) from \( O(g) \) is endothermic.
Answer: Oxygen and fluorine atoms are quite small and have many electrons packed closely together. When an extra electron is added to fluorine, it has to fit into a small space (the 2p orbital), but it results in a stable full shell, so energy is released. Hence, the formation of \( F^- \) from F is exothermic (releases energy).
In the case of oxygen, the formation of the \( O^{2-} \) ion from an oxygen atom needs energy (it's endothermic). This is because the first electron added to oxygen releases energy (exothermic), but the second electron added to form \( O^{2-} \) experiences strong repulsion from the already existing negative charge on \( O^- \). To overcome this repulsion and add the second electron, energy must be put in. However, the resulting \( O^{2-} \) ion gets a very stable, completely full 2p electron shell. The energy change when an atom gains an electron depends a lot on its size and how stable its new electron arrangement will be.
In simple words: When fluorine gains one electron, it releases energy because it becomes stable. But when oxygen tries to gain two electrons, adding the second one needs energy because of repulsion, even though the final ion is stable.

🎯 Exam Tip: Remember that while the first electron affinity is often exothermic, the second electron affinity is almost always endothermic due to electron-electron repulsion.

Question 45. What is the screening effect?
Answer: The screening effect, also called the shielding effect, happens when inner electrons push away (repulsion) outer (valence) electrons. This pushing away reduces how strongly the nucleus pulls on the outer electrons. So, the inner electrons work like a shield, blocking some of the nucleus's positive charge from reaching the outer electrons. This effect helps explain why elements behave differently even when they are in the same period or group.
In simple words: The screening effect is when inner electrons block the nucleus's pull on the outer electrons, making the outer electrons feel less attraction.

🎯 Exam Tip: The screening effect directly impacts an atom's effective nuclear charge, which in turn influences properties like atomic radius and ionization energy.

Question 46. Briefly give the basis for Pauling's scale of electronegativity.
Answer: Pauling's scale:
• Electronegativity is how much an atom pulls shared electrons towards itself when it's in a chemical bond.
• Pauling gave specific values for the electronegativity of hydrogen (2.2) and fluorine (4).
• Based on these values, we can find the electronegativity for other elements using this formula:
\( (X_A-X_B) = 0.182 \sqrt{E_{AB} - (E_{AA} E_{BB})} \)
Here, \( E_{AB}, E_{AA} \), and \( E_{BB} \) are the bond dissociation energies of the \( AB, A_2 \), and \( B_2 \) molecules, respectively. Pauling's scale is widely used because it provides a simple way to compare how strongly different atoms attract electrons in a bond.
In simple words: Pauling created a way to measure how strongly an atom attracts electrons in a bond. He used a formula based on bond energies and set values for hydrogen and fluorine to start.

🎯 Exam Tip: Understand that Pauling's scale is based on bond energies and provides a relative measure of electronegativity, not an absolute one.

Question 47. State the trends in the variation of electronegativity in groups and periods.
Answer: **Changes of Electronegativity in a period:**
As you move from left to right across a period, electronegativity generally increases. This is because atoms get smaller, and the nucleus pulls outer electrons more strongly. So, the ability to pull shared electrons closer gets stronger, and electronegativity increases.
**Changes of Electronegativity in a group:**
As you move down a group, electronegativity generally decreases. This happens because atoms get larger, and the nucleus pulls outer electrons less strongly due to increased shielding by inner electrons. Therefore, electronegativity decreases. Noble gases are considered to have zero electronegativity. Most p-block elements follow this pattern of decreasing electronegativity when moving down a group, except for groups 13 and 14. These trends are key to predicting how elements will bond with each other in chemical reactions.
In simple words: Electronegativity increases as you go across a period (left to right) because atoms get smaller and pull electrons harder. It decreases as you go down a group (top to bottom) because atoms get bigger, and the nucleus's pull is shielded by more inner electrons.

🎯 Exam Tip: Always explain electronegativity trends by referencing atomic size and effective nuclear charge (or nuclear attraction), as these are the primary influencing factors.

I. Very Short Question and Answers (2 Marks):

Question 1. State periodic law.
Answer: The Modern Periodic Law says that "The physical and chemical properties of the elements are the periodic functions of their atomic numbers." This law helped scientists organize elements in the periodic table based on their atomic number.
In simple words: The Modern Periodic Law states that element properties repeat in a pattern based on their atomic number.

🎯 Exam Tip: When stating a law, make sure to quote it accurately and briefly explain its significance.

Question 2. Write a note about Chancourtois classification.
Answer: In Chancourtois's system, elements whose atomic weights differed by 16, or multiples of 16, appeared almost on the same vertical line. Elements arranged directly under each other showed clear similarities. This was one of the very first periodic laws. Chancourtois's system was an early attempt to find patterns among elements, paving the way for the modern periodic table.
In simple words: Chancourtois arranged elements by atomic weight in a spiral. Elements with atomic weights differing by 16 (or multiples) lined up vertically and showed similar traits.

🎯 Exam Tip: Focus on the key idea of Chancourtois's classification: elements with atomic weight differences of 16 falling on vertical lines, showing similarities.

Question 3. What is Lavoiser's classification of elements?
Answer: Lavoisier sorted substances into four main types of elements: acid-making elements, gas-like elements, metallic elements, and earthy elements. This simple grouping was one of the first ways to organize chemical elements.
In simple words: Lavoisier classified elements into four simple groups: acid-makers, gases, metals, and earth-like elements.

🎯 Exam Tip: Remember the four broad categories Lavoisier used, as his classification was a foundational step in chemistry.

Question 4. State Mendeleev's periodic law.
Answer: Mendeleev's periodic law says that "The physical and chemical properties of elements are a periodic function of their atomic weights." Mendeleev's law was a major step in chemistry because it helped predict undiscovered elements.
In simple words: Mendeleev's law states that the properties of elements repeat in a pattern based on their atomic weights.

🎯 Exam Tip: Be precise with the wording of Mendeleev's law, specifically mentioning "atomic weights" as opposed to "atomic numbers" in the modern law.

Question 5. What are groups and periods?
Answer: In the modern periodic table, all elements are arranged in 18 upright columns and 7 flat rows. The vertical columns are called groups, and the horizontal rows are called periods. This arrangement helps us understand how elements are related and why they have similar properties.
In simple words: Groups are the vertical columns in the periodic table, and periods are the horizontal rows.

🎯 Exam Tip: Clearly distinguish between groups (vertical, similar properties) and periods (horizontal, gradual property change) in your definition.

Question 6. State modern periodic law.
Answer: The Modern Periodic Law says that, “The physical and chemical properties of the elements are periodic function of their atomic numbers." This law is fundamental to how we understand and organize all known elements today.
In simple words: The Modern Periodic Law states that element properties repeat in a pattern based on their atomic numbers.

🎯 Exam Tip: Emphasize "atomic numbers" as the basis for the modern periodic law, distinguishing it from Mendeleev's law based on atomic weights.

Question 7. What are p-block elements? Give their general electronic configuration.
Answer: The elements in groups 13 through 18 are called p-block elements or main group elements. Their general electronic configuration is \( ns^2, np^{1-6} \). P-block elements show a wide range of properties, from metals to non-metals.
In simple words: P-block elements are in groups 13-18 of the periodic table, and their outermost electrons fill the p-orbitals.

🎯 Exam Tip: Remember that the "p" in p-block refers to the p-orbital being filled by the valence electrons, and the groups involved are 13 to 18.

Question 8. Mention the names of the elements with atomic numbers 101, 102, 109, and 110.
Answer:
Z = 101 IUPAC name: Mendelevium
Z = 102 IUPAC name: Nobelium
Z = 109 IUPAC name: Meitnerium
Z = 110 IUPAC name: Darmstadtium
These elements are synthetic, meaning they are created in laboratories and are not found naturally.
In simple words: The elements are Mendelevium (101), Nobelium (102), Meitnerium (109), and Darmstadtium (110).

🎯 Exam Tip: When asked for IUPAC names of superheavy elements, ensure correct spelling and atomic number association.

Question 9. What are f-block elements? Give their properties.
Answer: The lanthanides and the actinides are called f-block elements. These elements are metals and melt at very high temperatures. Their compounds often form colored compounds, and these elements can have different positive charges when they react (variable oxidation states). These f-block elements are often placed in separate rows at the bottom of the periodic table for clarity.
In simple words: F-block elements are lanthanides and actinides. They are metals with high melting points, often form colorful compounds, and can have different charges.

🎯 Exam Tip: Recall that f-block elements are characterized by the filling of f-orbitals and are known for their metallic nature and variable oxidation states.

Question 10. Give the name and electronic configuration of elements of group 1 and 2 groups.
Answer:
• Elements of Group 1 are called alkali metals. Their electronic configuration is \( ns^1 \).
• Elements of Group 2 are called alkaline earth metals. Their electronic configuration is \( ns^2 \).
Both alkali metals and alkaline earth metals are very reactive because they easily lose their outermost electrons.
In simple words: Group 1 elements are alkali metals (\( ns^1 \)), and Group 2 elements are alkaline earth metals (\( ns^2 \)).

🎯 Exam Tip: Be sure to include both the common name of the group and its general valence electronic configuration.

Question 11. Define Atomic radius.
Answer: The atomic radius of an atom is defined as the distance from the center of its nucleus to its outermost electron shell. Atomic radius helps determine how tightly an atom holds onto its electrons and how it might bond with other atoms.
In simple words: Atomic radius is the distance from the nucleus to the outermost electron shell of an atom.

🎯 Exam Tip: A precise definition of atomic radius involves the distance from the nucleus to the valence electrons.

Question 12. Write any two characteristic properties of alkaline earth metals.
Answer:
• Alkaline earth metals easily give up their two outermost electrons to form ions with a +2 charge.
• As you go down the group, they become more like metals and more reactive.
These properties make them useful in various applications, such as magnesium in lightweight alloys.
In simple words: Alkaline earth metals easily form +2 ions by losing electrons. Their metallic character and reactivity increase as you move down their group.

🎯 Exam Tip: Focus on their tendency to form +2 ions and the trend in reactivity/metallic character down the group.

Question 13. Why is the covalent radius shorter than the actual atomic radius?
Answer: When a covalent bond forms, atomic orbitals (electron clouds) share space and overlap. This overlapping makes the distance between the two nuclei shorter than expected. Therefore, the covalent radius is always shorter than the actual atomic radius. This shortening happens because the shared electrons pull the nuclei closer together.
In simple words: Covalent radius is shorter than atomic radius because atoms overlap and share electrons when forming a bond, pulling their nuclei closer together.

🎯 Exam Tip: The key reason for the shorter covalent radius is the overlap of electron clouds during bond formation, leading to a decreased internuclear distance.

Question 14. Define metallic radius.
Answer: Metallic radius is defined as half the distance between the centers of two metal atoms that are touching each other in a solid metal crystal. This measurement helps chemists understand how metal atoms are arranged and interact within a solid.
In simple words: Metallic radius is half the distance between two touching metal atoms in a solid metal.

🎯 Exam Tip: Specify "adjacent" or "touching" atoms and "metallic crystal lattice" in the definition for accuracy.

Question 15. Halogens and chalcogens have highly negative electron gain enthalpies. Why?
Answer:
• Group 16 elements (chalcogens) and Group 17 elements (halogens) strongly want to gain one or two electrons respectively to get a stable electron arrangement like noble gases.
• Because they strongly want to do this, these elements release a lot of energy when they gain electrons, which is shown by their highly negative electron gain enthalpies.
This strong desire to gain electrons makes them very reactive non-metals.
In simple words: Halogens and chalcogens have very negative electron gain enthalpies because they eagerly gain one or two electrons to achieve a stable electron configuration, releasing much energy.

🎯 Exam Tip: The main reason for high negative electron gain enthalpy is the strong drive to achieve a stable noble gas configuration by gaining electrons.

Question 16. What is the effective nuclear charge? How is it approximated?
Answer: The effective nuclear charge is the overall positive charge that outer electrons feel from the nucleus. It is estimated by this formula: \( Z_{eff} = Z - S \).
Here, Z is the atomic number (total number of protons in the nucleus), and S is the screening constant, which is found using special rules called Slater's rules. Effective nuclear charge helps predict an atom's size and how strongly it holds onto its electrons.
In simple words: Effective nuclear charge is the net positive pull the outermost electrons feel from the nucleus. It's estimated by subtracting the shielding effect (S) from the atomic number (Z).

🎯 Exam Tip: Remember the formula \( Z_{eff} = Z - S \) and that S represents the shielding or screening constant due to inner electrons.

Question 17. Elements Zn, Cd, and Hg with electronic configuration \( (n-1)d^{10} ns^2 \) do not show most of the transition elements properties. Give reason.
Answer:
• Zinc (Zn), Cadmium (Cd), and Mercury (Hg) have full d-orbitals, meaning their \( d^{10} \) electron shell is complete.
• They do not have partially filled d-orbitals, unlike most other transition metals. Because of this, they do not act like typical transition elements and do not show many of their characteristic properties.
This stable electron configuration makes them behave more like main group elements than true transition metals.
In simple words: Zn, Cd, and Hg don't act like typical transition elements because their d-orbitals are completely full, making them very stable and less reactive.

🎯 Exam Tip: The defining characteristic of transition elements is having partially filled d-orbitals, so elements with full d-orbitals like Zn, Cd, Hg are often considered pseudo-transition elements.

Question 18. Define Ionisation energy.
Answer: Ionization energy is defined as the smallest amount of energy needed to take away the electron that is easiest to remove from the outermost electron shell of an atom that is alone and in its most stable energy state (ground state). This energy value helps chemists understand how easily an atom can form a positive ion.
In simple words: Ionization energy is the energy required to remove the outermost electron from a neutral, gaseous atom.

🎯 Exam Tip: Include "isolated neutral gaseous atom in its ground state" for a complete and accurate definition.

Question 19. Why d-block elements are called transition elements?
Answer: D-block elements connect the very reactive metals in the s-block with the less reactive elements in groups 13 and 14. They show varying properties between these two extremes. This is why they are called transition elements. Their ability to form this "bridge" also explains their varied chemical properties, like forming colored compounds.
In simple words: D-block elements are called transition elements because their properties gradually change, acting as a bridge between the highly reactive s-block metals and the less reactive p-block elements.

🎯 Exam Tip: The "bridge" concept, linking s-block and p-block elements in terms of properties, is key to explaining why d-block elements are called transition elements.

Question 20. Beryllium has higher ionization energy than Boron. Give reason.
Answer: Beryllium has a full and stable 2s electron orbital (\( 1s^2 2s^2 \)). This arrangement is more stable than Boron's. Boron, however, has a partially filled outer electron shell (\( 1s^2 2s^2 2p^1 \)), which is less stable. Because of Beryllium's more stable electron configuration, it takes more energy to remove an electron from Beryllium than from Boron. This highlights how the stability of electron configurations strongly influences an element's ionization energy.
In simple words: Beryllium has higher ionization energy than boron because beryllium's outer 2s orbital is completely full and stable, making it harder to remove an electron.

🎯 Exam Tip: Always compare the stability of the electron configurations (specifically full or half-full orbitals) of the elements when explaining exceptions to ionization energy trends.

Question 21. Write the electronic configuration of lanthanides and actinides?
Answer:
• The electronic configuration of lanthanides is \( 4f^{1-14} 5d^{0-1} 6s^2 \).
• The electronic configuration of actinides is \( 5f^{1-14} 6d^{0-1} 7s^2 \).
These elements are characterized by the filling of f-orbitals, which gives them unique magnetic and optical properties.
In simple words: Lanthanides have electron configuration \( 4f^{1-14} 5d^{0-1} 6s^2 \), and actinides have \( 5f^{1-14} 6d^{0-1} 7s^2 \).

🎯 Exam Tip: Remember the principal quantum numbers (4f for lanthanides, 5f for actinides) and the general structure for their f, d, and s orbitals.

Question 22. What is the effect of shielding on ionization energy?
Answer: As we move down a group, there are more inner electrons, and these electrons push away the outer (valence) electrons more strongly. This increased shielding effect reduces how much the nucleus pulls on the outer electron. Because of this, it takes less energy to remove an electron, so the ionization energy decreases. This shielding effect is why larger atoms in a group lose electrons more easily than smaller atoms.
In simple words: Shielding by inner electrons reduces the nuclear pull on outer electrons, making it easier to remove them. This means ionization energy decreases with increased shielding.

🎯 Exam Tip: Emphasize that increased shielding leads to a decrease in effective nuclear charge, which in turn lowers the ionization energy.

Question 23. Define electron affinity.
Answer: Electron affinity is defined as the amount of energy an atom gives off when it gains an electron and becomes a negative ion. This happens when an electron is added to the outermost electron shell of a single, uncharged gas atom in its lowest energy state. A high electron affinity means an atom easily accepts an extra electron, often leading to stable negative ions.
In simple words: Electron affinity is the energy released when a neutral atom gains an electron to form a negative ion.

🎯 Exam Tip: Note that electron affinity is typically an exothermic process (energy released), but exceptions exist, especially for elements with stable configurations or for adding a second electron.

Question 24. What are periodic properties? Give example.
Answer: The idea of periodicity means that elements show similar properties again and again at regular points as their atomic number increases in the periodic table.
Example:
• Atomic radii
• Ionization energy
• Electron affinity
• Electronegativity.
This repeating pattern of properties is why the periodic table is so useful for organizing elements.
In simple words: Periodic properties are characteristics of elements that show a repeating pattern as you move through the periodic table, like atomic size or how easily electrons are gained.

🎯 Exam Tip: When defining periodic properties, highlight the "regular intervals" and "atomic number" aspects, and provide a few common examples from the periodic table.

Question 25. Why is the electron affinity of Nitrogen almost zero?
Answer: Nitrogen has a stable electron arrangement with a half-filled 2p orbital (\( 1s^2, 2s^2, 2p^3 \)). Adding more electrons would upset this stable arrangement because the new electron would have to go into an already occupied 2p orbital, increasing electron-electron repulsion. This is why nitrogen has almost no desire to gain electrons (its electron affinity is almost zero). This stability makes nitrogen less likely to form negative ions by gaining electrons.
In simple words: Nitrogen has a stable half-filled electron shell, so it doesn't easily accept more electrons, which makes its electron affinity almost zero.

🎯 Exam Tip: The stability of the half-filled p-orbital in nitrogen is the key reason for its very low electron affinity. Avoid stating it is exactly zero, as it is very low but not precisely zero.

Question 26. The cationic radius is smaller than its corresponding neutral atom. Justify this statement.
Answer:
• When a neutral atom loses one or more electrons, it becomes a positive ion (cation). For example: \( Na \rightarrow Na^+ + e^- \)
• The size of this positive ion (\( r_{Na^+} \)) is smaller than its original atom (\( r_{Na} \)).
• When an atom becomes a positive ion, it still has the same number of protons, but now fewer electrons. This means the nucleus pulls the fewer remaining electrons in more tightly. This increased pull makes the electron cloud shrink. Hence, the positive ion's radius is smaller than its corresponding neutral atom. This size reduction makes cations behave differently in chemical reactions compared to their neutral atoms.
In simple words: A positive ion (cation) is smaller than its original atom. This is because it has lost electrons, so the nucleus pulls the remaining electrons closer, making the ion shrink.

🎯 Exam Tip: The two main reasons for a smaller cationic radius are the loss of the outermost shell and the increased effective nuclear charge on the remaining electrons.

Question 27. Define electronegativity.
Answer: Electronegativity is defined as how strongly an atom pulls shared electrons towards itself when it is part of a chemical bond (in a covalently bonded molecule). This property is crucial for understanding the polarity of chemical bonds.
In simple words: Electronegativity is an atom's ability to attract shared electrons in a bond.

🎯 Exam Tip: Emphasize "shared pair of electrons" and "covalently bonded molecule" to differentiate electronegativity from electron affinity.

Question 28. What are isoelectronic ions? Give example.
Answer: Isoelectronic ions are ions from different elements that have the exact same number of electrons. These ions often show similar properties because their electron configurations are identical.
Example: \( Na^+, Mg^{2+}, Al^{3+}, F^-, O^{2-}, N^{3-} \) (all have 10 electrons, like Neon)
In simple words: Isoelectronic ions are different ions that have the same total number of electrons. For example, \( Na^+ \) and \( F^- \) both have 10 electrons.

🎯 Exam Tip: Provide a clear example, showing multiple ions (cation and anion) that share the same electron count, usually matching a noble gas configuration.

Question 29. What is the variation of electronegativity in a group?
Answer: As we move down a group, the electronegativity generally decreases. This happens because the atomic radius increases, causing the nucleus's attractive force on the valence electrons to become weaker. This makes it less likely for an atom to attract shared electrons. Electronegativity is an important property that helps us understand how atoms bond with each other.
In simple words: Moving down a group, atoms get bigger, and the pull of the nucleus on outer electrons gets weaker. So, atoms become less able to attract electrons, and electronegativity goes down.

🎯 Exam Tip: Remember that increased atomic size and increased shielding effect are the main reasons for the decrease in electronegativity down a group.

Question 30. The ionization energy of beryllium is greater than the ionization energy of boron. Why?
Answer: Beryllium (Be, atomic number 4) has an electron configuration of \( 1s^2 2s^2 \), meaning its outermost 2s orbital is completely filled. This full subshell makes beryllium very stable. To remove an electron from this stable configuration, a lot of energy is needed, which results in a high first ionization energy. Boron (B, atomic number 5) has an electron configuration of \( 1s^2 2s^2 2p^1 \). Its outermost 2p orbital only has one electron. Removing this single 2p electron is easier than removing an electron from beryllium's stable, filled 2s orbital, even though boron has more protons. This makes boron's first ionization energy lower than beryllium's.
In simple words: Beryllium is very stable because its outer electron shell is completely full. Boron has one electron in a less stable outer shell, so it is easier to take that electron away. This means beryllium needs more energy to lose an electron than boron does.

🎯 Exam Tip: When comparing ionization energies, always check the electron configurations for filled or half-filled subshells, as these stable arrangements require extra energy to break.

Question 31. What is the diagonal relationship?
Answer: The diagonal relationship describes how elements in the second period sometimes show similar properties to elements diagonally placed below them in the third period. For example, lithium (Li) resembles magnesium (Mg), and beryllium (Be) resembles aluminum (Al). This similarity is not as strong as within a group but is still noticeable. This happens because crossing a period and moving down a group can balance out changes in atomic size and electronegativity. This unique relationship helps explain certain chemical behaviors.

In simple words: Some elements in the periodic table, like lithium and magnesium, share similar traits even though they are not in the same group, but are diagonally placed. This is called a diagonal relationship.

🎯 Exam Tip: Draw a simple diagonal relationship diagram to illustrate the concept. Common examples include Li-Mg and Be-Al.

Question 32. Define electron gain enthalpy or electron affinity. Give its unit.
Answer: Electron gain enthalpy, also known as electron affinity, is the amount of energy released when an electron is added to the valence shell of an isolated neutral gaseous atom. This process forms an anion (a negatively charged ion). For most elements, energy is released during this process, making the enthalpy change negative. This energy is a measure of an atom's ability to accept an electron. The unit for electron affinity is usually kilojoules per mole (\( \text{kJ mol}^{-1} \)) or electron volts (eV).
In simple words: Electron affinity is the energy given out when an atom takes in an extra electron to become a negative ion. Its unit is typically kJ per mole.

🎯 Exam Tip: Remember to specify "isolated neutral gaseous atom" in the definition for full accuracy, as the state of the atom affects the energy change.

II. Short Question and Answers (3 Marks):

Question 1. How Moseley determined the atomic number of an element using X-rays?
Answer: Henry Moseley studied the X-ray spectra of various elements by bombarding them with high-energy electrons. He found a direct relationship between the atomic number (Z) of an element and the frequency (\( \nu \)) of the X-rays it emitted. This relationship is given by the equation \( \sqrt{\nu} = a(Z - b) \), where 'a' and 'b' are constants. By measuring the frequency of the emitted X-rays, Moseley could accurately determine the atomic number of an unknown element. This work led to the modern periodic law, which arranges elements by atomic number rather than atomic weight.
In simple words: Moseley used X-rays to hit different elements. He found that the X-ray frequency was directly linked to the element's atomic number. This helped him find the atomic number of any element using a simple formula.

🎯 Exam Tip: Highlight Moseley's key formula \( \sqrt{\nu} = a(Z - b) \) and the significance of using atomic number as the basis for the periodic table.

Question 2. Write notes on Newlands classification of elements.
Answer: John Newlands proposed a classification system in 1865 called the Law of Octaves. He arranged the known elements in increasing order of their atomic weights. Newlands observed that the properties of every eighth element were similar to those of the first element, much like the notes in a musical octave. For example, sodium (Na) showed similar properties to lithium (Li), both being the eighth element after each other. This law worked well for lighter elements, up to calcium, but failed for heavier elements and newly discovered elements. It was an early attempt to find a pattern among elements. The concept of octaves was later recognized as a precursor to modern periodicity.
In simple words: Newlands arranged elements by weight and noticed every eighth element had similar properties, like music notes. This "Law of Octaves" worked for light elements but not for heavier ones.

🎯 Exam Tip: Focus on the "Law of Octaves," its basis on atomic weight, its success with lighter elements (up to calcium), and its limitations with heavier or newly discovered elements.

Question 3. Describe Mendeleev's periodic classification of elements.
Answer: Dmitri Mendeleev proposed the periodic law, stating that "The physical and chemical properties of the elements are the periodic functions of their atomic weights." In 1869, he created the first periodic table by arranging 70 known elements in vertical columns (groups) and horizontal rows (periods) based on increasing atomic weights. Mendeleev's table was revolutionary because he left blank spaces for undiscovered elements, predicting their properties. For example, he predicted the existence and properties of elements he called eka-aluminium (later gallium) and eka-silicon (later germanium), which were later found to match his predictions. This bold prediction helped gain acceptance for his periodic table. Mendeleev's table provided a logical framework for understanding elemental relationships.
In simple words: Mendeleev made the first periodic table, arranging elements by their atomic weights. He famously left gaps for new elements and correctly predicted their properties, proving his system was very clever.

🎯 Exam Tip: Emphasize Mendeleev's correct prediction of undiscovered elements (like Gallium and Germanium) as a key strength of his classification system.

Question 4. Write notes on Moseley's work.
Answer: Henry Moseley conducted significant work on the characteristic X-ray spectra of elements. By bombarding elements with high-energy electrons, he observed the X-rays they emitted. His experiments showed a linear relationship between the square root of the frequency (\( \sqrt{\nu} \)) of these X-rays and the atomic number (Z) of the element. This relationship is expressed by the equation \( \sqrt{\nu} = a(Z - b) \), where 'a' and 'b' are constants. Moseley's work provided a fundamental basis for arranging elements by atomic number rather than atomic weight, which led to the modern periodic law. This corrected the anomalies in Mendeleev's periodic table and proved that atomic number is a more fundamental property than atomic weight, reflecting the number of protons in an atom's nucleus.
In simple words: Moseley used X-rays to study elements and found a clear link between the X-ray frequency and the element's atomic number. This showed that atomic number, not atomic weight, is the best way to arrange elements in the periodic table.

🎯 Exam Tip: Clearly state the formula \( \sqrt{\nu} = a(Z - b) \) and its implication that atomic number (Z) is the primary factor determining an element's chemical properties, leading to the modern periodic law.

Question 5. What are the reasons behind Moseley's attempt in finding an atomic number?
Answer: Moseley focused on finding the atomic number for several key reasons. First, he observed that as the atomic number increases, the number of electrons also increases by the same amount. Second, changes in the atomic number directly affect the electronic structure of the atom. Third, it is the electrons in the outermost shell (valence electrons) that primarily determine an element's chemical properties. Mendeleev's periodic table, based on atomic weight, had some inconsistencies (e.g., tellurium and iodine). Moseley's work aimed to resolve these by identifying a more fundamental property, which he found to be the atomic number. This improved understanding helped correctly place elements in the periodic table, providing a more logical and consistent arrangement.
In simple words: Moseley wanted to find the atomic number because it matches the number of electrons, changes how electrons are arranged, and decides how elements react. This helped fix problems in the old periodic table.

🎯 Exam Tip: Emphasize that atomic number is more fundamental than atomic weight, as it directly relates to electron configuration and chemical behavior.

Question 6. What are s-block elements? Give their properties.
Answer: S-block elements are those in which the last electron enters the outermost s-orbital. These elements are found in Group 1 (alkali metals) and Group 2 (alkaline earth metals) of the periodic table. They typically have a general electronic configuration of \( ns^1 \) for Group 1 and \( ns^2 \) for Group 2.
Some properties of s-block elements include:
• They are soft metals with low melting and boiling points, and low ionization enthalpies.
• They are highly reactive and readily form ionic compounds by losing electrons.
• They are highly electropositive and tend to lose electrons easily.
• Many s-block elements and their salts impart characteristic colors to a flame test, like sodium's orange or potassium's lilac. This is due to the excitation of their electrons and subsequent emission of light. These properties are a direct result of their electron configurations.
In simple words: S-block elements are found in the first two groups. They are soft, very reactive metals that easily lose electrons to form ionic bonds, and they often give colors when heated in a flame.

🎯 Exam Tip: Remember the general electronic configurations (\( ns^1 \) and \( ns^2 \)) and the key characteristics: soft, reactive, electropositive, and form ionic compounds.

Question 7. Explain the classification of elements based on electronic configuration.
Answer: Elements are classified into blocks based on which orbital the last electron enters. This distribution of electrons into s, p, d, and f orbitals forms the basis of their electronic configuration.
• The main energy level, defined by the principal quantum number (n), determines the shell an electron occupies.
• An element's position in the periodic table is directly linked to its electronic configuration, specifically the quantum numbers of its last filled orbital.
• Elements are grouped into periods (horizontal rows, based on the outermost shell number) and groups (vertical columns, based on similar valence electron configurations).
• There are seven periods and 18 groups in the modern periodic table. For example, elements in Group 1 have a similar \( s^1 \) valence configuration, leading to similar properties. This systematic classification helps predict and explain chemical behavior.
In simple words: Elements are sorted by where their last electron goes (s, p, d, or f orbitals). This electron setup decides their place in the periodic table, whether in rows (periods) or columns (groups), because elements in the same group have similar outer electron arrangements.

🎯 Exam Tip: Emphasize that the "block" (s, p, d, f) corresponds to the orbital being filled, which is fundamental to predicting an element's properties and position.

Question 8. What are d-block elements? Give their properties.
Answer: D-block elements, also known as transition elements, are found in Groups 3 to 12 of the periodic table. In these elements, the last electron enters a d-orbital of the penultimate (n-1) shell. Their general valence shell electronic configuration is \( ns^{1-2} (n-1)d^{1-10} \). These elements are metallic in nature and serve as a bridge between the highly reactive s-block metals and the less active p-block elements.
Some key properties of d-block elements include:
• They show more than one oxidation state, meaning they can have different charges in compounds.
• They often form colorful ionic compounds.
• Many d-block elements act as catalysts, speeding up chemical reactions.
• They can form interstitial compounds (where small atoms fit into the gaps in their crystal lattice) and alloys (mixtures of metals).
• They are good conductors of heat and electricity. For instance, iron is a d-block element known for its multiple oxidation states and catalytic properties. These unique properties make them crucial in various industrial applications.
In simple words: D-block elements are transition metals in the middle of the periodic table. They have many uses and properties like forming colorful compounds, having different charges, and being good at speeding up reactions.

🎯 Exam Tip: Highlight their key characteristics: multiple oxidation states, formation of colored compounds, and catalytic properties, which differentiate them from s- and p-block elements.

Question 9. What is a covalent radius? Explain.
Answer: Covalent radius is defined as one-half of the internuclear distance between two identical atoms that are linked by a single covalent bond. This distance can be determined using advanced techniques like X-ray diffraction studies. For example, in a chlorine (\( \text{Cl}_2 \)) molecule, the internuclear distance is \( 1.98 \) Å. So, the covalent radius of a single chlorine atom is half of this, which is \( \frac{1.98}{2} = 0.99 \) Å. When atoms form a covalent bond, their orbitals overlap, which means the atoms get closer than they would be in their isolated state. This overlapping causes the covalent radius to be shorter than the actual atomic radius. This concept helps scientists understand the size of atoms in molecules and predict bond lengths.
In simple words: A covalent radius is half the distance between the centers of two identical atoms that are joined by a single chemical bond. It's usually shorter because the atoms share and overlap their electron clouds.

🎯 Exam Tip: Remember that covalent radius is specifically for *covalently bonded* atoms and is generally *shorter* than the atomic radius due to orbital overlap.

Question 10. How is the covalent radius of an individual atom can be calculated?
Answer: The covalent radius of an individual atom can be calculated when it is bonded to a different atom, say A and B, using their internuclear distance \( (d_{A-B}) \). A simple method proposed by Schomaker and Stevenson helps with this. Their formula is:
\( d_{A-B} = r_A + r_B - 0.09(X_A - X_B) \)
In this formula, \( r_A \) and \( r_B \) are the covalent radii of atoms A and B, respectively, and \( X_A \) and \( X_B \) are their electronegativities (in Pauling units). If \( X_A \) is greater than \( X_B \), the difference in electronegativity leads to a correction factor. This equation accounts for the extra shortening of bond length when atoms with different electronegativities form a bond, pulling the electron pair closer to the more electronegative atom. The radius is usually given in Angstroms (Å). This formula is especially useful for heteronuclear diatomic molecules.
In simple words: You can find an atom's covalent radius by using a special formula that includes the distance between two different atoms it's bonded to and how strongly each atom pulls electrons. This helps adjust for how different atoms share electrons.

🎯 Exam Tip: Learn the Schomaker-Stevenson formula and understand that the electronegativity difference is crucial for calculating covalent radii between dissimilar atoms.

Question 11. Write about the electronic configuration of the 1st and 2nd periods.
Answer:
**1st Period Electronic Configuration:** The first period contains only two elements: hydrogen and helium. This period starts by filling electrons into the first energy level (\( n=1 \)). The \( n=1 \) shell has only one orbital, the 1s orbital, which can hold a maximum of two electrons. Therefore, the 1st period has only two elements.
**2nd Period Electronic Configuration:** The second period contains eight elements, starting from lithium and ending with neon. This period begins with the filling of electrons into the second energy level (\( n=2 \)). The \( n=2 \) shell has four orbitals in total: one 2s orbital and three 2p orbitals. These four orbitals can accommodate a maximum of \( (1 \times 2) + (3 \times 2) = 2 + 6 = 8 \) electrons. Thus, the 2nd period has eight elements. The systematic filling explains the periodic properties.
In simple words: The first row of the periodic table has only two elements because its first energy level can hold just two electrons. The second row has eight elements because its second energy level can hold up to eight electrons in its s and p orbitals.

🎯 Exam Tip: Connect the number of elements in each period directly to the maximum number of electrons that can occupy the available orbitals in that principal energy level.

Question 12. Discuss the variation of electron affinity in the period.
Answer: As we move from left to right across a period, electron affinity generally increases. This means that more energy is typically released when an electron is added to an atom. This trend occurs because, across a period, the nuclear charge increases, and the atomic size decreases. Both factors lead to a stronger attraction between the nucleus and any incoming electron, making it easier for the atom to accept an electron and release more energy.
However, there are exceptions:
• Elements like beryllium (\( 1s^2 2s^2 \)) and nitrogen (\( 1s^2 2s^2 2p^3 \)) have stable electronic configurations (completely filled s-orbital or half-filled p-orbital). Adding an extra electron would disrupt this stability, so they have very low, or even zero, electron affinity.
• Noble gases have a very stable \( ns^2 np^6 \) configuration. Adding an electron to them is highly unfavorable and requires energy, so they have electron affinities close to zero or positive values.
• Halogens (Group 17) have an \( ns^2 np^5 \) configuration. They readily accept one electron to achieve a stable noble gas configuration (\( ns^2 np^6 \)), giving them very high electron affinities (large negative values). These exceptions highlight the importance of electronic stability.
In simple words: Moving across a period, atoms get smaller and their nucleus pulls stronger, so they attract new electrons more easily, releasing more energy. But some atoms, like noble gases, are already very stable, so they don't want extra electrons.

🎯 Exam Tip: Remember the general trend (increases across a period) but pay special attention to exceptions like noble gases, beryllium, and nitrogen due to their stable electron configurations.

Question 13. What are the two exceptions of block division in the periodic table?
Answer: The two main exceptions in the block division of the periodic table are hydrogen and helium.
1. **Helium (He):** Helium has an electronic configuration of \( 1s^2 \). Based on this, it should technically be placed in the s-block, as its outermost electron is in an s-orbital. However, due to its completely filled valence shell, helium exhibits properties similar to noble gases (Group 18), which belong to the p-block. Therefore, helium is placed with the noble gases in Group 18, not in the s-block.
2. **Hydrogen (H):** Hydrogen has only one s-electron (\( 1s^1 \)). This configuration suggests it could be placed in Group 1 (alkali metals) because it can lose one electron to form \( H^+ \). However, hydrogen can also gain one electron to achieve a stable duet configuration (like helium), forming \( H^- \) and behaving like halogens (Group 17). Due to these dual characteristics, hydrogen's position is unique and is usually placed separately at the top of the periodic table, not strictly within one block group. Its versatile chemistry reflects this dual nature.
In simple words: Hydrogen and helium are special. Helium should be in the 's' block but acts like 'p' block noble gases. Hydrogen can act like Group 1 or Group 17 elements, so it's placed alone at the very top.

🎯 Exam Tip: Focus on helium's stable noble gas properties despite its s-block configuration, and hydrogen's dual nature (can lose or gain an electron) as the reasons for their exceptional placements.

III. Long Question and Answers (5 Marks):

Question 1. Describe Moseley's work and Modern Periodic Law.
Answer: Henry Moseley conducted pioneering experiments in the early 20th century, where he studied the characteristic X-ray spectra emitted by different elements. By bombarding elements with high-energy electrons, he observed that the frequency of the emitted X-rays was directly related to the atomic number of the element. He formulated an empirical relationship known as Moseley's Law: \( \sqrt{\nu} = a(Z - b) \), where \( \nu \) is the frequency of the X-rays, \( Z \) is the atomic number, and \( a \) and \( b \) are constants. Plotting \( \sqrt{\nu} \) against \( Z \) yielded a straight line for all elements. This work demonstrated that atomic number, not atomic weight, is the fundamental property that dictates an element's chemical behavior.
This discovery led to the formulation of the **Modern Periodic Law**, which states: "The physical and chemical properties of the elements are periodic functions of their atomic numbers." This law corrected the anomalies present in Mendeleev's periodic table, where elements like iodine and tellurium were misplaced based on atomic weights. By arranging elements according to their increasing atomic numbers, the modern periodic table ensures that elements with similar properties recur at regular intervals, providing a more accurate and consistent classification system. This periodicity in properties, such as atomic size, ionization energy, and electronegativity, is a direct consequence of the regular repetition of similar outer electronic configurations. Moseley's work transformed the understanding of element organization.
In simple words: Moseley used X-rays to find that an element's atomic number is its most important feature. This led to the Modern Periodic Law, which says element properties repeat based on their atomic number. This made the periodic table better and more accurate.

🎯 Exam Tip: Clearly define Moseley's Law (\( \sqrt{\nu} = a(Z - b) \)) and the Modern Periodic Law. Explain how Moseley's work resolved issues in Mendeleev's table by establishing atomic number as the fundamental property.

Question 2. The first ionization enthalpy of magnesium is higher than that of sodium. On the other hand, the second ionization enthalpy of sodium is very much higher than that of magnesium. Explain.
Answer:
**First Ionization Enthalpy:** Magnesium (Mg) has a higher first ionization enthalpy than sodium (Na) because Mg has a higher nuclear charge (12 protons) compared to Na (11 protons), and its atomic radius is slightly smaller. Additionally, Na loses an electron from its \( 3s^1 \) configuration, leading to a stable noble gas configuration (Neon). Mg loses an electron from its \( 3s^2 \) configuration. The extra nuclear charge in Mg pulls its electrons more strongly, making it harder to remove the first electron compared to Na.
**Second Ionization Enthalpy:** The second ionization enthalpy of sodium (\( Na^+ \)) is much higher than that of magnesium (\( Mg^+ \)).
• After losing one electron, sodium becomes \( Na^+ \), which has the electron configuration of neon (\( 1s^2 2s^2 2p^6 \)). This is a very stable, completely filled noble gas configuration. Removing a second electron from this highly stable, full octet requires a very large amount of energy.
• Magnesium, after losing one electron, becomes \( Mg^+ \), with an electron configuration of \( 1s^2 2s^2 2p^6 3s^1 \). This \( Mg^+ \) ion still has one electron in its outermost 3s orbital. Removing this second electron is relatively easier than breaking sodium's stable noble gas core, leading to a much lower second ionization enthalpy for Mg compared to Na. The \( Mg^{2+} \) ion formed also achieves a stable noble gas configuration. This difference in stability profoundly impacts the energy required to remove subsequent electrons.
In simple words: Magnesium needs more energy to lose its first electron than sodium because it has a stronger nucleus. But to lose a second electron, sodium needs a lot more energy than magnesium because sodium becomes super stable after losing its first electron, like a noble gas. Magnesium still has an outer electron after losing its first, so losing the second is easier.

🎯 Exam Tip: When explaining ionization enthalpies, always consider both nuclear charge (Z) and electron configuration (especially stability of noble gas core or half-filled/full subshells).

Question 3. By using Pauling's method calculate the ionic radii of \( K^+ \) and \( Cl^- \) ions in the potassium chloride crystal. Given that \( d_{K^+ - Cl^-} = 3.14 \) Å.
Answer: Pauling's method allows us to calculate ionic radii based on interionic distance and effective nuclear charge.
Given: \( d_{K^+ - Cl^-} = 3.14 \) Å
This distance is the sum of the cationic and anionic radii:
\( d_{K^+ - Cl^-} = r_{K^+} + r_{Cl^-} = 3.14 \) Å (1)
Pauling's assumption states that for ions with noble gas configuration, the ionic radius is inversely proportional to the effective nuclear charge (Zeff) felt by the outermost electrons.
For \( Cl^- \) (isoelectronic with Ar, Z=17):
\( (Z_{eff})_{Cl^-} = Z - S = 17 - 10.9 = 6.1 \) (Using \( S_{Cl^-} = 10.9 \) from standard values or given context)
For \( K^+ \) (isoelectronic with Ar, Z=19):
\( (Z_{eff})_{K^+} = Z - S = 19 - 16.8 = 2.2 \) (Using \( S_{K^+} = 16.8 \) from standard values or given context)
According to Pauling's inverse proportionality:
\( \frac{r_{K^+}}{r_{Cl^-}} = \frac{(Z_{eff})_{Cl^-}}{(Z_{eff})_{K^+}} = \frac{6.1}{2.2} \approx 2.77 \)
So, \( r_{K^+} = 2.77 r_{Cl^-} \) (2)
Substitute equation (2) into equation (1):
\( 2.77 r_{Cl^-} + r_{Cl^-} = 3.14 \) Å
\( 3.77 r_{Cl^-} = 3.14 \) Å
\( r_{Cl^-} = \frac{3.14}{3.77} \approx 0.83 \) Å
Now, find \( r_{K^+} \) using equation (1):
\( r_{K^+} = 3.14 - r_{Cl^-} = 3.14 - 0.83 = 2.31 \) Å
Therefore, the ionic radius of the \( K^+ \) ion is \( 2.31 \) Å and the \( Cl^- \) ion is \( 0.83 \) Å. This method is crucial for understanding ion sizes within crystal lattices, as these sizes influence crystal structure and properties.
In simple words: We used Pauling's method to find the size of potassium and chloride ions. We first calculated how strongly their nucleus pulls electrons. Then, by knowing their total distance apart in a crystal, we divided that distance based on their relative pulls to find each ion's specific size.

🎯 Exam Tip: Carefully apply Slater's rules for calculating effective nuclear charge, as small errors in 'S' can significantly affect the final ionic radii values. Ensure the inverse proportionality is correctly applied.

Question 4. Discuss the variation of electronic configuration along the periods.
Answer: Electronic configuration changes systematically as we move across a period (horizontal row) in the periodic table. Each period represents the filling of electrons into a new principal energy shell.
• **First Period:** Starts with hydrogen (\( 1s^1 \)) and ends with helium (\( 1s^2 \)). Only the 1s orbital is filled, accommodating a maximum of two electrons.
• **Second Period:** Begins with lithium (\( [He]2s^1 \)) and ends with neon (\( [He]2s^2 2p^6 \)). The 2s and 2p orbitals are filled, accommodating a total of eight electrons.
• **Third Period:** Starts with sodium (\( [Ne]3s^1 \)) and ends with argon (\( [Ne]3s^2 3p^6 \)). The 3s and 3p orbitals are filled, also accommodating eight electrons.
• **Fourth Period:** This period is longer, starting with potassium (\( [Ar]4s^1 \)). After filling the 4s orbital, the 3d orbitals begin to fill (from scandium to zinc), followed by the 4p orbitals. These 10 d-block elements are known as the first transition series.
• **Subsequent Periods:** Similar patterns occur in later periods, where 4d, 5d, and 6d orbitals are filled for subsequent transition series. The sixth and seventh periods also include the filling of f-orbitals (4f and 5f respectively), leading to the inner-transition series (lanthanides and actinides), which are usually placed separately below the main table. This systematic filling of orbitals dictates the periodic properties of elements. The Aufbau principle guides this order of filling orbitals.
In simple words: As we go across a row in the periodic table, electrons fill up shells and subshells in a steady way. Each row adds electrons to a new main energy level, leading to changes in the elements' properties. Longer rows also include d and f orbitals filling up.

🎯 Exam Tip: Remember the specific orbitals filled in each period (1s, then 2s/2p, then 3s/3p, then 4s/3d/4p, etc.) and how this relates to the length of the period and the types of elements (s-block, p-block, d-block, f-block).

Question 5. Describe the nomenclature of elements with Atomic Number greater than 100.
Answer: For newly discovered elements with atomic numbers greater than 100, the International Union of Pure and Applied Chemistry (IUPAC) has a systematic nomenclature system. This method is used to assign a temporary name and symbol until a permanent name is officially approved, usually after a public discussion and consensus among discoverers.
The temporary name is derived directly from the atomic number using numerical roots for each digit. These roots are:
• 0: nil (n)
• 1: un (u)
• 2: bi (b)
• 3: tri (t)
• 4: quad (q)
• 5: pent (p)
• 6: hex (h)
• 7: sept (s)
• 8: oct (o)
• 9: enn (e)
These roots are put together in order of the digits in the atomic number, and the suffix "ium" is added at the end. For example, an element with atomic number 101 (un-nil-un-ium) is unnilunium. The symbol is derived from the first letter of each numerical root. Some minor adjustments are made for pronunciation, like omitting the final 'n' of 'enn' before 'nil' or 'un', and omitting the final 'i' of 'bi' or 'tri' before 'ium'. This system ensures that every newly synthesized element has a unique, systematic name immediately.
The following table illustrates the numerical roots: