Chapter 3. Classification of Elements and Periodicity in properties

Class 11 Chemistry Chapter 3. Classification of Elements and Periodicity in properties Internal Problems and Answers :

Problem 3.1 : What would be the IUPAC name and symbol for the element with atomic number 120?

Answer :  The element with atomic number 120 is known as "Unbinilium" according to the IUPAC naming system. Its symbol would be ‘ Ubn ’.

Problem 3.2 :  How would you justify the presence of 18 elements in the  period of the Periodic Table?

Answer : There are 18 elements in the 5th period because there are 9 orbitals available, and each orbital can hold a maximum of 2 electrons, resulting in a total capacity of 18 electrons and thus 18 elements.

Problem 3.3 : The elements Z = 117 and 120 have not yet been discovered. In which family/group would you place these elements and also give the electronic configuration in each case.

Answer :  Element with Z = 117:

Group: Element 117 would belong to Group 17, which is the Halogen family.

Predicted Electronic Configuration: The electronic configuration would be [Rn] 5f¹⁴6d¹⁰7s²7p⁵.

Element with Z = 120:

Group: Element 120 will be placed in Group 2, known as the Alkaline Earth Metals.

Predicted Electronic Configuration: The electronic configuration would be [Uuo] 8s².

Problem 3.4 : Considering the atomic number and position in the periodic table, arrange the following elements in the increasing order of metallic character : Si, Be, Mg, Na, P.

Answer : The order of increasing metallic character for the given elements is as follows:

P < Si < Be < Mg < Na

This order is determined by following the periodic trend that metallic character increases as you move down a group and decreases as you move from left to right along a period.

Problem 3.5 : Which of the following species will have the largest and the smallest size?   .

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

Therefore, the order of increasing atomic size is the reverse of this trend, with larger atoms being more metallic.

So, in terms of size:

Largest size: Na (sodium, an alkali metal)

Smallest size: Mg²⁺ (magnesium ion with a +2 charge)

Problem 3.6 : The first ionization enthalpy (∆i H ) values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ  . Predict whether the first  value for Al will be more close to 575 or 760 kJ  ? Justify your answer.

Answer : The first value for Al will be more close to 760 kJ/mol. This is because the effective shielding of 3p electrons by 3s electrons in aluminum reduces the attraction of the outermost electrons to the nucleus, making it slightly easier to remove an electron compared to magnesium.

Problem 3.7 :  Which of the following will have the most negative electron gain enthalpy and which the least negative?  P, S, Cl, F . Explain your answer.

Answer : The element with the most negative electron gain enthalpy is chlorine (Cl). It is on the right side of the period and closer to achieving a stable noble gas electron configuration, so adding an electron to its 3p-orbital is less repulsive.

The element with the least negative electron gain enthalpy is phosphorus (P), as it is on the left side of the period, and adding an electron to its 2p-orbital experiences greater electron repulsion.

Problem 3.8 : Using the Periodic Table, predict the formulas of compounds which might be formed by the following pairs of elements; (a) silicon and bromine (b) aluminium and sulphur.

Answer : (a) The formula of the compound formed between silicon (Si) and bromine (Br) would be  because silicon is a group 14 element with a valence of 4, and bromine is a halogen with a valence of 1.

(b) The formula of the compound formed between aluminum (Al) and sulfur (S) would be  because aluminum is a group 13 element with a valence of 3, and sulfur is a group 16 element with a valence of 2.

Problem 3.9 : Are the oxidation state and covalency of  in  same ?

Answer : No, the oxidation state and covalency of aluminum (Al) in are not the same. The oxidation state of aluminum is +3, indicating that it has lost 3 electrons. The covalency, on the other hand, is 6, which means aluminum is forming 6 covalent bonds with surrounding atoms, in this case, with chlorine and water molecules.

Problem 3.10 : Show by a chemical reaction with water that  is a basic oxide and  is an acidic oxide.

Answer : Reaction of with Water :

Sodium oxide is a basic oxide because it reacts with water to form a strong base, sodium hydroxide (NaOH), along with the liberation of heat . The chemical reaction is : 

Reaction of with Water :

Chlorine heptoxide () is an acidic oxide because it reacts with water to form a strong acid, perchloric acid (). The reaction is :

Class 11 Chemistry Chapter 3. Classification of Elements and Periodicity in properties Exercise Questions and Answers :

Q3.1 What is the basic theme of organisation in the periodic table?

Answer : The fundamental theme of organization in the periodic table is the arrangement of elements in order of increasing atomic number. This is known as the Modern Periodic Law. Elements are grouped by similar electronic configurations and properties, creating periodic patterns in their chemical behavior and physical characteristics. The table is organized into rows (periods) and columns (groups) to highlight these patterns.
Q3.2 Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?

Answer : Mendeleev classified elements in his periodic table primarily based on their atomic mass. While he initially arranged elements according to their atomic masses, he made exceptions when required to maintain chemical periodicity.

Q3.3 What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?

Answer : The basic difference is that Mendeleev's Periodic Law was based on atomic mass ,while the Modern Periodic Law is based on atomic number .

Q3.4 On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.

Answer : The sixth period of the periodic table should have 32 elements because the principal quantum number (n = 6) indicates the energy level, and the maximum number of electrons that can occupy this level is determined by the number of orbitals. In this case, there are 16 orbitals in the sixth period, each accommodating a maximum of 2 electrons, resulting in a total of 32 electrons and thus 32 elements to fill these electron slots.

Q3.5 In terms of period and group where would you locate the element with Z =114?

Answer : The element with atomic number 114 is Flerovium (Fl). It is located in period 7 (the seventh row) and in group 14 (the fourteenth column) of the periodic table.

Q3.6 Write the atomic number of the element present in the third period and seventeenth group of the periodic table.

Answer : The element present in the third period and seventeenth group of the periodic table is Chlorine, with an atomic number of 17.

Q3.7 Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group?

Answer :  (i) Lawrence Berkeley Laboratory: Lawrencium (Atomic number Lr is 103)

(ii) Seaborg's group: Seaborgium (Atomic number Sg is 106 )

Q3.8 Why do elements in the same group have similar physical and chemical properties?

Answer : Elements in the same group have similar properties because they share the same number of valence electrons, which determine an element's chemical behavior and physical characteristics. This common electron configuration leads to resemblances in reactivity and properties.

Q3.9 What does atomic radius and ionic radius really mean to you?

Answer : Atomic radius: The size of an atom, typically defined as the distance from the nucleus to the outermost electron cloud.

Ionic radius: The size of ions, which can differ from atomic radius due to the gain or loss of electrons. Cations are typically smaller, and anions are typically larger than their parent atoms.

Q3.10 How do atomic radius vary in a period and in a group? How do you explain the variation?

Answer : In a period, atomic radius generally decreases from left to right due to an increase in effective nuclear charge, which pulls electrons closer to the nucleus.

In a group, atomic radius generally increases as you move down due to the addition of energy levels, which results in greater electron-electron shielding, allowing outermost electrons to be farther from the nucleus.

Q3.11 What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
(i)       (ii)       (iii)      (iv) 

Answer : Isoelectronic species are atoms or ions that have the same number of electrons, resulting in identical electron configurations.

The isoelectronic species are :

 (i)   (Fluorine ion with a charge of – 1) is isoelectronic with Ne (neon), as they both have 10 electrons in their electron configuration.

(ii) Ar (Argon) is isoelectronic with (Potassium ion with a charge of + 1) and (Chlorine ion with a charge of – 1), as they all have 18 electrons in their electron configuration.

(iii)   (Magnesium ion with a charge of + 2) is isoelectronic with Ne (neon), as they both have 10 electrons in their electron configuration.

(iv)  (Rubidium ion with a charge of + 1) is isoelectronic with Kr (Krypton), as they both have 36 electrons in their electron configuration.

Q3.12 Consider the following species :   and
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.

Answer : (a) They are all isoelectronic with each other, and they share the same noble gas electron configuration. Each of these species has 10 electrons in their electron configuration, which is the same as the noble gas neon (Ne).

(b) Arrange them in the order of increasing ionic radii :   

Q3.13 Explain why cation are smaller and anions larger in radii than their parent atoms?

Answer : Cations are smaller than their parent atoms because they have lost electrons and the remaining electrons are pulled closer to the nucleus, while anions are larger than their parent atoms because they have gained electrons, causing increased electron-electron repulsion and an expanded electron cloud.

Q3.14 What is the significance of the terms - ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy? (Hint : Requirements for comparison purposes.)

Answer :  Isolated Gaseous Atom :

Ionization Enthalpy (IE) : Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom to form a singly positively charged ion. This condition is essential to eliminate any external influences or interactions, ensuring that the measured energy change is solely due to the removal of an electron from the atom.

Electron Gain Enthalpy (EGE) : Electron gain enthalpy is the energy change when an isolated gaseous atom gains an electron to form a singly negatively charged ion. Like IE, considering the atom in isolation in a gaseous state eliminates external factors that may affect the energy change during electron addition.

Ground State :

Ionization Enthalpy (IE) : Ionization enthalpy is typically defined for the ground state of an atom, which is its lowest energy state. This ensures that the electron being removed is in its most stable position within the atom, making the measurement consistent and applicable for comparison.

Electron Gain Enthalpy (EGE) : Electron gain enthalpy is also measured for atoms in their ground state, as it represents the most stable configuration for the atom when gaining an electron. The ground state ensures that the energy change observed is associated with the electron entering the lowest energy orbital.

Q3.15 Energy of an electron in the ground state of the hydrogen atom is  J. Calculate the ionization enthalpy of atomic hydrogen in terms of J  .
(Hint: Apply the idea of mole concept to derive the answer.)

Answer : The energy of an electron in the ground state of hydrogen atom is given as –  J.

Now, one mole of electrons is  electrons (Avogadro's number) .

Ionization Enthalpy  =

 

 J  

Ionization Enthalpy =  J

So, the ionization enthalpy of atomic hydrogen in terms of J is  J .

Q3.16 Among the second period elements the actual ionization enthalpies are in the order : Li < B < Be < C < O < N < F < Ne.
Explain why
(i) Be has higher  than B
(ii) O has lower  than N and F ?

Answer : (i) Beryllium (Be) has a higher second ionization enthalpy () than Boron (B) because Be's electron configuration is 1s² 2s². When an electron is removed from Be, it results in a fully filled 2s orbital, which is more stable, and thus, more energy is required to remove an electron compared to Boron, which has a less stable partially filled 2p orbital.

(ii) Oxygen (O) has a lower than Nitrogen (N) and Fluorine (F) because O has a half-filled 2p orbital (electron configuration 1s² 2s² 2p⁴). A half-filled orbital is relatively stable, and it takes less energy to remove an electron from such a configuration. In contrast, Nitrogen and Fluorine have partially filled 2p orbitals (N: 1s² 2s² 2p³, F: 1s² 2s² 2p⁵), which are less stable and require more energy to remove an electron.

Q3.17 How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?

Answer : The first ionization enthalpy of sodium is lower than that of magnesium due to sodium's 3s¹ electron configuration, which is closer to the nucleus and less tightly bound than magnesium's 3p¹ electron. However, sodium's second ionization enthalpy is higher because after losing one electron, it becomes a ion with a full 2p⁶ electron configuration, which is more stable and harder to remove. In contrast, magnesium still has a 3s² electron configuration, which is less stable and easier to remove, resulting in a lower second ionization enthalpy.

Q3.18 What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?

Answer : The ionization enthalpy of main group elements tends to decrease down a group primarily due to the following factors:

(i) As we move down a group, the atomic size increases, which results in a greater distance between the outermost electrons and the nucleus, reducing the effective nuclear charge on the valence electrons.

(ii) The increase in electron shells provides more inner electron shielding, reducing the attraction between the outermost electrons and the nucleus.

(iii) With additional electron shells, there is more electron-electron repulsion, making it easier to remove an outermost electron.

Q3.19 The first ionization enthalpy values (in kJ ) of group 13 elements are :
   B          Al        Ga        In        Tl
 801        577      579       558     589
How would you explain this deviation from the general trend ?

Answer : The deviation in the first ionization enthalpy values of group 13 elements can be explained by their electron configurations. Boron (B) has a relatively low ionization enthalpy due to the presence of one unpaired electron. Aluminum (Al) exhibits a lower ionization enthalpy as it has a stable half-filled 2p orbital. Gallium (Ga), Indium (In), and Thallium (Tl) all have stable electron configurations with fully or partially filled 2p orbitals, leading to lower ionization energies. These exceptions result from the stability of specific electron configurations, defying the general trend.

Q3.20 Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F (ii) F or Cl

Answer :  (i) The electron gain enthalpy for oxygen (O) would be more negative than for fluorine (F) because oxygen is in the same period as fluorine but has a smaller atomic size, leading to a stronger attraction for an additional electron.

(ii) The electron gain enthalpy for chlorine (Cl) would be more negative than for fluorine (F) because chlorine is in the same group as fluorine but has a larger atomic size, making it more favorable for chlorine to gain an additional electron.

Q3.21 Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.

Answer : The second electron gain enthalpy of oxygen (O) is expected to be more positive (less negative) than the first. After gaining the first electron, oxygen becomes negatively charged, making it less favorable for it to gain a second electron due to increased electron-electron repulsion. This results in a higher energy requirement and a less negative electron gain enthalpy for the second electron.

Q3.22 What is the basic difference between the terms electron gain enthalpy and electronegativity?

Answer : The basic differences between the terms electron gain enthalpy and electronegativity:

Electron Gain Enthalpy	Electronegativity
The energy change when an atom gains an electron to form a negative ion.	A measure of an element's ability to attract electrons in a chemical bond.
Generally increases from left to right across a period in the periodic table and decreases down a group.	Increases from left to right across a period and decreases down a group in the periodic table.
Provides a specific numerical value indicating the energy change associated with electron addition.	Provides a relative scale (e.g., Pauling scale) for comparing the electron-attracting ability of different elements.

Q3.23 How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?

Answer : The statement that the electronegativity of nitrogen (N) on the Pauling scale is 3.0 in all nitrogen compounds is incorrect. Electronegativity can vary in different compounds depending on the specific elements involved and the nature of the chemical bonds. Nitrogen's electronegativity is around 3.0, but it changes relative to other elements within different compounds, influencing the nature of the chemical bonds formed.

Q3.24 Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron

Answer :  (a) When an atom gains an electron, its size increases. The addition of an electron results in a higher electron count and more electron-electron repulsion. Since the number of protons (nuclear charge) remains the same, the effective nuclear charge decreases, causing the atom's radius to increase.

(b) When an atom loses an electron, its size decreases. The reduced electron count leads to decreased electron-electron repulsion, and with the nuclear charge remaining constant, the effective nuclear charge increases, causing the atom's radius to decrease.

Q3.25 Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.

Answer : The first ionization enthalpies for two isotopes of the same element should be the same. Ionization enthalpy depends on the number of electrons and the nuclear charge. Since isotopes of an element have the same number of protons and electrons, their ionization enthalpies will be identical.

Q3.26 What are the major differences between metals and non-metals?

Answer : The major differences between metals and non-metals are :

Metals	Non-metals
(i) Metals are typically solid at room temperature (with the exception of mercury, which is a liquid).	(i) Non-metals can exist in various states at room temperature, including solids, liquids, and gases.
(ii) They have a shiny and lustrous appearance.	(ii) They generally have a dull or non-lustrous appearance.
(iii) Metals are malleable and ductile, meaning they can be hammered into thin sheets (malleability) and drawn into wires (ductility).	(iii) Non-metals are typically brittle and not malleable or ductile.
(iv) Metals are good conductors of heat and electricity.	(iv) Non-metals are poor conductors of heat and electricity.
(v) They tend to have high density and high melting and boiling points.	(v) They tend to have lower density and lower melting and boiling points compared to metals.
(vi) In chemical reactions, metals tend to lose electrons and form cations (+ ions).	(vi) In chemical reactions, non-metals tend to gain, share, or hold onto electrons and form anions (- ions).

Q3.27 Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.

Answer : (a) An element with five electrons in the outer subshell is nitrogen (N).

(b) An element that would tend to lose two electrons is calcium (Ca).

(c) An element that would tend to gain two electrons is oxygen (O).

(d) Group 14 contains elements with a metal (carbon, tin), non-metal (carbon, silicon), liquid (tin), and gas (carbon and silicon) at room temperature.

Q3.28 The increasing order of reactivity among group 1 elements is Li < Na < K < Rb <Cs whereas that among group 17 elements is F > CI > Br > I. Explain.

Answer :  Group 1 (Alkali Metals): Reactivity increases down the group as metals due to decreasing ionization enthalpy. Larger atoms have weaker attraction to valence electron, facilitating its loss. The increasing order of reactivity among group 1 elements is Li < Na < K < Rb <Cs .

Group 17 (Halogens): Reactivity increases down the group as nonmetals due to decreasing electronegativity. Larger atoms more readily accept an electron to fill their valence shells, increasing reactivity. The increasing order of reactivity among group 17 elements is F > CI > Br > I .

Q3.29 Write the general outer electronic configuration of s-, p-, d- and f- block elements.

Answer : The general outer electronic configuration of elements in the s-, p-, d-, and f-blocks of the periodic table is as follows:

s-Block Elements:

The outer electronic configuration of s-block elements is  , where n = 2 to 7

p-Block Elements:

The outer electronic configuration of p-block elements is  , where n = 2 to 6

d-Block Elements:

The outer electronic configuration of d-block elements is  , where n = 4 to 7

f-Block Elements:

The outer electronic configuration of f-block elements is  , where n = 6 to 7

Q3.30 Assign the position of the element having outer electronic configuration
(i)   for n = 3 (ii)   for n = 4, and (iii)  for n = 6, in the periodic table.

Answer :  (i) The element with the outer electronic configuration for n = 3 belongs to the third period (n = 3) and has the electron configuration . This corresponds to the element sulfur (S), which is in Group 16 (also known as the oxygen group) and the third period of the periodic table.

(ii) The element with the outer electronic configuration for n = 4 belongs to the fourth period (n = 4) and has the electron configuration . This corresponds to the element titanium (Ti), which is in Group 4 and the fourth period of the periodic table.

(iii) The element with the outer electronic configuration for n = 6 belongs to the sixth period (n = 6) and has the electron configuration . This corresponds to the element gadolinium (Gd), which is in the lanthanide series and the sixth period of the periodic table.

Q3.31 The first () and the second () ionization enthalpies (in kJ ) and the () electron gain enthalpy (in kJ ) of a few elements are given below:

Elements
I	520	7300	– 60
II	419	3051	– 48
III	1681	3374	– 328
IV	1008	1846	– 295
V	2372	5251	+ 48
VI	738	1451	– 40

 Which of the above elements is likely to be :
(a) the least reactive element.
(b) the most reactive metal.
(c) the most reactive non-metal.
(d) the least reactive non-metal.
(e) the metal which can form a stable binary halide of the formula  (X = halogen).
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X = halogen)?

Answer : (a) The least reactive element: Element V

(b) The most reactive metal: Element II

(c) The most reactive non-metal: Element III

(d) The least reactive non-metal: Element V

(e) The metal which can form a stable binary halide of the formula  (X = halogen): Element VI

(f) The metal which can form a predominantly stable covalent halide of the formula MX (X = halogen): Element I .
Q3.32 Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.

(a) Lithium and oxygen  (b) Magnesium and nitrogen  (c) Aluminium and iodine   (d) Silicon and oxygen  (e) Phosphorus and fluorine  (f) Element 71 and fluorine

Answer :  The formulas of the stable binary compounds formed by the combination of the given pairs of elements:

(a) Lithium and oxygen:

(b) Magnesium and nitrogen:

(c) Aluminium and iodine:

(d) Silicon and oxygen:

(e) Phosphorus and fluorine:  or

(f) Element 71 (Lutetium) and fluorine: (Lutetium fluoride)

Q3.33 In the modern periodic table, the period indicates the value of :
(a) atomic number
(b) atomic mass
(c) principal quantum number
(d) azimuthal quantum number.

Answer :  (c) principal quantum number

In the modern periodic table, the period indicates the value of the principal quantum number (n). Each period represents a different energy level or principal quantum level, and as you move from one period to the next, the value of n increases by one.

Q3.34 Which of the following statements related to the modern periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (l ) for the last subshell that received electrons in building up the electronic configuration.

Answer : (b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.

This statement is incorrect. The d-block in the modern periodic table has 10 columns, not 8. This is because a maximum of 10 electrons can occupy all the orbitals in a d-subshell.

Q3.35 Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z )
(c) Nuclear mass
(d) Number of core electrons.

Answer :  (c) Nuclear mass

The factor that does not directly affect the valence shell and the chemistry of the element is nuclear mass. The nuclear mass represents the total mass of protons and neutrons in the nucleus, but it does not influence the arrangement or behavior of the electrons in the valence shell. The valence principal quantum number (n), nuclear charge (Z), and the number of core electrons are more directly related to the properties of the valence shell and the chemistry of the element.

Q3.36 The size of isoelectronic species -  , Ne and is affected by
(a) nuclear charge (Z )
(b) valence principal quantum number (n)
(c) electron-electron interaction in the outer orbitals
(d) none of the factors because their size is the same.

Answer :  (a) nuclear charge (Z).

The size of isoelectronic species is primarily affected by the nuclear charge (Z). In this case,   has a higher nuclear charge (Z = 9), which makes it smaller than Ne (Z = 10) and  (Z = 11), despite having the same number of electrons.

Q3.37 Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.

Answer :  (d) Removal of electron from orbitals bearing lower n value is easier than from orbitals having a higher n value.

In reality, removal of an electron from orbitals bearing lower n value (closer to the nucleus) is more difficult and requires more energy. Electrons in lower n value orbitals are closer to the nucleus and experience a stronger electrostatic attraction to the positively charged protons in the nucleus. As a result, it takes more energy to remove an electron from these inner orbitals. The ease of removal generally increases as you move to outer orbitals with higher n values. So, statement (d) is incorrect.

Q3.38 Considering the elements B, Al, Mg, and K, the correct order of their metallic character is :
(a) B > Al > Mg > K

(b) Al > Mg > B > K

(c) Mg > Al > K > B

(d) K > Mg > Al > B

Answer :  (d) K > Mg > Al > B

Potassium (K) is the most metallic element among the options, followed by magnesium (Mg), aluminum (Al), and boron (B). Metallic character generally increases as you move from right to left across the periodic table and from top to bottom within a group .

Q3.39 Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is :
(a) B > C > Si > N > F

(b) Si > C > B > N > F

(c) F > N > C > B > Si

(d) F > N > C > Si > B

Answer :  (c) F > N > C > B > Si

Fluorine (F) is the most non-metallic element among the options, followed by nitrogen (N), carbon (C), boron (B), and silicon (Si). Non-metallic character generally increases as you move from left to right across the periodic table and from top to bottom within a group.

Q3.40 Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is :
(a) F > Cl > O > N

(b) F > O > Cl > N

(c) Cl > F > O > N

(d) O > F > N > Cl

Answer : (b) F > O > Cl > N

Fluorine (F) is the most powerful oxidizing agent, followed by oxygen (O), chlorine (Cl), and nitrogen (N). This order is based on the tendency of these elements to gain electrons or attract electrons from other elements in chemical reactions.