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# Chapter 3 Classification of Elements and Periodicity in Properties

From this chapter, you will be able to appreciate the concept of grouping elements in accordance to their properties led to the development of the Periodic Table. You can also understand the Periodic Table and the significance of atomic number and electronic configuration. You can also name the elements according to IUPAC nomenclature and classify elements into s, p, d, f blocks. You will also be able to recognise the periodic trends in physical and chemical properties of elements. You can also compare the reactivity of elements and correlate it with their occurrence in nature. You can also explain the relationship between ionization enthalpy and metallic character.Use of scientific vocabulary appropriately to communicate ideas related to certain important properties of atoms are also known.

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### Exercise 1 ( Page No. : 99 )

• Q1 What is the basic theme of organisation in the periodic table?
Ans:

Before 18th century, only a few elements are known and it was quite east to study their chemical and physical properties. But with the passage of time, more and more elements were discovered and it was not possible to study and remember each of them individually. Because many of them have different properties. So, there arise the need of developing a table or chart in which all the elements can be classified according to similar properties ,so that there is less confusion and true information can be generated about the different elements present in periodic table.

Q2 Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?
Ans:

A Russian chemist, Dmitri Mendeleev was the first to develop a periodic table and he gave a law called Mendeleev periodic law which states that the physical and chemical properties of the elements are a periodic function of their atomic masses. On the basis of this law he developed a Mendeleev periodic table, where he  arranged the elements in his periodic table ordered by atomic weight or mass. He arranged the elements in periods and groups in order of their increasing atomic weight. He placed the elements with similar properties in the same group.

However, he did not stick to this arrangement for long. He found out that if the elements were arranged strictly in order of their increasing atomic weights, then some elements did not fit within this scheme of classification.

Therefore, he ignored the order of atomic weights in some cases. For example, the atomic weight of iodine is lower than that of tellurium. Still Mendeleev placed tellurium (in Group VI) before iodine (in Group VII) simply because iodine’s properties are so similar to fluorine, chlorine, and bromine.

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

A Russian chemist, Dmitri Mendeleev was the first to develop a periodic table and he gave a law called Mendeleev periodic law which states that the physical and chemical properties of the elements are a periodic function of their atomic masses. In his periodic table the elements are arranged in vertical rows called groups and horizontal rows known as periods.There are nine groups and seven periods. He arranged only 60 elements in periodic table.

A English physicist Moseley, in 1913 gave a modern periodic law which states that the physical and chemical properties of the elements are the periodic function of their atomic number. This law becomes the base of modern periodic law. In modern periodic table, there are 18 groups and 7 periods.

Q4 On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Ans:

Principal quantum number is number which determines the main energy level or shell in which the electron is present. It gives the average distance of the electron from the nucleus and specifies the value of the energy of the electron.

Azimuthal quantum number is the number which determines the sub shell in a principal energy shell to which an electron belongs.

In the periodic table of the elements, a period indicates the value of the principal quantum number (n) for the outermost shells. Each period begins with the filling of principal quantum number (n). The value of n for the sixth period is 6. For n = 6, azimuthal quantum number (l) can have values of 0, 1, 2, 3, 4.

According to Aufbau’s principle, electrons are added to different orbitals in order of their increasing energies. The energy of the 6d subshell is even higher than that of the 7s subshell.

In the 6th period, electrons can be filled in only 6s, 4f, 5d, and 6p subshells. Now, 6s has one orbital, 4f has seven orbitals, 5d has five orbitals, and 6p has three orbitals. Therefore, there are a total of sixteen (1 + 7 + 5 + 3 = 16) orbitals available. According to Pauli’s exclusion principle, each orbital can accommodate a maximum of 2 electrons. Thus, 16 orbitals can accommodate a maximum of 32 electrons.Hence, the sixth period of the periodic table should have 32 elements.

Q5 In terms of period and group where would you locate the element with Z =114?
Ans:

Elements with atomic numbers from Z = 87 to Z = 114 are present in the 7th period of the periodic table. Thus, the element with Z = 114 (Flerovium) with atomic weight 289 and a poor metal is present in the 7th period  and 14th group of the periodic table.

In the 7th period, first two elements with Z = 87 and Z= 88 are s-block elements, the next 14 elements excluding Z = 89 i.e., those with Z = 90 – 103 are f – block elements, ten elements with Z = 89 and Z = 104 – 112 are d – block elements, and the elements with Z = 113 – 118 are p – block elements. Therefore, the element with Z = 114 is the second p – block element in the 7th period.

Q6 Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Ans:

There are two elements in the 1st period and eight elements in the 2nd period., The third period starts with the element with Z = 11(sodium). Now, there are nine  elements in the third period. Thus, the 3rd period ends with the element with Z = 18(argon) i.e., the element in the 18th group of the third period has Z = 18. Hence, the element in the 17th group of the third period has atomic number Z = 17 ,chlorine(Cl) with atomic mass 35.453, which is a p block element.

Q7 Which element do you think would have been named by (i) Lawrence Berkeley Laboratory (ii) Seaborg’s group?
Ans:

(i) Lawrencium (Lr) with Z = 103  is an actinide which involves the filling of 5f orbital and Berkelium (Bk) with Z = 97  is an actinide which involves the filling of 5f orbital (both of them are f block elements).

(ii) Seaborgium (Sg) with Z = 106 is a d-block element present in 7th period and 6th column of periodic table.

Q8 Why do elements in the same group have similar physical and chemical properties?
Ans:

There are 18 groups in periodic table and each group is a  independent group. All the elements present in a group have same  electronic configuration of the atoms. The physical and chemical properties of elements depend on the number of valence electrons. Elements present in the same group have the same number of valence electrons. Therefore, elements present in the same group have similar physical and chemical properties.

Q9 What does atomic radius and ionic  radius really mean to you?
Ans:

Atomic radius and ionic radius are the periodic properties which are directly or indirectly related to the electronic configuration of their atoms and shows gradation on moving down a group or along a period.

Atomic radius is defined as the distance from the centre of the nucleus to the outer most shell containing the electrons. It measures the size of an atom. It is of 3 types:

A) Covalent radius- It is the one half of the distance between the centres of the nuclei of two adjacent similar atoms joined to each other by single covalent bond.

Covalent radius = inter nuclear distance in the bonded atoms/ 2

B) Metallic radius- It is defined as half the distance between the centres of the nuclei of two adjacent atoms in the metallic crystal.

C) Van der waal’s radius- It is defined as one half of the inter nuclear distance between 2 similar adjacent atoms belonging to the two neighbouring molecules of the same substance in the solid state.

Ionic radius means the radius of an ion (cation or anion). It is defined as the distance from the centre of the nucleus of the ion upto which it exerts its influence on the electron cloud. The ionic radii can be calculated by measuring the distances between the cations and anions in ionic crystals.

Since a cation is formed by removing an electron from an atom, the cation has fewer electrons than the parent atom resulting in an increase in the effective nuclear charge. Thus, a cation is smaller than the parent atom. For example, the ionic radius of Na+ ion is 95 pm, whereas the atomic radius of Na atom is 186 pm. On the other hand, an anion is larger in size than its parent atom. This is because an anion has the same nuclear charge, but more electrons than the parent atom resulting in an increased repulsion among the electrons and a decrease in the effective nuclear charge. For example, the ionic radius of F– ion is 136 pm, whereas the atomic radius of F atom is 64 pm.

Q10 How does atomic radius vary in a period and in a group? How do you explain the variation?
Ans:

Atomic radius of the elements generally decreases from left to the right in a period because on moving from left to right in a period the nuclear charge gradually increases by one unit and one electron is also added in the electron shell. Due to  this the electrons get attracted more and more towards the nucleus consequently the atomic radii decrease.

Atomic radius of the elements increases as we move downwords in a group because on moving down a group there is an increase in principal quantum number and thus, increase in the number of electron shells. Therefore the atomic size is expected to increase.

Q11 What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions. (i) F– (ii) Ar (iii) Mg2+ (iv) Rb+
Ans:

Isoelectronic species/ions/atoms are the species which have same number of electrons but different magnitude of nuclear charges and belongs to different atoms or ions.The isoelctronic ions with greater nuclear charge will have small size as compared to the ion with smaller nuclear charge.

(i) Fion has 9 + 1 = 10 electrons. Thus, the species isoelectronic with it will also have 10 electrons. Some of its isoelectronic species are Na+ ion (11 – 1 = 10 electrons), Ne (10 electrons), O2– ion (8 + 2 = 10 electrons), and Al3+ ion (13 – 3 = 10 electrons).

(ii) Ar has 18 electrons. Thus, the species isoelectronic with it will also have 18 electrons. Some of its isoelectronic species are S2– ion (16 + 2 = 18 electrons), Clion (17 + 1 = 18 electrons), K+ ion (19 – 1 = 18 electrons), and Ca2+ ion (20 – 2 = 18 electrons).

(iii) Mg2+ ion has 12 – 2 = 10 electrons. Thus, the species isoelectronic with it will also have 10 electrons. Some of its isoelectronic species are Fion (9 + 1 = 10 electrons), Ne (10 electrons), O2– ion (8 + 2 = 10 electrons), and Al3+ ion (13 – 3 = 10 electrons).

(iv) Rb+ ion has 37 – 1 = 36 electrons. Thus, the species isoelectronic with it will also have 36 electrons. Some of its isoelectronic species are Br ion (35 + 1 = 36 electrons), Kr (36 electrons), and Sr2+ ion (38 – 2 = 36 electrons).

Q12 Consider the following species: N3–, O2–, F–, Na+, Mg2+ and Al3+ (a) What is common in them? (b) Arrange them in the order of increasing ionic radii.
Ans:

(a) Each of the given species (ions) has the same number of electrons (10 electrons). Hence, the given species are isoelectronic, i.e

N3- has 7+3 = 10 electrons

O2-  has  8+2= 10 electrons

F- has 9+1 = 10 electrons

Na+ has 11-1 = 10 electrons

Mg2+  has 12-2 = 10 electrons

Al3+ has 13-3= 10 electrons

(b) The ionic radii of isoelectronic species increases with a decrease in the magnitudes of nuclear charge.

The arrangement of the given species in order of their increasing nuclear charge is as follows:

N3– < O2– < F < Na+ < Mg2+ < Al3+

Nuclear charge = +7 +8 +9 +11 +12 +13

Therefore, the arrangement of the given species in order of their increasing ionic radii is as follows:

Al3+ < Mg2+ < Na+ < F < O2– < N3–

Q13 Explain why cations are smaller and anions larger in radii than their parent atoms?
Ans:

Nucleus of an atom is positively charged and electrons are revolving around the nucleus in their respective orbitals.The more the cations or anions comes closer to the nucleus ,lesser will be their atomic radii. Therefore in the case of cations and anions are as follows:

A cation is a positively charged species that  has a fewer number of electrons than its parent atom, while its nuclear charge remains the same. As a result, the attraction of electrons to the nucleus is more in a cation than in its parent atom. Therefore, a cation is smaller in size than its parent atom.

On the other hand, an anion is a positively charged species that  has one or more electrons than its parent atom, resulting in an increased repulsion among the electrons and a decrease in the effective nuclear charge. As a result, the distance between the valence electrons and the nucleus is more in anions than in its parent atom. Hence, an anion is larger in radius than its parent atom.

Q14 What is the significance of the terms - ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy? Hint: Requirements for comparison purposes.
Ans:

Ionization enthalpy is the minimum amount of energy which is needed to remove the most loosly bound electron from a neutral isolated gaseous atom to form a cation.The cations are formed when the neutral atoms loses electrons. But for losing electrons they, should be in isolated gaseous form. Although the atoms are widely separated in the gaseous state, there are some amounts of attractive forces among the atoms. To determine the ionization enthalpy, it is impossible to isolate a single atom. But, the force of attraction can be further reduced by lowering the pressure. For this reason, the term ‘isolated gaseous atom’ is used in the definition of ionization enthalpy.

Ground state of an atom refers to the most stable state of an atom. If an isolated gaseous atom is in its ground state, then less amount energy would be required to remove an electron from it. Therefore, for comparison purposes, ionization enthalpy and electron gain enthalpy must be determined for an ‘isolated gaseous atom’ and its ‘ground state'.

Q15 Energy of an electron in the ground state of the hydrogen atom is –2.18 × 10–18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol–1.
Ans:

Ionization enthalpy is the minimum amount of energy which is needed to remove the loosely bound electron from a  isolated gaseous atom to form a cation.

It is given that the energy of an electron in the ground state of the hydrogen atom is –2.18 × 10–18 J.

Therefore, the energy required to remove that electron from the ground state of hydrogen atom is 2.18 × 10–18 J.

so Ionization enthalpy of atomic hydrogen = 2.18 × 10–18 J

Hence, ionization enthalpy of atomic hydrogen in terms of J mol–1 = 2.18 × 10–18 × 6.02 × 1023 J mol–1 = 1.31 × 106 J mol–1

Q16 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 ΔiH than B (ii) O has lower ΔiH than N and F?
Ans:

(i) Symmetry factor can be used for explaining higher ionization enthalpy of Be than B. The electronic configuration of Be is more symmetrical than that of B because both the occupied orbitals are filled in the case of Be while B has one half filled orbital in the 2p subshell. So during the process of ionization, the electron to be removed from beryllium atom is a 2s-electron, whereas the electron to be removed from boron atom is a 2p-electron. Now, 2s-electrons are more strongly attached to the nucleus than 2p-electrons and p orbital are also at slightly higher energy than the s orbital. Therefore, more energy is required to remove a 2s-electron of beryllium than that required to remove a 2p-electron of boron. Hence, beryllium has higher ΔiH than boron.

(ii)  According to symmetry factor, nitrogen has all the three 2p orbitals half filled while oxygen has one filled and two half filled orbitals in 2p subshell. In nitrogen, the three 2p-electrons of nitrogen occupy three different atomic orbitals. However, in oxygen, two of the four 2p-electrons of oxygen occupy the same 2p-orbital. This results in increased electron-electron repulsion in oxygen atom. As a result, the energy required to remove the fourth 2p-electron from oxygen is less as compared to the energy required to remove one of the three 2p-electrons from nitrogen. Hence, oxygen has lower ΔiH than nitrogen.

Fluorine contains one electron and one proton more than oxygen. As the electron is being added to the same shell, the increase in nuclear attraction (due to the addition of a proton) is more than the increase in electronic repulsion (due to the addition of an electron). Therefore, the valence electrons in fluorine atom experience a more effective nuclear charge than that experienced by the electrons present in oxygen. As a result, more energy is required to remove an electron from fluorine atom than that required to remove an electron from oxygen atom. Hence, oxygen has lower ΔiH than fluorine.

Q17 How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Ans:

Sodium is the 2nd member of group I (alkali metals) & magnesium is the 2nd member of group II (alkaline earth metals). The first ionization enthalpy of sodium is more than that of magnesium. This is primarily because  of smaller size & more symmetrical electronic configuration.For these reasons, the energy required to remove an electron from magnesium is more than the energy required in sodium. Hence, the first ionization enthalpy of sodium is lower than that of magnesium.

However, the second ionization enthalpy of sodium is higher than that of magnesium. This is because after losing 1  electron, sodium attains the stable noble gas configuration of neon (1s22s22p6) . On the other hand, magnesium, after losing 1 electron still has one electron in the 3s-orbital(1s2 2s22p63s1). In order to attain the stable noble gas configuration, it still has to lose one more electron. Thus, the energy required to remove the second electron in case of sodium is much higher than that required in case of magnesium. Hence, the second ionization enthalpy of sodium is higher than that of magnesium.

Q18 What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Ans:

The minimum amount of energy which is needed to remove the most loosely bound electron from a neutral isolated atom to form a cation is called as ionization energy or enthalpy.

The factors responsible for the ionization enthalpy of the main group elements to decrease down a group are listed below:

(i) Increase in the atomic size of elements: As we move down a group, the number of  main energy shells(n)  increases. As a result, the atomic size also increases gradually on moving down a group. As the distance of the valence electrons from the nucleus increases, the electrons are not held very strongly. Thus, they can be removed easily. Hence, on moving down a group, ionization energy decreases.

(ii) Increase in the shielding effect: The number of inner shells of electrons increases on moving down a group. Therefore, the shielding of the valence electrons from the nucleus by the inner core electrons increases down a group. As a result, the valence electrons are not held very tightly by the nucleus. Hence, the energy required to remove a valence electron decreases down a group.

Q19 The first ionization enthalpy values (in kJmol–1) of group 13 elements are : B Al Ga In Tl 801 577 579 558 589 How would you explain this deviation from the general trend?
Ans:

On moving down a group the ionization enthalpies generally decreses due to increase in atomic size and screening effect which is more than to compensate the effect of increase in nuclear charge. Consequently the electron becomes less and less tightly held by the nucleus as we move down the group.The sharp decrease in ionization enthalpy from B to Al is due to increase in size.In case of Ga ,there are 10 d electrons in its inner electronic configuration. Since the d electrons shield the nuclear charge less effectively than the s and p electrons,the outer electron are held strongly by the nucleus.As a result the ionization enthalpy increases slightly inspite of the increase in atomic size as we move from Al to Ga. The similar increase is also observed from In to Tl, which is due to presence of 14f electrons in inner electronic configuration of Tl which has poor shielding effect.

Q20 Which of the following pairs of elements would have a more negative electron gain enthalpy? (i) O or F (ii) F or Cl
Ans:

(i) Oxygen is a p block element present in 16th group and 2nd period while fluorine is also a p block element present in 17th group and 2nd period. An F atom has one proton and one electron more than O and as an electron is being added to the same shell, the atomic size of F is smaller than that of O. As F contains one proton more than O, its nucleus can attract the incoming electron more strongly in comparison to the nucleus of O atom. Also, F needs only one more electron to attain the stable noble gas configuration. Hence, the electron gain enthalpy of F is more negative than that of O.

(ii) Fluorine F and Cl belong to the same 17th group of the periodic table and they are p block elements with fluorine belonging to 2nd period and chlorine to 3rd group of periodic table. The electron gain enthalpy usually becomes less negative on moving down a group. However, in this case, the value of the electron gain enthalpy of Cl is more negative than that of F because of very small size of F atom. As a result there are strong interelectronic repulsions in the relatively small 2p subshell of F, thus the incoming electron does not feel much attraction. Therefore its electron affinity is small.

Q21 Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
Ans:

Oxygen belongs to group 16 of p block elements. It is the first member of group 16. It is a non metal and most abundant in earth crust(47%). When an electron is added to O atom to form O- ion, energy is released. Thus, the first electron gain enthalpy of O is negative. On the other hand, when an electron is added to O- ion to form O2- ion (superoxide ion), energy has to be given out in order to overcome the strong electronic repulsions between O- and e-. Thus, the second electron gain enthalpy of O is positive. Q22 What is the basic difference between the terms electron gain enthalpy and electronegativity?
Ans:
 Electron gain enthalpy Electronegativity It is the tendency of an atom to attract outside electrons It is the tendency of an atom to attract shared pair of electrons It is the property of an isolated atom It is the property of bonded atom It is the absolute electron attracting tendency of the atom It is the relative electron attracting tendency of an atom It has units like KJ/mol & Ev/atom It has no units

Q23 How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
Ans:

Linus Pauling described electronegativity as “the power of an atom in a molecule to attract electrons to itself.” Basically, the electronegativity of an atom is a relative value of that atom's ability to attract electron density towards itself when it bonds to another atom. The higher the electronegative an element, the more that atom will attempt to pull electrons towards itself and away from any atom it bonds to. The main properties of an atom dictate its electronegativity are its atomic number as well as its atomic radius.

Q24 Describe the theory associated with the radius of an atom as it (a) gains an electron (b) loses an electron
Ans:

Atomic radii- it is defined as the distance from the centre of the nucleus to the outer most shell containing the electrons.

(a) Along a period in periodic table ,the atomic radii of the elements generally decrease from left to right because the nuclear charges gradually increases on addition of electron in shell. As a result electrons are also getting attracted more and more towards nuclei.

(b) When an atom loses an electron, the number of electrons decreases by one while the nuclear charge remains the same. Therefore, the interelectronic repulsions in the atom decrease. As a result, the effective nuclear charge increases. Hence, the radius of the atom decreases.

Q25 Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Ans:

Ionization potential or enthalpy is the minimum amount of energy which is needed to remove the most loosely bound electron from a neutral isolated gaseous atom to form a cation also in gaseous state. The ionization enthalpy of an atom depends on the number of electrons and protons (nuclear charge) of that atom. Now, the isotopes of an element have the same number of protons and electrons. Therefore, the first ionization enthalpy for two isotopes of the same element should be the same.

Q26 What are the major differences between metals and non-metals?
Ans:
 METALS NON METALS Metals can lose electrons easily. Non-metals cannot lose electrons easily Metals are solids at room ( Exceptions – Hg, Ga ). temperature Non – metals may be solids, liquids or gases at room temperature Metals generally form ionic compounds. Non–metals generally form covalent compounds. Metals have low ionization enthalpies. Non–metals have high ionization enthalpies. Metals are less electronegative. They are rather electropositive elements. Non–metals are electronegative. Metals have a high reducing power. Non–metals have a low reducing power Metals have luster. They reflect light  from polished or freshly cut surface Non-metals do not have luster.( Exceptions – Diamond and Iodine ) Metals generally have high density Non-metals generally have low density They are good conductors of heat and  and electricity. They are usually bad conductors of heat electricity. (exception – carbon in the form of gas carbon and graphite ) Metals are malleable and ductile. They can be beaten into sheets and drawn into wires. Non-metals are not malleable and ductile They are brittle when solid.They can be crushed into powder They have a three dimensional crystal structure with metallic bonds They have different types of structures with covalent and van-der-Walls’ bonds Metals usually have high tensile  Strength Non- metals usually have low tensile strength Metals generally have 1 to 3 electrons in outermost shell of their atoms Non-metals generally have 4 to 8 in their outermost shell of the atom They generally form basic oxides. They generally form acidic oxides They act as reducing agents They act as oxidizing agents.

Q27 Use the periodic table to answer the following questions. (a) Identify an element with five electrons in the outer subshell. (b) Identify an element that would tend to lose two electrons. (c) Identify an element that would tend to gain two electrons. (d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.
Ans:

(a) The electronic configuration of an element having 5 electrons in its outermost subshell should be ns2 np5. This is the electronic configuration of the halogen group. Thus, the element can be F(2p5), Cl(3p5), Br(4p5), I(5p5), or At(6p5), or Uus(7p5).

(b) An element having two valence electrons will lose two electrons easily to attain the stable noble gas configuration. The general electronic configuration of such an element will be ns2. This is the electronic configuration of group 2 elements. The elements present in group 2 are Be(2s2), Mg(3s2), Ca(4s2), Sr(5s2), Ba(6s2), Ra (7s2).

(c) An element is likely to gain two electrons if it needs only two electrons to attain the stable noble gas configuration. Thus, the general electronic configuration of such an element should be ns2 np4. This is the electronic configuration of the oxygen family i.e O(2p4), S(3p4), Se(4p4), Te(5p4), Po(6p4), Uuh(7p4).

(d) Group 17 has metal, non–metal, liquid as well as gas at room temperature,i.e chlorine ,bromine  are non metals while iodine is a metal. Fluorides are liquids in state while chlorine, bromine & iodine are gases .

Q28 The increasing order of reactivity among group 1 elements is Li < Na < K < Rb CI > Br > I. Explain.
Ans:

These elements belongs to s block with general electronic configuration ns1 group 1 elements are called as alkali metals and are radioactive in nature.The elements present in group 1 have only 1 valence electron, which they tend to lose. Group 17 elements, on the other hand, need only one electron to attain the noble gas configuration. On moving down group 1, the ionization enthalpies decrease due to their very large atomic size, as a result of which the valence s electron can be readily removed .These values decrease down the group because of decrease in the magnitude of force of attraction with the nucleus on account of increased atomic radii and magnitude of screening effect. Thus, the increasing order of reactivity among group 1 elements is as follows:

Li < Na < K < Rb < Cs

They are called halogens (sea salt producers). In group 17, as we move down the group from Cl to I, the fluorine is the most reactive and iodine is least reactive because these halogens have low dissociation energies, as a result they easily dissociate into atoms and react with other substances. Also the halogens have high electron affinity and can easily gain an electron. Thus, the decreasing order of reactivity among group 17 elements is as follows:

F > Cl > Br > I

Q29 Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Ans:
 Element General outer electronic configuration s–block(alkali metals) ns1–2, where n = 2 – 7 p–block(metals & non metals) ns2np1–6, where n = 2 – 6 d–block(transition elements) (n–1) d1–10 ns0–2, where n = 4 – 7 f–block(inner transition elements) (n–2)f1–14(n–1)d0–10ns2, where n = 6 – 7

Q30 Assign the position of the element having outer electronic configuration (i) ns2 np4 for n = 3 (ii) (n - 1)d2 ns2 for n = 4, and (iii) (n - 2) f7 (n - 1)d1 ns2 for n = 6, in the periodic table.
Ans:

(i) Since n = 3, the element belongs to the 3rd period. It is a p–block element since the last electron occupies the p–orbital.

There are four electrons in the p–orbital. Thus, the corresponding group of the element

= Number of s–block groups (3s2) + number of d–block groups ([Ne]10 + number of p–electrons(3p4)

= 2 + 10 + 4

= 16

Therefore, the element belongs to the 3rd period and 16th group of the periodic table. Hence, the element is Sulphur ([Ne]103s23p4)

(ii) Since n = 4, the element belongs to the 4th period. It is a d–block element as d–orbitals are incompletely filled.

There are 2 electrons in the d–orbital.

Thus, the corresponding group of the element

= Number of s–block groups (4s2) + number of d–block groups (3d2)

= 2 + 2

= 4

Therefore, it is a 4th period and 4th group element. Hence, the element is Titanium ([Ar]183d24s2.

(iii) Since n = 6, the element is present in the 6th period. It is an f –block element as the last electron occupies the f–orbital. It belongs to group 3 of the periodic table since all f-block elements belong to group 3. Its electronic configuration is [Xe]54 4f7 5d1 6s2. Thus, its atomic number is 54 + 7 + 2 + 1 = 64. Hence, the element is Gadolinium.

Q31 The first (ΔiH1) and the second (ΔiH) ionization enthalpies (in kJ mol–1) and the (ΔegH) electron gain enthalpy (in kJ mol–1) of a few elements are given below: Elements ΔiH1 ΔiH ΔegH 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 MX2, (X=halogen). (f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?
Ans:

(a) Element V is likely to be the least reactive element. This is because it has the highest first ionization enthalpy (ΔiH1) and a positive electron gain enthalpy (ΔegH). It is a noble gas.

(b) Element II is likely to be the most reactive metal as it has the lowest first ionization enthalpy (ΔiH1) and a low negative electron gain enthalpy (ΔegH).The with lowest first ionization enthalpy is likely to be a reactive metal.

(c) Element III is likely to be the most reactive non–metal as it has a high first ionization enthalpy (ΔiH1) but less than the noble gas elements and the highest negative electron gain enthalpy (ΔegH).

(d) Element V is likely to be the least reactive non–metal since it has a very high first ionization enthalpy (ΔiH2) and a positive electron gain enthalpy (ΔegH).

(e) Element VI has a low negative electron gain enthalpy (ΔegH). Thus, it is a metal. Further, it has the lowest second ionization enthalpy (ΔiH2). Hence, it can form a stable binary halide of the formula MX2 (X=halogen).

(f) Element I has low first ionization energy and high second ionization energy. Therefore, it can form a predominantly stable covalent halide of the formula MX (X=halogen).

Q32 Predict the formula of the stable binary compounds that would be formed by the combination of the following pairs of elements. (a) Lithium and oxygen (b) Magnesium and nitrogen (c) Aluminium and iodine (d) Silicon and oxygen (e) Phosphorus and fluorine (f) Element 71 and fluorine
Ans:

CHEMICAL FORMULA- a brief representation of the molecule of a substance (element or compound) in terms of the symbols of the various elements present in it is called its chemical formula.

(a) LiO (Lithium dioxide)

(b) Mg3N2 (Magnesium nitrite)

(c) AlI3 (Aluminium triiodide)

(d) SiO2 (Silicon dioxide)

(e) PF3 or PF5 (Phosphorous trifluoride  or pentafluoride)

(f) The element with the atomic number 71 is Lutetium (Lu). It has valency 3. Hence, the formula of the compound is LuF3.(Lutetium trifluoride).

Q33 In the modern periodic table, the period indicates the value of: (a) Atomic number (b) Atomic mass (c) Principal quantum number (d) Azimuthal quantum number.
Ans:

The periodic table is a tabular arrangement of the chemical elements, organized on the basis of their atomic number (number of protons in the nucleus), electron configurations, and recurring chemical properties.

The value of the principal quantum number (n) for the outermost shell or the valence shell indicates a period in the Modern periodic table.

Q34 Which of the following statements related to the modern periodic table is incorrect? (a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell. (b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell. (c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell. (d) The block indicates value of azimuthal quantum number (l ) for the last subshell that received electrons in building up the electronic configuration.
Ans:

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

Option (b) is incorrect because the d-block has 10 columns  and maximum of 10 electrons can occupy all the orbitals in a d subshell. Since d sub shell can have a maximum of 5 orbitals amd 10 electrons, therefore, there are 10 vertical columns in d block.

Q35 Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell? (a) Valence principal quantum number (n) (b) Nuclear charge (Z) (c) Nuclear mass (d) Number of core electrons.
Ans:

Valence electrons are the electrons which are present in the outermost shell or valence shell of an atom and are responsible for the reactivity of any compound. Option (c) Nuclear mass does not affect the valence electrons because the nuclear mass is so small, it is considered as negligible.

Q36 The size of isoelectronic species — F–, Ne and Na+ is affected by (a) Nuclear charge (Z ) (b) Valence principal quantum number (n) (c) Electron-electron interaction in the outer orbitals (d) None of the factors because their size is the same.
Ans:

Isoelectronic species are the species belonging to different atoms or ions which have same number of electrons but different magnitudes of nuclear charges.

The size of an isoelectronic species increases with a decrease in the nuclear charge (Z). For example, the order of the increasing nuclear charge of F-, Ne, and Na+ is as follows:

F- < Ne < Na+

Z     9     10     11

Therefore, the order of the increasing size of F-, Ne and Na+ is as follows:

Na+ < Ne < F-

Q37 Which one of the following statements is incorrect in relation to ionization enthalpy? (a) Ionization enthalpy increases for each successive electron. (b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration. (c) End of valence electrons is marked by a big jump in ionization enthalpy. (d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Ans:

Ionization potential or enthalpy is the minimum amount of energy which is needed to remove the most loosely bound electron from a neutral isolated gaseous atom to form a cation also in gaseous state. Electrons in orbitals bearing a lower n value are more attracted to the nucleus than electrons in orbitals bearing a higher n value. Hence, the removal of electrons from orbitals bearing a higher n value is easier than the removal of electrons from orbitals having a lower n value.

Q38 Considering the elements B, Al, Mg, and K, the correct order of their metallic character is: (a) B > Al > Mg > K (b) Al > Mg > B > K (c) Mg > Al > K > B (d) K > Mg > Al > B
Ans:

Boron and aluminium belongs to p block elements and are group 13 elements whereas magnesium and potassium belongs to s block elements. Potassium belongs to group 1 and magnesium belongs to group 2. The metallic character of elements decreases from left to right across a period. Thus, the metallic character of Mg is more than that of Al.The metallic character of elements increases down a group. Thus, the metallic character of Al is more than that of B.

Considering the above statements, we get K > Mg.Hence, the correct order of metallic character is K > Mg > Al > B.

Q39 Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is: (a) B > C > Si > N > F b) Si > C > B > N > F (c) F > N > C > B > Si d) F > N > C > Si > B
Ans:

Boron is a p block element and is present in 13th group and 2nd period, carbon is a p block element and is present in 14th group and 2nd period, silicon is a p block element present in 14th group and 3rd period, nitrogen is p block element and belongs to 15th group and 2nd period while fluorine is also a p block element present in 17th group and 2nd period.The non-metallic character of elements increases from left to right across a period. Thus, the decreasing order of non-metallic character is F > N > C > B.Again, the non-metallic character of elements decreases down a group. Thus, the decreasing order of non-metallic characters of C and Si are C > Si. However, Si is less non-metallic than B i.e., B > Si. Hence, the correct order of their non-metallic characters is

F > N > C > B > Si.

Q40 Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is: (a) F > Cl > O > N (b) F > O > Cl > N (c) Cl > F > O > N (d) O > F > N > Cl
Ans:

All these elements are p block elements

The oxidizing character of elements increases from left to right across a period because of presence of vacant d orbitals in their valence shells. Thus, we get the decreasing order of oxidizing property as F > O > N.

Again, the oxidizing character of elements decreases down a group. Thus, we get F > Cl.

However, the oxidizing character of O is more than that of Cl i.e., O > Cl.

Hence, the correct order of chemical reactivity of F, Cl, O, and N in terms of their oxidizing property is F > O > Cl > N.