Chemistry Ncert Exemplar 2019 Solutions for Class 11 Science Chemistry Chapter 3 - Classification Of Elements And Periodicity In Properties

Answer:

Those species have the same number of electrons are called as isoelectronic species. Atomic and ionic size depend on the nuclear charge as we move left to right across the period, nuclear charge increases so the force of attraction applying from the nucleus to the outermost shell increases and atomic size decreases. In isoelectronic ionic species, ionic radius decreases for the species having high positive charge due to more nuclear charge as compared to the negative charge species (less nuclear charge).
In the given example Na⁺, Mg²⁺ having positive charge so the ionic radius is smaller as compared to F^– , O^2–. Mg²⁺  has more positive charge therefore it has least atomic radius as compared to Na⁺ while in negative charge containing species, O^2– has more negative charge and less nuclear charge so high ionic radius.
Mg²⁺ < Na⁺ < F^– < O^2–

Hence, the correct answer is option B.
 

Answer:

Terbium is not the actinoid since actinoids start from atomic number 90 and end to 103.

Hence, the correct answer is option D.

Answer:

Subshell which is near to the nucleus will show more shielding effect. so the order of the shielding effect is  s > p > d > f.

Hence, the correct answer is option A.

Answer:

Answer:

Gadolinium is a lanthanoid element having the electronic configuration [Xe] 4f⁷ 5d¹ 6s²

Hence, the correct answer is option C.

Answer:

For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals this is not the correct statement, actually, 3d-orbitals are filled with electrons after 3p-orbitals and 4s-orbitals.

Hence, the correct answer is option C.

Answer:

Generally, electron gain enthalpy decreases on moving down the group because of atomic size increases, but in the given elements fluorine has lesser electron gain enthalpy than chlorine due to electron-electron repulsion in fluorine. So the order is F < Cl > Br > I

Hence, the correct answer is option C.

Answer:

The period number in the long form of the periodic table is equal to the maximum Principal quantum number of any element of the period.

Hence, the correct answer is option C.
 

Answer:

The elements in which electrons are progressively filled in 4f-orbital are called lanthanoids.

Hence, the correct answer is option C.

Answer:

Positive charge containing species have more nuclear charge due to which they have smallest ionic radius as compared to the species having negative charge. So in the given species positive charge containing I⁺ has least size while I^– has largest size. Therefore, the order is I^– > I > I⁺

​Hence, the correct answer is option D.
 

Answer:

The process of formation of O^2– in gas phase is unfavourable even though O^2– is isoelectronic with neon because electron repulsion outweighs the stability gained by achieving noble gas configuration.

Hence, the correct answer is option C.

Answer:

(a) The element with atomic number 57 belongs to d-block.

Hence, the correct answer is option C.

(b) The last element of the p-block in 6^th period is represented by the outermost electronic configuration  4f¹⁴ 5d¹⁰ 6s² 6p⁶. This is a 

Description

Answer:

(A) 1s² 2s² 2p⁶ = Ne
(B) 1s² 2s² 2p⁴ = O
(C) 1s² 2s² 2p⁶ 3s¹ = Na
(D) 1s² 2s² 2p⁵ =  F

Fluorine has maximum electron gain enthalpy among the given elements due to the smaller size and neon has the least electron gain enthalpy due to completely filled configuration electron can not add easily. Out of oxygen and sodium, oxygen has more electron gain enthalpy due to lesser size than sodium. Therefore, the order is A < C < B < D.

Hence, the correct answer is option A.

Answer:

Elements that are present in the third period and after the carbon (covalency is four) they can show covalency greater than 4.
P and S both can show covalency greater than 4 because they have vacant d orbitals to expand the octet.

Hence, the correct answers are options B and C.

Answer:

Group 1 and group 2 elements impart the color due to low energy for the ionisation.

Hence, the correct answers are options A and C.

Answer:

In representative elements include s and p block elements having group no 1, 2, 13, 14, 15, 16, 17, 18.
 

Answer:

Sulphur and Chlorine have more electron gain enthalpy as compared to other elements of their group due to weaker electron-electron repulsion.

Hence, the correct answers are options A and D.

Answer:

Full filled configuration and lesser atomic size elements have more first ionisation enthalpy, therefore, Helium has the highest first ionisation enthalpy in the periodic table.

Mercury and bromine are liquids at room temperature due to low melting points.

In any period, the atomic radius of alkali metal is the highest due to the lesser nuclear charge.

Hence, the correct answers are options A, C and D.

Answer:

Isoelectronic ions are having the same total number of electrons.

Answer:

(i) Al³⁺ < Mg²⁺ < Na⁺ < F^– (increasing ionic size)
Ionic radius decreases on increasing nuclear charge. So F^–  has lesser nuclear charge due to which its size is more. The order provided in the question is correct.

(ii) B < C < N < O (increasing first ionisation enthalpy): Ionisation enthalpy increases on moving left to right across the period due to decrease the atomic size. Half filled and full filled configuration compounds also have more ionisation enthalpy. Nitrogen has half-filled configuration (1s²2s²2p³) that's why it's ionisation enthalpy is more as compared to oxygen. The order provided in the question is not the correct order.

(iii) I < Br < Cl < F (increasing electron gain enthalpy): Electron gain enthalpy of chlorine is more than fluorine due to more electron-electron repulsion in case of fluorine. The order provided in the question is not correct.

(iv) Li < Na < K < Rb (increasing metallic radius): Metallic radius increases on moving down the group due to increase in the number of shells. The order provided in the question is correct.

Hence, the correct answers are options B and C.

Answer:

Electron gain enthalpy and Ionisation enthalpy are having the unit of energy kJ/mol. Electronegativity and metallic character are the unitless terms.

Hence, the correct answers are options A and D.

Answer:

Ionic radius is inversely proportional to the effective nuclear charge and directly proportional to the screening effect. Screening effect is the repulsion between the inner electrons due to which size increases.

Hence, the correct answers are options A and C.

Answer:

Aluminium is an element, belongs to the 3^rd period and group-13 of the periodic table. It is a solid metal which is a â€‹good conductor of electricity.

Hence, the correct answers are options A and C.

Answer:

Electron gain enthalpy (EGE) is the amount of energy release on adding an extra electron in the outer most shell of an isolated gaseous atom. If electron gain enthalpy is more it means the extra added electron is strongly attracted by the nucleus due to which more negative energy value will be observed. Generally on moving left to right EGE increases while decreases on going down the group. 
So, Fluorine should have more EGE but instead of fluorine chlorine has more EGE in the respective group due to the small size of the fluorine electron density is more therefore, more repulsion is there between the extra added electron and the electrons already present in the fluorine atom. While this repulsion is not found up to the large extent in the case of chlorine hence, Chlorine has more electron gain enthalpy than fluorine. 

Answer:

d- block elements which have incompletely filled d- orbitals either in the neutral form or in any stable oxidation state of metal are not considered as transition elements. So that's why we say all transition elements are d-block elements, but all d-block elements are not transition elements.
Zn Cd and Hg are the elements which are not considered as transition elements.

Answer:

Group : 1
Valency : 1
Outermost electronic configuration = 8s¹
Formula of Oxide = M₂O

Answer:

Ionisation enthalpy order of second period elements is: Li
As we move from left to right across a period, the ionization enthalpy keeps on increasing due to increased nuclear charge and decrease in atomic radius. However, there are some exceptions given below

1. The first ionisation enthalpy of B is lower than that of Be. This is due to the presence of fully filled electronic configuration of Be [1s²2s² ] which is a stable electronic arrangement. Thus, higher energy is required to take out an electron from fully filled 2s orbital. While in B (1s²2s²2p¹) contains valence electrons in 2s and 2p-orbital. It can easily lose its one electron from 2p orbital in order to achieve noble gas configuration. Therefore, the first ionisation enthalpy of B is lower than that of Be.

2. The first ionisation enthalpy of N is higher than that of O because in case of N, an electron needs to be removed from the more stable half-filled electronic configuration (1s²2s²2p³) Which is not present in oxygen (1s²2s²2p⁴⁾. Therefore, the first ionisation enthalpy of N is higher than that of O.

Answer:

(i) Carbon has the highest first ionisation enthalpy because the atomic size of carbon among the given elements is smaller due to which ionisation enthalpy is highest.
(ii) Aluminium is the metal because it is a good conductor of heat and electricity, it is hard, shiny, malleable and ductile as well these all are the properties of metals.

Answer:

The four important characteristic properties of p-block elements are the following:
(a) p-Block elements include both metals and non-metals but the number of non-metals is much higher than that of metals. Further, the metallic character increases from top to bottom within a group and non-metallic character increases from left to right along a period in this block.
(b) Their ionization enthalpies are relatively higher as compared to s-block elements.
(c) They mostly form covalent compounds.
(d) Some of them show more than one (variable) oxidation state in their compounds. Their oxidizing character increases from left to right in a period and reducing character increases from top to bottom in a group.

Answer:

(i) 72, 160
Fluorine and Neon they both found in the same period extreme right in the periodic table. Generally, on moving left to right atomic size decreases due to increase the nuclear charge, but this is applicable only upto group 17 as we go to the group 18 elements then they have the highest size in the respective period because to calculate the atomic size of noble gases we use the concept of Vanderwall radius due which they show the highest radius.

Answer:

The oxidation state of an element is based on its electronic configuration. The various oxidation states of a transition metal are due to the involvement of unpaired electrons in (n-1)d and outer ns electrons (paired/unpaired) in bonding.
For example, Ti (22, electronic configuration [Ar]3d²4s²) can show three oxidation states (+2, +3 and +4) in various compounds like TiO₂ (+4), Ti₂O₃ (+3) and TiO (+2).
The non-transition elements, mainly the p-block elements generally show oxidation states from +n to (n – 8) where n is the number of electrons present in the outermost shell. For example, phosphorus can show -3, +3 and +5 oxidation states. 
 

Answer:

Electron gain enthalpy of nitrogen is positive because the outermost electronic configuration of nitrogen (2s² 2p³) is half-filled which is very stable, so high energy required to remove an electron and for the addition of an extra electron to any of the 2p orbital requires energy. Therefore, ionisation enthalpy of nitrogen is more and electron gain enthalpy is positive.
Oxygen (2s²2p⁴) has 4 electrons in 2p orbitals and acquires stable configuration 2p³ after removing one electron, so ionization enthalpy for oxygen will be lower than nitrogen, while for electron gain enthalpy electron can be easily added in the outer most shell of oxygen as compared to nitrogen and energy is released.

Answer:

First  member of each group of s and p block elements shows anomalous behaviour due to the following reasons:
(i) Small size
(ii) High ionization enthalpy
(iii) High electronegativity
(iv) Absence of d-orbitals
Examples: Li in the first group shows different properties from the rest of elements like the covalent nature of its compounds, formation of nitrides.
Similarly, beryllium, the first element of second group differs from its own group in the following ways:
• Beryllium carbide reacts with water to produce methane gas while carbides of other elements give acetylene.
• Beryllium shows a coordination number of four while other elements show a coordination number of six.

Answer:

In p-block, when we move from left to right in a period, the acidic character of the oxides increases due to an increase in electronegativity. For example,
(i)In 2nd period B₂O₃ < CO₂ < N₂O₃ (acidic nature increases). 
(ii) In 3rd period A1₂O₃ < SiO₂ < P₄O₁₀ < SO₃ < Cl₂O₇ (acidic character increases). 
On moving down the group, acidic character decreases and basic character increases, e.g.,
Nature of oxides of group 13 elements
B₂O₃ = Acidic
Al₂O₃, Ga₂O₃ = Amphoteric
In₂O₃, Tl₂O₃ = Basic
Nature of oxides of group 15 elements
N₂O₅, P₂O₅ = Acidic
As₄O₁₀, Sb₄O₁₀ = Amphoteric 
Bi₂O₃ = Basic
$$\begin{gathered} {\mathrm{B}}_2{\mathrm{O}}_3 + 3{\mathrm{H}}_2\mathrm{O}\rightleftharpoons 2{\mathrm{H}}_3{\mathrm{BO}}_3 \\ \mathrm{B}{\left(\mathrm{OH}\right)}_3 + {\mathrm{H}}_2\mathrm{O}\to [\mathrm{B}{\left(\mathrm{OH}\right)}_4{]}^{-}+{\mathrm{H}}^{+} \\ {\mathrm{Al}}_2{\mathrm{O}}_3 \mathrm{is} \mathrm{amphoteric} \mathrm{is} \mathrm{nature} \mathrm{and} \mathrm{is} \mathrm{insoluble} \mathrm{in} \mathrm{water}. \mathrm{But}, \mathrm{it} \mathrm{dissociates} \mathrm{to} \mathrm{react} \mathrm{with} \mathrm{acids}. \\ {\mathrm{Al}}_2{\mathrm{O}}_3 +2\mathrm{NaOH} \stackrel{∆}{\to }2{\mathrm{NaAlO}}_2 + {\mathrm{H}}_2\mathrm{O} \\ {\mathrm{Al}}_2{\mathrm{O}}_3 + 6\mathrm{HCl} \stackrel{∆}{\to }2{\mathrm{AlCl}}_3 + 3{\mathrm{H}}_2\mathrm{O} \\ {\mathrm{Tl}}_2\mathrm{O} \mathrm{is} \mathrm{very} \mathrm{basic} \mathrm{just} \mathrm{like} \mathrm{NaOH} \mathrm{due} \mathrm{to} \mathrm{its} \mathrm{lower} \mathrm{oxidation} \mathrm{state}. \\ {\mathrm{Tl}}_2\mathrm{O} + 2\mathrm{HCl} \to 2\mathrm{TlCl} + {\mathrm{H}}_2\mathrm{O} \\ {\mathrm{P}}_4{\mathrm{O}}_{10} \mathrm{reacts} \mathrm{with} \mathrm{water} \mathrm{to} \mathrm{form} \mathrm{orthophosphoric} \mathrm{acid}. \\ {\mathrm{P}}_4{\mathrm{O}}_{10} + 6{\mathrm{H}}_2\mathrm{O} \to 4{\mathrm{H}}_3{\mathrm{PO}}_4 \\ {\mathrm{Cl}}_2{\mathrm{O}}_7 \mathrm{is} \mathrm{strongly} \mathrm{acidic} \mathrm{in} \mathrm{nature} \mathrm{and} \mathrm{forms} \mathrm{perchloric} \mathrm{acid} \mathrm{in} \mathrm{water}. \\ {\mathrm{Cl}}_2{\mathrm{O}}_7 + {\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{HClO}}_4 \end{gathered}$$


 

Answer:

The electronic configurations are as follows:
Na - [Ne] 3s¹
Mg - [Ne] 3s²
Sodium has one valence electron in 3s¹ and by losing this electron it acquires stable configuration. Therefore the 1st ionization energy of Na is less than Mg. Therefore, the 1st ionization energy of Na is less than Mg. After removing an electron the Na acquires noble gas configuration whereas Mg has one electron left. To remove an electron from a noble gas configuration high energy is required.
Therefore, the second ionization enthalpy is higher than that of magnesium.

Answer:

Exothermic Reaction: A chemical reaction in which heat is given out is known as Exothermic reaction. 

                 $\mathrm{C} + {\mathrm{O}}_2 \to {\mathrm{CO}}_2 + \mathrm{Heat}$

Endothermic Reaction: A chemical reaction in which heat energy is absorbed is known as Endothermic reaction. 

                         ${\mathrm{N}}_2 + {\mathrm{O}}_2 \stackrel{\mathrm{Heat}}{\to } 2\mathrm{NO}$

 

Answer:

(i)  Ionisation enthalpy increases from left to right in a period and decreases down the group. N has higher ionisation enthalpy than O due to extra stability of half-filled orbitals. Similarly, P has higher ionisation enthalpy than S due to half-filled orbitals.

          S < P < N < O

(ii) Nonmetals tend to gain electrons in chemical reactions and have a high attraction for electrons within a compound. So the general trend which we follow in the periodic table is non-metallic character decreases down the group and increases from left to right in a period. 


            P < S < N < O

Answer:

As we move from left to right across a period, the ionization enthalpy keeps on increasing due to increased nuclear charge and decrease in atomic radius. However, there are some exceptions given below

1. The first ionisation enthalpy of B is lower than that of Be. This is due to the presence of fully filled electronic configuration of Be [1s²2s²] which is a stable electronic arrangement. Thus, higher energy is required to take out an electron from fully filled 2s orbital. While in B (1s²2s²2p¹) contains valence electrons in 2s and 2p-orbital. It can easily lose its one electron from 2p orbital in order to achieve noble gas configuration. Therefore, the first ionisation enthalpy of B is lower than that of Be.

2. The first ionisation enthalpy of N is higher than that of O because in case of N, an electron needs to be removed from the more stable half-filled electronic configuration (1s²2s²2p³) Which is not present in oxygen (1s²2s²2p⁴). Therefore, the first ionisation enthalpy of N is higher than that of O.

 

Answer:

(a) The electronegativity generally increases on moving left to right across a period due to decrease in atomic size and increase in effective nuclear charge. As a result of the increase in effective nuclear charge, the attraction between the outer electrons and the nucleus increases therefore, electronegativity increases.

(b) On moving down the group ionisation enthalpy decreases due to the lesser force of attraction on the outermost electrons which means more atomic size hence, lower energy is required to remove the outermost electron.

Answer:

Metals are elements which lose electrons to gain positive charge. So metallic character decreases on increasing ionisation enthalpy. Ionisation enthalpy increases on going left to right across the period therefore, the metallic character decreases on moving left to right.

Answer:

Radius of Na⁺ is less than Na because in Na 11 electrons are present and in Na⁺ only 10 electrons are present. So in Na 11 electrons experience the lesser force of attraction while in Na⁺ 10 electrons experience more force of attraction applying from the nucleus that is why Na⁺ has lesser radius than Na.

Answer:

On moving down the group, electronegativity decreases because of increases atomic size. Francium has the largest size, therefore it is least electronegative.

Answer:

Atomic size decreases due to increasing the effective nuclear charge on moving left to right across the period.

Answer:

(i) Most reactive non-metal: B, because it has high ionisation energy and electron gain enthalpy.

(ii) Most reactive metal: A, because it has the lesser first ionisation enthalpy and negative electron gain enthalpy.

(iii) Least reactive element: D, because it has the highest first ionisation enthalpy and electron gain enthalpy is positive.

(iv) Metal forming binary halide: C, it is an alkaline earth metal because it has low first ionisation enthalpy but higher than that of alkali metals.

Answer:

(i) 1s² 2s² 2p⁶  : D,  this is the Ne which is a noble gas and noble gas has positive electron gain enthalpy.

(ii) 1s² 2s² 2p⁶ 3s¹: A, this is the Na which is an alkali metal and which has lower electron gain enthalpy.

(iii) 1s² 2s² 2p⁵: B, this is the F which is a halogen and has more electron gain enthalpy.

(iv) 1s² 2s² 2p⁴: C, this is the O and has lesser electron gain enthalpy than the halogens.

Answer:

Ionisation enthalpy increases from left to right across the period due to decrease in atomic size. The electrons present within the subshell has almost the same effective nuclear charge. Therefore, Assertion and reason both are correct statements and reason is correct explanation of assertion.

Hence, the correct answer is option B. 

Answer:

Boron(1s²2s²2p¹) has smaller first ionisation enthalpy than beryllium(1s²2s²) because the penetration of 2s electron to the nucleus is more than the 2p electron hence, 2p electron is more shielded by the inner core of electrons than the 2s electrons, due to which lesser energy is required to remove an electron from 2p orbital(boron) than 2s orbital(beryllium). Therefore, assertion and reason both are correct statements and reason is the correct explanation for the assertion.

Hence, the correct answer is option C.

Answer:

Electron gain enthalpy is the amount of energy released when an electron is added in the isolated gaseous atom. So electron gain enthalpy is negative, which increases on moving left to right and decreases down to the group. Down the group size increases due to which added electron experience less force of attraction and lesser amount of energy will be released. Therefore, assertion is wrong statement but reason is correct statement.

Hence, the correct answer is option D.

Answer:

The magnitude of electron gain enthalpy depends on the following factors: 
(i) Size of the atom: Smaller the size of the atom, stronger is the attraction for the added electron towards nucleus. Thus, smaller the size of the atom, greater is the electron gain enthalpy.

(ii) Nuclear charge: Greater the nuclear charge, Stronger is the attraction for the added electron towards the nucleus. Thus, electron gain enthalpy increases as nuclear charge increases.

(iii) Electronic configuration: The elements having stable configuration (half-filled or fully filled valence subshells) have very little or no tendency to accept additional electron hence, electron gain enthalpies are low or zero in these cases.

Variation of electron gain enthalpies in periodic table generally, the electron gain enthalpy becomes less negative in going from top to bottom in a group and more negative from left to right in a period.  However, electron gain enthalpy does not show a regular trend along a period or a group.

Answer:

Ionisation enthalpy is the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom.
Ionisation enthalpy depends on the following factors:
(i) Size of the atom: The larger the atomic size, smaller is the value of ionisation enthalpy. In a larger atom, the outer electrons are far away from the nucleus and thus force of attraction with which they are attracted by the nucleus is less and hence can be easily removed.
Ionisation enthalpy ∝ (Atomic size)^-1

(ii) Screening effect: Higher the screening effect, the lesser is the value of ionisation enthalpy as the screening effect reduces the force of attraction towards nucleus and hence, the outer electrons can be easily removed.
Ionisation enthalpy ∝ (Screening effect)^-1

(iii) Nuclear charge: As the nuclear charge increases among atoms having same number of energy shells, the ionisation enthalpy increases because the force of attraction towards nucleus increases.
Ionisation enthalpy ∝ Nuclear charge

(iv) Half filled and fully filled orbitals: The atoms having half filled and fully filled orbitals are comparatively more stable hence, more energy is required to remove the electron from such atoms. The ionization enthalpy is rather higher than the expected value in such cases.
Ionisation enthalpy ∝ Stable electronic configuration

ln general, the ionisation energy decreases down the group due to increase in atomic size. On the other hand, the ionisation energy increases across the period from left to right due to decrease in atomic size.

Answer:

The meaning of the periodic function is that when the elements are arranged in order of their increasing atomic numbers in the same period or a group, there is gradual change (increase or decrease) in a particular property. In a period, the gradual change is due to the gradual change in atomic number or electronic configuration from member to member.
In a group, the chemical properties of the elements remain nearly the same due to the same valence shell configuration which occurs after the difference of 2 or 8 or 18 or 32 in the atomic number. There is a gradual change in the physical properties due to change in the size of atoms.
Examples:
(i) Metallic nature decreases gradually in a period.
(ii) Electronegativity increases gradually in a period.
(iii) Atomic radii decrease gradually in a period.
(iv) In the first group, the ionisation enthalpy decreases from top to bottom.
(v) In the carbon group, metallic nature increases from top to bottom.
(vii) In the halogen group, the electronegativity decreases from top to bottom. 

Answer:

Alkali elements are Lithium(Li), Sodium(Na), Potassium (K), Rubidium (Ru), Cesium (Cs) and Francium (Fr). 
The electronic configuration is given by ns¹. Since only one electron is present in the outermost shell of these elements therefore, they all are present in group 1.
 

Answer:

1. Position of hydrogen: Hydrogen is placed in group I. However, it resembles the elements of group I (alkali metals) as well as the elements of group VII (halogens). Therefore, the position of hydrogen in the periodic table is not correctly defined.

2. Anomalous pairs: In certain pairs of elements, the increasing order of atomic masses was not obeyed. In these cases, Mendeleev placed elements according to similarities in their properties and not in increasing order of their atomic masses. For example, argon (Ar, atomic mass 39.9) is placed before potassium (K, atomic mass 39.1). Similarly, cobalt (Co, atomic mass 58.9) is placed before nickel (Ni, atomic mass 58.6) and tellurium (Te, atomic mass 127.6) is placed before iodine (I, atomic mass 126.9). These positions were not justified.

3. Position of isotopes: Isotopes are the atoms of the same element having different atomic masses but same atomic number. Therefore, according to Mendeleev’s classification, these should be placed at different places depending upon their atomic masses. For example, isotopes of hydrogen with atomic masses 1,2 and 3 should be placed at three places. However, isotopes have not been given separate places in the periodic table.

4. Some similar elements are separated and dissimilar elements are grouped together: In the Mendeleev’s periodic table, some similar elements were placed in different groups while some dissimilar elements had been grouped together. For example, copper and mercury resembled in their properties but they had been placed in different groups. At the same time, elements of group IA such as Li, Na and K were grouped with copper (Cu), silver (Ag) and gold (Au), though their properties are quite different.

5. Cause of periodicity: Mendeleev did not explain the cause of periodicity among the elements.

6. Position of lanthanoids (or lanthanides) and actinoids (or actinides):
The fourteen elements following lanthanum (known as lanthanoids, from atomic number 58-71) and the fourteen elements following actinium (known as actinoids, from atomic number 90 – 103) have not been given separate places in Mendeleev’s table.

Answer:

Answer:

The ionization enthalpies decrease regularly as we move down the group from one element to the other in group 1 and group 17. This trend can be easily explained on the basis of increasing atomic size and screening effects as follows:

i) On moving down the group, the atomic size increases gradually due to the addition of one new shell, due to which the distance of the valence electrons from the nucleus increases. Consequently, the force of attraction by the nucleus for the valence electrons decreases and hence, ionisation enthalpy decreases.

ii) With the addition of new shells, the shielding or the screening effect increases. As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence, the ionsation enthalpy decreases.