Class 11 - Chemistry - Element Classify Periodicity

Question 3.1

What is the basic theme of organisation of periodic table?

Answer.

The basic theme of organization of elements in the periodic table is to simplify and systematize the

study of the properties of all elements and their compounds.

Elements are classified in periods and groups according to their properties.

Elements with similar properties are placed in same group.

 

 

Question 3.2

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

Answer.

Mendeleev used atomic weight as the basis of classification of elements in the periodic table.

He organized components in groups and periods according to the increasing atomic weight.

He did not stick to this arrangement for long.

He realized that when elements were arranged according to their increasing atomic weight,

some elements did not fit within this scheme of classification.

 

Question 3.3

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

Answer.

Mendeleev’ Periodic Law states that physical and chemical properties of elements are periodic function of their atomic weights.

Modern Periodic Law states that physical and chemical properties of elements are periodic functions of their atomic numbers.

 

 

Question 3.4

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

Answer.

The sixth period corresponds to sixth shell.

The orbitals present in this shell are 6s, 4f, 5p and 6d.

The maximum number of electrons present in these sub shells is 2 (s), 14 (f), 6 (p), 10 (d).

So, the maximum number of electrons present in these sub shells are 32.

Since the number of elements in a period corresponds to the number of electrons in the shells

so, sixth period should have maximum 32 elements.

 

 

Question 3.5

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

Answer.

Period – 7

Group – 14 (Block – p)

 

 

Question 3.6

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

Answer.

The element is chlorine (Cl) with atomic number, Z = 17.

 

 

Question 3.7

Which element do you think would have been named by

  • Lawrence Berkeley Laboratory
  • Seaborg’s group

Answer.

  • Lawrencium (Lr), Z = 103 and Berkelium (Bk) which has atomic number, Z = 97
  • Seaborgium (Sg), Z = 106

 

 

Question 3.8

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

Answer.

The physical and chemical properties of elements depend on the number of valence electrons.

Elements present in same group have same number of valence electrons.

So the elements present in the same group have similar physical and chemical properties.

 

 

Question 3.9

What does atomic radius and ionic radius really mean to you?

Answer.

Atomic radii is the distance from the centre of nucleus to the outermost shell of electrons in the atom of any element.

Ionic radii can be found out by measuring distances between cations and anions in ionic crystals.

 

 

Question 3.10

How to atomic radius vary in a period and in a group? How do you explain the variation?

Answer.

Atomic radii increase down the group because a continuous increase in the number of electronic shells or orbital numbers

in the structure of atoms of the elements down a group.

Atomic radii across a period generally decrease because effective nuclear charge increases from left to right.

 

 

Question 3.11

What do you understand by isoelectronic species?

Name a species that will be isoelectronic with each of the following atoms or ions.

  • Ar
  • F-
  • Mg2+
  • Rb+

Answer.

Isoelectronic species are those species which have same number of electrons.

  • K+

  • Na+ 
  • Na+

  • Sr2+

 

 

Question 3.12

Consider the following species:

N3-, O2-, F-, Na+, Mg2+ and Al3+

  • What is common in them?
  • Arrange them in the order of increasing ionic radii.

Answer.

  • All these are isoelectronic in nature having 10 electrons each.

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

Al3+ < Mg2+ < Na+ < F < O2– < N3–

 

 

Question 3.13

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

Answer.

A cation is smaller than the parent atom as it has lesser electrons while its nuclear charge remains the same.

Size of anion will be larger than that of parent atom as it has additional electrons as compared to parent atom.

These extra electrons will result in increased repulsion among the electrons and thus there is decrease in effective nuclear charge.

 

 

Question 3.14

What is the significance of the terms – ‘isolated gaseous atom’ and ‘ground state’ while defining

the ionization enthalpy and electron gain enthalpy?

Answer.

Ionization enthalpy is the energy that is required to expel an electron from an isolated gaseous atom in ground state.

Atoms are widely separated in the gaseous state, still there some attractive forces among the atoms.

In case we need to find out the ionization enthalpy, we cannot isolate a single atom but the

forces of attraction can be reduced. by lowering pressure.

That is the reason why isolated gaseous atom is used in definition of ionization enthalpy.

An atom is more stable in its ground state.

Less energy is required to expel an electron if the isolated gaseous atom is present in its ground state.

 

 

Question 3.15

Energy of an electron in the ground state of the hydrogen atom is -2.18 x 10-18 J.

Calculate the ionization enthalpy of atomic hydrogen in terms of J mol-1.

Answer.

Energy of electron of hydrogen = - 2.18 x 1018 J

This means -2.18 x 10-18 J amount of energy is required to expel an electron from ground state in H atom.

For H-atom ionization enthalpy = -2.18 x 1018 J

Ionization enthalpy of H atom in J mol-1 = -2.18 x 10-18 x 6.02 x 1023 J mol-1 = 1.31 x 106 J mol-1

 

 

Question 3.16

Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne.

Explain why

  • Be has higher ΔiH1 than B?
  • O has lower ΔiH1 than N and F?

Answer.

  • Electronic configuration of Be is 1s2 2s2

The outermost electron is present in 2s orbital.

Electronic configuration of B is 1s2 2s2 2p1

Outermost electron is present in 2p orbital.

Therefore, lesser amount of energy is required to knock out 2p-electron. 

ΔiH1  for Be is higher than B

  • Electronic configuration of N (Z-7) – 1s2 2p2 2p1x 2p1y 2p1z

Electronic configuration of O (Z-8) – 1s2 2p2 2p2x 2p1y 2p1z

In Nitrogen, 2p orbitals are exactly half filled, so it is difficult to remove an electron from N than from O.

 

 

Question 3.17

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.

Electronic configuration of Na and Mg is

Na – 1s2 2s2 2p6 3s1

Mg - 1s2 2s2 2p6 3s2

First electron in both cases has to be removed from 3s-orbital but nuclear charge of Na (11) is lower than Mg (12).

So the first ionization energy of sodium is lower than Mg.

After loss of first electron, electronic configuration of

Na+ - 1s2 2s2 2p6

Mg+ - 1s2 2s2 2p6 3s1

In case of sodium ion, now electron needs to be removed from inert gas configuration which is very stable.

So the removal of electron requires more energy as compared to Mg.

Second ionization enthalpy of sodium is higher than that of magnesium.

 

 

Question 3.18

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

Answer.

Factors affecting the ionization enthalpy:

  1. Increase in atomic size of the elements. As we move down in a group, number of shells goes on increasing, due to which, atomic size increases.
  2. As electrons move away from nucleus they are not held very strongly.
  3. Thus, they can be removed easily, so ionization energy decreases on moving down a group.

 

  1. Increase in shielding effect. On moving down a group, number of inner shells of electrons increases,
  2. so shielding effect of valence electrons from nucleus increases.
  3. Valence electrons are not held very tightly by the nucleus. Energy required to remove a valence electron decreases down a group.

 

 

Question 3.19

The first ionization enthalpy values in (kJ mol-1) of group 13 elements are:

B          Al    Ga    In    Tl

801      577 579   558    589

How would you explain this deviation from the general trend?

Answer.

The decrease in ΔiH1 value from B to Al is due to bigger size of Al.

In Ga, there are 10 electrons in 3d orbitals, shielding is not effective in this orbital.

So, the valence electrons of Ga experience greater effective nuclear charge than Al.

Moving from Ga to In, ionization enthalpy decreases again due to an increase in atomic size and shielding.

Then moving from In to Tl, ionization enthalpy increases again,

the shielding of electrons provided by electrons in these orbitals is not very effective. Hence, valence electrons are held quite strongly.

 

 

Question 3.20

Which of the following pairs of elements would have a more negative electron gain enthalpy?

  • O or F
  • F or Cl

Answer.

  • O and F lies in second period. As we move from O to F, atomic size decreases due to gain of one electron.
  • It only needs one more electron to gain stable noble gas configuration. Hence, electron gain enthalpy of F is more electronegative than O.
  • F and Cl are in the same group. As we move down in a group, electron gain enthalpy becomes less negative.
  • In this case, value of electron gain enthalpy of Cl is more electronegative than F because atomic size of F is smaller than Cl.

 

 

Question 3.21

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

Answer.

When an electron is added to O atom it form O-, energy is released.

O (g) + e- --> O- (g)

When an electron is added to O- to form O2- ion, there is a need to be given out in order to overcome the strong electronic repulsions.

Second electron gain enthalpy of O is positive.

O- (g) + e- --> O2- (g)

 

 

Question 3.22

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

Answer.

Electron gain enthalpy is the tendency of an isolated gaseous atom to accept an additional electron to form a negative ion.

Electronegativity is the tendency of an atom in a chemical compound to attract a shared par of electrons.

 

 

Question 3.23

Would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?

Answer.

Electronegativity of an element is a variable property.

Electronegativity of N on Pauling scale is 3.0 in all nitrogen compounds is incorrect.

Electronegativity of N is different in NH3 and NO2.

 

 

Question 3.24

Describe the theory associated with the radius of an atom as it

  • Gains an electron
  • Loses an electron

Answer.

  • Gain of an electron leads to the formation of anions.
  • The size of an anion will be larger than the parent atom because there is addition of one or more electrons.
  • This addition leads to repulsion among the electrons and decreases the effective nuclear charge.
  • Loss of electron leads to the formation of cation.
  • A cation is smaller than its parent because it has lesser electrons than the parent atom.
  • Nuclear charge remains same.

 

 

Question 3.25

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

Answer.

Ionization enthalpy depends on the electronic configuration and nuclear charge.

Isotopes of an element have same number of protons and electrons,

so the first ionization enthalpy of two isotopes of the same element should be same.

 

 

Question 3.26

What are the major differences between metals and non-metals?

Answer.

METALS

NON METALS

They lose electrons easily.

They cannot lose electrons easily.

They cannot gain electrons easily.

They can gain electrons easily.

Their oxides are basic in nature.

Their oxides are acidic in nature.

They have low ionization enthalpies.

They have high ionization enthalpies.

They have high reducing power.

They have a low reducing power.

They form ionic compounds.

They generally form covalent compounds.

 

 

Question 3.27

Use the periodic table to answer the following questions.

  • Identify an element with five electrons in the outer subshell.
  • Identify an element that would tend to lose two electrons.
  • Identify an element that would tend to gain two electrons.
  • Identify the group having metal, non-metals, liquid as well as gas at the room temperature.

Answer.

  • Element belongs to Nitrogen family that is Group 15.
  • Element belongs to Alkaline Earth family that is Group 2.
  • Element belongs to Oxygen family that is Group 16.
  • Group 17 has metal, non-metal, liquid as well as gas at room temperature.

 

 

Question 3.28

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

Explain.

Answer.

Elements of group I have only one electron in their valence shells and thus, they have a strong tendency to lose their electron.

This tendency to lose electrons depends upon the ionization enthalpy.

As we go down in a group, ionization enthalpy decreases, so reactivity goes on increasing.

While taking of Group 17, they have 7 electrons in their valence shell,

they have strong tendency to accept one or more electron to make their electronic configuration stable.

But now, electron gain enthalpy and electronegativity decreases down the group, reactivity goes also decreases.

 

 

Question 3.29

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

Answer.

s-block elements have general outer electronic configuration ns1-2 where n = 2 – 7.

p-block elements have general outer electronic configuration ns2 np1-6 where n = 2 – 6

d-block elements have general outer electronic configuration (n – 1) d1-10 ns0-2 where n = 4 – 7

f-block elements have general outer electronic configuration (n – 2) f0-14 (n – 1) d0-1 ns2 where n = 6 - 7

 

 

Question 3.30

Assign the position of the elements having outer electronic configuration

  • ns2 np4 for n = 3
  • (n -1) d2ns2 for n = 4
  • (n – 2) f7 (n-1) d1ns2 for n = 6, in the periodic table.

Answer.

  • n = 3

Element belong to third period, p-block element.

Valence shell contains 6 electrons, group number is 10 + 6 = 16.

Electronic configuration is 1s2 2s2 2p6 3s2 3p4.

Element is sulphur.

  • n = 4

Electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d2 4s2

Element is Titanium.

  • n = 6

Group is 3.

Element is gadolinium (Z = 64)

Electronic configuration is [Xe] 4f7 5d1 6s2.

 

 

 

Question 3.31

The first (Δ1H1) and the second (Δ1H2) ionization enthalpies (in kJ mol-1) and the (ΔeqH)

electron gain enthalpy (in kJ mol-1) of a few elements are given below:

Elements           ΔH1       ΔH2     ΔeqH

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:

  • The least reactive element
  • The most reactive element
  • The most reactive non-metal
  • The least reactive non-metal
  • The metal which can form a stable binary halide of the formula MX2 (X = halogen)
  • The metal which can form a predominantly stable covalent halide of the formula MX (X = halogen)?

Answer.

  • Element II has lowest first ionization enthalpy and have negative electron affinity, hence, it is least reactive among all.
  • Element V has highest first ionization enthalpy and have positive electron affinity, hence it is most reactive.
  • Element III has high first ionization enthalpy and have negative electron affinity, hence it is most reactive non-metal.
  • Element V has high first ionization enthalpy and have positive electron affinity, hence, it is least reactive non-metal.
  • Element VI has low second ionization enthalpy and have a negative electron affinity. Hence, it is a metal. Thus, metal can easily form stable binary halide and has formula MX2.
  • Element I has low first ionization enthalpy and high second ionization enthalpy. Metal can easily form predominantly stable covalent halide with formula MX.

 

 

Question 3.32

Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of the elements.

  • Lithium and oxygen
  • Magnesium and nitrogen
  • Aluminium and iodine
  • Silicon and oxygen
  • Phosphorous and fluorine
  • Element 71 and fluorine

Answer.

  • LiO2 Lithium oxide
  • Mg3N2 Magnesium nitride
  • AlI3 Aluminium iodide
  • SiO2 Silicon dioxide
  • Phosphorous and fluorine
  • Z – 71 Element Lutetium has atomic number 71. It has valency 3. Formula is LuF3.

 

 

Question 3.33

In the modern periodic table, the period indicates the value of

  • atomic number
  • atomic mass
  • principal quantum number
  • azimuthal quantum number

Answer.

Correct option is (c)

Period in Modern periodic table indicates the value of ‘n’ that is principal quantum number.

 

 

Question 3.34

Which of the following statements related to the modern periodic table is incorrect?

  • The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
  • The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
  • Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
  • The block indicates values of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration.

Answer.

Correct option is (b)

d-block has 10 columns as a maximum of 10 electrons can occupy all the orbitals in a subshell.

 

 

Question 3.35

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?

  • Valence principal quantum number (n)
  • Nuclear charge (Z)
  • Nuclear mass
  • Number of core electrons

Answer.

Correct option is (c)

 

 

Question 3.36

The size of isoelectronic species – F-, Ne and Na+ is affected by

  • Nuclear charge (Z)
  • Valence principal quantum number (n)
  • Electron-electron interaction in the outer orbitals
  • None of the factors because their size is the same

Answer.

Correct option is (a)

The atomic size decreases with increase in nuclear charge. The arrangement according to increasing order of nuclear charge is F- < Ne < Na+.

Arrangement according to increasing order of atomic size is Na+ < Ne < F-.

 

 

Question 3.37

Which one of the following statements is incorrect in relation to ionization enthalpy?

  • Ionization enthalpy increases for each successive electron.
  • The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
  • End of valence electrons is marked by a big jump in ionization enthalpy.
  • Removal of electron from orbitals bearing lower n value is easier than from orbital having n value.

Answer.

Correct option is (d)

 

 

Question 3.38

Considering the elements B, Al, Mg and K the correct order of their metallic character is:

  • B > Al > Mg > K
  • Al > Mg > B > K
  • Mg > Al > K > B
  • K > Mg > Al > B

Answer.

Correct option is (d)

As we move from left to right in a period, the metallic character decreases. (Mg > Al)

As we move down in a group, metallic character also decreases. (Al > B)

Correct order will be K > Mg > Al > B

 

 

Question 3.39

Considering the elements B, C, N, F and Si, the incorrect order of their non-metallic character is

  • B > C > Si > N > F
  • Si > C > B > N > F
  • F > N > C > B > Si
  • F > N > C > Si > B

Answer.     

Correct option is (c).

As we move from left to right in a period, non-metallic character decreases. (F > N > C > B > Si)

As we move down the group metallic character decreases. (C > Si)

Correct sequence will be F > N > C > B > Si

 

 

Question 3.40

Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is

  • F > Cl > O > N
  • F > O > Cl > N
  • Cl > F > O > N
  • O > F > N > Cl

Answer.

Correct option is (a)

As we move from left to right in a period non-metallic character of the elements increases and as we move down in a group metallic character of elements decreases.

F > O > N (period)

F > Cl (group)

O is having higher oxidizing characteristics than Cl. (O > Cl)

So the correct order of chemical reactivity is F > Cl > O > N.

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