RBSE Solutions for Class 11 Chemistry Chapter 3 Periodic Table

December 17, 2021

Rajasthan Board RBSE Class 11 Chemistry Chapter 3 Periodic Table

RBSE Class 11 Chemistry Chapter 3 Text Book Questions

RBSE Class 11 Chemistry Chapter 3 Multiple Choice Questions

Question 1.
The basis of Mendeleevs Periodic Law was:
(a) Valency
(b) Atomic weight
(c) Atomic number
(d) Atomic volume
Answer:
(b) Atomic weight

Question 2.
The element with electronic configuration (Ar) 3d² 4s² belongs to which block?
(a) s-block
(b) p-block
(c) d-block
(d) f-block
Answer:
(c) d-block

Question 3.
Which of the following element has highest ionisation enthalpy?
(a) Boron
(b) Carbon
(c) Nitrogen
(d) Oxygen
Answer:
(c) Nitrogen

Question 4.
The element from atomic number 58 to 71 are called:
(a) Transition elements
(b) Lanthanoids
(c) Actinoids
(d) Alkali Metals
Answer:
(b) Lanthanoids

Question 5.
Which of the following has highest atomic radius?
(a) F^–
(b) Cl^–
(c) Br^–
(d) I^–
Answer:
(d) I^–

RBSE Class 11 Chemistry Chapter 3 Very Short Answer Type Questions

Question 1.
What is the basis of Mendeleev periodic Law?
Answer:
According to Mendeleev’s periodic law, the physical and chemical properties of elements are the periodic function of their atomic weights. Thus elements are grouped according to their atomic weight.

Question 2.
What is the basis of modern periodic law?
Answer:
According to the Modern periodic law, the properties of the element and their compound are a periodic function of their atomic numbers. When element are placed according to atomic number, then similar electronic configuration are repeated after regular interval. Hence element are placed in periodic table according to increasing atomic.

Question 3.
What is the name given to elements of 17th group?
Answer:
The name of 17th group’s elements is halogens.

Question 4.
In a periodic table, the penultimate shell of which of the elements is incomplete?
Answer:
Transition elements because last two shells of these elements namely outermost and penultimate shells are incomplete.

Question 5.
Why there are 14 elements in 4f series?
Answer:
The lanthanides are f-block elements which contain seven f-orbitals. The maximum electrons can be filled in f-orbitals are 14. Hence, lanthanides have total 14 elements.

Question 6.
What are transuranic elements?
Answer:
The transuranic elements (also known as transgranic elements) are the chemical elements with atomic numbers greater than 92 (the atomic number of uranium). All of these elements are unstable and decay radioactively into other elements.

Question 7.
What is the IUPAC name and symbol of elements with atomic number 120?
Answer:
The IUPAC name for atomic number 120 is Unbinilium and symbol is Ubn.

Question 8.
Element X is placed at first position in 15 group. Write outer most electronic configuration of this element.
Answer:
Nitrogen is placed at first position in 15 groups. Atomic number is N = 7
Electronic configuration is 1s², 2s², 2p³, so outer most electronic configuration is [2s², 2p³]

Question 9.
Write two de-merits of Mendeleev’s periodic table.
Answer:
De-merits of Mendeleev’s periodic table are

Couldn’t explain position of Hydrogen.
Position of isotopes was not clear.

Question 10.
Write two demerits of Modern periodic table.
Answer:
Two demerits of Modern Periodic table are :
(a) Hydrogen resembles both the alkali metals and halogens, But it has been placed with the alkali metals.
(b) The lanthanides and actinides have not been placed in the main body of the table.

Question 11.
Write atomic number of element with symbol Uub.
Answer:
The atomic number of element Uub is 112.

Question 12.
Which element in a period has highest atomic size?
Answer:
Atomic radii decrease from left to right across a period. Therefore, first element in a period has highest atomic size.

Question 13.
Arrange in increasing order S^2-,Cl^–, Ar, K⁺, Ca⁺², Sc⁺³.
Answer:
It is an iso-electronic series. Isoelectronic series have same number of electrons but different numbers nuclear charge. All have 18 electrons. In an iso-electronic series, more the positive charge, smaller will be the size. Thus, increasing order will be. Sc⁺³ < Ca²⁺ < K⁺ < Cl^– < S^2-

Question 14.
What is Vander Waal’s radius?
Answer:
Vander Waals radius is defined as half of the internuclear distance of two non-bonded atoms of the same element on their closest possible approach and is denoted by r_v.

RBSE Solutions for Class 11 Chemistry Chapter 3 Periodic Table 1

Question 15.
Which of the following has smallest radius – Na⁺, Mg²⁺ and Al³⁺?
Answer:
On moving left to right in a period, with increasing atomic number, effective nuclear charge increases and atomic radius decreases. In isoelectronic series, with increase in positive charge ionic radius decreases. Thus Al³⁺ has smallest radius.

Question 16.
Which of the following has smallest radius- cation or parent atom?
Answer:
The cation has smallest radius. The reason for the smaller size of cation are:
1. A cation is formed by loss of one or more electrons from the parent atom. Generally the whole of the outermost shell of electrons is removed so that the resulting cation is smaller in size.
2. Also, in the formation of cation, electrons are removed but the magnitude of nuclear charge remains the same i.e., same nuclear charge will work on less number of electrons. In other words, effective nuclear charge per electron increases and as a result, electrons are strongly attracted and pulled towards the nucleus. This causes decrease in size.

Question 17.
How the ionization enthalpy changes on moving left to right in a period?
Answer:
As we move from left to right in a period, the atomic number of elements increases which means that the number of protons and electrons in the atoms increases (the extra electrons being added to the same shell). Due to large positive charge on the nucleus, the valence electrons are pulled more strongly by the nucleus and it becomes more and more difficult for the atoms to lose electrons. Thus, on moving from left to right in a period, the tendency of atoms to lose electrons decreases. Hence, the ionization energy increases across the period.

Question 18.
Which of the element has lowest ionization enthalpy in alkali metals?
Answer:
Caesium (376 kJ/mol) has lowest ionization enthalpy in alkali metals.

Question 19.
Which element has highest electron gain enthalpy in periodic table?
Answer:
Chlorine has highest electron gain enthalpy in periodic table. As we know that elctron gain enthalpy increases across period and decreases along group. Hence halogen has highest electron gain enthalpy. Among halogen fluorine has lower electron gain enthalpy than chlorine due to small size. In this way chlorine has highest electron gain enthalpy in periodic table.

Question 20.
Which of the following has highest second ionization enthalpy Na or Mg?
Answer:
Na has the highest second ionization enthalpy.

Question 21.
Which is the most electronegative element in a periodic table?
Answer:
Fluorine is the most electronegative elements in the periodic table.

Question 22.
What is the relation between Mulliken and Pauling Scale for measurement electronegativity?
Answer:
According to Mullikan scale :
Electronegativity = Ionisation enthalpy + Electron gain enthalpy / 2
When IE and electron gain enthalpy are expressed in eV.
Electronegativity = Ionisation enthalpy + Electron gain enthalpy / 540 When ionization enthalpy and electron gain enthalpy are expressed in kJ/mol.
χ_pauling = χ_mullikan/2.8
χ_pauling = Ionisation enthalpy + Electron gain enthalpy /5.6

Question 23.
Write the formula for calculation of percentage ionic character.
Answer:
Every ionic compound have some percentage of covalent character according to Fajan’s rule. The
percentage of ionic character in a compound having some covalent character can be calculated by the
following equation.
Percentage ionic character $=\frac{\text { Obsereed dipole moment }}{\text { Calculated dipole moment }} \times 100$

Question 24.
What is the Valency?
Answer:
Combining capacity of an element is known as Valency.

Question 25.
What is the valency of alkaline earth metal?
Answer:
Alkaline earth metals have a valency of 2. Alkaline earth metals are placed in group 2 of the periodic table, which implies that they each have 2 electrons in their outermost shells.

RBSE Class 11 Chemistry Chapter 3 Short Answer Type Questions

Question 1.
What is the necessity of classification of elements?
Answer:
The reason for necessity of classification of elements are:

Its help the systematic study, of the properties of elements.
Easy to understand and remember the properties of elements.
Its help us in learning efficiently the regular change that occurs in a group and the relation that present between its element.

Question 2.
In transition elements, on moving left to right, there is less change in atomic size. Why?
Answer:
Generally atomic size decreases on moving left to right across a period. But this trend is less systematic in the case of transition elements. As the number of electrons increases going across a period, there is more pull of these electrons toward the nucleus. However, with the d-electrons, there is some added electron- electron repulsion. Atomic and ionic radii increases from 3d series to 4d series but radii of third series elements are nearly same as those of the corresponding member of the second series because of lanthanoid contraction and poor shielding effect of 4f.

Question 3.
The radius of anion is always larger than parent atom. Why?
Answer:
The reasons for larger size of anion are :

The negative ion is formed by gain of one or more electrons in neutral atom and the number of electrons increases while the magnitude of nuclear charge remains the same.
As a result, the same nuclear charge acts on larger number of electrons than were present in the neutral atom. In other words, effective nuclear charge per electron is reduced and the electron cloud is held tightly by the nucleus. This results in increase in size.
Example : Chloride ion is larger in size than chlorine atom.

Question 4.
Cs⁺ from Cs is obtained easily as compared to Na⁺ from Na. Explain
Answer:
On moving down the group, ionisation energy decreases due to increase in size. As Cs is larger than Na, thus, ionisation energy of Cs is lower than Na. Therefore, electrons can be easily removed by Cs as compared to Na. This is the reason why Cs⁺ from Cs is obtained easily as compared to Na⁺ from Na.

Question 5.
Na₂O is basic whereas Cl₂O₇ is acidic. Explain with reason.
Answer:
Sodium oxide (Na₂O) is a strongly basic oxide. It reacts with water to form NaOH which is a strong base.
Na₂O + H₂O ➝ 2NaOH
Cl₂O₇ reacts with water to from HClO₄, which is a strong acid. Therefore, Cl₂O₇ is acidic.

Question 6.
Why Alkali metals do not form di-positive ions?
Answer:
Alkali metals have only one electron in their outermost energy levels, so (they can loose one electron to form a mono positive ion only. That’s why alkali metal do not form dispositive ions.

Question 7.
The ionisation enthalpy of noble gases is very high. Explain
Answer:
As noble gas posses complete valence shell configuration. Hence energy required to remove electron from valence shell of noble is highest in a period.

Question 8.
The ionization enthalpy of N is more than Oxygen. Why ?
Answer:
Nitrogen has configuration of 1s², 2s², 2p³, while oxygen has configuration is 1s², 2s², 2p⁴. In nitrogen p- sub shell is half filled. According to Hund’s half-filled and full filled orbitals are more stable. Hence it is difficult to remove an electron from 2p, so more amount of energy is required to remove such electrons. Therefore ionisation energy of nitrogen is higher than Oxygen.

Question 9.
Alkaline Earth Metal do not show variable oxidation states. Why?
Answer:
Alkaline earth metals do not show varible oxidation state. It possess only +2 oxidation state because after removal 2 electron from valence shell alkaline earth metal acquire stable configuration
For example ➝ ₁₂Mg ➝ 2s² 2s² 2p⁶ 3s²
Mg⁺² ➝ 1s² 2s² sp⁶

Question 10.
Calculate percentage of ionic character in CsF. The electronegativity of Cs and F are 0.7 and 0.4 respectively
Answer:
The electronegativity of Cs = 0.7 and F = 0.4
Percentage of ionic character :
= 16(χ_A – χ_B) + 3.5(χ_A – χ_B)²
Here, χ_A and χ_B are electronegativity’s of two atoms A and B
Percentage of ionic character
= 16(0.7 – 0.4) + 3.5 (0.7 – 0.4)²
= 16 (0.3) + 3.5 (0.09)
= 4.8 + 0.315
= 48 + 0.315
Percentage of ionic character = 5.115%

Question 11.
Explain the reason that electron gain enthalpy of F is less than Cl.
Answer:
Fluorine is the most electronegative element but still its electron gain enthalpy is less than that of chlorine .It is because of the small size of fluorine. The electron gain enthalpy of an element is the energy given off when a neutral atom in the gas phase gains an extra electron to form a negatively charged ion. A fluorine atom in the gas phase, for example, gives off energy when it gains an electron to form a fluoride ion.

Electron gain enthalpy of fluorine is less than that of chlorine because the size of the fluorine is less than the chlorine and due to this adding electron will only increase the electronic repulsion and thus will disfavour the process whereas adding electron to chlorine produces energy more as compare to fluorine.

Question 12.
Ionisation enthalpy of boron is less than Beryllium. Explain with reason.
Answer:
The electronic configuration of B and Be is
Be : 1s², 2s²
B : 1s², 2s², 2p¹
Here Be has a stable electron configuration, which means that more , energy is required to remove an electron from it. On the other hand, B has one electron in p-orbital.

Since p-orbitals are higher in energy than s-orbital, therefore less stable, it is easier for an electron to be removed from a p-orbital. Electrons located in p-orbitals are farther away from the nucleus, which means that they can withstand its effective nuclear charge better, more easy to remove from the atom.

Losing an electron from its outermost shell will make B more stable, Hence, it will be lower in energy.

Question 13.
Second electron gain enthalpy of Oxygen is greater than first. Explain with reason.
Answer:
When the first electron is added to the gaseous atom, it forms a uni-negative ion and enthalpy change during the processes is called first electron gain enthalpy. If an electron added to the uni-negative ion, it experiences a repulsive force from the anion. The energy to be supplied to overcome the repulsive force. Thus, When an electron is added to O atom to form O^– ion, the energy is released, so the first electron gain enthalpy is negative. But when second electron is added to O^– ion, due to electron-electron repulsions, energy is to be added. Hence, the second electron gain enthalpy is positive.
O_g + e^– ➝ O(g) (H)₁ = -14 kJ
O_(g)+ ^– ➝ O^2-_g (H)₂ = +780 kJ

Question 14.
The bond angle of NH₃ is greater than NF₃. Explain with reason.
Answer:
Electronic repulsion for bonded electrons—
l.p.-l.p. > l.p-b.p > b.p-b.p.
RBSE Solutions for Class 11 Chemistry Chapter 3 Periodic Table img 2
Nitrogen is more electronegative than Hydrogen, it attracts shared pair of electron and electron density accumulates near N, to minimize repulsion between bond pairs, angle increases
For NF₃ :

Bond pairs are near fluorine, this would lead to more b.p.l.p. repulsion and to minimize it, bond angle decreases. Although; NH₃ and NF₃ have same hybridization but F is more electronegative than H, the amount of p character in the N-F bonds in NF₃, will be higher than in N-H bonds in NH₃. Hence, bond angle in NF₃ is smaller.

RBSE Class 11 Chemistry Chapter 3 Long Answer Type Questions

Question 1.
Write a short note on long form of periodic table. Write characteristics of the periodic Table. What are it’s de-merit?
Answer:
Modem Periodic Lows:
In 1913, Moseley showed that when an electron moving with a high speed strikes a metal surface, then square root of frequency of X-rays produced is directly proportional to nuclear charge of atom of the metal. A plot of √Va against atomic number (Z) gave a straight line and not the plot of √v vs atomic weight.
RBSE Solutions for Class 11 Chemistry Chapter 3 Periodic Table img 4
Since the elements can not have same atomic number. Thus, he considered atomic number as fundamental property of element and gave modern periodic law by modifying Mandeleev’s periodic law. According to this law” The physical and chemical properties of elements are periodic function of their atomic numbers”. This means that if elements are arranged in order of this increasing atomic number, elements of similar properties resources at regular intervals or periodicity. The modern periodic law revealed important analogies among the 94 naturally occuring elements.

Advantages of Long Form of Periodic Table
The important features of modern periodic table are given below

In this periodic table elements are arranged in the order of increasing atomic number
In periodic table, there is non sub group
In modern periodic table, metals and non-metals are placed separately. Metals are placed on left hand side while non-metals and metalloids are placed on right hand side.
All isotopes are placed in a certain place because isotopes have similar atomic number.
In this periodic table the position of lanthanides and actinides is more clear. These are placed in downward of periodic table
In Mendeleev’s periodic table elements which were not following the order of increasing atomic mass are now arranged automatically in order of increasing atomic numbers.
It is easy to understand the electronic configuration of elements.
Division of elements in s, p, d and f block could helpful to understand their properties.

Shortcomings in the Long Form of Periodic Table:
Although long form of periodic table was better than Mendeleev’s periodic table in many aspects, however there are many shortcomings in this table in which some are given below

Position of hydrogen is not clear. It is similar to Mendeleev’s periodic table.
No, proper place is given to lanthanides and actinides in periodic table.
Some similar elements are placed at different positions for example Ba and Pb, Cu and Hg

Question 2.
Define atomic radius. Explain the factors affecting it and its periodicity.
Answer:
Atomic radius is the radius of an atom. Atomic radius is defined as the distance from the center of the nucleus to the outer most shell containing the electrons. It measures the size of an atom. If the atom is a metal, atomic radius states the metallic radius while if atom is a non-metal, atomic radius states the covalent radius. Factors affecting atomic Radius :

Effective nuclear charge : The net charge of nuclear experienced by outer most electron known as effective nuclear charge .It is represented by Z_eff with increase in Zeff atomic radius decreases.
Number of shells : With increase in number of shell .or principal quantum number, atomic radius increases.
Shielding or screening Effect : In multi-electron atoms, the outer-most electrons are shielded or screened from nucleus by the inner electrons. This is called shielding or screening effects. As the result the outer most electrons does not feel the full charge of the nucleus. The actual charge felt by electron is termed as effective nuclear charge. Hence with increase in shielding effect, atomic radius increases.
Number of Bonds : Covalent radius depends on number of bonds. With increase in number of covalent bonds, atomic radius decreases. Atomic radius is inversely proportional to number of bonds.

Variation of atomic radius in a period.
Atomic radius generally decreases from left to right in a period. This is due to the fact that on moving from left to right in a period, electrons are added one at a time to the same outermost shell. A proton is also added one at a time. Thus, the effective nuclear charge per electron increases. The outermost electrons experience increasing strong nuclear attraction, so the electrons come closer to the nucleus and more tightly bound to it. This results in decreasing the atomic radius.

Variation of Atomic radius in a group :
Atomic radius increases on moving down the group. This is due to the fact that on moving down the group number of orbits keeps on increasing along with the number of protons. The space required to accommodate the extra orbits takes prevalence and therefore the atomic size increases.

Question 3.
What is Ionization enthalpy of an element? Explain the factors affecting it and it’s periodicity?
Answer:
Ionisation enthalpy is the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom so as to convert it into a gaseous cation. It is represented as IE. The process can be represented as:
M(g) ➝ M⁺ (g) + e^–
Where, M(g) and M⁺ (g) are gaseous atom and resultant gaseous cation. Unit of ionisation enthalpy is kJ mol^-1. The value of ionisation enthalpy is always positive. Factors affecting ionization enthalpy are :
1. Size of atom : The Ionization enthalpy depends upon the size of atom, because with increase in size, the distance between nucleus and valence electrons increases and hence force of attraction , between nucleus and valence electron decreases. As a result of which the valence electrons are loosely held and smaller energy is required to remove an electron. Thus Ionization enthalpy decreases with increase in size and increase with decrease in size.

2. Magnitude of nuclear charge: As the nuclear charge increases, the force of attraction between nucleus and valence electrons increases and hence it is difficult to remove an electron from valence shell. Thus with increase in nuclear charge, the Ionization enthalpy increases.

3. Screening effect of the inner electron : The inner electrons present in shells between nucleus and valence shell reduce the attraction between nucleus and the outermost electrons. This shielding effect depends upon the number of inner electrons. Larger the number of electrons in the inner shells, the greater is the screening effect. More the screening effect, easier will be to remove an electron and hence lesser will be ionization enthalpy.

4. Penetration effect of the electron : s-electrons are closer to the nucleus than p-electrons, which is closer than d-electron and these in turn are closer than f-electrons of the same principal energy level. Hence s-electrons experience more attraction from the nucleus than p, d and f-electrons. Thus Ionization enthalpy to remove an electron from a given energy level decrease in order s > p > d > f.

5. Electronic Configuration: The atom having a more stable configuration has fewer tendencies to lose the electron and consequently, has high value of ionization enthalpy. For example : The noble gases have stable configuration. They have highest ionization enthalpy in their respective periods.

Periodicity:

Ionisation enthalpy decreases from top to bottom in a group due to increased atomic size and screening effect.
Ionisation enthalpy increases from left to right due to increased nuclear charge and smaller atomic radius.
Exception : I.E. of fulfilled sub-shell > I.E. of half-filled sub-shell > I.E. of partially filled sub-shell

Question 4.
Write applications of electronegativity and electron gain enthalpy.
Answer:
Electronegativity, which was initially defined by Pauling in 1932 as “the power of an atom in a molecule to attract electrons to itself,” is one of the most important concepts in chemistry, physics, and materials science..

Application of electronegativity:
1. Nature of bond : The concept of electronegativity can be used to predict whether the bond between similar or dissimilar atoms is non-polar covalent bond, polar covalent bond (or) ionic bond.
(a) When χ_A = χ_B i.e. χ_A – χ_B = 0, then A-B bond is non-polar covalent bond or simply covalent bond and is represented as A-B.
(b) When χ_A is slightly greater than χ_B, i.e. χ_A – χ_B is small, the A-B bond is polar covalent bond and is represented as A^δ- – B^δ+.
(c) When χ_A > χ_B, i.e., χ_A – χ_B is very large, A-B bond is more ionic or polar bond and is represented as A – B, Since χ_A >> χ_B.

2. Percentage of ionic character in a polar covalent bond : It χ_A – χ_A > 1.7 then covalent character is less then 50%. while ionic character more than 50%.
1. When (x_A – x_a) = 1.7 then bond is 50% ionic and 50% covalent.
2. When (x_A – x_B) < 1.7 then covalent character is more than 50%. While ionic character is less then 50%.

Application of electron gain enthalpy:

Nature of Bonds : If the difference between ionization enthalpy and electron gain enthalpy of two elements is less than lattice energy, then they form ionic bond. Whereas if the difference is more than lattice energy, the bond formed will be covalent.
Reactivity of Elements : The elements whose electron gain enthalpy is high are generally more reactive. In a period, halogens have highest electron gain enthalpy and are highly reactive.
Oxidising Tendency : As the value of electron gain enthalpy increases, oxidising tendency increases.

Question 5.
Define electron gain enthalpy. Explain the factors affecting it and its periodicity.
Answer:
Amount of energy released when an electron is added to an isolated gaseous neutral atom is called electron gain enthalpy. It is denoted ∆_egH
A (g) + e^– ➝ A^– (g) ; ∆H = ∆_egH
Electron gain enthalpy is measured in electron volts per atom or kJ per mole. Higher the energy released in the process of taking an extra electron, the higher will be electron gain enthalpy. Higher the value of electron gain enthalpy of an atom, the more is the tendency to change into anion.

Factors Affecting Electron Gain Enthalpy :

Effective Nuclear Charge : Electron gain enthalpy is directly proportional to nuclear charge. With increase in effective nuclear charge, electron gain enthalpy increases.
Atomic size: Electron gain enthalpy is inversely proportional to atomic size, with increase in atopiic size, value of electron gain enthalpy decreases.
Shielding Effect : Electron-gain enthalpy is inversely proportional to the shielding effect. As shielding effect increases, effective nuclear charge decreases and electron gain enthalpy decreases.
Half Filled and Fully Filled orbitals: Half-filled and fully filled orbitals are stable. Therefore, their electron gain enthalpy is low.

Periodicity:

Electron gain enthalpy increases from left to right across a period due to decrease in atomic size and increase in ionisation enthalpy.
Electron enthalpy decreases from top to bottom in going down the group due to increased atomic size.

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