Classification of Elements and Periodicity in Properties

Assertion (A): Electron gain enthalpy becomes less negative as we go down a group.
Reason (R): Size of the atom increases on going down the group and the added electron would be farther from the nucleus.
(a) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
(b) Assertion and reason both are correct statements and reason is correct explanation of assertion.
(c) Assertion and reason both are wrong statement.
(d) Assertion is wrong statement but reason is correct statement.

(b) As one moves down the group, the electron gain enthalpy decreases because the atomic size grows and the new electron is further away from the nucleus.

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Assertion (A): Boron has a smaller first ionization enthalpy than beryllium. Reason (R): The penetration of a 2s electron to the nucleus is more than the 2p electron, hence, 2p electron is more shielded by the inner core of electrons that the 2s electrons.
(a) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
(b) Assertion is correct statement but reason is wrong statement.
(c) Assertion and reason both are correct statements and reason is correct explanation of assertion.
(d) Assertion and reason both are wrong statement.

(c) Because beryllium (1s2 2s2) has a fully filled, boron (1s2 2s2 2p1) has a lower initial ionisation enthalpy than beryllium (1s2 2s2). s-subshell. When compared to 2s-electrons, 2s-electrons are...

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Assertion and Reason Type Questions
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given.
Choose the correct option out of the choices given below each question.
Assertion (A): Generally, ionization enthalpy increases from left to right in a period.
Reason (R): When successive electrons are added to the orbitals in the same principal quantum level, the shielding effect of inner core of electrons does not increase very much to compensate for the increased attraction of the electron to the nucleus.
(a) Assertion is correct statement and reason is wrong statement.
(b) Assertion and reason both are correct statements and reason is correct explanation of assertion.
(c) Assertion and reason both are wrong statements.
(d) Assertion is wrong statement and reason is correct statement.

(b) As atomic size decreases, the ionization enthalpy increases from left to right over time. The effective nuclear charge of the electrons in the subshell is about the same.

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The statement that is not correct for periodic classification of elements is:
(i) The properties of elements are periodic function of their atomic numbers.
(ii) Non-metallic elements are less in number than metallic elements.
(iii) For transition elements, the 3d-orbitals are filled with electrons after3p-orbitals and before 4s-orbitals.
(iv) The first ionisation enthalpies of elements generally increase withincrease in atomic number as we go along a period.

Option (iii) is the answer. The Aufbau principle describes how electrons first fill low-energy orbitals (near to the nucleus) before moving on to higher-energy orbitals. They fill the orbitals...

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Ionisation enthalpies of elements of the second period are given below: Ionisation enthalpy/ kcal mol–1: 520, 899, 801, 1086, 1402, 1314, 1681, 2080. Match the correct enthalpy with the elements and complete the graph given in Fig. 3.1. Also, write symbols of elements with their atomic number.

Solution: N has a higher first ionisation enthalpy than O, despite the fact that O has a higher nuclear charge. This is because the electron in N must be removed from a more stable, exactly...

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An element belongs to the 3rd period and group-13 of the periodic table. Which of the following properties will be shown by the element?
(i) Good conductor of electricity
(ii) Liquid, metallic
(iii) Solid, metallic
(iv) Solid, non-metallic

Option (i) and (iii) are the answers. The element belonging to 3rd period and 13th group is aluminium which is a metal. Hence, it is solid, metallic and good conductor of electricity.

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Ionic radii vary in
(i) inverse proportion to the effective nuclear charge.
(ii) inverse proportion to the square of effective nuclear charge.
(iii) direct proportion to the screening effect.
(iv) direct proportion to the square of screening effect.

Option (i) and (iii) are the answers. Ionic radii decreases as the effective nuclear charge increases due to inverse proportional relation. Also, ionic radii increases as the screening effect...

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In which of the following options order of arrangement does not agree with
the variation of property indicated against it?
(i) Al3+ < Mg2+ < Na+ < F– (increasing ionic size)
(ii) B < C < N < O (increasing first ionisation enthalpy)
(iii) I < Br < Cl < F (increasing electron gain enthalpy)
(iv) Li < Na < K < Rb (increasing metallic radius)

Option (ii) and (iiii) are the answers. For increasing first ionization enthalpy, the order should be: B < C < O < N For increasing electron gain enthalpy, the order should be: I < Br...

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Which of the following statements are correct?
(i) Helium has the highest first ionisation enthalpy in the periodic table.
(ii) Chlorine has less negative electron gain enthalpy than fluorine.
(iii) Mercury and bromine are liquids at room temperature.
(iv) In any period, the atomic radius of alkali metal is the highest.

Option (i), (iii) and (iv) are the answers. Because of its larger size and lower electronic repulsion, chlorine has a higher negative electron gain enthalpy than fluorine.  

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Those elements impart colour to the flame on heating in it, the atoms of which require low energy for the ionisation (i.e., absorb energy in the visible region of the spectrum). The elements of which of the following groups will impart colour to the flame?
(i) 2
(ii) 13
(iii) 1
(iv) 17

Option (i) and (iii) are the answers. Ionization enthalpies are low in group 1 (alkali metals) and group 2 (alkaline earth metals). As a result, they give flame colour.

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Electronic configurations of four elements A, B, C and D are given below :

    \[\left( \mathbf{A} \right)\text{ }\mathbf{1}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{p}}^{\mathbf{6}}}~\]


    \[\left( \mathbf{B} \right)\text{ }\mathbf{1}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{p}}^{\mathbf{4}}}\]


    \[\left( \mathbf{C} \right)\text{ }\mathbf{1}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{p}}^{\mathbf{6}}}~\mathbf{3}{{\mathbf{s}}^{\mathbf{1}}}~\]


    \[\left( \mathbf{D} \right)\text{ }\mathbf{1}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{p}}^{\mathbf{5}}}\]


Which of the following is the correct order of increasing tendency to gain electron :
(i) A < C < B < D
(ii) A < B < C < D
(iii) D < B < C < A
(iv) D < A < B < C

Option (i) is the answer. (a) A – Is2 2s2 2p6 – Noble gas configuration B -1s2 2s2 2p4 – 2 electrons short of stable configuration C – 1s2 2s2 2p6 3.?1 – Requires one electron to complete 5-orbital...

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Comprehension given below is followed by some multiple-choice questions.
Each question has one correct option. Choose the correct option.
In the modern periodic table, elements are arranged in order of increasingatomic numbers which are related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic the table has been divided into four blocks, viz, s, p, d and f. The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Aufbau principle, the seven periods (1 to 7) have 2, 8, 8, 18, 18, 32 and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of f-block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table.
(i) The element with atomic number 57 belongs to
(a) s-block
(b) p-block
(c) d-block
(d) f-block
(ii) The last element of the p-block in 6th period is represented by the outermost electronic configuration.
(a) 7s2 7p6
(b) 5f 14 6d10 7s 2 7p 0
(c) 4f 14 5d10 6s2 6p6
(d) 4f 14 5d10 6s 2 6p 4
(iii) Which of the elements whose atomic numbers are given below, cannot be accommodated in the present set up of the long form of the periodic table?
(a) 107
(b) 118
(c) 126
(d) 102
(e) The electronic configuration of the element which is just above the element with atomic number 43 in the same group is ________.
(a) 1s2 2s2 2p6 3s2 3p6 3d5 4s2
(b) 1s2 2s2 2p6 3s2 3p6 3d5 4s3 4p6
(c) 1s2 2s2 2p6 3s2 3p6 3d6 4s2
(d) 1s2 2s2 2p6 3s2 3p6 3d7 4s2
(v) The elements with atomic numbers 35, 53 and 85 are all ________.

(a) noble gases
(b) halogens
(c) heavy metals
(d) light metals

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The formation of the oxide ion, O2- (g), from oxygen atom requires first an exothermic and then an endothermic step as shown below:

    \[\mathbf{O}\text{ }\left( \mathbf{g} \right)\text{ }+\text{ }\mathbf{e}\text{ }\to \text{ }{{\mathbf{O}}^{-}}\left( \mathbf{g} \right)\text{ };\text{ };\text{ }\text{ }\mathbf{HV}\text{ }=\text{ }\text{ }\mathbf{141}\text{ }\mathbf{kJ}\text{ }\mathbf{mol}\mathbf{1}\]


    \[{{\mathbf{O}}^{-}}\left( \mathbf{g} \right)\text{ }+\text{ }\mathbf{e}\text{ }\to \text{ }{{\mathbf{O}}^{\mathbf{2}-}}\left( \mathbf{g} \right)\text{ };\text{ }\text{ }\mathbf{HV}\text{ }=\text{ }+\text{ }\mathbf{780}\text{ }\mathbf{kJ}\text{ }\mathbf{mol}\mathbf{1}\]


Thus the process of formation of O2– in the gas phase is unfavourable even though O2- is isoelectronic with neon. It is due to the fact that
(i) oxygen is more electronegative.
(ii) addition of electron in oxygen results in larger size of the ion.
(iii) electron repulsion outweighs the stability gained by achieving a noble gas configuration.
(iv) O- ion has a comparatively smaller size than an oxygen atom.

Option (iii) is the answer. This is due to the fact that when an electron is introduced to a negatively charged ion, it is repelled rather than attracted. As a result, the addition of the second...

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The period number in the long form of the periodic table is equal to
(i) magnetic quantum number of any element of the period.
(ii) an atomic number of any element of the period.
(iii) maximum Principal quantum number of any element of the period.
(iv) maximum Azimuthal quantum number of any element of the period.

Option (iii) is the answer. Period number = maximum n of any element where 'n' stands for the principle quantum number. It determines the element's period number. Mg, for example, has a maximum main...

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The order of screening effect of electrons of s, p, d and f orbitals of a given shell of an atom on its outer shell electrons is:
(i) s > p > d > f
(ii) f > d > p > s
(iii) p < d < s > f
(iv) f > p > s > d

Option (i) is the answer. In every atom with more than one electron shell, this effect, known as the screening effect, describes the decrease in attraction between an electron and the nucleus. The...

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