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Classification of Elements and Periodicity in Properties

Madhya Pradesh Board · Class 11 · Chemistry

NCERT Solutions for Classification of Elements and Periodicity in Properties — Madhya Pradesh Board Class 11 Chemistry.

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EXERCISES

3.1What is the basic theme of organisation in the periodic table?Show solution
The basic theme is arrangement of elements in order of increasing atomic number so that elements with similar properties come together in groups.

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3.2Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?Show solution
Mendeleev used atomic weights as the main property to classify elements in his periodic table. However, he did not follow atomic weight strictly in every case. Where atomic-weight order conflicted with similar properties, he ignored the order and placed elements with similar properties together; for example, he placed iodine with fluorine, chlorine and bromine instead of after tellurium.

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3.3What is the basic difference in approach between the Mendeleev's Periodic Law and the Modern Periodic Law?Show solution
The basic difference is that Mendeleev’s Periodic Law was based on atomic weights: properties of elements are periodic functions of atomic weights. The Modern Periodic Law is based on atomic numbers: properties of elements are periodic functions of atomic numbers. So modern classification uses atomic number as the fundamental basis, not atomic mass.

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3.4On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.Show solution
For the sixth period, n=6n=6. The available subshells are 6s6s, 4f4f, 5d5d and 6p6p. Their number of orbitals is:

- 6s6s: 1 orbital
- 4f4f: 7 orbitals
- 5d5d: 5 orbitals
- 6p6p: 3 orbitals

Total orbitals =1+7+5+3=16=1+7+5+3=16.

Each orbital can hold 2 electrons, so maximum electrons =16×2=32=16\times 2=32.

Hence the sixth period should have 32 elements.

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3.5In terms of period and group where would you locate the element with Z=114Z=114 ?Show solution
Element with Z=114Z=114 is flerovium (Fl). From the periodic table in the chapter, it lies in Group 14 and Period 7.

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3.6Write the atomic number of the element present in the third period and seventeenth group of the periodic table.Show solution
The element in the third period and seventeenth group is chlorine (Cl). Its atomic number is 17.

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3.8Why do elements in the same group have similar physical and chemical properties?Show solution
Elements in the same group have similar physical and chemical properties because they have the same valence shell electronic configuration. Since the number and distribution of electrons in the outermost orbitals are the same, they show similar chemical behaviour.

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3.9What does atomic radius and ionic radius really mean to you?Show solution
Atomic radius means the size of an atom, usually taken as half the distance between two bonded atoms in a molecule for non-metals, or half the distance between adjacent metal atoms in a crystal for metals. Ionic radius means the size of an ion, estimated from distances between ions in an ionic crystal. A cation has a smaller radius than its atom, while an anion has a larger radius than its atom.

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3.10How do atomic radius vary in a period and in a group? How do you explain the variation?Show solution
Across a period, atomic radius generally decreases from left to right because the effective nuclear charge increases while electrons are added to the same shell, so electrons are pulled closer to the nucleus. Down a group, atomic radius increases because new shells are added and shielding increases, placing the outer electrons farther from the nucleus.

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3.11What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.Show solution
Isoelectronic species are atoms or ions having the same number of electrons. From the chapter examples:

- F−F^- is isoelectronic with Ne
- ArAr is isoelectronic with K^+$? No; the direct match from the chapter’s idea is a species with 18 electrons such as Cl^-
- Mg2+Mg^{2+} is isoelectronic with Ne
- Rb+Rb^+ is isoelectronic with Kr

Using standard school-level knowledge, the closest one-electron matches are those species with the same electron count.

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3.12Consider the following species : N3−N^{3-} , O2−O^{2-} , F−F^- , Na+Na^+ , Mg2+Mg^{2+} and Al3+Al^{3+}Show solution
All the given species are isoelectronic and each has 10 electrons.

Their ionic radii increase as nuclear charge decreases. So the order of increasing ionic radii is:

Al3+<Mg2+<Na+<F−<O2−<N3−Al^{3+} < Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}

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3.13Explain why cation are smaller and anions larger in radii than their parent atoms?Show solution
A cation is smaller than its parent atom because it has lost one or more electrons, so the same nuclear charge attracts fewer electrons more strongly and the electron–electron repulsion decreases. An anion is larger than its parent atom because added electrons increase repulsion among electrons and reduce the effective attraction per electron, so the ion expands.

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3.14What is the significance of the terms — 'isolated gaseous atom' and 'ground state' while defining the ionization enthalpy and electron gain enthalpy?Show solution
The term isolated gaseous atom means the atom is considered alone in the gas phase so that its ionization enthalpy or electron gain enthalpy can be compared without interference from other atoms. Ground state means the atom is in its lowest energy state. These conditions are used for a standard, well-defined comparison of energies.

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3.15Energy of an electron in the ground state of the hydrogen atom is −2.18×10−18J-2.18 \times 10^{-18} J . Calculate the ionization enthalpy of atomic hydrogen in terms of Jmol−1J mol^{-1} .Show solution
Energy needed to remove one electron from one hydrogen atom is 2.18×10−18 J2.18\times 10^{-18}\,J.

For 1 mole:

2.18×10−18×6.022×1023=1.313×106 J mol−1 2.18\times 10^{-18}\times 6.022\times 10^{23} = 1.313\times 10^6\,J\,mol^{-1}

=1313 kJ mol−1 = 1313\,kJ\,mol^{-1}

So the ionization enthalpy of atomic hydrogen is about **1312 kJ mol−1^{-1}**.

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3.16Among the second period elements the actual ionization enthalpies are in the order Li<B<Be<C<O<N<F<NeLi < B < Be < C < O < N < F < Ne .Explain whyShow solution
The actual order is due to small differences in subshell energies and electron pairing.

- Be has higher ionization enthalpy than B because Be has a filled 2s22s^2 configuration, while boron loses a 2p2p electron, which is easier to remove.
- O has lower ionization enthalpy than N because oxygen has paired electrons in a 2p2p orbital, so electron–electron repulsion makes removal easier.
- Ne has the highest ionization enthalpy because it has a very stable closed shell.

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3.17How 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?Show solution
The first ionization enthalpy of sodium is lower than magnesium because Na has the electronic configuration 3s13s^1, so its outermost electron is easier to remove than Mg’s 3s23s^2 electron. But the second ionization enthalpy of Na is higher than Mg because after losing one electron, Na becomes Na+Na^+ with a noble-gas configuration, and removing another electron from this stable configuration requires much more energy. Magnesium after losing one electron is still not a noble-gas ion, so its second electron is removed more easily than sodium’s second.

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3.18What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?Show solution
Ionization enthalpy of main group elements tends to decrease down a group because:

- Atomic size increases, so the outer electron is farther from the nucleus.
- Shielding effect increases due to more inner shells.
- The nucleus attracts the valence electron less strongly, so less energy is needed to remove it.

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3.19The first ionization enthalpy values (in kJ mol⁻¹) of group 13 elements are :Show solution
The group 13 ionization enthalpies given are:

- B: 801 kJ mol−1^{-1}
- Al: 577 kJ mol−1^{-1}
- Ga: 579 kJ mol−1^{-1}
- In: 558 kJ mol−1^{-1}
- Tl: 589 kJ mol−1^{-1}

The deviation from smooth decrease is due to poor shielding by inner d and f electrons and the inert pair effect in heavier elements. These effects make Ga and Tl show higher values than expected.

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3.20Which of the following pairs of elements would have a more negative electron gain enthalpy?Show solution
Electron gain enthalpy becomes more negative across a period, and within group 17 it becomes less negative down the group. The chapter states that chlorine has a more negative electron gain enthalpy than fluorine because the added electron goes into the larger 3p3p level in chlorine and suffers less repulsion than in fluorine’s compact 2p2p level. So among the listed pairs, the more negative value is for chlorine in the comparison F or Cl.

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3.21Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.Show solution
The second electron gain enthalpy of oxygen would be positive. The first electron addition forms O−O^-, but adding a second electron to the negatively charged ion means the incoming electron is repelled by the anion. Energy must be supplied to overcome this repulsion, so the second electron gain enthalpy is positive.

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3.22What is the basic difference between the terms electron gain enthalpy and electronegativity?
3.23How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
3.24Describe the theory associated with the radius of an atom as it
3.25Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
3.26What are the major differences between metals and non-metals?
3.27Use the periodic table to answer the following questions.
3.28The 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.
3.29Write the general outer electronic configuration of s-, p-, d- and f- block elements.
3.30Assign the position of the element having outer electronic configuration (i) ns²np⁴ for n=3 (ii) (n-1)d²ns² for n=4, and (iii) (n-2) f⁷ (n-1)d¹ns² for n=6, in the periodic table.
3.31The first (ΔH1)(\Delta H_1) and the second (ΔH2)(\Delta H_2) ionization enthalpies (in kJ mol−1^{-1}) and the (ΔeqH)(\Delta_{eq}H) electron gain enthalpy (in kJ mol−1^{-1}) of a few elements are given below:
3.32Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.
3.33In the modern periodic table, the period indicates the value of :
3.34Which of the following statements related to the modern periodic table is incorrect?
3.35Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
3.36The size of isoelectronic species — F⁻, Ne and Na⁺ is affected by
3.37Which one of the following statements is incorrect in relation to ionization enthalpy?
3.38Considering the elements B, Al, Mg, and K, the correct order of their metallic character is :
3.39Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is :
3.40Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is :

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