Brief Questions and Answers: Chemistry Periodic Classification of Elements

Periodic Classification of Elements | Chemistry

Answer the following questions

24. Define modern periodic law.

The modern periodic law was developed which states that "the physical and chemical properties of the elements are periodic functions of their atomic numbers". 

25. What are isoelectronic ions? Give examples.

Ions with same number of electrons are called isoelectronic ions. Eg.Na⁺, F, Mg²⁺, Ne(10e⁻), Ar,Cl⁻, K⁺(18e⁻).

26. What is effective nuclear charge ?

The net nuclear charge experienced by valence electrons in the outermost shell is called the effective nuclear charge. It is approximated by the below mentioned equation.

Z_eff = Z-S Here, Z - atomic number,

S - Screening constant. 

27. Is the definition given below for ionisation enthalpy is correct?

"Ionisation enthalpy is defined as the energy required to remove the most loosely bound electron from the valence shell of an atom"

Correct. Ionisation enthalpy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.

M_(g)+ IE₁ → M⁺_(g)+ 1e⁻

Where IE₁ represents the first ionisation energy. 

28. Magnesium loses electrons successively to form Mg+, Mg2+ and Mg3+ ions. Which step will have the highest ionisation energy and why?

The effective nuclear charge of the cation is higher than the corresponding neutral atom. Thus the successive ionisation energies, always increases in the following order

Ie₁ < IE₂ < IE₃ < …………

Mg²⁺ has stable inert gas configuration. Removing and electron from it need more energy.

Thus Mg²⁺ + IE₃ → Mg³⁺ step has highest ionisation enthalpy

29. Define electronegativity.

It is defined as the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself.

30. How would you explain the fact that the second ionisation potential is always higher than first ionisation potential?

IE₂ > IE₁. Because the electron must be removed against net positive charge on the metal ion.

31. Energy of an electron in the ground state of the hydrogen atom is -2.8 x 10-18 J. Calculate the ionisation enthalpy of atomic hydrogen in terms of kJ mol-1.

Energy of an electron in the ground state of the hydrogen atom (E₁) = −2.8 × 10⁻¹⁸ J

Energy of an electron at infinity (E_∞) = 0

the ionisation enthalpy of atomic hydrogen = E_∞ − E₁ = 0 − (−2.8 × 10⁻¹⁸ J)

 = 0 + 2.8×10⁻¹⁸

 = 2.8 × 10⁻¹⁸ × 10⁻³ × 6.023 ×10²³ KJ mol⁻¹

ionisation enthalpy of atomic hydrogen in terms of kJ mol⁻¹ = 1.686 × 10³ KJ mol⁻¹

 (or) = 1686 KJ mol⁻¹

32. The electronic configuration of atom is one of the important factor which affects the value of ionisation potential and electron gain enthalpy. Explain

● It is expected that boron(B) has higher ionisation energy than beryllium since it has higher nuclear charge.

● However, the actual ionisation energies of beryllium and boron are 899 and 800 kJ mol⁻¹ respectively contrary to the expectation.

● It is due to the fact that beryllium with completely filled 2s orbital, is more stable than partially filled valence shell electronic configuration of boron. (2s²,2p¹)

● In case of elements such as beryllium (1s²,2s²), nitrogen (1s², 2s², 2p³) the addition of extra electron will disturb their stable electronic configuration and they have almost zero electron affinity.

● Noble gases have stable ns², np⁶ configuration, and the addition of further electron is unfavourable and requires energy. Halogens having the general electronic configuration of ns², np⁵ readily accept an electron to get the stable noble gas electronic configuration (ns²,np⁶), and therefore in each period the halogen has high electron affinity. (High negative values)

33. In what period and group will an element with Z = 118 will be present?

The electronic configuration is (1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f¹⁴ 5d² 5p⁶ 5d¹⁰ 5f¹⁴ 6s² 6p⁶ 6d¹⁰ 7s² 7p⁶ ). Therefore, this element belongs to period No.7 and group No.18 along with inert gases.

34. Justify that the fifth period of the periodic table should have 18 elements on the basis of quantum numbers.

For 5^th period, n = 5,

l =0 5s has 2 electrons = 2 elements

l =1 5p has 6 electrons = 6 elements

l =2 4d has 10 electrons = 10 elements

Total number of orbitals = 1+3+5 = 9

Maximum no of electrons in 9 orbitals = 9×2 = 18 = 18 elements

35. Elements a, b, c and d have the following electronic configurations:

a: 1s2, 2s2, 2p6

b: 1s2, 2s2, 2p6, 3s2, 3p1

c: 1s2, 2s2, 2p6, 3s2, 3p6

d: 1s2, 2s2, 2p1

which elements among these will belong to the same group of periodic table

a and c – inert gases – 18^th group

b and d – boron group – 3^rd group

36. Give the general electronic configuration of lanthanides and actinides?

Lanthanides : 4f ¹⁻¹⁴ 5d⁰⁻¹ 6s²

Actinides : 5f ⁰⁻¹⁴ 6d⁰⁻² 7s²

37. Why halogens act as oxidising agents?

Halogens having the general electronic configuration of ns², np⁵ readily accept an electron to get the stable noble gas electronic configuration (ns², np⁶), and therefore in each period the halogen has high electron affinity. (high negative values). Hence they act as Oxidising agents.

38. Mention any two anomalous properties of second period elements.

● The elements of the same group show similar physical and chemical properties.

● However, the first element of each group differs from other members of the group in certain properties.

● For example, lithium and beryllium form more covalent compounds, unlike the alkali and alkaline arth metals which predominantly form ionic compounds.

● The elements of the second period have only four orbitals (2s & 2p) in the valence shell and have a maximum co-valence of 4, whereas the other members of the subsequent periods have more orbitals in their valence shell and shows higher valences. For example, boron (B) forms BF₄ and aluminium forms A1F₆³⁻_.

39. Explain the pauling method for the determination of ionic radius.

● Ionic radius of uni-univalent crystal can be calculated using Pauling's method from the inter ionic distance between the nuclei of the cation and anion.

● Pauling assumed that ions present in a crystal lattice are perfect spheres, and they are in contact with each other

therefore, d = r_C++ r_A^- ………. (1)

● Where d is the distance between the centre of the nucleus of cation C⁺ and anion A− and r_C+ , r_A− are the radius of the cation and anion respectively.

Pauling also assumed that the radius of the ion having noble gas electronic configuration (Na⁺ and F⁻ having 1s², 2s², 2p⁶ configuration) is inversely proportional to the effective nuclear charge felt at the periphery of the ion.

i.e. r_C+ ∝  1 / (Z_{eff )} C⁺ ………… (2)

 r_A- ∝ 1 / (Z_eff)_A ………… (3) and

Where Z_eff is the effective nuclear charge and Z_eff = Z−S

Dividing the equation 2 by 3

r_C+ / r_A− = (Z_eff A⁻) / (Z_eff C⁺)      ………… (4)

By solving equation (1) & (4), r_C+ and r_A- can be calculated

40. Explain the periodic trend of ionisation potential.

Periodic variation in period:

● The ionisation energy usually increases along a period with few exceptions.When we move from left to right along a period, the valence electrons are added to the same shell, at the same time protons are added to the nucleus. This successive increase of nuclear charge increases the electrostatic attractive force on the valence electron and more energy is required to remove the valence electron resulting in high ionisation energy.

Periodic variation in group:

● The ionisation energy decreased down a group. As we move down a group, the valence electron occupies new shells, the distance between the nucleus and the valence electron increases.

● So, the nuclear forces of attraction on valence electron decreases and hence ionisation energy also decreases down a group.

41. Explain the diagonal relationship.

On moving diagonally across the periodic table, the second and third period elements show certain similarities. Even though the similarity is not same as we see in a group, it is quite pronounced in the following pair of elements.

The similarity in properties existing between the diagonally placed elements is called 'diagonal relationship'.

42. Why the first ionisation enthalpy of sodium is lower that that of magnesium while its second ionisation enthalpy is higher than that of magnesium?

Size of Mg is smaller than Na. Therefore the IE₁ of Na is less than Mg. Na loses an electron to become Na⁺, which acquires stable inert gas configuration. From a stable electronic configuration removal of electron requires more energy. Thus second ionisation enthalpy of Na is higher than that of magnesium.

43. By using paulings method calculate the ionic radii of K+ and Cl- ions in the potassium chloride crystal. Given that dK+-Cl- = 3.14 Ǻ

 d_k-cl- = r_k+ + r_cl- = 3.14 Ǻ               ……………..(1)

Electronic configuration of K⁺ & Cl^-

(1s²)(2s²2s⁶) (3s²3p⁶)

 Z_eff = Z − S

Z_eff K⁺ = 19 − [(0.35×7) + (0.85×8) + (1.00×2)]

 = 19 − 11.25 = 7.75

Z_eff Cl⁻ = 17 − 11.25 = 5.75

 rk⁺ / rcl⁻ = (Z_eff )Cl⁻ / (Z_eff)K⁺ = 5.75 / 7.75 = 0.74

 r_k+ = 0.74 r_cl−         ………….. (2)

Substituting (2) in (1)

 d_k+ − d_cl- = 0.74 r_cl- + r_cl- = 3.14 Ǻ

 1.74 r_cl-= 3.14 Ǻ

 r_cl-= 3.14 / 1.74 = 1.81 Ǻ

 r_k+ = 0.74 r_cl- = 0.74 × 1.81

 r_k+ = 1.33 Ǻ

r_k+ = 1.33 Ǻ

r_cl- = 1.81 Ǻ

44. Explain the following, give appropriate reasons.

i. Ionisation potential of N is greater than that of O

ii. First ionisation potential of C-atom is greater than that of B atom, where as the reverse is true is for second ionisation potential.

iii. The electron affinity values of Be, Mg and noble gases are zero and those of N (0.02 eV) and P (0.80 eV) are very low

iv. The formation of F- (g) from F(g) is exothermic while that of O2-(g) from O (g) is endothermic.

Reasons :

i) Nitrogen with 1s², 2s², 2p³ electronic configuration has higher ionisation energy (1402 kJ mol⁻¹) than oxygen (1314 kJ mol^-l). Since the half filled electronic configuration is more stable, it requires higher energy to remove an electron from 2p orbital of nitrogen. Whereas the removal one 2p electron from oxygen leads to a stable half filled configuration. This makes comparatively easier to remove 2p electron from oxygen

ii) C (Z=6) electronic configuration : 1s², 2s², 2p²

B(Z=5) electronic configuration : 1s², 2s², 2p¹

Though the nuclear charge of C is more than B, by losing one electron B acquires completely filled stable configuration. From a stable configuration removal of electron requires more energy. Hence, the second IE of B > C

iii) It is due to the fact that beryllium with completely filled 2s orbital, is more stable. The electronic configuration of beryllium (Z=4) in its ground state is 1s², 2s² nitrogen with (Z = 7) has 1s², 2s², 2p³ - half filled electronic configuration - more stable, Magnesium has 1s², 2s², 2p⁶ 3s² completely filled stable configuration phosphorus has 1s², 2s², 2p⁶ 3s² 2p³ - half filled stable configuration, It requires higher energy to remove an electron from stable half filled and completely filled electronic configuration

(iv) F_(g) + e⁻ → F_(g)⁻ + Energy

Fluorine readily accepts an electron to aquire stable inert gas configuration. Hence it is exothermic.

O_(g) + e⁻ → O ²⁻

This is endothermic. The electron is added to a negative ion which is very small and has high electron density. The additive electron leads to electron - electron repulsion.

45. What is screening effect? 

● In addition to the electrostatic forces of attraction between the nucleus and the electrons, there exists repulsive forces among the electrons.

● The repulsive force between the inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus.

● Thus, the inner shell electrons act as a shield between the nucleus and the valence electrons. This effect is called shielding effect.

46. Briefly give the basis for pauling's scale of electronegativity.

● Electro-negativity is not a measurable quantity. However, a number of scales are available to calculate its value.

● One such method was developed by Pauling's. He assigned arbitrary value of electro-negativities for hydrogen and fluorine as 2.1 and 4.0 respectively.

● Based on this the electro-negativity values for other elements can be calculated using the following expression

(χ_A − χ_B) = 0.182 √E_AB-(E_AA*E_BB)^1/2

Where E_AB, E_AA and E_BB are the bond dissociation energies of AB, A₂ and B₂ molecules respectively.

46. State the trends in the variation of electronegativity in group and periods.

Variation in periods:

● The electronegativity generally increases across a period from left to right. The atomic radius decreases in a period, as the attraction between the valence electron and the nucleus increases.

● Hence the tendency to attract shared pair of electrons increases. Therefore, electronegativty decreases.

Variation in group:

As we down the group, the electronegativity decreases.

Reason:

1) Atomic radius Increases

2) The attraction between the nucleus and valence electrons decreases.