Classification of Elements and Periodicity in Properties - 1 - JEE MCQ

Generally as we move from left to right in a period, there is regular decrease in atomic radii and in a group as the atomic number increases the atomic radii also increases. Thus the atomic radius of Sn should be less than lanthanides. La > Sn. But due to lanthanide contraction, in case of lanthanides there is a continuous decrease in size with increase in atomic number. Hence the atomic radius follow the given trend:

Ce > Sn > Yb > Lu. 

With increase in atomic number i.e. in moving down a group, the number of the principal shell increases and therefore, the size of the atom increases. But in case of f ­block elements there is a steady decrease in atomic size with increase in atomic number due to lanthanide contraction. As we move through the lanthanide series, 4f electrons are being added one at each step.

The mutual shielding effect of f electrons is very little. This is due to the shape of the f ­orbitals. The nuclear charge, however increases by one at each step. Hence, the inward pull experienced by the 4f electrons increases. This causes a reduction in the size of the entire 4fⁿ shell.

The electronic configurations are as follows;

V:(Ar)3d³4s²
Cr:(Ar)3d⁵4s¹
Mn:(Ar)3d⁵4s²
Fe:(Ar)3d⁶4s²

The second ionization of Cr means removal of electron from the stable configuration of 3d⁵.
The highest second ionization enthalpy is Cr
Hence a is the correct answer.

Isoelectronic species are those which have same number of electrons.

K⁺ = 19 – 1 = 18 ; Ca²⁺ = 20 – 2 = 18

Sc³+ = 21 – 3 = 18 ; Cl^– = 17 + 1 = 18

Thus all these ions have 18 electrons in them.

The Lanthanide Contraction is caused by a poor shielding effect of the 4f electrons. Gd because as atomic number increases, the atomic radius decreases. These electrons are do not shield good, causing a greater nuclear charge.

For isoelectronic species, as the z/e decreases, ionic radius increases
such as ionic radii ∝ 1/z.

Among isoelectronic species, ionic radii increase as the charge increases.

Order of ionic radii:

The number of electrons remains the same, but the nuclear charge increases with an increase in the atomic number, causing a decrease in size. 

(a): Among isoelectronic ions, the ionic radii of anions is more than that of cations.
Further, the size of the anion increases with an increase in negative charge, and the size of the cation decreases with an increase in positive charge.

Pauling introduced the electronegativity concept. He introduced the idea that the ionic character of a bond varies with the difference in electronegativity. A large difference in electronegativity leads to a bond with a high degree of polar character, i.e., the bond is predominantly ionic or vice versa.

(d): Among options (a), (c), and (d), option (d) has the highest ionisation energy because of extra stability associated with a half-filled 3p-orbital.
In option (b), the presence of 3d¹⁰ electrons offers a shielding effect, as a result, the 4p electrons do not experience much nuclear charge and hence the electrons can be removed easily.

Effective nuclear charge (i.e. Z/e ratio) decreases from F⁻ to N³⁻, hence the radii follows the order:
F^- < O^2- < N^3-

Z/e ratio calculations:

(A) For more polarity electronegativity difference should be  more E.N. (P) ≈ E.N. (H)

N–H > As–H > Sb–H > P–H is the order, (B) and (C) more is the size is gaseous phase less is hydration thus less is

hydrated size ∴ more is ionic mobility.

(D) Lattice energy depends on charge of cation and anion and internuclear distance thus order

MgO > SrO > NaF

(i) Ions having half-filled or fully filled orbitals have extra stability.
(ii) Larger the size of the cation, the more stable it will be.

Pb²⁺ (5d¹⁰ 6s²) has the most stable +2 oxidation state because its d-orbital is completely filled, making it more stable than Fe²⁺ (3d⁶).
Ag⁺ (4d¹⁰) is also more stable due to the completely filled d-orbital, and Ag²⁺ is not easily obtained.
Pb²⁺ is more stable than Sn²⁺ (4d¹⁰ 5s²) due to its larger size.

(i) Noble gases do not have covalent radii. They have only van der Waal’s radii.
(ii) Covalent radii are always smaller than corresponding van der Waal’s radii.
The atomic radius of neon, being a van der Waal’s radius, is larger than that of fluorine, which in fact represents its covalent radius.