Test: JEE Previous Year Questions- Classification of Elements & Periodicity in Properties - 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. 

Yb³⁺ < Pm³⁺< Ce³⁺<La³⁺

Due to the lanthanide contraction size of +3 ions decreases from La³⁺ to Lu³⁺

According to modern periodic law, the properties of the elements are repeated after certain regular intervals when these elements are arranged in order of their increasing atomic numbers. 

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.

Isoelectronic species are the neutral atoms, cations or anions of different elements which have the same number of electrons but different nuclear charge. N3^-, F^– and Na^+ each have 10 electr

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.

This can be explained on the basis of z/e, where as z/e ratio increases, the size decreases and when z/e ratio decreases the size increases.

z is atomic number and e is number of electrons.

For Li⁺, z/e = 3/2 = 1.5

For F^-, z/e = 9/10 = 0.9

For O^2-, z/e = 8/10 = 0.8

For B³⁺, z/e = 5/2 = 2.5

The correct option is A.

While moving along a group from top to bottom acidic nature of oxides decreases and along a period left to right, acidic nature increases. 
Al, Si, P, S

This process is unfavourable in the gas phase because the resulting increase in electron-electron repulsion overweights the stability gained by achieving the noble gas configuration.

Nitrogen has a stable configuration.

So, ionization enthalpy of nitrogen is greater than that of Oxygen, so the correct order is (B)B<C<N<O: Increasing first ionization energy

H₃O⁺, NH₃ and CH₃^- are not only isoelectronic but are also iso-structural with pyramidal shape.

However CH₃⁺ is trigonal planar with one empty orbital perpendicular to this plane.

In each vertical column of transition elements, the elements of second and third transition series resemble each other more closely than the elements of first and second transition series on account of lanthanide contraction.

The pairs of elements such as Zr-­Hf, Mo-­W, Nb-­Ta, etc possess almost the same properties.

Lanthanide contraction is due to poor shielding of one of 4f electron by another in the sub-shell. 

In general as we move from left to right in a period, the ionisation enthalpy increases with increasing atomic number. The ionisation enthalpy decreases as we move down a group. P (1s² 2s² 2p⁶ 3s² 3p³) has a stable half-filled electronic configuration than S (1s² 2s² 2p⁶ 3s² 3p⁴). For this reason, ionization enthalpy of P is higher than S.

The correct option is D.

Each of K⁺, Cl^-, Ca²⁺, Sc³⁺ has 18 electrons.

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.

Isoelectronic species are those species which have the same number of electrons. In the given set, option b has the same no. of electrons in all given molecules(14 e^-).

On moving down the group, the basic tendency of hydrides of group 15 decreases.

O^2-, F^-, Na⁺, Mg²⁺,Al³⁺ are isoelectronic species (10 electrons each)
Cation having a positive charge is smaller in size than neutral atom whereas anion having a negative charge is larger in size than neutral atom.

While moving from left to right in periodic table basic character of oxide of elements will decrease.

And while descending in the group basic character of corresponding oxides increases.

Gadolinium is a chemical element with symbol Gd and atomic number 64. It is a silvery-white, malleable and ductile rare-earth metal. It is found in nature only in combined form.

The outer electron configuration of Gd (Atomic No.: 64) is: Xe 4f⁷ 5d¹ 6s²

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