Test: JEE Previous Year Questions- d & f-Block Elements - NEET MCQ

The correct answer is Option A.
Atomic and ionic radii of Lanthanides decrease from La to Lu.
Their order of ionic radii:  
Yb³⁺ < Pm3⁺< Ce³⁺ < La³⁺

The correct answer is Option B.
 
A solution of potassium chromate, when treated with an excess of dilute nitric acid doesn't oxidize nitric acid as N is already present in its maximum oxidation state (+5). The products formed are Cr2O₇^2- and H₂O.

The correct answer is Option D.

Due to lanthanide contraction, the ionic radii of Ln3+ (lanthanide ions) decreases from La3+ to Lu3+. Thus the lowest value (here 0.85Å) is the ionic radius of Lu3+.

The incorrect statement about the Cerium is only a C option.
C: The +4 oxidation state of cerium is not known in solutions [ Incorrect ]
As +4 oxidation state is common for the Cerium.
The correct statement about the Cerium are as follows:
A: The common oxidation states of Cerium are +3, and +4.
B : The +3 oxidation state of the Cerium is more stable than the +4 oxidation state.
D: Cerium (IV) acts as an oxidising agent

The correct answer is option B
It is given that an excess of KI reacts with CuSO₄​ solution and then Na2​S2​O3​ is added to it. This is hypo test and reactions are as follows:
2CuSO_4​ + 4KI → 2K₂​SO₄ ​+ 2CuI+I₂​
2Na₂​S₂​O₃​ + I2​→ Na₂​S_4​O₆ ​+ 2NaI
CuI2​ is not formed according to the following reaction.
2CuI_2​(unstable) → 2CuI+I₂​
Hence, the statement in option B is incorrect.
 

The correct answer is option A
When calomel (H_g2​Cl_2​) reacts with NH4​OH solution, the compound formed is [HgO.Hg(NH₂​)Cl]. It is black in colour.
Hg₂​Cl_2​+2NH₄​OH→[HgO.Hg(NH₂​)Cl]↓+NH₄​Cl+2H₂​O

The correct answer is Option C
Due to poor shielding effect of 4f electrons the nucleus experiences more effective nuclear charge and results in contraction which is known as lanthanide contraction. Due to lanthanide contraction 5th and 6th period transition series have approximately the same size. Therefore, lanthanide contraction is responsible for Zr and Hf having the same size.

The elements in the Lanthanide series are La, Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb and Lu.
In Lanthanides (Ce−Lu), the atomic and ionic radii decrease steadily. This steady decrease in atomic and ionic radii is known as Lanthanide Contraction. 
The contraction is due to the fact that f−orbitals are not capable of providing effective shielding for the valence electrons from the nuclear attraction.

The correct answer is option D
4f and 5f belongs to different energy levels, hence the shielding effect on them is not the same. Shielding of 4f is more than 5f.
 

The correct answer is option D
Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d and 6s orbitals is quite large. On the other hand, the energy difference between 5f, 6d and 7s orbitals is very less. Hence, actinoids display a large number of oxidation states.

The correct answer is option A
As oxidation state increases, electronegativity increases thus acidic characteristic increases not basic.
 

The correct answer is option  B
Ln(III) compounds are coloured due to the f−f transitions in the lanthanide but the colour is fainter as compared to the d−d transitions in transition metals.
 

The correct answer is Option D.

Ferrous oxide is more basic in nature than ferric oxide. Ferrous compounds are relatively more ionic than the corresponding ferric compounds. Ferrous compounds are less volatile than the corresponding ferric compounds as ferrous compounds are more ionic. 
 
But ferric compounds are more easily hydrolyzed than the corresponding ferrous compounds due to their more positive charge.