Retro (Past 13 Year) JEE Main (D Block Elements) - JEE MCQ

(a) V²⁺ = 3 unpaired electrons, Cr²⁺ = 4 unpaired electrons
Mn²⁺ = 5 unpaired electrons, Fe²⁺ = 4 unpaired electrons
Hence, the order of paramagnetic behaviour should be V²⁺ < Cr²⁺ < Fe²⁺ < Mn²⁺
(b) Ionic size decreases from left to right in the same period.
(c) (As per data from NCEFIT)
Co³⁺/Co²⁺ = 1.97; Fe³⁺/Fe²⁺ = 0.77;Cr³⁺/Cr²⁺ =-0.41, Sc³⁺ is highly stable (it does not show + 2).
(d) The oxidation state increases as we go from group 3 to group 7 in the same period.

SRP value normally increases from left to right in the period of cf-block elements. Some SRP value are exceptionally higher due to stability of product ion, e.g.

(a) FeO > Fe₂O₃ (basic). Thus, it is correct statement.
(b) FeCI₂ > FeCI₃ (ionic), larger the charge, greater the polarising power. Thus, greater the covalent nature. Thus, the given statement is correct.
(c) Fe²⁺ salts are more ionic thus less volatile than Fe³⁺ salts. Thus, given statement is correct.
(d) Greater the covalent nature, more easily they are hydrolysed, Thus, FeCI₃ is more hydrolysed than FeCI₂. Thus, statement (d) is incorrect.

(a) There is a gradual decrease in the radii of the lanthanoids with increasing atomic number a case of lanthanide contraction, thus true.
(b) Ionisation potential for the formation of Lu³⁺ is comparatively low, hence +3 state is favourable, thus true,
(c) Due to lanthanoid contraction, Zr and Hf; Nb and Ta,
Mo and W have the same size and thus, similar properties and thus separation is notfwsy, thus true.
(d) Formation of +4 state requires very high energy, thus incorrect.

All the electrons in 47-orbital are unpaired, hence stable. Thus,
Gd (64) has EC as [Xe]₅₄ 4f⁷, 5d\ 6s² instead of [Xe]₅₄ 4f⁸, 6s².

Usually across the first transition series, the negative values for standard electrode potential decrease except for Mn due to stabled⁵-configuration. So, correct order: Mn > Cr > Fe > Co

The most common oxidation state of lanthanoid is +3. Lanthanoids in +3 oxidation state usually have unpaired electron in f-subshell and impart characteristic colour in solid as well as in solution state due to f-f transition.

(Except lanthanum and lutetium)

As oxidation state increases, electronegativity increases thus acidic character increases not basic.

Lanthanoids : [Xe] f^1-145d^0-16s²
Actinoids : [Rn]5f^1-14 6d^0-17s²
Lanthanoids and actinoids use core d and f-orbitals also to show higher oxidation state. As actinoids have comparatively low energy difference between f and d-orbitals, thus they show more number of oxidation states.

4f-orbital is nearer to nucleus as compared to 5f-orbital. Therefore, shielding of 4f is more than 5f-orbital.

The actinoids (5f-element) exhibit more number of oxidation states in general than the lanthanoids because 5f-orbitals extend farther from the nucleus than the 4f-orbitals.
Thus, the attraction on the 5f-electrons in actinoid series decreases. Hence, they can easily participate in bonding and therefore, exhibits more number of oxidation states.

Imperfect shielding on outer electrons by 4f-electrons from the nuclear charge is due to the diffused large shape of 4f-orbitals. This imperfect shielding is unable to counterbalance the effect of the increased nuclear charge. Hence, the net result shows contraction in size.

Following reaction takes place during Bessemerisation
2Cu₂O + CU₂S → 6Cu + SO₂

+3 and +4 states are shown by Ce in aqueous solution. Thus, statement (c) is incorrect.

Cu, Ag, Au are called coinage metals because in the early times, they were used in making coins.

Metal oxides lying below Hg in electrochemical series decompose to form metal.

Mercury (Hg) solidifies after glass, so used mercury for making good quality mirrors.

Ammonia can dissolve precipitate of AgCI due to the formation of complex as given below.
AgCI + 2NH₃ → [Ag(NH₃)₂CI]

Hgl₂ becomes soluble in water, when Kl is added to it due to the formation of K₂Hgl₄ complex which is soluble in water.
Hgl₂ + 2KI → K₂Hgl₄(Soluble)
But when Hgl₂ is heated, Hgl₂ decomposes to give violet coloured fumes of I₂ that deposits on the cooler parts of the test tube.

Cerium, Ce₅₈ = [Xe] 4f¹5d¹6s². Its most stable oxidation state is +3 but +4 is also existing.

Larger the number of unpaired electrons, greater the magnetic moment, 
where, n = Number of unpaired electrons

Number of unpaired electrons = 5

Number of unpaired electrons = 4

Number of unpaired electrons = 2


Number of unpaired electrons = 4
Hence, Mn²⁺: [Ar] 3d⁵ with five unpaired electrons have maximum magnetic moment.