31 Year NEET Previous Year Questions: The d & f-Block Elements - 1 - NEET MCQ

La (OH)₃ is more basic than Li (OH)₃. In lanthanides the basic character of hydroxides decreases as the ionic radius decreases.

The Lanthanides are transition metals from Atomic numbers 58 (Ce) to 71(Lu).
Hence the electron configulation becomes : (n –2) f ^{1– 14} (n – 1) s²p⁶ d^{0 – 1} ns².

Mn : [Ar] 3d⁵  4s²
Shows +2, +3, +4, +5, +6 & +7 oxidation states

The basic character of the transition metal monoxide is TiO > VO > CrO > FeO because basic character of oxides decrease with increase in atomic number. Oxides of transitional metals in low oxidation state i.e., + 2 and + 3 are generally basic except Cr₂O₃.

Transition metals have the ability to utilize (n –1)d -orbitals. Thus variable oxidation states enables the transition element to associate with the reactants in different forms.

In lanthanide series, there is a regular decrease in the atomic as well as ionic radii of trivalent ions (M³⁺) as the atomic number increases. Although the atomic radii do show some irregularities but ionic radii decrease from La(103 pm) to Lu (86pm).

The electronic configuration of  different species  given in the question are :

(a) 

(b)  

(c) 

(d)  

Thus options (a) and (c) are discarded; now

let us observe the second point of difference. Thus option (b) is discarded

Lanthanides are 4 f-series elements starting from cerium  (Z= 58) to lutetium ( Z = 71).
These are placed in the sixth period and in third group.

In Sc³⁺ there as no unpaired electron So the aqueous solution of Sc³⁺ will be colourless.

The cause of showing different oxidation is due to the fact that there is only a small difference between the energies of electron in the ns orbitals and (n – 1)d orbitals with the result both ns as well (n – 1)d electrons can be used for compound formation. Lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals result in larger no. of oxidation state.

For third ion ization enthalpy last configuration of

For third Ionization enthalpy Mn has stable configuration due to half filled d-orbital.

Copper sulphate on treatment with excess of KCN forms complex K₃[Cu(CN)₄] or  

Sc³⁺ : 1s², 2s²p⁶, 3s²p⁶d⁰, 4s⁰; no unpaired electron.
Cu⁺ : 1s², 2s²p⁶, 3s2p⁶d¹⁰, 4s⁰; no unpaired electron.
Ni²⁺: 1s², 2s²p⁶, 3s²p⁶d⁸, 4s⁰; unpaired electron present.
Ti³⁺ : 1s², 2s²p⁶, 2s²p⁶d¹, 4s⁰; unpaired electron present Co²⁺ : 1s², 2s²p⁶, 3s²p⁶d⁷, 4s⁰; unpaired electron present So from the given option the only correct combination is Ni²⁺ and Ti³⁺

All the group IV elements show tetravalency which can be explained on the basis of shifting of one e^– of ns to the vacant np orbital.These four orbitals give rise to four sp3 hybrid orbitals.
When ns² electrons of the outermost shell do not participate in bonding it is called inert pair and the effect is called inert pair effect. The last three elements of group 4 i.e. Ge, Sn and Pb, have a tendency to form divalent as well as tetravalent ions. Since the inert pair effect increases from Ge to Pb, the stability of tetravalent ion decreases and that of divalent ion increases that is why Pb⁺⁺ is more stable than Pb⁴⁺ and Sn⁴⁺ ions is more stable than Sn²⁺.

For chromium ion + 3oxidation state is most stable.

There is a steady decrease in the radii as the atomic number of the lanthanide elements increases. For every additional proton added in nucleus the corresponding electron goes to 4f subshell.
The shape of f -orbitals is very much diffused and they have poor shielding effect. The effective nuclear charge increases which causes the contraction in the size of electron charge cloud. This contraction in size is quite regular and known as Lanthanoid contraction.
Since the change in the ionic radii in the lanthanide series is very small, thus their chemical properties are similar.

Ti ; Z (22) is 1s²2s²2p⁶3s²3p⁶4s²3d²
V ; Z (23) is 1s²2s²2p⁶3s²3p⁶4s²3d³
Cr ; Z (24) is 1s²2s²2p⁶3s²3p⁶3d⁵4s¹
Mn ; Z (25) is 1s²2s²2p⁶3s²3d⁵4s²

 The second electron in all the cases (except Cr) is taken out from 4s-orbital and for Cr it is an electron from completely half filled 3d-orbital. The force required for removal of second electron will be more for Mn (except for Cr) having more positive charge.
Based on this we find the correct order is Mn > V > Ti.  Cr > Mn > V > Ti. So correct answeris (a).

The no. of various oxidation states possible are +  2, + 3, + 4, (+ 5), + 6, + 7.

The colour exhibited by transition metal ions is due to the presence of unpaired electrons in -orbitals which permits the  - excitation of electrons.

 Ti is in + 4 O.S. ; 3d⁰ = colourless

 Co is in + 3 O.S ; 3d⁵ = coloured

In Cu₂Cl₂– Cu is in +1 O.S. ; 3d¹⁰ – colourless

  Ni is in + 2 O.S ; 3d⁸ – coloured

In general, the atomic and ionic radii increases on moving down a group. But the elements of second transition series (eg, Zr, Nb, Mo etc) have the almost same radii as the elements of third transition series (eg Hf, Ta, W etc.) This is because of lanthanide contraction i.e imperfect shielding of one 4f- electron by another.

Correct choice : (a)

The green colour appears due to the formation of Cr⁺⁺⁺ion

Na₂Cr₂O₇ is hygroscopic.

The melting points of the transition element first rise to a maximum and then fall as the atomic number increases manganese have abnormally low melting points.

The transition metals and their compounds are used as catalysts. Because of the variable oxidation states, due to this, they easily absorb and re-emit wide range of energy to provide the necessary activation energy.

Ac (89) = [Rn] [6 ¹] [7 ²]