If valence shell electron configuration of vanadium is 3d34s2 and that of uranium is 5f36d17s2, then both [vo2]+ and (uo2)2+ will have :

The answer is same number of d-electrons, with \([vo_{2}]^{+}\) bent and \([uo_{2}]^{2+}\) linear.In VO2+ vanadium is +5, which removes all five valence electrons (3d34s2): d0. In UO22+ uranium is +6, which removes all six (5f36d17s2): also no d electrons. So both have zero d electrons.Their shapes differ. The dioxovanadium(V) ion is bent (cis-dioxo), as usual for d0 transition-metal dioxo species, which use d orbitals that favour a cis arrangement. The uranyl ion is strictly linear (O=U=O), because the 5f orbitals let uranium form strong pi bonds to two trans oxygens. That rules out 'bent for both' and 'linear for both', and the oxidation states are irrelevant to the comparison asked.

Lanthanides and Actinides Questions and Answers

Q: Lanthanides and actinides are not known to form neutral carbonyl complexes because :

The keyed reason is They lack electrons in d orbitals.Stable metal carbonyls depend on synergic bonding: CO donates a lone pair to the metal (σ) and the metal pushes d-electron density back into the empty π* orbitals of CO (π back-bonding). Without back-donation the M-CO bond is too weak to hold.The lanthanides and actinides have their valence electrons in 4f and 5f orbitals, with few or no d electrons available, and the f orbitals are too contracted and buried to overlap with CO π*. So there is nothing to back-donate and neutral carbonyls are not formed under normal conditions. Their size is not the problem, and they do not hydrate CO.

Q: R-S term symbol of Eu2+ (At. No. = 63) is :

The term symbol is 8S7/2.Europium (Z = 63) is [Xe]4f76s2; Eu2+ loses the two 6s electrons and keeps a half-filled 4f7 shell.Seven electrons in seven f orbitals, all with parallel spin, give S = 7/2, so the multiplicity is 2S + 1 = 8. One electron in each orbital, ml = +3 to −3, gives L = 0, an S term. Then J = |L ± S| = 7/2.7F6 and 7F0 are terms of f6 or f8, the ground term of Eu3+ being 7F0. 6H5/2 is the ground term of f5, as in Sm3+.

Q: The recovery of two lanthanoids easier than the others due to their redox chemistry. These lanthanoids are :

The two lanthanoids are cerium and europium, recovered as Ce4+ and Eu2+.Almost all lanthanoids are stable only as Ln3+, with very similar radii, which is why they are so hard to separate. Cerium can be oxidised to Ce4+ (an empty 4f0 shell), whose chemistry differs sharply from Ln3+: it hydrolyses and precipitates or is extracted selectively. Europium can be reduced to Eu2+ (a half-filled 4f7 shell), which behaves like an alkaline-earth ion and can be removed, for example as insoluble EuSO4.Pr4+, Tb4+, Nd4+ and Dy4+ exist only in a few solid fluorides and oxides and are not usable in solution, and 'Ln' is the general symbol for any lanthanoid, so options A, C and D do not describe a practical separation.

Q: In Frost diagram, NE° values for four different oxidation states of Thorium is found to be Th(II) \(\gt\) Th(I) \(\gt\) Th(III) \(\gt\) Th(IV), then :

The correct conclusion is that Th(II) can disproportionate to Th(I) and Th(III).In a Frost diagram, nE° (proportional to the free energy of a species relative to the element) is plotted against oxidation state. The species with the lowest nE° is the most stable, and a species whose point lies above the line joining its two neighbours is unstable to disproportionation into them.Th(II) has the highest nE° of all four, higher than both its neighbours Th(I) and Th(III), so it lies above the line joining them and will disproportionate. Th(II) is therefore the least stable state, not the most stable; Th(IV), with the lowest value, is the most stable, in line with thorium's real chemistry. Option B would need Th(III) to lie above the line joining Th(I) and Th(IV), which the given order does not establish, and the order alone does not fix the comproportionation in option D either.

Q: Correct statement regarding the electronic transitions of f-block elements is :

The correct statement is that there is little coupling of electronic transitions with molecular vibrations.The 4f orbitals of the lanthanides lie deep inside the atom, shielded by the filled 5s and 5p shells, so they interact only weakly with the ligands. The f-f transitions are therefore hardly affected by ligand vibrations, and their bands are sharp and line-like, almost like those of free ions.That same shielding makes the other statements wrong. f-f bands are narrow, not broad. Mixing of f with d orbitals is small. And f-f transitions are Laporte forbidden, since they occur within one subshell, which is why the colours of lanthanide ions are pale.

21.

Lanthanides and actinides are not known to form neutral carbonyl complexes because :

A.They lack electrons in d orbitals
B.Their d orbitals do not overlap with \(\pi\) orbitals of CO
C.They tend to hydrate CO under ambient conditions
D.Their large metal atoms cannot interact with smaller diatomic ligands

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The keyed reason is They lack electrons in d orbitals.

Stable metal carbonyls depend on synergic bonding: CO donates a lone pair to the metal (σ) and the metal pushes d-electron density back into the empty π* orbitals of CO (π back-bonding). Without back-donation the M-CO bond is too weak to hold.

The lanthanides and actinides have their valence electrons in 4f and 5f orbitals, with few or no d electrons available, and the f orbitals are too contracted and buried to overlap with CO π*. So there is nothing to back-donate and neutral carbonyls are not formed under normal conditions. Their size is not the problem, and they do not hydrate CO.

Discuss in Forum

22.

R-S term symbol of Eu²⁺ (At. No. = 63) is :

A.\({}^{8}\mathrm{S}_{7/2}\)
B.\({}^{7}\mathrm{F}_{6}\)
C.\({}^{6}\mathrm{H}_{5/2}\)
D.\({}^{7}\mathrm{F}_{0}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The term symbol is ⁸S_7/2.

Europium (Z = 63) is [Xe]4f⁷6s²; Eu²⁺ loses the two 6s electrons and keeps a half-filled 4f⁷ shell.

Seven electrons in seven f orbitals, all with parallel spin, give S = 7/2, so the multiplicity is 2S + 1 = 8. One electron in each orbital, m_l = +3 to −3, gives L = 0, an S term. Then J = |L ± S| = 7/2.

⁷F₆ and ⁷F₀ are terms of f⁶ or f⁸, the ground term of Eu³⁺ being ⁷F₀. ⁶H_5/2 is the ground term of f⁵, as in Sm³⁺.

Discuss in Forum

23.

The recovery of two lanthanoids easier than the others due to their redox chemistry. These lanthanoids are :

A.Pr\(^{4+}\), Tb\(^{4+}\)
B.Ce\(^{4+}\), Eu\(^{2+}\)
C.Nd\(^{4+}\), Ln\(^{2+}\)
D.Dy\(^{4+}\), Ln\(^{4+}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The two lanthanoids are cerium and europium, recovered as Ce⁴⁺ and Eu²⁺.

Almost all lanthanoids are stable only as Ln³⁺, with very similar radii, which is why they are so hard to separate. Cerium can be oxidised to Ce⁴⁺ (an empty 4f⁰ shell), whose chemistry differs sharply from Ln³⁺: it hydrolyses and precipitates or is extracted selectively. Europium can be reduced to Eu²⁺ (a half-filled 4f⁷ shell), which behaves like an alkaline-earth ion and can be removed, for example as insoluble EuSO₄.

Pr⁴⁺, Tb⁴⁺, Nd⁴⁺ and Dy⁴⁺ exist only in a few solid fluorides and oxides and are not usable in solution, and 'Ln' is the general symbol for any lanthanoid, so options A, C and D do not describe a practical separation.

Discuss in Forum

24.

In Frost diagram, NE° values for four different oxidation states of Thorium is found to be Th(II) \(\gt\) Th(I) \(\gt\) Th(III) \(\gt\) Th(IV), then :

A.Th(II) is the most stable oxidation state
B.Th(III) can disproportionate to Th(I) and Th(IV)
C.Th(II) can disproportionate to Th(I) and Th(III)
D.Th(II) and Th(IV) can consproportionate to Th(III)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The correct conclusion is that Th(II) can disproportionate to Th(I) and Th(III).

In a Frost diagram, nE° (proportional to the free energy of a species relative to the element) is plotted against oxidation state. The species with the lowest nE° is the most stable, and a species whose point lies above the line joining its two neighbours is unstable to disproportionation into them.

Th(II) has the highest nE° of all four, higher than both its neighbours Th(I) and Th(III), so it lies above the line joining them and will disproportionate. Th(II) is therefore the least stable state, not the most stable; Th(IV), with the lowest value, is the most stable, in line with thorium's real chemistry. Option B would need Th(III) to lie above the line joining Th(I) and Th(IV), which the given order does not establish, and the order alone does not fix the comproportionation in option D either.

Discuss in Forum

25.

If valence shell electron configuration of vanadium is 3d³4s² and that of uranium is 5f³6d¹7s², then both [vo₂]⁺ and (uo₂)²⁺ will have :

A.Same number of d-electrons and bent shape for both
B.Same number of d-electrons with bent shape for \([vo_{2}]^{+}\) and linear shape for \([uo_{2}]^{2+}\)
C.Same number of d-electrons and linear shape for both
D.Different oxidation states with linear shape for both

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The answer is same number of d-electrons, with \([vo_{2}]^{+}\) bent and \([uo_{2}]^{2+}\) linear.

In VO₂⁺ vanadium is +5, which removes all five valence electrons (3d³4s²): d⁰. In UO₂²⁺ uranium is +6, which removes all six (5f³6d¹7s²): also no d electrons. So both have zero d electrons.

Their shapes differ. The dioxovanadium(V) ion is bent (cis-dioxo), as usual for d⁰ transition-metal dioxo species, which use d orbitals that favour a cis arrangement. The uranyl ion is strictly linear (O=U=O), because the 5f orbitals let uranium form strong pi bonds to two trans oxygens. That rules out 'bent for both' and 'linear for both', and the oxidation states are irrelevant to the comparison asked.

Discuss in Forum

26.

Correct statement regarding the electronic transitions of f-block elements is :

A.f-f transition bands are broad
B.There is large d-f orbital mixing
C.The transitions are Laporte allowed
D.There is a little coupling of electronic transitions with molecular vibrations

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The correct statement is that there is little coupling of electronic transitions with molecular vibrations.

The 4f orbitals of the lanthanides lie deep inside the atom, shielded by the filled 5s and 5p shells, so they interact only weakly with the ligands. The f-f transitions are therefore hardly affected by ligand vibrations, and their bands are sharp and line-like, almost like those of free ions.

That same shielding makes the other statements wrong. f-f bands are narrow, not broad. Mixing of f with d orbitals is small. And f-f transitions are Laporte forbidden, since they occur within one subshell, which is why the colours of lanthanide ions are pale.

Lanthanides and Actinides Questions and Answers

Topics

Other Subjects

Tags: Lanthanides and Actinides Questions and Answers

About Us

Lanthanides and Actinides Questions and Answers

Topics

Other Subjects

Tags: Lanthanides and Actinides Questions and Answers

About Us

Stay informed about all important Dates & Notifications