D & F Block Elements MCQs for NEET — Chemistry Questions with Answers

The contraction is, however, greater from element to element in this series resulting from poor shielding by 5f electrons. The statement that actinoid contraction is more significant than lanthanoid contraction due to more effective shielding by 5f electrons is incorrect. The context states that 'The contraction is, however, greater from element to element in this series resulting from poor shielding by 5f electrons.' This implies that poor shielding, not effective shielding, is the cause, and while the contraction per element is greater, the overall significance is often attributed to lanthanoid contraction's impact on subsequent elements, as stated: 'The lanthanoid contraction is more important because...'

The text states: 'The irregularities in the electronic configurations of the actinoids, like those in the lanthanoids are related to the stabilities of the $f^0$, $f^7$ and $f^{14}$ occupancies of the 5f orbitals. Thus, the configurations of Am and Cm are [Rn] $5f^7 7s^2$ and [Rn] $5f^7 6d^1 7s^2$.' Americium (Am) clearly shows a $5f^7$ configuration, indicating stability due to a half-filled orbital.

The text states: 'There is a greater range of oxidation states, which is in part attributed to the fact that the 5f, 6d and 7s levels are of comparable energies.' This allows for the participation of more electrons in bonding.

The text states: 'Hydrochloric acid attacks all metals but most are slightly affected by nitric acid owing to the formation of protective oxide layers; alkalies have no action.' Thus, hydrochloric acid attacks all actinoid metals.

The text states: 'The fourteen electrons are formally added to 5f, though not in thorium (Z = 90) but from Pa onwards the 5f orbitals are complete at element 103.' This clearly indicates that thorium (Th) does not have 5f orbital occupancy in its ground state configuration as explained here.

Referring to Table 4.11 and the accompanying text: '...the maximum oxidation state increases from +4 in Th to +5, +6 and +7 respectively in Pa, U and Np...' The highest oxidation state listed is +7 for Neptunium (Np).

The text states: 'The actinoids are radioactive elements and the earlier members have relatively long half-lives, the latter ones have half-life values ranging from a day to 3 minutes for lawrencium (Z =103). The latter members could be prepared only in nanogram quantities. These facts render their study more difficult.'

The text mentions: 'The magnetic properties of the actinoids are more complex than those of the lanthanoids.' 'The chemistry of the actinoids is more complex in view of their ability to exist in different oxidation states.' 'The structural variability is obtained due to irregularities in metallic radii which are far greater than in lanthanoids.' Electronegativity trend is not described as complex or varied.

The text states: 'All the actinoids are believed to have the electronic configuration of $7s^2$ and variable occupancy of the 5f and 6d subshells.' This corresponds to $5f^{1-14}6d^{0-1}7s^2$ when expanded for the 5f and 6d subshells, with n=7, so (n-2)f = 5f, (n-1)d = 6d, ns = 7s. The option '$(n-2)f^{1-14}(n-1)d^{0-1}ns^2$' is the general definition for f-block, but 5f, 6d, 7s refers to the specific series.

The text states: 'The 5f electrons, will therefore, be more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids. Because the outer electrons are less firmly held, they are available for bonding in the actinoids.' Wait, the text actually says: 'when 5f orbitals are beginning to be occupied, they will penetrate less into the inner core of electrons. The 5f electrons, will therefore, be more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids. Because the outer electrons are less firmly held, they are available for bonding in the actinoids.' This means more effective shielding leads to less firmly held electrons and lower ionization enthalpy. So options are reversed. Let's re-evaluate. 'This is quite reasonable since it is to be expected that when 5f orbitals are beginning to be occupied, they will penetrate less into the inner core of electrons. The 5f electrons, will therefore, be more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids. Because the outer electrons are less firmly held, they are available for bonding in the actinoids.' Effective shielding means less attraction to the nucleus, hence easier to remove. So 'Weaker shielding of 5f electrons by the nuclear charge' is incorrect if it leads to lower ionization enthalpy.