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

The transition elements form interstitial compounds in which small atoms of light elements such as H,C and N occupy the voids in their lattices.

The NCERT states, 'There is an increase in ionisation enthalpy along each series of the transition elements from left to right due to an increase in nuclear charge which accompanies the filling of the inner d orbitals.' It also mentions that the 'nuclear charge increases from scandium to zinc but electrons are added to the orbital of inner subshell, i.e., 3d orbitals. These 3d electrons shield the 4s electrons from the increasing nuclear charge somewhat more effectively than the outer shell electrons can shield one another. Therefore, the atomic radii decrease less rapidly. Thus, ionization energies increase only slightly along the 3d series.'

The NCERT states, 'The first ionisation enthalpy, in general, increases, but the magnitude of the increase in the second and third ionisation enthalpies for the successive elements, is much higher along a series.' It further explains that 'A general trend of increasing values of second ionisation enthalpy is expected as the effective nuclear charge increases because one d electron does not shield another electron from the influence of nuclear charge because d-orbitals differ in direction.'

The NCERT mentions, 'The irregular trend in the first ionisation enthalpy of the metals of 3d series, though of little chemical significance, can be accounted for by considering that the removal of one electron alters the relative energies of 4s and 3d orbitals.'

The NCERT explicitly states, 'You have learnt that when d-block elements form ions, ns electrons are lost before (n – 1) d electrons.'

The NCERT lists, 'The three terms responsible for the value of ionisation enthalpy are attraction of each electron towards nucleus, repulsion between the electrons...'. Nuclear charge is inherently part of the attraction of electrons towards the nucleus. Magnetic properties are not listed as a direct factor for ionization enthalpy.

The NCERT states, 'In beryllium, the electron removed during the ionization is an s-electron whereas the electron removed during ionization of boron is a p-electron. The penetration of a 2s-electron to the nucleus is more than that of a 2p-electron; hence the 2p electron of boron is more shielded from the nucleus by the inner core of electrons than the 2s electrons of beryllium. Therefore, it is easier to remove the 2p-electron from boron compared to the removal of a 2s- electron from beryllium. Thus, boron has a smaller first ionization enthalpy than beryllium.'

The NCERT states, 'However, the trend of steady increase in second and third ionisation enthalpy breaks for the formation of Mn$^{2+}$ and Fe$^{3+}$ respectively. In both the cases, ions have d$^{5}$ configuration.' For iron, the third ionization enthalpy breaks the trend because Fe$^{3+}$ achieves a stable d$^{5}$ configuration. For second ionization enthalpy, the break is typically observed before or at elements gaining stability in their +2 oxidation state. However, the context is specific about the break for Fe$^{3+}$ in the third ionization enthalpy. The question asks for a 'similar break' for second ionization enthalpy. Mn is the primary example for a break in the second ionization enthalpy due to the stability of Mn$^{2+}$ (d$^{5}$). If the question intends a similar type of stability, Fe would be a good choice if considering a subsequent stable electron configuration. Let's re-evaluate. The text says 'the trend of steady increase in second and third ionisation enthalpy breaks for the formation of Mn$^{2+}$ and Fe$^{3+}$ respectively.' This means Mn$^{2+}$ (d$^{5}$) causes a break in the second ionization enthalpy trend. So, the question asks for an element where a similar break occurs for its second ionization enthalpy. Mn is the direct example for the second IE break (Mn to Mn$^{2+}$ is d$^{5}$). Sc is d$^{0}$, V is d$^{3}$, Co is d$^{7}$. Fe as Fe$^{2+}$ is d$^{6}$. The stability of d$^{5}$ for Mn$^{2+}$ makes its second IE relatively lower than expected. The question is a bit tricky, but based on the direct statement 'breaks for the formation of Mn2+ and Fe3+ respectively', Mn is the direct answer for the second IE break. If it meant 'similar reason' then Fe is not for second IE. Let's provide Mn as it is explicitly linked to 2nd IE break. My previous thought of 'subsequent stable electron configuration' was leading to a wrong path.

The NCERT states, 'Energy is always required to remove electrons from an atom and hence ionization enthalpies are always positive. The second ionization enthalpy will be higher than the first ionization enthalpy because it is more difficult to remove an electron from a positively charged ion than from a neutral atom. In the same way the third ionization enthalpy will be higher than the second and so on.' This confirms that $IE_1 < IE_2 < IE_3$.

The NCERT states, 'To understand these trends, we have to consider two factors : (i) the attraction of electrons towards the nucleus, and (ii) the repulsion of electrons from each other.'