NEET Practice Questions (MCQs) with Answers & Solutions

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.'

The NCERT explains, 'Another “anomaly” is the smaller first ionization enthalpy of oxygen compared to nitrogen. This arises because in the nitrogen atom, three 2p-electrons reside in different atomic orbitals (Hund’s rule) whereas in the oxygen atom, two of the four 2p-electrons must occupy the same 2p-orbital resulting in an increased electron-electron repulsion. Consequently, it is easier to remove the fourth 2p-electron from oxygen than it is, to remove one of the three 2p-electrons from nitrogen.'

The NCERT problem 3.6 directly addresses this: 'The first ionization enthalpy ($\Delta_i H$) values of the third period elements, Na, Mg and Si are respectively 496, 737 and 786 kJ mol⁻¹. Predict whether the first $\Delta_i H$ value for Al will be more close to 575 or 760 kJ mol⁻¹ ? Justify your answer.' The solution provided is: 'It will be more close to 575 kJ mol⁻¹. The value for Al should be lower than that of Mg because of effective shielding of 3p electrons from the nucleus by 3s-electrons.'