Chemistry MCQs for NEET — Practice Questions with Answers

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.'

The NCERT states, '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 explicitly 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.' While d$^{10}$ and d$^{0}$ are stable, the context specifically mentions the d$^{5}$ configuration of Mn$^{2+}$ causing a 'break' in the steady increasing trend of ionization enthalpy. The stability of Sc$^{3+}$ (d$^{0}$) means its value for the $E°(M^{3+}/M^{2+})$ reduction potential is low, reflecting its stability, but not described as an anomaly in the general trend of increasing ionization enthalpies at that point.

Lithium is an alkali metal (minima on the graph in Fig. 3.5), known for its low ionization enthalpy due to its larger atomic radius, a single loosely bound valence electron in the outermost s-orbital, and the associated high reactivity. All factors contribute to its lower ionization enthalpy compared to its period neighbors.

According to the NCERT text, 'A hydrocarbon is termed saturated if it contains only carbon-carbon single bonds.' This means all carbon valencies are satisfied by single bonds with other carbon or hydrogen atoms.