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

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.

Referring to Table 4.4 and 4.2, the standard electrode potential (E$^{\text{o}}$) for the reduction of M$^{2+}$ to M are: Ti (-1.63 V), V (-1.18 V), Cr (-0.90 V), Mn (-1.18 V), Fe (-0.44 V), Co (-0.28 V), Ni (-0.25 V), Cu (+0.34 V), Zn (-0.76 V). Titanium has the most negative E$^{\text{o}}$ value of -1.63 V, indicating the strongest tendency to get oxidized (i.e., its M$^{2+}$ ion is hardest to reduce).

A high positive standard electrode potential for the reduction of M$^{3+}$ to M$^{2+}$ (like Co$^{3+}$ + e$^{-}$ $\to$ Co$^{2+}$ with E$^{\text{o}}$ = +1.81 V or +1.97 V) indicates that the M$^{3+}$ ion has a strong tendency to gain an electron and be reduced. Therefore, Co$^{3+}$ acts as a strong oxidizing agent.

According to the text, 'The comparatively low value for V is related to the stability of V$^{2+}$ (half-filled t$_{2g}$ level, Unit 5).' A low E$^{\text{o}}$ for M$^{3+}$/M$^{2+}$ means that M$^{2+}$ is relatively stable and does not readily get oxidized to M$^{3+}$, or M$^{3+}$ is easily reduced to M$^{2+}$. In this case, the stability of V$^{2+}$ makes it less prone to oxidation to V$^{3+}$. (V$^{2+}$ is d$^3$ in octahedral field, having t$_{2g}^3$ configuration, which is a half-filled t$_{2g}$ level - this needs knowledge from Unit 5 as mentioned in NCERT).

The context states: 'The highest value for Zn is due to the removal of an electron from the stable d$^{10}$ configuration of Zn$^{2+}$'. This implies that oxidizing Zn$^{2+}$ (d$^{10}$) to Zn$^{3+}$ is very difficult, leading to a very high positive reduction potential for Zn$^{3+}$/Zn$^{2+}$ (if it were to exist), or more practically, it means Zn$^{2+}$ is exceedingly stable. The M$^{3+}$/M$^{2+}$ potential is actually listed in Table 4.2 for transition elements, and Zn is not listed here, but the statement refers to the difficulty of removing an electron from stable Zn$^{2+}$ which would correspond to a very high positive reduction potential if Zn$^{3+}$ were involved.

The text explicitly states: 'whereas E$^{\text{o}}$ for Ni is related to the highest negative $\Delta_{\text{hyd}}H^{\text{o}}$'. A highly negative hydration enthalpy helps stabilize the M$^{2+}$ ion in solution, making the overall reduction process from M$^{2+}$ to M less favorable, contributing to a more negative E$^{\text{o}}$.