The question asks why the second ionization enthalpy is always greater than the first. This relates to the energy required to remove successive electrons from an atom. We need to consider the changes in electrostatic attraction and electron configuration after the first electron is removed.
B) Because removing an electron from a positively charged ion is more difficult than from a neutral atom — This statement accurately describes why the second ionization enthalpy is greater. After the first electron is removed, the atom becomes a positive ion. The remaining electrons are held more tightly by the nucleus due to a higher effective nuclear charge, making it harder to remove the next electron.
The first ionization enthalpy is the energy required to remove the outermost electron from a neutral gaseous atom. We need to compare the electronic configurations and shielding effects for Boron (B) and Beryllium (Be) to understand why Boron's first ionization enthalpy is lower.
Because the \(2p\) electron in Boron is at a higher energy level and is better shielded, it requires less energy to remove it compared to the \(2s\) electron in Beryllium. Therefore, the first ionization enthalpy of Boron is less than that of Beryllium.
C) The 2p electron of B is more shielded and at higher energy than the 2s electron of Be. This statement accurately explains why the first ionization enthalpy of Boron is lower than that of Beryllium, considering both the higher energy of the \(2p\) orbital and the increased shielding experienced by the \(2p\) electron.
The question asks why the first ionization enthalpy of oxygen is less than that of nitrogen. This can be understood by examining their electronic configurations and the stability associated with them, particularly Hund's rule and the concept of half-filled and fully-filled orbitals.
B) Nitrogen has half-filled stable 2p³ configuration; in oxygen, two electrons in same 2p orbital cause repulsion, easing electron removal. This option correctly identifies the key reasons: the stability of nitrogen's half-filled 2p subshell and the electron-electron repulsion in oxygen's 2p subshell, which makes electron removal easier for oxygen.
The question asks to identify the element with the highest first ionization enthalpy. Ionization enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom. Key factors influencing ionization enthalpy are nuclear charge, atomic size, shielding effect, and electron configuration.
D) Helium has the highest first ionization enthalpy. It is the smallest atom with a completely filled valence shell (\(1s^2\)), leading to maximum nuclear attraction for its electrons and thus the highest energy required to remove an electron.
The question asks about the general trends of first ionization enthalpy across a period and down a group in the periodic table. Ionization enthalpy is the energy required to remove the most loosely bound electron from an isolated gaseous atom. Understanding the factors affecting ionization enthalpy, such as atomic size, nuclear charge, and shielding effect, is key to determining these trends.
A) increases; decreases — Based on the analysis, first ionization enthalpy generally increases across a period and decreases down a group. This option correctly reflects both trends.
The question asks to explain the anomalous trend in the first ionization enthalpy (IE₁) values for Group 13 elements, specifically the increase from Indium (In) to Thallium (Tl). We need to consider factors that influence ionization enthalpy, such as nuclear charge, atomic size, and shielding effect.
A) Poor shielding of nuclear charge by intervening 4f and 5d electrons (lanthanide contraction effect) — This option correctly identifies the reason for the anomalous increase in ionization enthalpy from Indium to Thallium. The presence of 4f and 5d electrons in Tl, which are poor shielders, leads to an increased effective nuclear charge on the valence electrons, making them harder to remove.