Chemistry MCQs for NEET — Practice Questions with Answers

In the atomic spectrum of hydrogen, the Balmer series corresponds to the electronic transitions from higher energy levels (n ≥ 3) to the second energy level (n=2). Therefore, the spectral lines pertaining to the transition from n=4 to n=2 belong to the Balmer series.

The second ionization energy of an element is the energy required to remove an electron from the singly ionized atom. In the case of helium, the first ionization removes one electron, leaving He+ (singly ionized helium). The second ionization energy removes the remaining electron, giving He2+. Since helium has two electrons, the second ionization energy is simply twice the first ionization energy of hydrogen, which is 13.6 eV. Therefore, the second ionization energy of helium is 54.4 eV.

The maximum number of electrons that can be accommodated in s, p, and d subshells are 2, 6, and 10, respectively. This is based on the Aufbau principle and the Pauli exclusion principle.

The $d_{xy}$ orbital can accommodate a maximum of 2 electrons with opposite spins, following the Pauli exclusion principle. This is true for all the five d-orbitals ($d_{xy}$, $d_{yz}$, $d_{xz}$, $d_{x^2-y^2}$, and $d_{z^2}$).

The total number of electrons present in any main energy level (n) can be calculated from the formula 2n^2. This is based on the Aufbau principle, which states that electrons occupy orbitals in the order of increasing energy.

The two sets of quantum numbers X (3, 2, -2, +1/2) and Y (3, 0, 0, +1/2) have different values for the subsidiary quantum number (l) and the magnetic quantum number (m), indicating that they represent electrons with different energy levels and orientations.