Structure Of Atom – NEET UG ChemistryPractice Questions & PYQs

Energy emission during an electronic transition occurs when an electron shifts from a higher energy shell to a lower one, releasing a photon with energy equal to the difference between those two states. The energy level of an electron in a specific shell $n$ for a hydrogen atom is defined by $E_n = \frac{-2.18 \times 10^{-18}}{n^2} \text{ J}$. To find the energy released when an electron transitions from $n=4$ to $n=1$, the magnitude of the difference between these levels is calculated as $\Delta E = 2.18 \times 10^{-18} \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \text{ J}$. This simplifies to $\Delta E = 2.18 \times 10^{-18} \times (1 - 0.0625) \text{ J}$, which equals $2.04 \times 10^{-18} \text{ J}$. Applying the relation $\nu = \frac{\Delta E}{h}$ with Planck's constant $h = 6.625 \times 10^{-34} \text{ J s}$, the frequency is determined by calculating $\frac{2.04 \times 10^{-18} \text{ J}}{6.625 \times 10^{-34} \text{ J s}}$. This division yields a frequency of approximately $3.08 \times 10^{15} \text{ s}^{-1}$.

The wavelength of a light wave is determined by the relationship $\lambda = \frac{c}{\nu}$. By substituting the velocity of 3.0 × 10^8 m/s and the frequency of 8 × 10^15 s^-1 into the equation $\lambda = \frac{3.0 \times 10^8}{8 \times 10^{15}}$, the result is 0.375 × 10^-7 m. Converting this figure to nanometers by multiplying by 10^9 yields 37.5 nm, which is the value closest to 4 × 10^1 nm.

The fundamental relationship governing the energy states of an electron in a hydrogen atom dictates that the kinetic energy is always equal in magnitude but opposite in sign to the total energy of that orbit. Mathematically, this is expressed as $K.E. = -E_n$. Since the total energy for the first excited state is given as $-3.4 \text{ eV}$, calculating the kinetic energy for this orbit simply requires applying this relationship, which results in $K.E. = -(-3.4 \text{ eV}) = +3.4 \text{ eV}$.

Bonding in diatomic molecules depends on how the electron clouds of atomic orbitals align with the internuclear axis, which is designated here as the $z$-axis. Since $p_{x}$ and $p_{y}$ orbitals possess lobes that are oriented perpendicular to this axis, they lack the head-on approach required for $\sigma$ bonding. Instead, these orbitals interact via sidewise overlap, which is the defining characteristic of $\pi$ molecular orbitals.

Every unique set of quantum numbers designates one specific orbital in an atom. The principal quantum number $n=3$ sets the energy shell, while $l=2$ identifies the subshell as a $d$-type. Because $m=+2$ limits the spatial orientation to a single specific position, these parameters describe exactly one of the five available 3d orbitals.

The energy of a photon is inversely proportional to its wavelength as defined by the relation $E = \frac{hc}{\lambda}$. By rearranging this equation to isolate the wavelength, we get $\lambda = \frac{hc}{E}$. Substituting the constants $h = 6.626 \times 10^{-34}$ J s, $c = 3 \times 10^8$ m/s, and the energy $E = 3.03 \times 10^{-19}$ J results in the calculation $\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.03 \times 10^{-19}}$. Evaluating this provides a wavelength of $6.56 \times 10^{-7}$ meters, which is equivalent to 656 nm.

Isoelectronic species are defined as atoms, molecules, or ions that contain the exact same total number of electrons. In the set comprising ${CO}$, ${CN}^{-}$, ${NO}^{+}$, and ${C}_{2}^{2-}$, every single species contains exactly 14 electrons when you sum the atomic numbers and adjust for the corresponding ionic charges. Since they share this identical electron count, they satisfy the requirement for being isoelectronic.