Structure Of Atom – NEET UG ChemistryPractice Questions & PYQs

Radial nodes occur where the radial wavefunction $\psi_r$ intersects the horizontal axis, excluding the points at $r=0$ and infinity, and the number of such nodes is governed by the expression $n - l - 1$. A plot with two radial nodes must exhibit exactly two zero-crossings across the horizontal axis. Figure D displays two distinct intersections, distinguishing it from the other plots which show only zero or one.

In Bohr's theory of the hydrogen atom, the energy of an electron is constrained by the principal quantum number $n$, which defines the specific energy level of each allowable orbit. Theoretical derivation of these stationary states demonstrates that the binding energy is inversely proportional to the square of the quantum number, leading to the expression $E_n = -\frac{13.6}{n^2} \text{ eV}$. This inverse square dependence explains why energy levels become progressively less negative and closer together as the electron occupies higher, more distant orbits.

The physics governing this scenario relies on the relationship between total power output and individual quantum energy. A bulb rated at 150 watts consumes energy at a rate of 150 joules per second, but only 8% of this power is actually transformed into light. By calculating 8% of the total input power, we find that the effective light output is 12 joules per second. Because every photon carries a specific amount of energy, the number of photons emitted each second is determined by dividing the light output power by the energy of a single photon:

$n = \frac{P_{\text{light}}}{E_{\text{photon}}}$

Substituting the known values into this relationship gives the expression $n = \frac{12}{4.42 \times 10^{-19}}$. Performing this division yields $n = 2.71 \times 10^{19}$ photons released every second.

Spherical nodes, often referred to as radial nodes, are specific regions within an atomic orbital where the electron density vanishes, and their count is defined by the expression $n - l - 1$. In this formula, $n$ represents the principal quantum number indicating the energy level, while $l$ stands for the azimuthal quantum number corresponding to the subshell shape. For a $3p$ orbital, the shell is defined by $n = 3$ and the $p$ subshell type requires $l = 1$. Substituting these values into the calculation $3 - 1 - 1$ yields a total of one spherical node.

The principle stating that no two electrons within a single atom can share the exact same set of all four quantum numbers-namely the principal ($n$), azimuthal ($l$), magnetic ($m_l$), and spin ($m_s$) quantum numbers-is defined as Pauli's Exclusion Principle. This fundamental rule implies that since an orbital can hold a maximum of two electrons, these two electrons must necessarily possess opposite spins to distinguish their total quantum states. Because this restriction prevents multiple electrons from occupying an identical state, it governs how electrons populate orbitals throughout the entire electronic structure of an atom.

The azimuthal quantum number $l$ dictates the shape and type of the subshell, with $l=3$ specifically identifying the f-subshell. Within any given subshell, the number of orbitals is determined by $2l + 1$, and since each orbital is restricted to a maximum of two electrons, the total capacity is found by doubling the number of orbitals. Applying this to the f-subshell, we find seven orbitals which accommodate a total of 14 electrons.

The energy level of an electron in an orbital is determined by the sum of its principal and azimuthal quantum numbers, represented as $(n+l)$. By calculating this value for each set, we find that the sets correspond to (a) $3+2=5$, (b) $4+2=6$, (c) $4+1=5$, and (d) $5+0=5$. Comparing these results, the sum of 6 is the highest, which identifies the orbital described in option (b) as having the greatest energy.

According to the Aufbau principle, electrons occupy atomic orbitals in increasing order of energy, which would theoretically predict a configuration ending in $4s^2 3d^9$ for an atom with atomic number 29. Nature, however, favors the extra stability associated with fully filled and half-filled subshells, which arises from symmetric charge distribution and increased exchange energy. To achieve this stability, a single electron from the $4s$ orbital shifts to the $3d$ subshell to complete the set. This shift results in the stable ground-state configuration of $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$.

When evaluating the electronic structure of transition metal ions, the distribution of electrons across principal energy levels dictates their chemical characteristics. The copper ion $Cu^+$ is formed by the removal of one electron from the neutral copper atom, resulting in an electronic configuration of $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10}$. Within this arrangement, the third energy level contains the fully occupied $3 s^2$, $3 p^6$, and $3 d^{10}$ subshells, which collectively account for a total of $2 + 6 + 10 = 18$ electrons in the outermost shell.

The absorption of light in a Hydrogen atom occurs as an electron transitions from a lower energy level to a higher one, with the wavelength of the absorbed photon being inversely related to the energy difference between those states, as defined by $\Delta E = \frac{hc}{\lambda} = R_H (\frac{1}{n_{initial}^2} - \frac{1}{n_{final}^2})$. Because the photon's energy must precisely match the difference between levels, we calculate the inverse wavelengths for each specific transition to determine their final ratio.

For the transition from $n=2$ to $n=3$, the inverse wavelength $\frac{1}{\lambda_1}$ is proportional to $R_H (\frac{1}{4} - \frac{1}{9})$, which simplifies to $\frac{5R_H}{36}$, making $\lambda_1 = \frac{36}{5R_H}$. For the transition from $n=4$ to $n=6$, the inverse wavelength $\frac{1}{\lambda_2}$ is proportional to $R_H (\frac{1}{16} - \frac{1}{36})$, which simplifies to $R_H (\frac{9-4}{144})$ or $\frac{5R_H}{144}$, making $\lambda_2 = \frac{144}{5R_H}$. Taking the ratio of these two wavelengths yields:
$\frac{\lambda_1}{\lambda_2} = \frac{36/5R_H}{144/5R_H} = \frac{36}{144} = \frac{1}{4}$

The Bohr model relates the energy and radius of an electron in a single-electron species to the atomic number $Z$ and the principal quantum number $n$. Higher nuclear charges result in a more negative potential energy because the electron is more strongly attracted to the nucleus, while simultaneously causing the orbital radius to decrease. The relationships between these variables are defined by:

$E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2} \text{ J}$
$r_n = 52.9 \times \frac{n^2}{Z} \text{ pm}$

For the $\mathrm{Li}^{2+}$ ion, which has a nuclear charge of $Z=3$ and resides in the first orbit $n=1$, the energy calculation using the formula yields -19.62 $\times$ 10$^{-18}$ J. The corresponding orbital radius, calculated by dividing the Bohr radius by the atomic number, is 17.6 pm.

Similarly, for the $\mathrm{He}^{+}$ ion, the nuclear charge is $Z=2$ and the principal quantum number is $n=1$. Applying the same principles, the energy is computed as -8.72 $\times$ 10$^{-18}$ J, while the orbital radius is found to be 26.4 pm.

The energy of an electron in a hydrogen-like species is governed by the expression $E_n = -R_H \left( \frac{Z^2}{n^2} \right)$, where $Z$ is the atomic number and $n$ represents the energy level. For the $\mathrm{He}^+$ ion, which has an atomic number of 2, the energy of the $n=1$ state is $-x$, satisfying the equation $-x = -R_H \left( \frac{2^2}{1^2} \right) = -4R_H$, which confirms that $R_H = \frac{x}{4}$. Substituting this value into the energy formula for the $\mathrm{Be}^{3+}$ ion, which has an atomic number of 4, the energy at state $n=2$ is given by $E_2 = - \left( \frac{x}{4} \right) \left( \frac{4^2}{2^2} \right)$. This calculation simplifies to $E_2 = - \left( \frac{x}{4} \right) \left( \frac{16}{4} \right) = -x$ J, indicating that the final energy is identical to that of the initial ground state.

The energy level of an electron is determined by the $(n+l)$ rule, which states that higher $(n+l)$ values correspond to greater energy, and for orbitals with identical $(n+l)$ sums, the one with the higher principal quantum number $n$ holds more energy. Calculating these values, we find $(n+l)=6$ for electron I (4d), $(n+l)=5$ for both electron II (3d) and III (4p), and $(n+l)=4$ for electron IV (3p). Since electron I has the largest sum and electron IV the smallest, and electron III possesses a larger $n$ value than electron II despite their shared $(n+l)$ sum, the decreasing energy order is $I > III > II > IV$.

Quantum numbers provide a structured framework to uniquely identify the characteristics of an electron orbiting a nucleus. The principal quantum number $n$ fundamentally establishes the size of an orbital, acting as the primary indicator of the energy level. Meanwhile, the azimuthal quantum number $l$ determines the orbital's geometric shape, distinguishing between different subshells. The magnetic quantum number $m_l$ describes the specific spatial orientation of these orbitals within a three-dimensional coordinate system, whereas the spin quantum number $m_s$ indicates the orientation of the electron's own intrinsic spin.

The Rydberg formula for the hydrogen spectrum is expressed as $\overline{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. For the Balmer series, the transition involves electrons falling to the $n_1 = 2$ energy level, and the lowest energy photon corresponds to the transition from the nearest orbit, $n_2 = 3$. Substituting these values gives $\overline{\nu} = \frac{5 R_H}{36}$ cm$^{-1}$, which confirms that the first statement is true. Regarding blackbody radiation, increasing the temperature causes the peak intensity of the emission curve to shift toward shorter wavelengths, validating the second statement as well.

The magnetic quantum number $m$ defines the specific orientation of orbitals, and for any given azimuthal quantum number $l$, the total number of permissible values $n_m$ is expressed by the relationship $n_m = 2l + 1$. To determine the azimuthal quantum number in terms of the total number of magnetic values, we rearrange this equation by subtracting $1$ from both sides to obtain $n_m - 1 = 2l$. Dividing both sides of this resulting expression by $2$ leads directly to the formula $l = \frac{n_m - 1}{2}$.

The magnetic quantum number $m_l$ determines the spatial orientation of an orbital and is strictly required to include all integer values ranging from $-l$ to $+l$, zero inclusive. When the azimuthal quantum number is $l=2$, the valid set of magnetic quantum numbers must be $-2, -1, 0, +1, +2$. Because the set provided in the second option omits zero, it fails to satisfy the mathematical definition of allowed values for an orbital within a subshell.

In quantum mechanics, the wave function, $\Psi$, serves as a mathematical description of an electron's state and is explicitly defined as a function of the spatial coordinates, meaning its value changes based on the electron's position within the atom. While this function contains all necessary information about the system, it is not directly proportional to the probability of finding the electron; rather, the probability density is given by the square of the wave function, $|\Psi|^2$.

Atoms represent the fundamental building blocks of matter, characterized by specific structural components and historical models. An atom consists of three primary particles, namely electrons, protons, and neutrons, with the electron specifically weighing $9.10939 \times 10^{-31} \text{ kg}$. Because chemical behavior depends on electronic configuration, all isotopes of a given element display identical chemical properties, a concept that aligns with Dalton's atomic theory regarding the atom as the ultimate particle of matter. Since protons and neutrons are grouped together as nucleons within the nucleus, the statements validating the electron's mass, isotope behavior, and Dalton's theory constitute the accurate observations.

The principal quantum number $n$ indicates the main energy shell, while the azimuthal quantum number $l$ determines the subshell shape, specifically where $l = 0$ denotes an $s$ orbital, $l = 1$ represents a $p$ orbital, and $l = 2$ signifies a $d$ orbital. By applying these definitions, we can identify $n = 2, l = 1$ as $2p$, $n = 3, l = 2$ as $3d$, $n = 3, l = 0$ as $3s$, and $n = 2, l = 0$ as $2s$. This direct relationship allows us to map each quantum set to its specific orbital designation.