Structure Of Atom – JEE Main ChemistryPractice Questions & PYQs

The energy of an orbital is governed by the $(n + l)$ rule, where orbitals with the same $(n + l)$ value are ranked by their principal quantum number $n$, such that a lower $n$ results in lower energy. In multielectron systems, increasing the atomic number raises the effective nuclear charge, which causes the energy of orbitals within a given subshell to decrease rather than increase, making the second statement incorrect. Orbital size is determined by the principal quantum number $n$, and since $n=2$ for the $2p_x$ orbital and $n=3$ for the $3p_x$ orbital, the $2p_x$ orbital is physically smaller. Finally, radial nodes are calculated using the expression $(n - l - 1)$, which yields $1$ node for $5f$, $5$ nodes for $6s$, $1$ node for $4d$, $3$ nodes for $5p$, and $2$ nodes for $5d$; since $5d$ possesses two radial nodes, the assertion that none of these orbitals have two radial nodes is false.

The energy of an electron in a stationary state for a hydrogen-like system is defined by the formula $E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2} \, \mathrm{J}$, where $Z$ is the atomic number and $n$ is the principal quantum number. For the second orbit of $\mathrm{He}^+$ ($Z=2$, $n=2$), the energy is $-2.18 \times 10^{-18} \times \frac{2^2}{2^2} = -2.18 \times 10^{-18} \, \mathrm{J}$. Calculating for the first orbit of $\mathrm{He}^+$ ($Z=2$, $n=1$) yields $-2.18 \times 10^{-18} \times \frac{2^2}{1^2} = -8.72 \times 10^{-18} \, \mathrm{J}$. For the second orbit of the H atom ($Z=1$, $n=2$), the energy is $-2.18 \times 10^{-18} \times \frac{1^2}{2^2} = -0.545 \times 10^{-18} \, \mathrm{J}$. For the third orbit of the $\mathrm{Li}^{2+}$ ion ($Z=3$, $n=3$), the energy is $-2.18 \times 10^{-18} \times \frac{3^2}{3^2} = -2.18 \times 10^{-18} \, \mathrm{J}$, which confirms the value provided in the final statement.

The energy of a spectral line in the hydrogen spectrum is determined by the formula $\Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) eV$, where $n_1$ and $n_2$ are the principal quantum numbers of the lower and higher energy levels, respectively. For the first line of the Paschen series ($n_1=3, n_2=4$), the energy factor is $\frac{1}{3^2} - \frac{1}{4^2} = \frac{7}{144} \approx 0.0486$. For the second line of the Balmer series ($n_1=2, n_2=4$), the energy factor is $\frac{1}{2^2} - \frac{1}{4^2} = \frac{3}{16} = 0.1875$. For the third line of the Paschen series ($n_1=3, n_2=6$), the energy factor is $\frac{1}{3^2} - \frac{1}{6^2} = \frac{3}{36} \approx 0.0833$. For the fourth line of the Brackett series ($n_1=4, n_2=8$), the energy factor is $\frac{1}{4^2} - \frac{1}{8^2} = \frac{3}{64} \approx 0.0469$. Comparing these values, $0.0469 < 0.0486 < 0.0833 < 0.1875$, which corresponds to the sequence $D < A < C < B$.

The Heisenberg uncertainty principle applies inherently to all microscopic particles, such as electrons, because their wave-particle duality makes it impossible to define both position and momentum with absolute precision. Orbital size is dictated primarily by the principal quantum number, so the $2p_x$ orbital, with its lower principal quantum number, is inherently smaller than the $3p_x$ orbital. Unlike hydrogen, where orbital energy depends solely on the principal quantum number, multi-electron atoms like lithium possess unique effective nuclear charges that cause the energy of their 2s orbitals to differ from that of hydrogen. Lastly, chromium exhibits an electronic configuration of $[\text{Ar}] 3d^{5} 4s^{1}$, an arrangement favored because the half-filled 3d subshell provides greater stability than the expected filling pattern.

The energy of spectral lines in the hydrogen spectrum is given by the Rydberg formula $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$, where $n_f$ is the final energy level and $n_i$ is the initial energy level. For the first line of the Lyman series ($L_1$), the transition occurs from $n_i = 2$ to $n_f = 1$, resulting in $\Delta E_{L_1} = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \times \frac{3}{4} \text{ eV}$. The first line of the Balmer series ($B_1$) involves a transition from $n_i = 3$ to $n_f = 2$, yielding $\Delta E_{B_1} = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \times \frac{5}{36} \text{ eV}$. Calculating the ratio $x = \frac{\Delta E_{L_1}}{\Delta E_{B_1}}$ gives $x = \frac{3/4}{5/36} = \frac{3}{4} \times \frac{36}{5} = \frac{27}{5} = 5.4$. Expressing this result in the required format $5.4 = 54 \times 10^{-1}$ leads to the value 54.

The radius of an electron in a hydrogen-like species follows the relationship $r = a_0 \frac{n^2}{Z}$, where $a_0$ represents the Bohr radius of a hydrogen atom, $n$ is the principal quantum number, and $Z$ is the atomic number. Dividing the specified radius of 70.53 pm by the provided hydrogen Bohr radius of 52.9 pm reveals the required ratio of $\frac{n^2}{Z}$ to be $\frac{70.53}{52.9} = \frac{4}{3}$. Evaluating the given options shows that for a $\text{Li}^{2+}$ ion, which has an atomic number $Z=3$, the second energy state where $n=2$ results in the ratio $\frac{2^2}{3} = \frac{4}{3}$, matching the calculated value precisely.

The wave number of spectral lines in the Hydrogen atom is governed by the Rydberg equation, where the value is proportional to the difference between the inverse squares of the transition energy levels. The first line of the Balmer series involves an electron falling from $n_2 = 3$ to $n_1 = 2$, whereas the first line of the Brackett series involves a transition from $n_2 = 5$ to $n_1 = 4$.

Dividing the Balmer term by the Brackett term involves the following calculation:
$\frac{\frac{1}{2^2} - \frac{1}{3^2}}{\frac{1}{4^2} - \frac{1}{5^2}} = \frac{5/36}{9/400}$
Evaluating this expression yields $500/81$, which is equivalent to $5 : 0.81$.

The nuclear notation ${}_{Z}^{A}\mathrm{X}$ defines $Z$ as the number of protons and $A$ as the mass number (protons plus neutrons), meaning ${}_{12}^{24}\mathrm{Mg}$ comprises 12 protons and $24 - 12 = 12$ neutrons, rendering statement A incorrect.

The wavelength $\lambda$ is related to frequency $\nu$ by the formula $\lambda = c/\nu$. Substituting the speed of light $c = 3 \times 10^8 \mathrm{m/s}$ and frequency $\nu = 4.5 \times 10^{15} \mathrm{s}^{-1}$:
$\lambda = \frac{3 \times 10^8}{4.5 \times 10^{15}} = 6.67 \times 10^{-8} \mathrm{m}$
This calculation confirms that statement B is correct.

Energy $E$ is inversely proportional to wavelength $\lambda$ according to the relation $E = hc/\lambda$, which implies the ratio $E_1/E_2 = \lambda_2/\lambda_1$. Substituting $\lambda_1 = 900 \mathrm{nm}$ and $\lambda_2 = 300 \mathrm{nm}$ yields:
$\frac{E_1}{E_2} = \frac{300 \mathrm{nm}}{900 \mathrm{nm}} = \frac{1}{3}$
Since the calculated ratio is $1:3$ rather than $3:1$, statement C is incorrect.

The total number of photons $n$ required to provide energy $E$ is given by $n = E\lambda / hc$. Converting $\lambda = 2000 \mathrm{pm}$ to $2 \times 10^{-9} \mathrm{m}$ and substituting $E = 1 \mathrm{J}$, $h = 6.626 \times 10^{-34} \mathrm{Js}$, and $c = 3 \times 10^8 \mathrm{m/s}$:
$n = \frac{1 \times 2 \times 10^{-9}}{(6.626 \times 10^{-34}) \times (3 \times 10^8)} = 1.006 \times 10^{16}$
This result confirms that statement D is correct.

The wavelength of spectral lines in hydrogen-like atoms is determined by the Rydberg formula, where the transition between energy levels dictates the energy of the emitted photon. The shortest wavelength in any spectral series corresponds to an electronic transition from an infinite energy level to the series limit, whereas the longest wavelength corresponds to a transition between the two closest adjacent energy levels within that series.

For the hydrogen atom Lyman series, the shortest wavelength $x$ occurs as an electron falls from infinity to the first orbital, where the atomic number $Z = 1$. This relationship, $\frac{1}{x} = R \times 1^2 \times (\frac{1}{1^2} - \frac{1}{\infty ^2})$, simplifies directly to $\frac{1}{x} = R$. For the Balmer series of the helium ion $\mathrm{He}^{+}$, where $Z = 2$, the longest wavelength $\lambda$ is achieved by an electron transitioning from the third energy level to the second. The mathematical relationship is expressed as:
$\frac{1}{\lambda} = R \times 2^2 \times \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$
This simplifies to $\frac{1}{\lambda} = 4R \times (\frac{1}{4} - \frac{1}{9})$, which reduces to $\frac{1}{\lambda} = 4R \times \frac{5}{36}$. Substituting $R = \frac{1}{x}$ into this expression results in $\frac{1}{\lambda} = \frac{5}{9x}$, which leads to the final wavelength of $\lambda = \frac{9x}{5}$.

The energy of a photon emitted during an electronic transition in a hydrogen atom is proportional to the difference of the inverse squares of the principal quantum numbers, given by $\Delta E \propto \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$, where $n_1 = 2$ for the Balmer series. The first Balmer line corresponds to the transition from $n_2 = 3$ to $n_1 = 2$, and the second Balmer line corresponds to the transition from $n_2 = 4$ to $n_1 = 2$.

Substituting these values, the energy of the first line is $x \propto \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = \frac{5}{36}$, while the energy of the second line is $E_2 \propto \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = \frac{3}{16}$. Calculating the ratio of the energy of the second line to the first yields $\frac{E_2}{x} = \frac{3/16}{5/36} = \frac{3}{16} \times \frac{36}{5} = \frac{27}{20} = 1.35$. Thus, the energy of the second Balmer line is $1.35\,x$.

According to Bohr's theory, the radius of an orbit in a hydrogen-like species is determined by the formula $r_n = a_0 \frac{n^2}{Z}$, where $n$ represents the principal quantum number and $Z$ denotes the atomic number. Calculating the radius for each case, hydrogen in the first orbit results in $a_0 \frac{1^2}{1} = a_0$, whereas the helium ion in its first orbit yields $a_0 \frac{1^2}{2} = 0.5 a_0$. Continuing the analysis, the helium ion in its second orbit gives $a_0 \frac{2^2}{2} = 2 a_0$, while the lithium ion in the first orbit computes to $a_0 \frac{1^2}{3} \approx 0.33 a_0$. Finally, the beryllium ion in its second orbit results in $a_0 \frac{2^2}{4} = a_0$, which confirms that both hydrogen in the first orbit and beryllium in the second orbit share the same radius of $a_0$.

The energy of a photon is given by the equation $E = h\nu = \frac{hc}{\lambda} = hc\bar{\nu}$, where energy is directly proportional to frequency and wave number, and inversely proportional to wavelength. To compare the energies, express all values as frequency ($\nu$) using the speed of light $c \approx 3 \times 10^8 \text{ m/s}$.

For photon A, $\nu_A = \frac{c}{\lambda_A} = \frac{3 \times 10^8 \text{ m/s}}{400 \times 10^{-9} \text{ m}} = 7.5 \times 10^{14} \text{ s}^{-1}$. The frequency of photon B is given directly as $\nu_B = 10^{16} \text{ s}^{-1}$, which is equivalent to $100 \times 10^{14} \text{ s}^{-1}$. For photon C, given a wave number $\bar{\nu}_C = 10^4 \text{ cm}^{-1} = 10^6 \text{ m}^{-1}$, the frequency is $\nu_C = c \bar{\nu}_C = (3 \times 10^8 \text{ m/s})(10^6 \text{ m}^{-1}) = 3 \times 10^{14} \text{ s}^{-1}$. Comparing these magnitudes, $100 \times 10^{14} > 7.5 \times 10^{14} > 3 \times 10^{14}$, which results in the energy order $B > A > C$.

The nodal surface is defined as the region where the electron probability density $|\psi|^2$ is zero, which mathematically corresponds to the point where the wave function $\psi$ intersects the radial distance axis. In the plot of the wave function against distance shown in Figure 2, point B represents this intersection where the value of $\psi$ transitions from positive to negative, identifying it as the radial node of the 2s orbital.

The ionization energy of a hydrogen-like species is determined by the square of its atomic number, as the increased nuclear charge binds the electron more tightly than in a standard hydrogen atom. For a single-electron system such as $\mathrm{Li}^{2+}$, the atomic number is $Z = 3$, and the energy required to remove the electron from the ground state is given by the relation $E = E_H \times Z^2$. Substituting the value of $E_H = 2.18 \times 10^{-18}$ J/atom and $Z = 3$ into this equation gives $E = 2.18 \times 10^{-18} \times 3^2$ J/atom, which simplifies to $E = 1.962 \times 10^{-17}$ J/atom.

Einstein's photoelectric equation relates the maximum kinetic energy of liberated electrons $K$ to the incident photon energy $E$ and the work function $\Phi$ through the expression $K = E - \Phi$. Representing the work functions as $\Phi_A = \Phi$ and $\Phi_B = 2\Phi$ satisfies the given $1:2$ ratio.

The kinetic energies are expressed as $K_A = 6 - \Phi$ and $K_B = 6 - 2\Phi$. Substituting these into the ratio $\frac{K_A}{K_B} = 2.642$ yields $\frac{6 - \Phi}{6 - 2\Phi} = 2.642$. This results in $6 - \Phi = 15.852 - 5.284\Phi$, which simplifies to $4.284\Phi = 9.852$, giving $\Phi \approx 2.3$ eV. Thus, the work functions are $\Phi_A = 2.3$ eV and $\Phi_B = 4.6$ eV.

The wave number of a spectral line in the hydrogen spectrum is determined by the Rydberg formula $\bar{\nu} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$, where the Balmer series corresponds to electron transitions ending at the second energy level, meaning $n_1 = 2$ and $n_2 = 3, 4, 5, \dots$. By substituting $n_1 = 2$ and integer values for $n_2$ into the formula $\bar{\nu} = R \left( \frac{1}{4} - \frac{1}{n_2^2} \right)$, the validity of the provided spectral lines can be verified.

For $n_2 = 3$, the calculation yields $\bar{\nu} = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36}$. For $n_2 = 4$, the result is $\bar{\nu} = R \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{3R}{16}$. Finally, for $n_2 = 5$, the value is $\bar{\nu} = R \left( \frac{1}{4} - \frac{1}{25} \right) = \frac{21R}{100}$. As these three values directly correspond to the Balmer series formula for consecutive transitions, this set represents spectral lines belonging to the series.

The de Broglie wavelength of a charged particle accelerated through a potential difference $V$ is given by the relation $\lambda = \frac{h}{\sqrt{2mqV}}$, where $h$ is Planck's constant, $m$ is the mass, and $q$ is the charge of the particle. For two particles with equal charges accelerated through the same potential difference, the wavelength is inversely proportional to the square root of the mass, expressed as $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2}{m_1}}$. Substituting the given values $m_1 = 1 \text{ amu}$ and $m_2 = 4 \text{ amu}$ into this expression results in $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{4}{1}} = 2$, which signifies that $\lambda_1 = 2\lambda_2$, leading to $x = 2$.

The energy required for an electronic transition in a hydrogen atom is calculated using the Bohr formula $\Delta E = R_H (\frac{1}{n_1^2} - \frac{1}{n_2^2})$, where the Rydberg constant is $R_H = 2.18 \times 10^{-18} \text{ J}$ (converted from $2.18 \times 10^{-11} \text{ ergs}$). For an excitation from the first Bohr orbit ($n_1 = 1$) to the second Bohr orbit ($n_2 = 2$), the energy change per atom is $\Delta E = 2.18 \times 10^{-18} \times (1 - \frac{1}{4}) = 2.18 \times 10^{-18} \times 0.75 = 1.635 \times 10^{-18} \text{ J}$. To determine the energy required per mole, this value is multiplied by Avogadro's number ($N_A \approx 6.022 \times 10^{23} \text{ mol}^{-1}$), yielding $1.635 \times 10^{-18} \times 6.022 \times 10^{23} = 9.835 \times 10^5 \text{ Jmol}^{-1}$.

The Rydberg formula for hydrogen-like ions is expressed as $\frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$, where $Z$ is the atomic number, $R_H \approx 109677 \ \text{cm}^{-1}$ is the Rydberg constant, and $n_1$ and $n_2$ are the principal quantum numbers of the transition levels. Solving the system of equations $n_1 + n_2 = 4$ and $n_1 - n_2 = 2$ yields $n_1 = 3$ and $n_2 = 1$. Substituting $Z = 3$ for the $\mathrm{Li}^{2+}$ ion into the formula gives $\frac{1}{\lambda} = 109677 \times 3^2 \times \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 109677 \times 9 \times \left( \frac{8}{9} \right) = 877416 \ \text{cm}^{-1}$. Calculating the inverse results in $\lambda \approx 1.14 \times 10^{-6} \ \text{cm}$.

The Aufbau principle dictates that electrons fill atomic orbitals in order of increasing energy, beginning with the lowest energy levels. For chromium with an atomic number of 24, the electronic configuration is $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{1} 3d^{5}$. Adding these up, the first 18 electrons occupy the shells through the 3p subshell, placing the 19th electron into the 4s subshell. Because this electron resides in a 4s orbital, the principal quantum number $n=4$ and the azimuthal quantum number $\ell=0$. Since the azimuthal value is 0, the magnetic quantum number $m=0$, and using the standard spin convention $s=+\frac{1}{2}$, we arrive at the set $n=4, \ell=0, m=0, s=+\frac{1}{2}$.