Structure Of Atom – JEE Main ChemistryPractice Questions & PYQs

The energy of an orbital in a multi-electron atom is determined by the ($n+\ell$) rule, which states that orbitals with lower ($n+\ell$) values possess lower energy; if the ($n+\ell$) sums are equal, the orbital with the lower principal quantum number $n$ occupies a lower energy state. Applying this rule to the given orbitals, the ($n+\ell$) values are $3+2=5$ for (A), $4+0=4$ for (B), $6+1=7$ for (C), $5+1=6$ for (D), and $2+1=3$ for (E). Ordering these calculated sums from lowest to highest yields the sequence $3 < 4 < 5 < 6 < 7$, which corresponds to the orbital arrangement $E < B < A < D < C$.

For case (A), the principal quantum number $n=5$ allows for subshells with azimuthal quantum numbers $l = 0, 1, 2, 3, 4$. The condition $m_l = -1$ is satisfied in any subshell where $l \geq |m_l|$, which includes $l=1, 2, 3, 4$ (the $5p, 5d, 5f,$ and $5g$ subshells). Each of these four subshells contains exactly one orbital with the magnetic quantum number $m_l = -1$, and because each orbital holds a maximum of 2 electrons with opposing spins, the total number of electrons is $4 \times 2 = 8$.

For case (B), the quantum numbers $n=3, l=2, m_l=-1, m_s=+\frac{1}{2}$ fully specify a single quantum state within the $3d$ subshell. According to the Pauli exclusion principle, which states that no two electrons in an atom can have the same set of four quantum numbers, this specific combination defines the state of exactly $1$ electron.

Atomic emission spectra arise because electrons in excited atoms return to lower energy states, emitting radiation with energy corresponding to the difference between these states, characterized by discrete frequencies defined by the Rydberg formula $\bar{\nu} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$. Since the frequency of electromagnetic radiation is directly proportional to the wavenumber, comparing transitions requires evaluating the factor $Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$. For the second line of the Balmer series where $n_1=2$ and $n_2=4$ in $He^+$ with $Z=2$, the factor is $2^2 (\frac{1}{2^2} - \frac{1}{4^2}) = 4 (\frac{1}{4} - \frac{1}{16}) = 4 (\frac{3}{16}) = \frac{3}{4}$. For the first line of the Lyman series where $n_1=1$ and $n_2=2$ in a hydrogen atom with $Z=1$, the factor is $1^2 (\frac{1}{1^2} - \frac{1}{2^2}) = 1 (1 - \frac{1}{4}) = \frac{3}{4}$. Because both calculations yield the same factor, the frequencies of these transitions are identical.

To understand the distribution of nodes in an atomic orbital, we apply the rules that the number of radial nodes is given by $n - \ell - 1$ and the number of nodal planes is equal to the azimuthal quantum number $\ell$. For the 2s orbital, where $n=2$ and $\ell=0$, the calculation results in 1 radial node and zero nodal planes, which matches entry IV. Evaluating the 3s orbital with $n=3$ and $\ell=0$ gives 2 radial nodes and zero nodal planes, corresponding to entry III. Similarly, the 3p orbital with $n=3$ and $\ell=1$ yields 1 radial node and 1 nodal plane, aligning with entry II, while the 4d orbital with $n=4$ and $\ell=2$ results in 1 radial node and 2 nodal planes, corresponding to entry I.

The phenomenon where incident radiation ejects electrons from a metal surface is governed by Einstein's photoelectric equation, which dictates that the energy of an incoming photon must equal the sum of the metal's work function and the maximum kinetic energy of the ejected electron. Mathematically, this is expressed as $\frac{hc}{\lambda} = \Phi + K.E.$, where the incident energy depends on the wavelength $\lambda$.

Before performing the calculation, all values must be expressed in consistent SI units. The work function of 2.3 eV is converted to Joules by multiplying by the provided factor of $1.60 \times 10^{-19} \text{ J/eV}$, resulting in $3.68 \times 10^{-19} \text{ J}$. Adding this to the kinetic energy of $2.8 \times 10^{-20} \text{ J}$, which is equivalent to $0.28 \times 10^{-19} \text{ J}$, yields a total incident energy of $3.96 \times 10^{-19} \text{ J}$.

Using the relationship $\frac{hc}{\lambda} = E$, we substitute the constants $h = 6.6 \times 10^{-34} \text{ J s}$ and $c = 3.0 \times 10^8 \text{ m/s}$ to solve for $\lambda$. This gives the equation $\frac{19.8 \times 10^{-26}}{\lambda} = 3.96 \times 10^{-19}$, which simplifies to $\lambda = 5 \times 10^{-7} \text{ m}$. Converting this to nanometers gives 500 nm, and since the target format is $x \times 10^2 \text{ nm}$, the value of x is 5.

The enhanced stability of atoms with half-filled subshells arises from several key electronic factors that create a more favorable energy state. Electrons adopt a highly symmetrical distribution within the subshell, which inherently lowers the coulombic repulsion energy and reduces the shielding effect of electrons upon one another. Furthermore, the configuration facilitates a greater number of potential electron exchanges, leading to larger exchange energy that significantly contributes to the overall stability of the system.

The energy of an electron in a hydrogen-like species is determined by the expression $E_n = -13.6 \frac{Z^2}{n^2}$, which indicates that for a constant principal quantum number $n$, the energy varies directly with the square of the atomic number $Z$ but carries a negative sign. This functional relationship, represented as $E_n \propto -Z^2$, describes an inverted parabola passing through the origin because the energy value becomes increasingly negative as the atomic number increases. Consequently, the graph shows a curve that drops further below the x-axis, illustrating that higher nuclear charges result in lower, more stable energy states for any given orbit.

Heisenberg's Uncertainty Principle mandates that a particle's position and momentum cannot be simultaneously specified with arbitrary precision, confirming Statement I is true. Applying the standard inequality $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$ under the given condition that the uncertainties are equal, $\Delta x = \Delta p$, transforms the expression into $(\Delta p)^2 \ge \frac{h}{4\pi}$, which simplifies to $\Delta p \ge \frac{1}{2} \sqrt{\frac{h}{\pi}}$. Substituting $m \Delta v$ for $\Delta p$ leads to $m \Delta v \ge \frac{1}{2} \sqrt{\frac{h}{\pi}}$, and dividing by $m$ confirms the value $\Delta v \ge \frac{1}{2m} \sqrt{\frac{h}{\pi}}$ provided in Statement II is also true.

Energy levels in a multielectron atom are governed by the sum of the principal quantum number $n$ and the azimuthal quantum number $l$. When this sum ($n+l$) is identical for two orbitals, they occupy the same energy state in the absence of external electric or magnetic fields. Applying this rule to the given sets, the sum for A is 1, for B is 2, and for C is 3. Since both D and E result in a sum of 5, they represent the pair of orbitals with equivalent energy.

The Bohr model treats the electron as a particle restricted to a definite circular path around the nucleus, an assumption that mimics planetary motion. Modern quantum mechanics rejects this concept because it violates the Heisenberg uncertainty principle, which dictates that it is impossible to define both the precise position and momentum of a subatomic particle simultaneously. Instead of fixed trajectories, the quantum model defines the electron's behavior through wave functions that map out probability distributions known as orbitals.

The radius of an orbit in the Bohr model of the hydrogen atom scales with the square of its principal quantum number, following the proportionality $r \propto n^2$. Comparing the fourth orbit ($n=4$) to the second ($n=2$), the ratio of their radii is $\frac{16}{4} = 4$, confirming the fourth orbit is four times larger. Similarly, the ratio of the eighth orbit ($n=8$) to the fourth ($n=4$) is $\frac{64}{16} = 4$, and the ratio of the third ($n=3$) to the first ($n=1$) is $\frac{9}{1} = 9$. When applying this to the sixth orbit ($n=6$) and the fourth ($n=4$), the ratio of the radii is $\frac{36}{16} = \frac{9}{4} = 2.25$, which demonstrates that the statement claiming the sixth orbit is three times larger than the fourth is inaccurate.

In a single-electron system like the hydrogen atom, the energy of an orbital is determined solely by its principal quantum number, $n$, where higher values of $n$ correspond to higher orbital energies. By examining the given orbitals, we find that $4s$ and $4p_z$ are associated with $n=4$, while $3p_x$, $3d_{x^{2-}y^2}$, and $3d_{z^2}$ all correspond to $n=3$. Because $n=3$ is the lowest principal quantum number among the provided choices, the orbitals $3p_x$, $3d_{x^{2-}y^2}$, and $3d_{z^2}$ represent the lowest energy levels.

The ground state electronic configuration of chromium ($Z=24$) is $1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$ due to the stability of half-filled subshells. An azimuthal quantum number of $l=1$ represents $p$-subshells, which consist of $2p^6$ and $3p^6$, yielding a total of $6 + 6 = 12$ electrons. An azimuthal quantum number of $l=2$ represents $d$-subshells, which consist of the $3d^5$ subshell, resulting in 5 electrons.

The Bohr quantization condition states that the circumference of an electron's orbit must be an integral multiple of its de-Broglie wavelength, expressed as $2 \pi r_n = n \lambda$. For a hydrogen atom, the radius of the $n^{th}$ orbit is given by $r_n = \frac{a_0 n^2}{Z}$ where $Z=1$, and substituting this radius into the quantization condition results in $\frac{2 \pi a_0 n^2}{Z} = n \lambda$. Simplifying this expression by substituting $Z=1$ leads to $2 \pi a_0 n^2 = n \lambda$, which isolates the wavelength as $\lambda = 2 \pi a_0 n$. Evaluating this relation for the second orbit where $n=2$ results in a wavelength of $4 \pi a_0$, which corresponds to the expression $\frac{8 \pi a_0}{n}$ evaluated for the same orbit.

The subsidiary quantum number $I$ fundamentally identifies the shape of an orbital, and the provided diagram correctly represents the $2p_x$ orbital. In contrast, the positive and negative signs appearing in wave functions denote the mathematical phase rather than electrical charge, and the nodal plane of the $2p_x$ orbital, where the wave function is zero, is the $yz$ plane rather than the $xy$ plane.

With an atomic number of 9, the electronic configuration for fluorine is $1s^2 2s^2 2p^5$, where the presence of five electrons with $m_s = +\frac{1}{2}$ and four electrons with $m_s = -\frac{1}{2}$ confirms the first statement is correct. The final electron resides in a $2p$ orbital, characterized by $n=2$ and $l=1$, which verifies the third statement. Other assertions are incorrect because the sum of angular nodes for the occupied orbitals is 3, obtained by summing the $l$ values of $0, 0, 1, 1,$ and $1$ for the $1s, 2s, 2p_x, 2p_y,$ and $2p_z$ orbitals, while the $p_z$ orbital can accommodate up to two electrons rather than being restricted to one.

The relationship between the orbit radius of a hydrogen-like ion and the fundamental Bohr radius is defined by the expression $r = a_0 \frac{n^2}{z}$, where $n$ represents the principal quantum number and $z$ is the atomic number. For a helium ion, the atomic number is $2$, and the first excited state corresponds to the second energy level, meaning $n = 2$. Substituting these specific values into the formula yields $r = a_0 \frac{(2)^2}{2}$, which simplifies to $r = 2a_0$.

The total energy of an electron in a stationary state like the $1\text{ s}$ orbital of a hydrogen atom is constant and independent of its instantaneous position relative to the nucleus. Because the electron exists in a quantized state with a fixed energy level determined by the principal quantum number $n$, asserting that the total energy varies and reaches a maximum at a distance $r = a_0$ is fundamentally incorrect according to quantum mechanical principles.

Henry Moseley's experimental work on X-ray spectra proved that the frequency of emitted radiation depends on the atomic number ($Z$) rather than the atomic mass. The specific relationship follows Moseley's law, expressed as $\sqrt{v} = a(Z - b)$, where $a$ and $b$ are constants. Since the linear dependency strictly applies to the square root of the frequency plotted against the atomic number, both statements are flawed: the first incorrectly cites atomic mass, and the second fails to identify the square root relationship required to obtain a straight line.

The energy levels of a hydrogen atom depend solely on the principal quantum number, meaning that all orbitals sharing the same principal shell, such as the $2s$ and $2p$ subshells, possess identical energy. The correct hierarchy follows the sequence
$1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f$
Statements that imply an energy difference between orbitals within the same principal shell, or that incorrectly order orbitals across different shells, contradict this principle of degeneracy. Consequently, any arrangement proposing that $2s < 2p$ or that $4s < 3d$ is inherently incorrect.