Structure Of Atom – NEET UG ChemistryPractice Questions & PYQs

Quantum numbers define the characteristics of atomic orbitals, where the principal quantum number $n$ represents the main energy shell and the azimuthal quantum number $l$ determines the subshell shape. Each value of $l$ corresponds to a specific letter designation: $l=0$ denotes an $s$ orbital, $l=1$ denotes a $p$ orbital, $l=2$ denotes an $d$ orbital, and $l=3$ denotes an $f$ orbital. By combining the given shell $n$ with the subshell type for $l$, we identify that the set $n=2, l=1$ results in a 2p orbital, $n=4, l=0$ results in a 4s orbital, $n=5, l=3$ results in a 5f orbital, and $n=3, l=2$ results in a 3d orbital. This sequence links A to II, B to III, C to IV, and D to I.

The radius of an electron in a hydrogen-like ion is defined by the expression $r_n = r_H \frac{n^2}{Z}$, where $r_H$ is the Bohr radius for hydrogen, $n$ represents the principal quantum number, and $Z$ is the atomic number. For any atom in its ground state, $n$ is 1, which means the radius is inversely proportional to the atomic number.

With the ground-state radius for hydrogen given as 0.53 \AA and the atomic number for lithium as 3, the calculation is straightforward. Applying these values, we determine the ion's radius by dividing 0.53 \AA by 3, which results in 0.17 \AA.

In the Bohr model of the hydrogen atom, the radius of an orbit increases proportionally with the square of the principal quantum number. This relationship is defined by the formula $r_n = n^2 \times r_1$, where $r_n$ represents the radius of the orbit and $r_1$ corresponds to the radius of the ground state.

Since the ground state radius for $n=1$ is 0.530 Å, the radius of the first excited state at $n=2$ is obtained by squaring the principal quantum number and multiplying it by the initial radius. Utilizing the expression $r_2 = 2^2 \times 0.530 = 2.12$, the resulting radius for the second orbit is 2.12 Å.

While Niels Bohr originally described electrons as moving in fixed circular paths, this model proved insufficient for explaining the fine spectral lines observed in atomic experiments. To account for these complexities, Arnold Sommerfeld extended Bohr's theory by proposing that electrons can also travel along elliptical orbits, providing a more detailed understanding of electron movement around the nucleus.

Heisenberg's Uncertainty Principle dictates that the product of the uncertainties in a particle's position and momentum must be at least $\frac{h}{4 \pi}$. Since momentum is defined as the product of mass and velocity, the uncertainty relationship is rearranged to $\Delta x = \frac{h}{4 \pi m \Delta v}$ to determine the minimum uncertainty in position.

The absolute uncertainty in velocity is derived from the stated $0.001\%$ accuracy of the $3 \times 10^{4} \text{cm/s}$ speed, resulting in $\Delta v = 3 \times 10^{4} \times \frac{0.001}{100} = 0.3 \text{cm/s}$. By substituting the provided mass $m = 9.1 \times 10^{-28} \text{g}$ and Planck's constant $h = 6.626 \times 10^{-27} \text{erg s}$ into the uncertainty expression, the calculation is structured as:

$\Delta x = \frac{6.626 \times 10^{-27}}{4 \times \pi \times 9.1 \times 10^{-28} \times 0.3}$

Performing this division results in an uncertainty in the electron's position of $1.93 \text{cm}$.

The longest wavelength in any spectral series corresponds to the transition that releases the minimum amount of energy, which occurs between the two energy levels that are closest to each other. For the Balmer series, electrons transition to the $n_1 = 2$ state, meaning the smallest energy drop occurs during the transition from the $n_2 = 3$ level. By substituting these quantum numbers and the Rydberg constant $R_H = 109677 \text{ cm}^{-1}$ into the Rydberg formula:

$\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = 109677 \left(\frac{1}{4} - \frac{1}{9}\right) = 15233 \text{ cm}^{-1}$

Taking the reciprocal of this value gives the wavelength $\lambda \approx 6.56 \times 10^{-5} \text{ cm}$, which converts to $656 \text{ nm}$.

Atomic spectra are primarily determined by the specific electronic configuration of a species, meaning that atoms or ions with the same number of electrons will exhibit comparable spectral characteristics. Neutral helium possesses 2 electrons, so we must identify which option shares this exact electronic count. Although hydrogen ($H$) and the $He^+$ ion contain only 1 electron and sodium ($Na$) contains 11, the $Li^+$ ion effectively loses one electron from the neutral lithium atom to maintain a stable 2-electron configuration, making its spectrum analogous to that of helium.

A neutral nitrogen atom has 7 electrons configured as $1s^2 2s^2 2p^3$. To form the $N^{2+}$ ion, two electrons are removed from the $2p$ subshell, resulting in the electronic configuration $1s^2 2s^2 2p^1$. With all electrons in the lower $1s$ and $2s$ orbitals being fully paired, the single electron residing in the $2p$ orbital is left unpaired.

Chemical species are classified as isoelectronic when they possess an identical total number of electrons. The neutral molecule $CO$ contains 6 electrons from carbon and 8 from oxygen, summing to 14 electrons in total. Similarly, the $CN^{-}$ ion derives 6 electrons from carbon, 7 from nitrogen, and 1 from its negative charge, also reaching a total of 14 electrons and making it isoelectronic with $CO$.

Each subshell contains a specific number of orbitals determined by the azimuthal quantum number $l$, where the total count of these orbitals is always calculated as $2l+1$. Since the Pauli Exclusion Principle limits each individual orbital to a maximum of two electrons with opposite spins, the total electron capacity of the subshell is simply twice the number of orbitals. Applying this rule leads to the expression $2(2l+1)$, which simplifies to $4l+2$ as the final formula for the maximum number of electrons.

Species are defined as isoelectronic when they possess the exact same number of electrons. An analysis of the ionic species shows that ${O}^{2-}$, ${N}^{3-}$, ${F}^{-}$, and ${Na}^{+}$ each contain 10 electrons, while the ${Tl}^{+}$ ion contains 80 electrons. Because of this distinct difference in electron quantity, ${Tl}^{+}$ does not share the same electronic configuration as the others.

The radius of an electron orbit in a hydrogen atom depends on the principal quantum number $n$ according to the Bohr model. Since the orbital radius is proportional to the square of this quantum number, the radius of the $n^{th}$ orbit is calculated as $r_n = r_0 n^2$, where $r_0$ is the radius of the first orbit. This quadratic dependence implies that the distance of the electron from the nucleus increases significantly as $n$ increases.

Heisenberg's uncertainty principle establishes a fundamental limit on the precision with which we can simultaneously know a particle's position and momentum, defined by the relationship:

$\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$

Because the uncertainty in position is given as 1.0 nm for both the electron and the helium atom, the product of position and momentum uncertainty is governed by the same physical constraint. Since the position uncertainty is identical for both particles, the required minimum uncertainty in momentum must also be equivalent to satisfy this inequality. Consequently, the minimum uncertainty in the measurement of the helium atom's momentum is equal to the given value for the electron, which is $\Delta p = 5.0 \times 10^{-26} \,\text{kg} \,\text{m} \, \text{s}^{-1}$.

The de Broglie wavelength describes the wave-like character of matter by relating a particle's wavelength to its mass and velocity through the equation $\lambda = \frac{h}{mv}$. To determine the wavelength for this particle, we convert the mass of 1 g into $10^{-3}$ kg to maintain consistency with the SI units of the Planck constant, which is $6.63 \times 10^{-34}$ J s. Plugging the mass and the velocity of 100 m/s into the formula gives

$\lambda = \frac{6.63 \times 10^{-34}}{10^{-3} \times 100} \text{ m}$

which simplifies to a final result of $6.63 \times 10^{-33}$ m.

Heisenberg's uncertainty principle asserts that there is a fundamental limit to the precision with which certain pairs of physical properties, such as position and momentum, can be known simultaneously. This inherent constraint is expressed by the mathematical inequality $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.

Calculating the minimum uncertainty in position given a momentum uncertainty of $1 \times 10^{-5} \text{ kg m/s}$ requires rearranging the principle to solve for $\Delta x = \frac{h}{4 \pi \Delta p}$. Substituting the value of Planck's constant $h = 6.62 \times 10^{-34} \text{ kg m}^2/\text{s}$ and the provided momentum uncertainty into this relationship results in the expression $\Delta x = \frac{6.62 \times 10^{-34}}{4 \times 3.14 \times 1 \times 10^{-5}}$. Completing this calculation yields a position uncertainty of $5.27 \times 10^{-30} \text{ m}$.

The Aufbau principle dictates that electrons populate atomic orbitals in order of increasing energy, a hierarchy determined by the $(n+l)$ rule where lower sums fill first and ties are resolved by the principal quantum number $n$. Following this rule, the $4s$ orbital fills first with $n+l = 4$, followed by the orbitals with $n+l = 5$ in increasing order of $n$, which are $3d$, $4p$, and $5s$. Finally, the $4d$ orbital is filled because its higher energy corresponds to $n+l = 6$.

Calcium has an atomic number of 20, indicating that a neutral atom contains 20 electrons in total. We simplify the configuration by using the symbol for the preceding noble gas, argon, which accounts for the first 18 electrons as $[Ar]$. According to the Aufbau principle, the remaining two electrons occupy the $4s$ orbital, resulting in the final electronic configuration of $[Ar] 4s^2$.

Bohr's model dictates that an electron's energy in a hydrogen atom depends on its principal quantum number, with energy increasing as the electron transitions to higher orbits. To determine the energy required for a jump from $n=1$ to $n=3$, we first calculate the energy for each state using the expression $E_n = -\frac{1312}{n^2} \text{ kJ mol}^{-1}$.

For the ground state at $n=1$, the energy is $E_1 = -1312 \text{ kJ mol}^{-1}$. When the electron moves to $n=3$, its energy becomes $E_3 = -\frac{1312}{3^2} = -145.7 \text{ kJ mol}^{-1}$. The difference in energy is $\Delta E = E_3 - E_1 = -145.7 - (-1312) = 1166.3 \text{ kJ mol}^{-1}$, confirming that energy is absorbed during this transition.

To express this value per atom, we divide the molar energy by Avogadro's number, $6.023 \times 10^{23} \text{ mol}^{-1}$, which yields $193.6 \times 10^{-23} \text{ kJ}$. Converting this to joules by multiplying by $1000$ gives $193.6 \times 10^{-20} \text{ J}$. Finally, converting the result to ergs using the conversion factor $1 \text{ J} = 10^7 \text{ ergs}$ results in $193.6 \times 10^{-13} \text{ ergs}$, which is equivalent to $0.1936 \times 10^{-10} \text{ ergs}$.

The principal quantum number $n=2$ paired with the azimuthal quantum number $l=1$ identifies the $2p$ subshell, which consists of three distinct orbitals. Since each individual orbital can accommodate a maximum of two electrons based on the Pauli exclusion principle, the entire $2p$ subshell holds a total of $3 \times 2 = 6$ electrons.

Heisenberg's uncertainty principle, formulated mathematically as $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$, establishes that a particle's position and momentum cannot be simultaneously determined with exact precision. This concept directly contradicts the foundation of Bohr's atomic model, which relies on the assumption that electrons follow well-defined, circular orbits where their precise location and velocity can be known. Because Bohr's theory assumes a deterministic path for electrons, it does not incorporate the inherent uncertainty regarding simultaneous measurements that this principle dictates.