Structure Of Atom – NEET UG ChemistryPractice Questions & PYQs

According to Bohr's theory of hydrogen-like atoms, the radius of an orbit depends on the principal quantum number $n$ and the atomic number $Z$ through the relationship $r \propto \frac{n^2}{Z}$. To find the radius of the third orbit of $\mathrm{Li}^{2+}$, where $n=3$ and $Z=3$, we compare it to the known radius of the second orbit of $\mathrm{He}^{+}$, which is $105.8\,\text{pm}$ with $n=2$ and $Z=2$. Setting up the proportion $\frac{r(\mathrm{Li}^{2+})}{105.8\,\text{pm}} = \frac{3^2 / 3}{2^2 / 2}$ simplifies the calculation to a factor of $\frac{3}{2}$. Multiplying the reference value by this ratio of $1.5$ yields a final radius of $158.7\,\text{pm}$.

The size of an orbital is determined by the principal quantum number, while its geometric shape is governed by the azimuthal quantum number, meaning that all $d$-orbitals share the same general shape regardless of the shell they occupy, even as their size increases from the $3d$ to $4d$ and $5d$ levels. Although all five $3d$ orbitals are degenerate in an atom in a free state, they do not all exhibit the same shape. Specifically, the $d_{x^{2}-y^{2}}$ and $d_{z^{2}}$ orbitals possess unique and distinct geometries, making the assertion that they are similar to one another incorrect.

The Einstein photoelectric effect describes how the energy of an incident photon is distributed between the work function, representing the minimum energy required to extract an electron, and the resulting kinetic energy, expressed as $E = \phi + K.E.$ The energy for one mole of photons is determined using the relation $E = \frac{N_A hc}{\lambda}$ with the provided constants $N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}$, $h = 6.626 \times 10^{-34} \, \text{Js}$, $c = 3 \times 10^8 \, \text{ms}^{-1}$, and $\lambda = 300 \times 10^{-9} \, \text{m}$, yielding $3.975 \times 10^5 \, \text{J} \, \text{mol}^{-1}$. Isolating the work function leads to the subtraction $\phi = E - K.E.$, or $\phi = (3.975 - 1.68) \times 10^5 \, \text{J} \, \text{mol}^{-1}$ using the given kinetic energy of $1.68 \times 10^5 \, \text{J} \, \text{mol}^{-1}$. This calculation results in $2.295 \times 10^5 \, \text{J} \, \text{mol}^{-1}$, which aligns with $2.31 \times 10^5 \, \text{J} \, \text{mol}^{-1}$.

Electromagnetic waves propagate at a constant speed that is the product of their frequency and wavelength. The governing principle relates the speed of light, $c$, the frequency, $\nu$, and the wavelength, $\lambda$, through the equation $\lambda = \frac{c}{\nu}$.

Converting the broadcast frequency of 1,368 kHz to hertz yields 1,368 x 10^3 Hz. Using the speed of light value of 3.0 x 10^8 m/s, the wavelength calculation is performed as $\lambda = \frac{3.0 \times 10^8}{1,368 \times 10^3} = 219.3 \text{ m}$. This result represents the wavelength of the electromagnetic radiation emitted by the transmitter.

The quantization condition for a hydrogen atom dictates that an electron's orbit must accommodate an integer number of de Broglie wavelengths along its circumference, which is expressed by the relationship $n \lambda = 2 \pi r$. Since the radius of the $n$-th orbit is defined by the formula $r = \frac{n^2}{Z} a_0$, substituting this into the wavelength equation allows us to derive the expression $\lambda = \frac{2 \pi n a_0}{Z}$.

Applying this to the second Bohr orbit of a hydrogen atom involves setting $n$ to 2, $Z$ to 1, and using the given Bohr radius $a_0$ of 52.9 pm. Placing these values into our derived expression results in $\lambda = \frac{2 \pi \times 2 \times 52.9}{1}$ pm, which simplifies the final wavelength to $\lambda = 211.6 \pi$ pm.

To determine the relative energies of atomic orbitals, we use the Aufbau Principle, where energy depends on the sum of the principal quantum number $n$ and the azimuthal quantum number $l$. An orbital with a higher $(n+l)$ sum possesses greater energy, and if the sums are equal, the orbital with the lower principal quantum number $n$ has the lower energy.

Calculating these values, $5f$ results in $n+l = 5+3 = 8$, $6p$ results in $n+l = 6+1 = 7$, $5p$ results in $n+l = 5+1 = 6$, and $4d$ results in $n+l = 4+2 = 6$. Comparing these sums, $5f$ clearly has the highest energy, followed by $6p$. Because $5p$ and $4d$ both have an $(n+l)$ value of $6$, we compare their principal quantum numbers; since $5 > 4$, $5p$ is higher in energy than $4d$. Organizing these orbitals from highest to lowest energy provides the sequence $5f > 6p > 5p > 4d$.

The number of angular nodes corresponds directly to the azimuthal quantum number $\ell$, which means $\ell = 3$. Since the total number of nodes is defined by the expression $n - 1$, we can find the principal quantum number by setting this expression equal to the given count of 3:
$n - 1 = 3$
Solving this equation provides $n = 4$, and pairing this principal quantum number with an azimuthal value of $\ell = 3$, which signifies an $f$ subshell, identifies the orbital as 4f.

The electromagnetic spectrum of a hydrogen atom is organized into distinct series based on the final energy level to which an electron transitions, where each series corresponds to a specific portion of the electromagnetic spectrum. Among these, the Balmer series consists of electron transitions specifically ending at the $n=2$ energy level, releasing photons that possess the precise wavelength range detectable by the human eye as visible light.

Hund's Rule of Maximum Multiplicity dictates that for degenerate orbitals like the p-subshell, electrons must occupy each orbital singly before any pairing occurs. For a nitrogen atom with seven electrons, this principle requires that the three 2p electrons be distributed one by one into the available orbitals, making any configuration that prematurely pairs these electrons physically impossible. The remaining statements are correct because the orbital angular momentum of an s-orbital is zero since its azimuthal quantum number satisfies $l=0$, resulting in a magnitude of $\sqrt{l(l+1)}\hbar$ equal to zero. Furthermore, an orbital is defined by three quantum numbers whereas an electron requires four, and the magnetic quantum number is conventionally assigned a value of $m=0$ for the $d_{z^2}$ orbital.

In hydrogen-like atoms, which contain only one electron, the energy of an orbital depends solely on the principal quantum number ($n$). Because both the 2s and 2p orbitals share the same principal quantum number of $n=2$, they are considered degenerate, meaning they possess identical energy levels. Therefore, claiming that the 2s orbital has lower energy than the 2p orbital contradicts the fundamental principle that orbital energy is independent of the azimuthal quantum number in single-electron systems.

D-orbitals are divided into two sets based on their spatial orientation relative to the Cartesian axes, with the $d_{xy}$, $d_{yz}$, and $d_{xz}$ orbitals having electron density concentrated in the planes between the axes. In contrast, the $d_{x^2-y^2}$ orbital has lobes directed along the x and y axes, and the $d_{z^2}$ orbital has lobes oriented along the z axis. Consequently, these specific orbitals are the only ones with electron density positioned directly along the axes.

The capacity of an electron shell is determined by the angular quantum number $l$ and the fundamental Pauli Exclusion Principle. An angular quantum number of $l = 1$ identifies a $p$ subshell, which consists of a set of orbitals defined by the formula $2l + 1$. Substituting the given value into this expression yields $2(1) + 1 = 3$ individual orbitals. Since each orbital can accommodate a maximum of 2 electrons, the total electron capacity for this subshell is $3 \times 2 = 6$ electrons.

The Pauli Exclusion Principle states that no two electrons in an atom can have an identical set of all four quantum numbers. When two electrons occupy the same orbital, they share the same principal, azimuthal, and magnetic quantum numbers, which define the orbital's specific energy, shape, and orientation in space. To satisfy the principle, these two electrons must be distinguished by their spin quantum number, taking values of $+\frac{1}{2}$ and $-\frac{1}{2}$ to represent opposite electron spins.

The orbital angular momentum of an electron is determined by its azimuthal quantum number $l$ using the formula $\sqrt{l(l+1)} \hbar$. Since an electron residing in a d-orbital is assigned an azimuthal quantum number of $l = 2$, we substitute this value directly into the expression. This leads to the calculation $\sqrt{2(2+1)} \hbar$, which simplifies to $\sqrt{6} \hbar$.

The Aufbau principle dictates that electrons occupy atomic orbitals based on increasing energy levels, which are determined by the sum of the principal and azimuthal quantum numbers. With an atomic number of 22, the Titanium atom exhibits the electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2$. By evaluating these specific orbitals within the sequence, the energy levels follow a distinct progression starting from 3s, continuing to 3p, then 4s, and finally 3d.

To determine the number of d-electrons in the ferrous ion, we first examine the neutral iron atom ($Z=26$), which has the electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2$. When forming the $\mathrm{Fe}^{2+}$ ion, the two electrons are removed from the $4s$ orbital, leaving the $3d$ subshell with exactly 6 electrons.

Evaluating the options based on this count of 6, we find that neutral Chlorine ($Z=17$) has the configuration $1s^2 2s^2 2p^6 3s^2 3p^5$, which contains 6 electrons in the $2p$ subshell and 5 in the $3p$ subshell, resulting in a total of 11 p-electrons. This value does not equal 6, whereas the other choices all align with the iron ion; neutral iron itself retains 6 d-electrons, Neon ($Z=10$) has 6 p-electrons within its $2p^6$ configuration, and Magnesium ($Z=12$) possesses 6 s-electrons distributed across its $1s$, $2s$, and $3s$ orbitals.

An orbital is uniquely defined by a specific set of three quantum numbers: the principal quantum number $n$, the azimuthal quantum number $l$, and the magnetic quantum number $m$. When you define these values as $n=3$, $l=1$, and a specific magnetic value $m=a$, you are selecting one unique orientation in space, which corresponds to the $3p_z$ orbital. Since each distinct combination of these three quantum numbers identifies a single, precise orbital, the maximum number that can be identified with this exact set is one.

The energy of a photon is fundamentally tied to its wavelength through the Planck-Einstein relation, which shows that energy increases as wavelength decreases. Expressing the given wavelength of 45 nm in standard meters is a necessary first step for consistency with the SI units used for the speed of light and Planck's constant, resulting in $45 \times 10^{-9}$ m.

Utilizing the formula $E = \frac{hc}{\lambda}$, we substitute the provided constants into the expression to find the result. Multiplying Planck's constant of $6.63 \times 10^{-34}$ Js by the speed of light of $3 \times 10^8$ m/s and dividing the product by the wavelength of $45 \times 10^{-9}$ m follows the operation:
$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{45 \times 10^{-9}} = 4.42 \times 10^{-18} \text{ J}$

Two species are isoelectronic when they possess the exact same number of electrons. The $\mathrm{Be}^{2+}$ ion contains 2 electrons, which is determined by taking the atomic number of beryllium, 4, and subtracting the two electrons lost during ionization. This count perfectly matches the $\mathrm{Li}^{+}$ ion, which also holds 2 electrons because the neutral lithium atom, with an atomic number of 3, loses one electron to achieve its charge.

The set of quantum numbers where $n = 3$, $l = 1$, and $m = -1$ specifies one particular orbital within the $3p$ subshell. By the Pauli exclusion principle, any individual orbital is limited to holding a maximum of two electrons with opposite spins. Since the magnetic quantum number $m = -1$ isolates a single orbital, the total capacity for this set is two electrons.