Structure Of Atom – JEE Main ChemistryPractice Questions & PYQs

Ionization enthalpy represents the energy required to remove an electron completely from the ground state of an atom, effectively moving it to an infinite distance from the nucleus. Since the energy levels within a hydrogen atom are determined by the inverse square of the principal quantum number, the excitation energy needed to move an electron between any two specific levels is proportional to the difference between the inverse squares of those quantum numbers. To find the energy required for a transition from $n=1$ to $n=2$, we simply apply this fractional change to the given ionization enthalpy of $1.312 \times 10^{6}$ J/mol.

Calculating the difference between the inverse squares of the initial and final states gives us $1 - 1/4$, which is a factor of $3/4$. Multiplying the ionization enthalpy by this factor provides the excitation energy:
$\Delta E = 1.312 \times 10^{6} \times \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 1.312 \times 10^{6} \times \frac{3}{4} = 9.84 \times 10^{5}\,\mathrm{J}\,\mathrm{mol}^{-1}$

Isoelectronic species are defined by having the exact same total number of electrons. Within the group comprising $\mathrm{NO}^{+}$, $\mathrm{C}_2^{2-}$, $\mathrm{CN}^{-}$, and $\mathrm{N}_2$, each species possesses exactly 14 electrons, which is confirmed by the individual electron counts of $7+8-1$ for $\mathrm{NO}^{+}$, $12+2$ for $\mathrm{C}_2^{2-}$, $6+7+1$ for $\mathrm{CN}^{-}$, and $14$ for $\mathrm{N}_2$. Because every member in this set shares this identical electron count, they are classified as isoelectronic.

The energy level of an orbital in an atom is governed by the sum of its principal quantum number $n$ and azimuthal quantum number $l$, with higher $(n + l)$ values corresponding to higher energy. Calculating this sum for each of the given sets yields results of $3+0=3$, $3+1=4$, $3+2=5$, and $4+0=4$. Because the set $n=3, l=2$ produces the maximum sum of $5$, it represents the orbital with the highest energy among the choices provided.

Bohr's atomic theory asserts that an electron in a stable orbit possesses quantized angular momentum defined by the equation $mvr = \frac{nh}{2\pi}$, where $n$ is the principal quantum number. By substituting $n = 5$ for the fifth orbit into this formula, the calculation becomes $mvr = \frac{5h}{2\pi}$. Dividing the numerator 5 by 2 yields 2.5, which simplifies the final expression to $\frac{2.5h}{\pi}$.

Isoelectronic species are defined as atoms or ions that share an identical total number of electrons, a condition determined by adjusting the atomic number according to the charge of each ion. When examining the set containing $K^+$, $Cl^-$, $Ca^{2+}$, and $Sc^{3+}$, we find that potassium, with an atomic number of 19, has $19 - 1 = 18$ electrons, while chlorine, with an atomic number of 17, has $17 + 1 = 18$ electrons. Calcium, possessing an atomic number of 20, yields $20 - 2 = 18$ electrons, and scandium, with an atomic number of 21, arrives at $21 - 3 = 18$ electrons, meaning all these ions consistently contain exactly 18 electrons and thus form a perfect isoelectronic series.

The Heisenberg Uncertainty Principle establishes that it is impossible to simultaneously measure the exact position and momentum of a particle with infinite precision, as defined by the relation $\Delta x \cdot m \Delta v \geq \frac{h}{4\pi}$. Given that the electron moves at $300 \text{ ms}^{-1}$ with an accuracy of $0.001\%$, the uncertainty in its velocity is $0.001\%$ of the total speed, which simplifies to $\Delta v = 3 \times 10^{-3} \text{ ms}^{-1}$.

Rearranging the principle to solve for the uncertainty in position yields the formula:
$\Delta x = \frac{h}{4 \pi m \Delta v}$
Substituting the constants $h = 6.63 \times 10^{-34} \text{ Js}$ and $m = 9.1 \times 10^{-31} \text{ kg}$, along with the determined uncertainty in velocity, gives the following expression:
$\Delta x = \frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 3 \times 10^{-3}}$
Completing this calculation confirms that the uncertainty in the position of the electron is $1.92 \times 10^{-2} \text{ m}$.

Isoelectronic species are defined as atoms, ions, or molecules that possess an identical total number of electrons. In the set consisting of $\mathrm{SO}_3^{2-}$, $\mathrm{CO}_3^{2-}$, and $\mathrm{NO}_3^{-}$, we find that the sulfite ion $\mathrm{SO}_3^{2-}$ has a total of $16 + 24 + 2 = 42$ electrons. Because the carbonate ion $\mathrm{CO}_3^{2-}$ contains $6 + 24 + 2 = 32$ electrons and the nitrate ion $\mathrm{NO}_3^{-}$ contains $7 + 24 + 1 = 32$ electrons, this group does not possess the same electron count.

The energy of orbitals in a multi-electron atom is governed by the $(n+l)$ rule, which states that orbital energy increases as the sum of the principal quantum number $n$ and the azimuthal quantum number $l$ increases. Because external electric and magnetic fields are absent, the magnetic quantum number $m$ does not affect orbital energy, meaning any orbitals sharing the same $(n+l)$ value will be degenerate. Calculating these sums for the provided options, we find for (A) $n+l = 1+0 = 1$, for (B) $n+l = 2+0 = 2$, for (C) $n+l = 2+1 = 3$, and for both (D) and (E) $n+l = 3+2 = 5$. Since orbitals (D) and (E) result in the same sum of 5, they possess equal energy levels.

Isoelectronic species are defined by having an identical total number of electrons. While $\mathrm{CH}_3^{+}$ contains 8 electrons, $\mathrm{H}_3\mathrm{O}^{+}$, $\mathrm{NH}_3$, and $\mathrm{CH}_3^{-}$ all contain 10 electrons. This uniform count of 10 electrons confirms that $\mathrm{H}_3\mathrm{O}^{+}$, $\mathrm{NH}_3$, and $\mathrm{CH}_3^{-}$ are isoelectronic with each other.

The orbital notation 4f indicates that the principal quantum number $n=4$, while the f subshell corresponds to an azimuthal quantum number $l=3$. Because the magnetic quantum number $m$ must satisfy the condition $-l \leq m \leq l$, valid values for this orbital are any integers from $-3$ to $3$, and the spin quantum number $s$ is strictly limited to $s=+\frac{1}{2}$ or $s=-\frac{1}{2}$. Evaluating these requirements against the given options, the set $n=4$, $l=3$, $m=+1$, and $s=+\frac{1}{2}$ represents a valid state for an electron in the 4f orbital.

The azimuthal quantum number defines the subshell type, where $l=1$ denotes $p$ orbitals and $l=2$ denotes $d$ orbitals. Chromium, with $Z=24$, follows the electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$. The total number of electrons with $l=1$ is found by summing those in the $2p$ and $3p$ subshells, which provides $6 + 6 = 12$ electrons. For the azimuthal quantum number $l=2$, we focus on the $3d$ subshell, which holds 5 electrons.

Isoelectronic species are identified by having an identical total number of electrons. The set $\mathrm{K}^{+}$, $\mathrm{Ca}^{2+}$, $\mathrm{Sc}^{3+}$, and $\mathrm{Cl}^{-}$ consists of ions that each possess 18 electrons, determined by the calculations $19 - 1 = 18$, $20 - 2 = 18$, $21 - 3 = 18$, and $17 + 1 = 18$. Because all members of this group have the same electron count, they represent a collection of isoelectronic species.

The spectral lines for a hydrogen atom are determined by the Rydberg formula, which relates the wavelength of emitted radiation to the energy transition between quantum levels. For an electron falling from an infinite energy level to the ground state, defined by $n_1 = 1$ and $n_2 = \infty$, the relationship is expressed as:

$\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

For a hydrogen atom, the atomic number is $Z = 1$ and the Rydberg constant is given as $R = 1.097 \times 10^7 \text{ m}^{-1}$. Substituting these parameters into the equation yields:

$\frac{1}{\lambda} = 1.097 \times 10^7 \times 1^2 \left( \frac{1}{1^2} - \frac{1}{\infty ^2} \right)$

Because any finite value divided by infinity is zero, the term $\frac{1}{\infty ^2}$ vanishes, simplifying the expression to $\frac{1}{\lambda} = 1.097 \times 10^7 \text{ m}^{-1}$. Solving for $\lambda$ gives $\lambda = \frac{1}{1.097 \times 10^7} \text{ m}$, which results in $9.11 \times 10^{-8} \text{ m}$. Converting this figure to nanometers by multiplying by $10^9$ provides a wavelength of 91.1 nm, which is effectively 91 nm.

The visible portion of the hydrogen spectrum is defined by the Balmer series, which involves electron transitions landing on the principal quantum level $n=2$. Because spectral lines are ordered by increasing energy, the red end represents the lowest energy transition in the series, starting from $n=3$. Counting sequentially from this first line, the third line requires starting from an energy level of $n=5$. Consequently, the electron emits energy as it moves from the $n=5$ state down to the $n=2$ state.

The orbital angular momentum of an electron depends on the azimuthal quantum number $l$, and for an s-electron, this value is defined as $l=0$. By substituting this into the formula $\sqrt{l(l+1)} \frac{h}{2\pi}$, the calculation becomes

$\sqrt{0(0+1)} \frac{h}{2\pi} = \sqrt{0} \frac{h}{2\pi} = 0$

which indicates that the orbital angular momentum for an s-electron is zero.

Species are classified as isoelectronic when they possess an identical total number of electrons, regardless of their individual atomic numbers. Determining this involves adjusting the neutral atomic number by the ion's charge, where anions gain electrons and cations lose them. For instance, $\mathrm{N}^{3-}$ ($Z=7$) contains $7+3=10$ electrons, $\mathrm{F}^{-}$ ($Z=9$) holds $9+1=10$ electrons, and $\mathrm{Na}^{+}$ ($Z=11$) results in $11-1=10$ electrons. Because each member of this specific group shares a total of 10 electrons, they satisfy the requirements for an isoelectronic series.

The wave-like behavior of a moving object is defined by its de Broglie wavelength, which is inversely proportional to its momentum. For a tennis ball with a mass of 60 g and a velocity of 10 m/s, we must first express the mass in the standard unit of 0.06 kg. Applying the relationship $\lambda = \frac{h}{mv}$ with Planck's constant $h = 6.63 \times 10^{-34} \, \text{J}\cdot\text{s}$, the calculation is set up as follows:

$\lambda = \frac{6.63 \times 10^{-34}}{0.06 \times 10}$

Simplifying the denominator gives 0.6, which allows the expression to be written as $\lambda = \frac{6.63 \times 10^{-34}}{0.6}$. By adjusting the power of ten to maintain precision, this becomes $\frac{6.63 \times 10^{-33}}{6}$, resulting in a value approximately equal to $10^{-33}$ meters.

The energy levels of a hydrogen atom are determined by the principal quantum number $n$, following the relationship $E_n = -13.6 \times \frac{Z^2}{n^2}$ eV. While the ground state occurs at $n=1$, the second excited state corresponds to $n=3$ because each successive excited state adds one to the quantum number. For a hydrogen atom where $Z=1$, substituting $n=3$ into the energy equation provides the calculation:

$E = -13.6 \times \frac{1^2}{3^2} = -13.6 \times \frac{1}{9}$

Performing the arithmetic confirms that the energy of the electron in this state is -1.51 eV.

Heisenberg's Uncertainty Principle establishes a trade-off in measurement precision, stating that the product of position uncertainty and momentum uncertainty cannot be less than the limit $\frac{h}{4\pi}$. Since momentum uncertainty is the product of mass and velocity uncertainty, we express the relationship as $\Delta x \cdot m \cdot \Delta v \geq \frac{h}{4\pi}$. We first ensure all values are in the standard SI system, converting the mass of 25 g into 0.025 kg to align with the units of Planck's constant.

Rearranging the expression to isolate the uncertainty in velocity, we obtain $\Delta v = \frac{h}{4\pi \cdot m \cdot \Delta x}$. Substituting the provided values, we divide 6.6 × 10^-34 by the product of 4, 3.14, 0.025 kg, and 10^-5 m. The denominator becomes 3.14 × 10^-6. Dividing the numerator by this denominator results in an uncertainty in velocity of 2.1 × 10^-28 m s^-1.

The energy of a photon is inversely proportional to its wavelength according to the relation $E = \frac{hc}{\lambda}$, meaning that for two distinct radiations, the ratio of their energies is equivalent to the inverse ratio of their corresponding wavelengths. By setting up the ratio of the energies for these radiations, the constants $h$ and $c$ cancel out, leaving the expression:
$\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$
Using the provided values of $\lambda_1 = 2000 \text{ Å}$ and $\lambda_2 = 6000 \text{ Å}$, the ratio is calculated as $\frac{6000}{2000}$, which simplifies to 3.