Compare the energies of following sets of quantum numbers for multielectron system.
(A) n = 4, 1 = 1
(B) n= 4, l = 2
(C) n = 3, l = 1
(D) n = 3, l = 2
(E) n = 4, 1 = 0
Choose the correct answer from the options given below :
(A) n = 4, 1 = 1
(B) n= 4, l = 2
(C) n = 3, l = 1
(D) n = 3, l = 2
(E) n = 4, 1 = 0
Choose the correct answer from the options given below :
\((B)>(A)>(C)>(E)>(D)\)
\((E)>(C)<(D)<(A)<(B)\)
\((E)>(C)>(A)>(D)>(B)\)
\((C)<(E)<(D)<(A)<(B)\)
The Correct Option is D
Approach Solution - 1
To determine the order of energy levels for the given sets of quantum numbers in a multi-electron atom, we use the concept of the n + l rule. According to this rule, the energy of an electron is determined first by the sum of the principal quantum number \( n \) and the azimuthal (angular momentum) quantum number \( l \). The higher the sum of \( n + l \), the higher the energy. If two orbitals have the same \( n + l \) value, the one with the lower \( n \) value is lower in energy.
Let's evaluate each given set:
- (A) \( n = 4, l = 1 \) → \( n + l = 4 + 1 = 5 \)
- (B) \( n = 4, l = 2 \) → \( n + l = 4 + 2 = 6 \)
- (C) \( n = 3, l = 1 \) → \( n + l = 3 + 1 = 4 \)
- (D) \( n = 3, l = 2 \) → \( n + l = 3 + 2 = 5 \)
- (E) \( n = 4, l = 0 \) → \( n + l = 4 + 0 = 4 \)
Now, let's order the orbitals by energy:
- Sets (C) and (E) both have the smallest \( n + l \) values (4), but set (C) has a lower \( n \) (\(n = 3\) compared to \(n = 4\)), so set (C) is lower in energy than set (E).
- Sets (A) and (D) have the same \( n + l \) values (5), but set (D) has a lower \( n \), making it lower in energy than set (A).
- Set (B) has the highest \( n + l \) value (6), so it is the highest in energy.
Thus, the order of energy levels is: (C) < (E) < (D) < (A) < (B)
The correct answer is: \((C)<(E)<(D)<(A)<(B)\)
Approach Solution -2
In multielectron systems, the energy of an electron in an orbital depends on both the principal quantum number (\(n\)) and the azimuthal quantum number (\(l\)). The energy increases as the value of \(n + l\) increases. For orbitals with the same \(n + l\), the one with the lower \(n\) has lower energy.
- (A) \(n + l = 4 + 1 = 5\)
- (B) \(n + l = 4 + 2 = 6\)
- (C) \(n + l = 3 + 1 = 4\)
- (D) \(n + l = 3 + 2 = 5\)
- (E) \(n + l = 4 + 0 = 4\)
Order by \(n + l\):
\((C) = (E) < (D) < (A) < (B)\)
For orbitals with the same \(n + l\), compare \(n\):
\((C) < (E)\), as \(n = 3\) for (C) and \(n = 4\) for (E).
Final Answer: \((C) < (E) < (D) < (A) < (B)\).
Top Questions on Atomic Structure
- Consider the following spectral lines for atomic hydrogen :
A. First line of Paschen series
B. Second line of Balmer series
C. Third line of Paschen series
D. Fourth line of Bracket series
The correct arrangement of the above lines in ascending order of energy is :- JEE Main - 2026
- Chemistry
- Atomic Structure
- The energy of first (lowest) Balmer line of H atom is \(x\) J. The energy (in J) of second Balmer line of H atom is:
- JEE Main - 2026
- Chemistry
- Atomic Structure
- The energy required by electrons, present in the first Bohr orbit of hydrogen atom, to be excited to second Bohr orbit is ________ J mol\(^{-1}\). Given: \(R_H = 2.18 \times 10^{-11}\) ergs.
- JEE Main - 2026
- Chemistry
- Atomic Structure
- Consider two Group IV metal ions X$^{2+}$ and Y$^{2+}$. A solution containing 0.01 M X$^{2+}$ and 0.01 M Y$^{2+}$ is saturated with H$_2$S. The pH at which the metal sulphide YS will form as a precipitate is ___. (Nearest integer)
Given:
$K_{sp}(\mathrm{XS}) = 1 \times 10^{-22}$ at 25$^\circ$C
$K_{sp}(\mathrm{YS}) = 4 \times 10^{-16}$ at 25$^\circ$C
$[\mathrm{H_2S}] = 0.1$ M
$K_{a1} \times K_{a2} (\mathrm{H_2S}) = 1.0 \times 10^{-21}$
$\log 2 = 0.30,\ \log 3 = 0.48,\ \log 5 = 0.70$
- JEE Main - 2026
- Chemistry
- Atomic Structure
- X and Y are the number of electrons involved, respectively during the oxidation of I$^-$ to I$_2$ and S$^{2-}$ to S by acidified K$_2$Cr$_2$O$_7$. The value of X + Y is ___.
- JEE Main - 2026
- Chemistry
- Atomic Structure
Questions Asked in JEE Main exam
A laser beam has intensity of $4.0\times10^{14}\ \text{W/m}^2$. The amplitude of magnetic field associated with the beam is ______ T. (Take $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2/\text{N m}^2$ and $c=3\times10^8\ \text{m/s}$)
- JEE Main - 2026
- Electromagnetic waves
- When $1$ g of compound $(X)$ is subjected to Kjeldahl's method for estimation of nitrogen, $15$ mL of $1$ M $\mathrm{H_2SO_4}$ was neutralized by ammonia evolved. The percentage of nitrogen in compound $(X)$ is:
- JEE Main - 2026
- Quantitative Analysis
- Let $A = \{2, 3, 5, 7, 9\}$. Let $R$ be the relation on $A$ defined by $xRy$ if and only if $2x \le 3y$. Let $l$ be the number of elements in $R$, and $m$ be the minimum number of elements required to be added in $R$ to make it a symmetric relation. Then $l + m$ is equal to :
- JEE Main - 2026
- Functions
- Let $z$ be the complex number satisfying $|z - 5| \le 3$ and having maximum positive principal argument. Then $34 \left| \frac{5z - 12}{5iz + 16} \right|^2$ is equal to :
- JEE Main - 2026
- Complex numbers
- If the line $ax + 4y = \sqrt{7}$, where $a \in \mathbb{R}$, touches the ellipse $3x^2 + 4y^2 = 1$ at the point $P$ in the first quadrant, then one of the focal distances of $P$ is :
- JEE Main - 2026
- Conic sections