For hydrogen like species, which of the following graphs provides the most appropriate representation of E vs Z plot for a constant n? [E: Energy of the stationary state, Z: atomic number, n: principal quantum number]

For a constant principal quantum number n, the energy E is directly proportional to -Z2. This represents a quadratic relation where the curve is a downward-opening parabola starting from the origin in the negative energy region. Since energy values are inherently negative for bound states, as Z increases, E becomes rapidly more negative following a parabolic curve. Class 11 Chemistry: Structure of Atom

For hydrogen like species, which of the following graphs provides the most appropriate representation of E vs Z plot for a constant n? [E: Energy of the stationary state, Z: atomic number, n: principa

For a constant principal quantum number n, the energy E is directly proportional to -Z2. This represents a quadratic relation where the curve is a downward-opening parabola starting from the origin in the negative energy region. Since energy values are inherently negative for bound states, as Z increases, E becomes rapidly more negative following a parabolic curve. Class 11 Chemistry: Structure of Atom

For hydrogen like species, which of the following graphs provides the most appropriate representation of E vs Z plot for a constant n? [E: Energy of the stationary state, Z: atomic number, n: principal quantum number]

For a constant principal quantum number n, the energy E is directly proportional to -Z2. This represents a quadratic relation where the curve is a downward-opening parabola starting from the origin in the negative energy region. Since energy values are inherently negative for bound states, as Z increases, E becomes rapidly more negative following a parabolic curve. Class 11 Chemistry: Structure of Atom

For hydrogen like species, which of the following graphs provides the most appropriate representation of E vs Z plot for a constant n? [E: Energy of the stationary state, Z: atomic number, n: principa

For a constant principal quantum number n, the energy E is directly proportional to -Z2. This represents a quadratic relation where the curve is a downward-opening parabola starting from the origin in the negative energy region. Since energy values are inherently negative for bound states, as Z increases, E becomes rapidly more negative following a parabolic curve. Class 11 Chemistry: Structure of Atom

For hydrogen like species, which of the following graphs provides the most appropriate representation of E vs Z plot…

Year	2026	2025	2024	Total
Questions	14	15	9	38

Q jee_main_2024_29_january_evening Hydrogen Spectral Series

Match List I with List II:

List I (Spectral Series for Hydrogen)	List II (Spectral Region/Higher Energy State)
A. Lyman	I. Infrared region
B. Balmer	II. UV region
C. Paschen	III. Infrared region
D. Pfund	IV. Visible region

Choose the correct answer from the options given below :

A. A-II, B-III, C-I, D-IV
B. A-I, B-III, C-II, D-IV
C. A-II, B-IV, C-III, D-I
D. A-I, B-II, C-III, D-IV

Solution

Related Formula

(1)/(λ) = RH ( (1)/(n₁²) - (1)/(n₂²) )

Core Logic

Analyzing electromagnetic spectrum regions for each hydrogen electronic decay series:

Lyman Series ( $n₁=1$ $n_1=1$) emits photons in the high-energy Ultraviolet (UV) region (II).
Balmer Series ( $n₁=2$ $n_1=2$) falls cleanly into the Visible region spectrum (IV).
Paschen Series ( $n₁=3$ $n_1=3$) transitions sit inside the standard Near-Infrared region (III).
Pfund Series ( $n₁=5$ $n_1=5$) shifts sit even deeper in the Infrared region zone (I).

Step 1: Resolution

Thus, compiling match mappings yields configuration: A-II, B-IV, C-III, D-I.

Pattern Recognition

Lyman is always UV; Balmer is visible; all subsequent higher level series ( $n₁ \geq 3$ $n_1 \ge 3$) belong strictly to infrared segments.

Chapter Mix

Class 11 Chemistry: Atomic Structure

Q jee_main_2024_27_jan_morning Quantum Numbers and Electron Capacity

The number of electrons present in all the completely filled subshells having

n=4

and

s=+(1)/(2)

is

. (Where

n=

principal quantum number and

s=

spin quantum number)

Numerical Answer. Answer: 16 to 16

Solution

Step 1: Identify all available subshells within the $n=4$ $n=4$ energy shell

For principal quantum level $n=4$ $n=4$, the allowed values of azimuthal quantum numbers ( $l$ $l$) are:

$4s$ $4\text{s}$ ( $l=0$ $l=0$) $arrow 1 orbital arrow 2 electrons capacity$
$4p$ $4\text{p}$ ( $l=1$ $l=1$) $arrow 3 orbitals arrow 6 electrons capacity$
$4d$ $4\text{d}$ ( $l=2$ $l=2$) $arrow 5 orbitals arrow 10 electrons capacity$
$4f$ $4\text{f}$ ( $l=3$ $l=3$) $arrow 7 orbitals arrow 14 electrons capacity$

Step 2: Filter capacity using spin values

Every single spatial orbital holds exactly 2 electrons maximum; one with spin $s=+(1)/(2)$ $s=+\frac{1}{2}$ and one with spin $s=-(1)/(2)$ $s=-\frac{1}{2}$. Total number of orbitals across $n=4$ $n=4$ is:

1 + 3 + 5 + 7 = 16 orbitals

Thus, the total count of electrons featuring spin value $s=+(1)/(2)$ $s=+\frac{1}{2}$ across these completely filled configurations is exactly $16$ $16$.

Pattern Recognition

Total orbitals in shell $n$ $n$ is $n²$ $n^2$. Since each orbital contributes exactly 1 electron with $s=+(1)/(2)$ $s=+\frac{1}{2}$, capacity is simply $n² = 4² = 16$ $n^2 = 4^2 = 16$.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q65 jee_main_2024_27_jan_morning Electronic Configuration and Magnetic Moment

Which of the following electronic configuration would be associated with the highest magnetic moment?

A. $[Ar] 3d⁷$ $[Ar] 3d^7$
B. $[Ar] 3d⁸$ $[Ar] 3d^8$
C. $[Ar] 3d³$ $[Ar] 3d^3$
D. $[Ar] 3d⁶$ $[Ar] 3d^6$

Solution

Related Formula

Spin-only magnetic moment formula:

μ = \sqrt(n(n+2)) BM

where $n$ $n$ is the number of unpaired electrons.

Step 1: Audit configurations and count unpaired electrons

Configuration	Unpaired $e^-$ $e^-$ ( $n$ $n$)	Magnetic Moment (BM)
$[Ar] 3d⁷$ $[Ar] 3d^7$	3	$\sqrt(15)$ $\sqrt{15}$
$[Ar] 3d⁸$ $[Ar] 3d^8$	2	$\sqrt(8)$ $\sqrt{8}$
$[Ar] 3d³$ $[Ar] 3d^3$	3	$\sqrt(15)$ $\sqrt{15}$
$[Ar] 3d⁶$ $[Ar] 3d^6$	4	$\sqrt(24)$ $\sqrt{24}$

Step 2: Conclusion

Since $[Ar] 3d⁶$ $[Ar] 3d^6$ contains the maximal count of 4 unpaired electrons, it yields the highest spin-only magnetic moment value.

Pattern Recognition

Maximal unpaired configuration in high-spin $3dⁿ$ $3d^n$ series yields the highest $μ$ $\mu$. Check $n$ $n$ systematically.

Chapter Mix

Class 11 Chemistry: Structure of Atom Class 12 Chemistry: d-and f-Block Elements

Q66 jee_main_2024_29_jan_morning Quantum Numbers

The correct set of four quantum numbers for the valence electron of rubidium atom

(Z = 37)

is:

A. $5, 0, 0, +(1)/(2)$ $5, 0, 0, +\frac{1}{2}$
B. $5, 0, 1, +(1)/(2)$ $5, 0, 1, +\frac{1}{2}$
C. $5, 1, 0, +(1)/(2)$ $5, 1, 0, +\frac{1}{2}$
D. $5, 1, 1, +(1)/(2)$ $5, 1, 1, +\frac{1}{2}$

Solution

Core Logic

Rubidium ( $Rb$ $Rb$) has the atomic number $Z = 37$ $Z = 37$.

The noble gas core preceding it is Krypton ( $Kr$ $Kr$, $Z = 36$ $Z = 36$). The electronic configuration is:

Rb arrow [Kr] 5s¹

Thus, the valence electron resides in the $5s$ $5s$ orbital.

Step 1: Assigning Quantum Numbers

For a $5s$ $5s$ electron:

Principal quantum number, $n = 5$ $n = 5$ (from the shell number).
Azimuthal quantum number, $l = 0$ $l = 0$ (for an s-orbital).
Magnetic quantum number, $m = 0$ $m = 0$ (since $m$ $m$ ranges from $-l$ $-l$ to $+l$ $+l$, and $l=0$ $l=0$).
Spin quantum number, $s = +(1)/(2)$ $s = +\frac{1}{2}$ or $-(1)/(2)$ $-\frac{1}{2}$.

The correct set matching this is $5, 0, 0, +(1)/(2)$ $5, 0, 0, +\frac{1}{2}$.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q90 jee_main_2024_30_january_evening Atomic Spectra

Number of spectral lines obtained in

He^+

spectra, when an electron makes transition from fifth excited state to first excited state will be

Numerical Answer. Answer: 10 to 10

Solution

Related Formula

Number of spectral lines = (Δ n (Δ n + 1))/(2)

where $Δ n = n₂ - n₁$ $\Delta n = n_2 - n_1$

Core Logic

Identify the principal quantum numbers ( $n$ $n$) for the given states:

Fifth excited state means $n₂ = 5 + 1 = 6$ $n_2 = 5 + 1 = 6$.
First excited state means $n₁ = 1 + 1 = 2$ $n_1 = 1 + 1 = 2$.

Calculate the difference:

Δ n = n₂ - n₁ = 6 - 2 = 4

Step 1: Calculate Total Spectral Lines

Substitute $Δ n = 4$ $\Delta n = 4$ into the formula:

Maximum number of spectral lines = (4(4 + 1))/(2)

= (4 \times 5)/(2)

$= 10$ $= 10$

Chapter Mix

Class 11 Chemistry: Structure of Atom

For hydrogen like species, which of the following graphs provides the most appropriate representation of E vs Z plot for a constant n? [E: Energy of the stationary state, Z: atomic number, n: principal quantum number]

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Structure of Atom Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Resolution

Pattern Recognition

Chapter Mix

Solution

Step 1: Identify all available subshells within the n=4$n=4$ energy shell

Step 2: Filter capacity using spin values

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Audit configurations and count unpaired electrons

Step 2: Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Assigning Quantum Numbers

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Total Spectral Lines

Chapter Mix

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Step 1: Identify all available subshells within the $n=4$ $n=4$ energy shell