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 :

Related Formula

Core Logic

From the de-Broglie equation: λ ∝ (1)/(p)$\lambda \propto \frac{1}{p}$ ⇒ λ p = h (constant)$\Rightarrow \lambda p = h \text{ (constant)}$

This represents the equation of a rectangular hyperbola (xy = c$xy = c$).

Step 1: Graph Identification

The plot of λ$\lambda$ versus p$p$ will be a rectangular hyperbola curve in the first quadrant. Graph 1 correctly depicts this hyperbolic relationship.

Pattern Recognition

Inverse proportionality y = (k)/(x)$y = \frac{k}{x}$ always graphs as a rectangular hyperbola in the positive quadrant.

Chapter Mix

Class 11 Chemistry: Structure of Atom

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

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:

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

Related Formula

Spin-only magnetic moment formula:

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

Step 1: Audit configurations and count unpaired electrons

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

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:

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