The four quantum numbers for the electron in the outer most orbital of potassium (atomic no. 19) are

Q: The four quantum numbers for the electron in the outer most orbital of potassium (atomic no. 19) are

### Core Logic The atomic number of potassium (K) is 19. Its electronic configuration is: 1s2, 2s2, 2p6, 3s2, 3p6, 4s1. The outermost orbital of potassium is the 4s orbital. For the 4s orbital: Principal quantum number, n = 4 Azimuthal quantum number, l = 0 (for s-subshell) Magnetic quantum number, m = 0 (since l = 0) Spin quantum number, s = pm (1/2) ### Step 1: Final Matching Comparing these values with the options, option (2) provides n = 4, l = 0, m = 0, s = +(1/2), which perfectly describes

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Q27 jee_main_2025_29_jan_evening Bohr Model for Hydrogen-like Species

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]

A. Graph (1)
B. Graph (2)
C. Graph (3)
D. Graph (4)

Solution

Related Formula

Eₙ = -13.6 (Z²)/(n²) eV

Core Logic

For a constant principal quantum number $n$ $n$, the energy $E$ $E$ is directly proportional to $-Z²$ $-Z^2$. This represents a quadratic relation where the curve is a downward-opening parabola starting from the origin in the negative energy region.

Pattern Recognition

Since energy values are inherently negative for bound states, as $Z$ $Z$ increases, $E$ $E$ becomes rapidly more negative following a parabolic curve.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q40 jee_main_2025_29_jan_evening Heisenberg's Uncertainty Principle

Given below are two statements: Statement (I): It is impossible to specify simultaneously with arbitrary precision, both the linear momentum and the position of a particle. Statement (II) If the uncertainty in the measurement of position and uncertainty in measurement of momentum are equal for an electron, then the uncertainty in the measurement of velocity is

gesqrt(h)/(pi)\times(1)/(2m)

In the light of the above statements, choose the correct answer from the options given below:

A. Statement I is true but Statement II is false.
B. Both Statement I and Statement II are true.
C. Statement I is false but Statement II is true.
D. Both Statement I and Statement II are false.

Solution

Related Formula

Delta x cdot Delta p ge (h)/(4pi)

Core Logic

Statement I is a verbatim definition of Heisenberg's Uncertainty Principle, hence it is completely true.

For Statement II, we are given that $Delta x = Delta p$ $Delta x = Delta p$:

Delta p cdot Delta p ge (h)/(4pi) implies (Delta p)² ge (h)/(4pi)

Delta p ge sqrt(h)/(4pi) = (1)/(2)sqrt(h)/(pi)

Since $Delta p = m cdot Delta v$ $Delta p = m cdot Delta v$:

m cdot Delta v ge (1)/(2)sqrt(h)/(pi) implies Delta v ge (1)/(2m)sqrt(h)/(pi)

This perfectly matches Statement II, so it is also true.

Pattern Recognition

When solving inequality bounds for identical uncertainties, always substitute directly to obtain a clean quadratic form before taking the square root.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q38 jee_main_2025_28_jan_morning Quantum Numbers and Orbital Energies

In a multielectron atom, which of the following orbitals described by three quantum numbers with have same energy in absence of electric and magnetic fields? A.

n = 1, l = 0, ml = 0

B.

n = 2, l = 0, ml = 0

C.

n = 2, l = 1, ml = 1

D.

n = 3, l = 2, ml = 1

E.

n = 3, l = 2, ml = 0

Choose the correct answer from the options given below:

A. $A and B only$ $\text{A and B only}$
B. $B and C only$ $\text{B and C only}$
C. $C and D only$ $\text{C and D only}$
D. $D and E only$ $\text{D and E only}$

Solution

Core Logic

In multi-electron systems, the energy profile depends on both the primary shell ( $n$ $n$) and azimuthal subshell ( $l$ $l$) quantum values via the $(n+l)$ $(n+l)$ rule. Orbitals sharing the exact same $n$ $n$ and $l$ $l$ values are energy-degenerate as long as external field metrics are zero. Let us map each set:

A: $1s$ $1s$
B: $2s$ $2s$
C: $2p$ $2p$
D: $3d$ $3d$ ( $ml = 1$ $m_l = 1$)
E: $3d$ $3d$ ( $ml = 0$ $m_l = 0$)

Since both D and E describe subcomponents of the same $3d$ $3d$ subshell ( $n=3, l=2$ $n=3, l=2$), they share identical energies, establishing perfect degeneracy.

Pattern Recognition

Sees: Orbitals matching energy level in multi-electron space. Trap: Confusing single-electron hydrogen atoms (where energy depends solely on $n$ $n$) with multielectron atoms.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q26 jee_main_2025_03_april_morning Bohr Model of Atom

Which of the following postulate of Bohr's model of hydrogen atom is not in agreement with quantum mechanical model of an atom?

A. An atom in a stationary state does not emit electromagnetic radiation as long as it stays in the same state
B. An atom can take only certain distinct energies $E₁, E₂, E₃,$ $E_1, E_2, E_3,$ etc. These allowed states of constant energy are called the stationary states of atom
C. When an electron makes a transition from a higher energy stationary state to a lower energy stationary state, then it emits a photon of light
D. The electron in a H atom's stationary state moves in a circle around the nucleus

Solution

Related Formula

Eₙ = -(13.6 \cdot Z²)/(n²) eV

Δ E = E₂ - E₁ = hν

Core Logic

According to quantum mechanics (wave mechanics and Heisenberg's uncertainty principle), electrons do not move in fixed two-dimensional circular orbits around the nucleus. Instead, their positions are represented by three-dimensional probability density functions called orbitals.

Step 1: Evaluation of Options

Statements 1, 2, and 3 are consistent with both models (quantized energy states and non-radiating stationary states).

Statement 4 asserts that electrons move in well-defined circular paths, which violates quantum mechanical principles where electron motion is three-dimensional and probabilistic.

Pattern Recognition

Bohr model = 2D planar circular orbits. Quantum Mechanical model = 3D spatial electron clouds (orbitals). Fixed circular path is strictly a Bohr concept not supported by quantum mechanics.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q38 jee_main_2025_04_april_evening Quantum Numbers and Electronic Configuration

Consider the ground state of chromium atom

(Z = 24)

. How many electrons are with Azimuthal quantum number

l = 1

and

l = 2

respectively?

A. 12 and 4
B. 16 and 4
C. 12 and 5
D. 16 and 5

Solution

Related Formula

Azimuthal Quantum Number: l=1 p-orbital, l=2 d-orbital

Core Logic

The ground-state electronic configuration of Chromium ( $Z=24$ $Z=24$) is anomalous due to half-filled stability:

Cr: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹

Now count the electrons in specific subshells:

For $l = 1$ $l = 1$ (p-electrons): present in $2p⁶$ $2p^6$ and $3p⁶ 6 + 6 = 12$ $3p^6 \implies 6 + 6 = 12$ electrons.
For $l = 2$ $l = 2$ (d-electrons): present in $3d⁵ 5$ $3d^5 \implies 5$ electrons.

Thus, the counts are 12 and 5 respectively.

Pattern Recognition

Always remember the $3d⁵ 4s¹$ $3d^5 4s^1$ exception for Chromium ( $Z=24$ $Z=24$) and $3d¹⁰ 4s¹$ $3d^{10} 4s^1$ for Copper ( $Z=29$ $Z=29$). Forgetting the half-filled d-subshell exception will lead to the wrong answer (12 and 4).

Chapter Mix

Class 11 Chemistry: Structure of Atom

The four quantum numbers for the electron in the outer most orbital of potassium (atomic no. 19) are

Solution & Explanation

Core Logic

Step 1: Final Matching

Chapter Mix

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Solution

Related Formula

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