Q.Explain, giving reasons, which of the following sets of quantum numbers are not possible.

Concept: Quantum Number Constraints

Each electron in an atom is described by four quantum numbers that must obey strict rules derived from solving the Schrödinger equation.

The allowed ranges are:

• Principal quantum number: $n \geq 1$ (positive integer)
• Azimuthal quantum number: $0 \leq l \leq n-1$ (integer)
• Magnetic quantum number: $-l \leq m_l \leq +l$ (integer)
• Spin quantum number: $m_s = \pm\tfrac{1}{2}$ only

Checking each set:

(a) $n=0$ violates the rule $n \geq 1$. Energy levels start at $n=1$.

(b) All conditions satisfied: $n=1$, $l=0$ (allowed since $0 \leq 0 \leq 0$), $m_l=0$ (allowed since $-0 \leq 0 \leq 0$), $m_s=-\tfrac{1}{2}$. This describes a 1s electron.

(c) $l=1$ requires $l \leq n-1$, so $1 \leq 0$, which is false. The first shell ($n=1$) has only an s-subshell ($l=0$). …

Why quantum numbers have rules

The four quantum numbers describe an electron's state in an atom, and each arises from solving the Schrödinger equation under specific physical constraints. The principal quantum number $n$ emerges from boundary conditions requiring finite energy, so $n$ must be a positive integer. The angular momentum quantum number $l$ is bounded by $n$ because higher angular momentum states require more energy. The magnetic quantum number $m_l$ represents the projection of angular momentum along an axis, so it cannot exceed $l$ in magnitude. The spin quantum number $m_s$ is an intrinsic property with only two possible values.

These aren't arbitrary rules—they're consequences of the mathematics of quantum mechanics and the physics of bound states.

Now let's examine each set systematically.

Checking each set of quantum numbers

1. Set (a): $n = 0, l = 0, m_l = 0, m_s = +\tfrac{1}{2}$

The principal quantum number $n$ must be at least 1 because it represents the energy level of the electron, and $n = 0$ would correspond to an electron at the nucleus with infinite negative energy—a physically meaningless state. The ground state of hydrogen has $n = 1$, not $n = 0$.

This set is impossible.

2. Set (b): $n = 1, l = 0, m_l = 0, m_s = -\tfrac{1}{2}$

Here $n = 1$ is valid. For $n = 1$, the allowed values of $l$ are $0, 1, \ldots, (n-1) = 0$, so $l = 0$ works. When $l = 0$, the only possible value of $m_l$ is $0$. The spin $m_s = -\tfrac{1}{2}$ is one of the two allowed spin states. This describes an electron in the 1s orbital with spin down.

This set is possible.

3. Set (c): $n = 1, l = 1, m_l = 0, m_s = +\tfrac{1}{2}$

While $n = 1$ is valid, the angular momentum quantum number $l$ must satisfy $l \leq n - 1$. For $n = 1$, we have $l \leq 0$, so only $l = 0$ is allowed. The value $l = 1$ would correspond to a p-orbital, but p-orbitals first appear at $n = 2$.

This set is impossible.

4. Set (d): $n = 2, l = 1, m_l = 0, m_s = -\tfrac{1}{2}$

With $n = 2$, the allowed values of $l$ are $0$ and $1$, so $l = 1$ is fine. For $l = 1$, $m_l$ can be $-1, 0,$ or $+1$, so $m_l = 0$ is allowed. The spin is valid. This describes an electron in a 2p orbital.

This set is possible.

5. Set (e): $n = 3, l = 3, m_l = -3, m_s = +\tfrac{1}{2}$ …