QJEEM24Chd01e: JEE Main 2024 Chemistry - Class 11 JEEM24Chd01e Solution: In an atom, total number of electrons having quantum numbers…

To find the total number of electrons with the given quantum numbers, let's first understand what each quantum number represents:

• n (Principal Quantum Number): This specifies the energy level or shell of the electron in the atom. Here, $n = 4$ means we are dealing with the fourth energy level.
• l (Azimuthal Quantum Number or Orbital Angular Momentum Quantum Number): This defines the shape of the orbital and is dependent on the value of n. It can take values from 0 to $n-1$ . For $n = 4$ , $l$ can be 0 (s orbital), 1 (p orbital), 2 (d orbital), or 3 (f orbital).
• $m_l$ (Magnetic Quantum Number): This describes the orientation of the orbital in space and can take values from $-l$ to $+l$ , including zero. The condition $\left|\mathrm{~m}_l\right|=1$ implies $m_l = -1$ or $+1$ .
• $m_s$ (Spin Quantum Number): This indicates the spin direction of the electron, with possible values of $+\frac{1}{2}$ or $-\frac{1}{2}$ . Here, $\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}$ specifies electrons with a downward spin.

Given $\left|\mathrm{~m}_l\right|=1$ , this directly implies that we are not considering s orbitals (which have $l = 0$ and hence $m_l = 0$ ). We are considering p, d, and f orbitals which can have $m_l = 1$ or $m_l = -1$ .

For $l = 1$ (p orbitals):

• This corresponds to p orbitals, where $\left|\mathrm{~m}_l\right|=1$ can be achieved with $m_l = -1$ or $m_l = +1$ .
• With $\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}$ , we only take half of the available m_l values since each orbital can host two electrons with different spin quantum numbers.
• So, we get 2 electrons for p orbitals with the specified conditions.

For $l = 2$ (d orbitals):

• This corresponds to d orbitals, where again $\left|\mathrm{~m}_l\right|=1$ indicates $m_l = -1$ or $m_l = +1$ .
• Similarly, we consider only electrons with a spin quantum number of $-\frac{1}{2}$ , leading us to again have 2 electrons.

For $l = 3$ (f orbitals):

• f orbitals also can have $m_l = -1$ or $m_l = +1$ satisfying the $\left|\mathrm{~m}_l\right|=1$ .
• With $\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}$ , like before, this limits us to 2 electrons.

Adding up all the electrons from the p, d, and f orbitals that meet the given criteria, we get:

• $2\ (\mathrm{p\ orbitals}) + 2\ (\mathrm{d\ orbitals}) + 2\ (\mathrm{f\ orbitals}) = 6\ \mathrm{electrons}$

Therefore, the total number of electrons having quantum numbers with $\mathrm{n}=4, \left|\mathrm{~m}_l\right|=1$ and $\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}$ is 6.