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)

Solution

Chemistry ma’am · 4 steps

Fill the whole fourth shell first, then keep only the spin-up half of its electrons.

1. List the subshells. For n = 4, l can be 0, 1, 2 or 3. That gives four subshells: 4s, 4p, 4d and 4f.
2. Count when full. A subshell holds 2(2l + 1) electrons, so full 4s, 4p, 4d and 4f hold 2 + 6 + 10 + 14 = 32 electrons. This is the same as 2n² = 2 × 4² = 32.
3. Pick one spin. In a completely filled orbital, one electron has s = +1/2 and the other has s = -1/2. The fourth shell has n² = 16 orbitals, so exactly half of the 32 electrons are spin-up.

   Electrons with s = +1/2 = 2n²/2 = n² = 16

4. So. The number of electrons with n = 4 and s = +1/2 is 16.

The common slip: 32. Ignoring the spin condition and counting every electron in the full shell gives 32. Only half of those have s = +1/2.

The idea to keep. Shell n has n² orbitals and 2n² electrons when full. Fixing the spin picks one electron per orbital, so the count drops back to n².

Answer from the NTA final answer key. Solution by PrepWiser.