Question

Instead of Principal quantum number $\left(n\right)$, Azimuthal quantum number $\left(l\right)$ and Magnetic quantum number $\left(m\right)$, a set of new quantum number $S$, $t$ and $u$ was introduced with similar logic but different values as defined below:
$S=1,2,3\dots \dots$ (all positive integral values)
$t=(S^2-1^2),(S^2-2^2),(S^2-3^2)\dots \dots$ No negative values
$u=-\frac{(t+1)}{2}$ to $\frac{(t+1)}{2}$ in integral steps.
Each orbital can have maximum four electrons. $(S+t)$ rule is defined similar to $(n+l)$ rule.
Now answer the following questions:

Number of electrons belongs to $S=2$, $t=3$ for an element with atomic number $Z=24$ is:

• A. 8
• B. 4
• C. \N
• D. None of these

Answer 1

The correct option is C \N

i) For $S=1$
$t=0$ two orbitals -1/2 and +1/2
ii) For $S=2$,
$t=3,0$
number of orbitals corresponding to $t=3$ are five as per below:
$u=-\frac{(3+1)}{2}$ to $\frac{(3+1)}{2}$ i.e. $-2,-1,0,1,2$
Similarly for $t=0$, total number of orbitals is 2 i.e. $-\frac{1}{2},+\frac{1}{2}$
Since one orbital can accomodate maximum of four electrons hence total of 28 $e^{-}$ belong to $S=2$.

iii) For $S=3$,
number of subshells $t=8,5,0$
$u_{\left(8\right)}=-\frac{9}{2},...,+\frac{9}{2}\to 10\text{orbitals}$
$u_{\left(5\right)}=-3,...,+3\to 7\text{orbitals}$
$u_{\left(0\right)}=-\frac{1}{2},+\frac{1}{2}\to 2\text{orbitals}$
according to (S+t) rule order of filling before $S=2andt=3$:
$S,t,numberof{e}^{-}$
$1,0,8$
$2,0,8$
$3,0,8$
$4,0,8$
So, there are no electrons in S=2 and t=3.