Question

If possible values of spin quantum numbers are $3$ i.e. $\frac{1}{2},0,+\frac{1}{2}$. The permissible values of other quantum numbers and rules for filling of orbitals remains unchanged, then number of elements in $4th$ period is:

• A. $27$
• B. $18$
• C. $9$
• D. $54$

Answer 1

Answer: (A)

$27$

Now, each orbital can hold $3e^{-}$s each, so, total member of electrons is $3n^2$. So, $75$ electrons totally. The configuration of fifth group is $5s4d5p$. Hence, total $27$ elements will be here in $5$th period.