Question

How many elements are possible for the $1^{st}$ period of periodic table if azimuthal quantum number can have integral values from $0$ to $(n+1)$. $[n=$ shell number and other rules are remaining same to form period table?

• A. 4
• B. 6
• C. 2
• D. 8

Answer 1

Answer: (D)

8

Value of azimuthal quantum number (as per the question) is from 0 to (n+1)

Therefore for,

$n=1,l=0,1,2$

$n=2,l=0,1,2,3,$

$n=3,l=0,1,2,3,4$

Now applying $(n+1)$ rule we get subshell filling order like this

$1s,1p,2s,2p.....$

Since after $1p,$ second shell starts filling only the elements having electrons present in $1s$ and $1p$ subshells belong to the first period.

And the total number of such elements possible is $2$ (capacity of 1s subshell) $+6$ (Capacity of 1p subshell)

Which is equal to a total of $8$ elements.

So, the correct option is $D$