Question
Assign the position of the element having outer electronic configuration.
$n s^2 n p^4$ for $n=3$
$n s^2 n p^4$ for $n=3$
✓
Answer
Since $\mathrm{n}=3$, the element belongs to the $3^{\text {rd }}$ period. It is a p -block element since the last electron occupies the p-orbital.
There are four electrons in the p-orbital. Thus, the corresponding group of the element
$=\text { Number of s-block groups }+ \text { number of d-block groups }+ \text { number of p-electrons }$
$=2+10+4$
$=16$
Therefore, the element belongs to the $3^{\text {rd }}$ period and $16^{\text {th }}$ group of the periodic table. Hence, the element is Sulphur.
There are four electrons in the p-orbital. Thus, the corresponding group of the element
$=\text { Number of s-block groups }+ \text { number of d-block groups }+ \text { number of p-electrons }$
$=2+10+4$
$=16$
Therefore, the element belongs to the $3^{\text {rd }}$ period and $16^{\text {th }}$ group of the periodic table. Hence, the element is Sulphur.
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