Assign the position of the element having an outer electronic configuration in the periodic table.
ns2 np4 for n = 3
Since n = 3, the element belongs to the 3rd 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
= Number of s–block groups + number of d–block groups + number of p–electrons
= 2 + 10 + 4
Therefore, the element belongs to the 3rd period and 16th group of the periodic table. Hence, the element is Sulphur.
n = 3
Thus element belongs to 3rd period, p-block element.
Since the valence shell contains = 6 electrons, group No = 10 + 6 = 16 configuration
=1s2 2s2 2p6 3s23p4 element name is sulphur
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