Question

Assign the position of the element having outer electronic configuration

1. $ns^ 2 np^ 4$ for $n$ = $3$
2. $(n - 1)d^ 2 \;ns^ 2$ for $n$ =$4$, and
3. $(n - 2) f ^7 (n - 1)d ^1 ns^ 2$ for $n$ = $6$, in the periodic table.

Answer 1

(i) Since $n$ = $3$, the element belongs to the $3 rd$ 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
= Number of $s-block$ groups + number of $d-block$ groups + number of $p-electrons$
= $2 + 10 + 4$
= $16$
Therefore, the element belongs to the $3 ^{rd}$ period and $16 ^{th}$ group of the periodic table. Hence, the element is $Sulphur$.

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(ii) Since $n = 4$, the element belongs to the $4^{ th}$ period. It is a $d-block$ element as $d-orbitals$ are incompletely filled.
There are $2$ electrons in the $d-orbital$.
Thus, the corresponding group of the element
= Number of $s-block$ groups + number of $d-block$ groups
= $2 + 2$
= $4$
Therefore, it is a $4 ^{th }$ period and $4 ^{th}$ group element. Hence, the element is $Titanium$.

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(iii) Since $n = 6$, the element is present in the $6^{ th}$ period. It is an $f -block$ element as the last electron occupies the $f-orbital$. It belongs to group $3$ of the periodic table since all $f-block$ elements belong to group $3$. Its electronic configuration is $[Xe]$ $4f^7$ $5d ^1$ $6s ^2$ . Thus, its atomic number is $54 + 7 + 2 + 1 = 64$.
Hence, the element is $Gadolinium$.