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Modern Periodic Table — Chemistry STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceChemistryModern Periodic Table6 Marks
Question
Outer electronic configurations of a few elements are given below. Explain them and identify the period, group and block in the periodic table to which they belong.
$_2He: 1s^2, _{54}Xe: 5s^25p^6, _{16}S: 3s^23p^4, _{79}Au: 6s^15d^{10}$
✓

Answer

$i. \  _2He: 1s^2$
Here, $n = 1.$ Therefore, $_2He$ belongs to the $1^{st}$ period.
The shell $n = 1$ has only one subshell, namely $1s.$ The outer electronic configuration $1s^2$ of $‘He \ ’$ corresponds to the maximum capacity of $1s,$ the complete duplet. Therefore, He is placed at the end of the $1^{st}$ period in the group $18$ of inert gases. So, $‘He \ ’$ belongs to $p-$block.
$ii. \ _{54}Xe: 5s^25p^6$
Here, $n = 5.$ Therefore, $_{54}Xe$ belongs to the $5^{th}$ period.
The outer electronic configuration. $5s^25p^6$ corresponds to complete octet. Therefore, $_{54}Xe$ is placed in group $18$ and belongs to $p-$block.
$iii. \  _{16}S: 3s^23p^4$
Here, $n = 3.$ Therefore, $_{16}S$ belongs to the $3^{rd}$ period. The $3p$ subshell in $‘S \ ’$ is partially filled and short of completion of octet by two electrons. Therefore, $‘S \ ’$ belongs to $(18 – 2) = 16^{th}$ group and $p-$block.
$iv. \ _{79}AU: 6s^15d^{10}$
Here $n = 6.$ Therefore, $‘Au \ ’$ belongs to the $6^{th}$ period.
The sixth period begins with filling of electron into $6s$ and then into $5d$ orbital.
The outer configuration of $‘Au \ ’: 6s^1 5d^{10}$ implies that $(1 + 10) = 11$ electrons are filled in the outer orbitals to give $‘Au \ ’.$ Therefore $‘Au \ ’$ belongs to the group $11.$
As the last electron has entered $‘d \ ’$ orbital $‘Au \ ’$ belongs to the $d-$block.

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