Question 3.30
NCERT Class XI Chemistry
Assign the position of the element having outer electronic configuration (i) ns2np4 for n=3 (ii) (n-1)d 2ns2 for n=4, and (iii) (n-2) f 7 (n-1)d 1ns2 for n=6, in the periodic table.
Solution in Detail
(video solution below this)
$\displaystyle \underline{\underline{3s^23p^4}}$
Compare with the general configuration of p-block elements $\displaystyle ns^2np^{1-6}$
n = 3 gives period as 3
$\displaystyle p^4$ gives group as 16 - chalcogens - O [period 2], S[period 3], Se[period 4] . . .
hence, the element is sulphur (S)
$\displaystyle \underline{\underline{4s^23d^2}}$
General config of d-block: $\displaystyle ns^{0-2}(n - 1)d^{1-10}$
$\displaystyle 4s^2$ gives n = period = 4
$\displaystyle 4s^23d^2$ gives group as (GII) + 2 = Group IV (see video for explanation)
Period 4, Group 4 $\displaystyle \implies$ Z = [Ar] + 4 = 22 titanium (Ti)
$\displaystyle \underline{\underline{6s^24f^75d^1}}$
Period is 6 because of 6s2
$\displaystyle 4f^75d^1 \implies$ 7-th lanthanide
therefore,Z = 57 [La] + 7 = 64, Gadolinium (Gd)
Video Explanation
Please watch this youtube video for a quick explanation of the solution:
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