Question 3.29
NCERT Class XI Chemistry
Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Solution in Detail
(video solution below this)
$\displaystyle \underline{\underline{\text{s-block}}}$
Group I has $\displaystyle ns^1$ and Group II has $\displaystyle ns^2$
hence, general config is $\displaystyle ns^{1-2}$
$\displaystyle \underline{\underline{\text{d-block}}}$
The general configuration is $\displaystyle ns^{0-2} (n - 1)d^{1-10}$
The d-subshell starts filling after $\displaystyle ns^2$. The expected configuration is $\displaystyle ns^2 (n - 1)d^{1-10}$. But due to exchange energy stabilization, s electrons [0, 1 or both 2] move to the d sub-shell so that d is fully filled or half-filled. Hence, the general configuration becomes as per written above.
Ex. 1: [Z = 29] Cu $\displaystyle \text{[Ar]}4s^13d^{10}$.
Ex. 2: [Z = 46] Palladium $\displaystyle \text{[Kr]}5s^04d^{10}$.
$\displaystyle \underline{\underline{\text{f-block}}}$
The general configuration is $\displaystyle ns^2 (n - 2)f^{1-14} (n - 1)d^{0-2}$
$\displaystyle \underline{\underline{\text{p-block}}}$
The $\displaystyle ns^2$ subshell, followed by the inner $\displaystyle (n-2)f^{14}$, $\displaystyle (n-1)d^{10}$ subshells [wherever available] are fully filled, and now np -subshell starts filling.
G13 $\displaystyle ns^2np^1$ . . . G18 $\displaystyle ns^2np^{6}$
hence, the general configuration is $\displaystyle ns^{2}np^{1-6}$
Video Explanation
Please watch this youtube video for a quick explanation of the solution:
This Blog Post/Article "(solved)Question 3.29 of NCERT Class XI Chemistry Chapter 3" by Parveen is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.