Co Ordination Compounds Question 91
Question: Which one of the following complexes will most likely absorb visible light? (At nos. Sc=21, Ti=22, V=23, Zn=30)
Options:
A) $ {{[Sc{{(H_2O)}_6}]}^{3+}} $
B) $ {{[Ti{{(NH_3)}_6}]}^{4+}} $
C) $ {{[V{{(NH_3)}_6}]}^{3+}} $
D) $ {{[Zn{{(NH_3)}_6}]}^{2+}} $
Show Answer
Answer:
Correct Answer: C
Solution:
The absorption of visible light and hence coloured nature of the transition metal cation is due to the promotion of one or more unpaired - d - electron from a lower to higher level withing same d-subshell.
Hence higher will be the number of unpaired electron higher will be the absorpion in visible light.
The electronic configuration of the given elements is $ S{c^{3+}}(18)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{0}4s^{0}- $ no Unpaired $ {e^{-}} $ .
$ T{i^{4+}}(18)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{0}4s^{0} $ - no Unpaired $ {e^{-}} $ .
$ {V^{3+}}( 20 )=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{2}4s^{0} $ - Two Unpaired $ {e^{-}} $ .
$ Z{n^{2+}}(28)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{0} $ - no Unpaired $ {e^{-}} $ .
hence $ {{[V{{(NH_3)}_6}]}^{3+}} $ will most likely absorb visible light.
BETA
