Co-Ordination Compounds Question 171
Question: Select the correct statement from the following.
Options:
A) $ {{[Sc{{(H_2O)}_6}]}^{3+}} $ and $ {{[Ti{{(H_2O)}_6}]}^{3+}} $ both are colourless.
B) $ [Co{{(NH_3)}_4}Br_2]Cl $ shows ionization isomers and geometrical isomers.
C) $ [Pd{{(NO_2)}_2}{{(NH_3)}_2}] $ is square planar and shows geometrical as well as linkage isomers.
D) Both [b] and [c] are correct.
Show Answer
Answer:
Correct Answer: D
Solution:
(1) $ S{c^{3+}}-{{[Ar]}^{18}}3d^{0}4s^{0}; $ there is no unpaired electron in d-orbitals so no d-d transition takes place and the complex is colourless.
$ T{i^{3+}}-{{[Ar]}^{18}}3d^{1}4s^{0}; $ it has one unpaired electron so d-d transition of electron form $ t_{2g} $ level to empty level takes place and thus the complex is colored.
(2) $ [Co{{(NH_3)}_4}Br_2]CI- $
$ [Co{{(NH_3)}_4}BrCI]Br $ (Ionisation isomers)
(3) Pd has $ 4d^{8} $ configuration which has higher CFSE and thus the complex is square planar and diamagnetic.
In some ligands, like ambidenate ligands, there are two possible coordination sites.
In such cases, linkage isomerism exist, e.g., $ NO_2 $ group can be bonded to metal ions through nitrogen $ (-NO_2) $ or through oxygen $ (-ONO).$
