Neet Solved Paper 2015 Question 14
Question: Which of these statements about $ {{[Co{{(CN)}_6}]}^{3-}} $ is true?
Options:
A) $ {{[Co{{(CN)}_6}]}^{3-}} $ has no unpaired electrons and will be in a low-spin configuration.
B) $ {{[Co{{(CN)}_6}]}^{3-}} $ has four unpaired electrons and will be in a low-spin configuration.
C) $ {{[Co{{(CN)}_6}]}^{3-}} $ has four unpaired electrons and will be in a high-spin configuration.
D) $ {{[Co{{(CN)}_6}]}^{3-}} $ has no unpaired electrons and will be in a high-spin configuration.
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{[Co{{(CN)}_6}]}^{3-}} $ $ C{o^{3+}}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6} $ $ C{N^{-}} $ is a strong field ligand and as it approaches the metal ion, the electrons must pair up.
The splitting of the d-orbitals into two sets of orbitals in an octahedral $ {{[Co{{(CN)}_6}]}^{3-}} $ may be represented as
Here, for $ d^{6} $ ions, three elections first enter orbitals with parallel spin put the remaining may pair up in $ t _{2g} $ orbital giving rise to low spin complex (strong ligand) field.
$ \therefore \ {{[Co{{(CN)}_6}]}^{3-}} $ has no unpaired electron and will be in a low spin configuration.
BETA
