NEET Solved Paper 2017 Question 34
Question: Pick out the correct statement with respect $ {{[Mn{{(CN)}_6}]}^{3-}} $
Options:
A) It is $ dsp^{2} $ hybridised and square planar
B) It is $ sp^{3}d^{2} $ hybridised and octahedral
C) It is $ sp^{3}d^{2} $ hybridised and tetrahedral
D) It is $ d^{2}sp^{3} $ hybridised and octahedral
Show Answer
Answer:
Correct Answer: D
Solution:
- $ {{[Mn{{(CN)}_6}]}^{3-}} $ $ Mn(III)=[Ar]3d^{4} $ $ C{N^{-}} $ being strong field ligand forces pairing of electrons This gives $ t_2^{4}e_g^{0} $
$ \therefore $ Mn(III) = [Ar] ∵ Coordination number of Mn = 6
$ \therefore $ Structure = octahedral $ {{[Mn{{(CN)}_6}]}^{3-}}= $
BETA
