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Pick out the correct statement with respect to [ [ [ Mn(CN) 6 ] 3 ā . _6]^{3-}. 6 ā ] 3 ā .
medium
Coordination Compounds
2017
chemistry
It is s p 3 d 2 sp^3d^2 s p 3 d 2 hybridized and tetrahedral.
It is d 2 s p 3 d^2sp^3 d 2 s p 3 hybridized and octahedral.
Explanation To solve this problem, we need to determine the hybridization and geometry of the complex [ [ [ Mn(CN) 6 ] 3 ā . _6]^{3-}.
\newline 6 ā ] 3 ā . Step-by-step analysis:
\newline 1. Determine the oxidation state of Mn:
\newline
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It is d s p 2 dsp^2 d s p 2 hybridized and square planar.
It is s p 3 d 2 sp^3d^2 s p 3 d 2 hybridized and octahedral.
ā¢ The complex ion is
Mn(CN)
6 ] 3 ā . _6]^{3-}.
\newline 6 ā ] 3 ā . ā¢ Let the oxidation state of Mn be
ā¢ Each CN ligand has a charge of
ā¢ The overall charge of the complex is
ā¢ Therefore, the equation is:
x + 6 ( ā 1 ) = ā 3. x ā 6 = ā 3 x = + 3 x + 6(-1) = -3.
\newline x - 6 = -3
\newline x = +3
\newline x + 6 ( ā 1 ) = ā 3. x ā 6 = ā 3 x = + 3 ā¢ Mn is in the +3 oxidation state.
2. Determine the electronic configuration of Mn in the +3 state:
ā¢ The atomic number of Mn is 25.
ā¢ The electronic configuration of Mn is
Ar
] ā 3 d 5 ā 4 s 2 . ] \, 3d^5 \, 4s^2.
\newline ] 3 d 5 4 s 2 . ā¢ In the +3 oxidation state, Mn loses 3 electrons: 2 from 4s and 1 from 3d.
ā¢ The configuration becomes
Ar
] ā 3 d 4 . ] \, 3d^4.
\newline ] 3 d 4 . 3. Determine the hybridization:
ā¢ CN is a strong field ligand and causes pairing of electrons in the 3d orbitals.
ā¢ After pairing, the configuration is
t 2 g 4 ā e g 0 . t_{2g}^4 \, e_g^0.
\newline t 2 g 4 ā e g 0 ā . ā¢ The hybridization involves the use of 2 paired 3d orbitals, 1 4s orbital, and 3 4p orbitals.
ā¢ This results in
hybridization.
4. Determine the geometry:
ā¢ The
hybridization corresponds to an octahedral geometry.
Conclusion:
The correct statement is: It is
hybridized and octahedral.
This corresponds to Option 2.