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Diamonds and graphite are both made of pure carbon — just arranged differently.
Among the following complexes, the one which shows zero crystal field stabilization energy (CFSE) is
medium
Coordination Compounds
2014
chemistry
[ M n ( H 2 O ) 6 ] 3 + [Mn(H_2O)_6]^{3+} [ M n ( H 2 O ) 6 ] 3 +
[ F e ( H 2 O ) 6 ] 3 + [Fe(H_2O)_6]^{3+} [ F e ( H 2 O ) 6 ] 3 +
[ C o ( H 2 O ) 6 ] 2 + [Co(H_2O)_6]^{2+} [ C o ( H 2 O ) 6 ] 2 +
[ C o ( H 2 O ) 6 ] 3 + [Co(H_2O)_6]^{3+} [ C o ( H 2 O ) 6 ] 3 +
Explanation To solve this problem, we need to determine which complex shows zero crystal field stabilization energy (CFSE).
\newline CFSE depends on the distribution of electrons in the d-orbitals of the metal ion in the complex.
\newline Let's analyze each option:
\newline • Option 1: [ M n ( H 2 O ) 6 ] 3 + [Mn(H_2O)_6]^{3+}
\newline [ M n ( H 2 O ) 6 ] 3 +
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-
has an electronic configuration of
- In an octahedral field, the electrons will be distributed as
t 2 g 3 e g 1 . t_{2g}^3 e_g^1.
\newline t 2 g 3 e g 1 . - This results in a non-zero CFSE.
• Option 2:
[ F e ( H 2 O ) 6 ] 3 + [Fe(H_2O)_6]^{3+}
\newline [ F e ( H 2 O ) 6 ] 3 + -
has an electronic configuration of
- In an octahedral field, the electrons will be distributed as
t 2 g 3 e g 2 . t_{2g}^3 e_g^2.
\newline t 2 g 3 e g 2 . - This results in zero CFSE because the stabilization from
is balanced by the destabilization from
• Option 3:
[ C o ( H 2 O ) 6 ] 2 + [Co(H_2O)_6]^{2+}
\newline [ C o ( H 2 O ) 6 ] 2 + -
has an electronic configuration of
- In an octahedral field, the electrons will be distributed as
t 2 g 5 e g 2 . t_{2g}^5 e_g^2.
\newline t 2 g 5 e g 2 . - This results in a non-zero CFSE.
• Option 4:
[ C o ( H 2 O ) 6 ] 3 + [Co(H_2O)_6]^{3+}
\newline [ C o ( H 2 O ) 6 ] 3 + -
has an electronic configuration of
- In an octahedral field, the electrons will be distributed as
t 2 g 4 e g 2 . t_{2g}^4 e_g^2.
\newline t 2 g 4 e g 2 . - This results in a non-zero CFSE.
Therefore, the complex that shows zero CFSE is
[ F e ( H 2 O ) 6 ] 3 + . [Fe(H_2O)_6]^{3+}.
\newline [ F e ( H 2 O ) 6 ] 3 + . This corresponds to Option 2.