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Need explanation for: - p- d- and f-block elements - BITSAT

The C.F.S.E. for the complex K_{4}[Fe(CN)_{6}] is:

  • Option 1)

    0.6 \Delta_{o}

  • Option 2)

    -3.6 \Delta_{o}

  • Option 3)

    -2.4 \Delta_{o}

  • Option 4)

    -0.4 \Delta_{o}

 
Answers (1)
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As we learnt in 

CFSE in octahedral complex -

CFSE =\left [- \frac{2}{5}\left ( No. of\;electron\;in\;t_{2}g \right ) +\frac{3}{5}\left ( No. of\;\acute{e}s\;in\;eg \right ) \right ]\bigtriangleup \circ

-

 

Electronic Configuration of Fe^{2+} in K_{4}\left [ Fe\left ( CN \right )_{6} \right ]

Fe^{2+}: 3d^{6} & 4s^{0},  as CN- is a strong ligand.

Therefore, CFSE in Octahedral Complexes. 

= \left [ \frac{-2}{5}\left ( no. \: of\: \: electrons\: \: in\: \: t_{2}g \right ) +\frac{3}{5}\left ( no. \: of\: \: electrons\: \: in\: \: eg \right ) \right ]\Delta _{0}

= \left [ \frac{-2}{5}\times 6 +\frac{3}{5}\left ( 0 \right ) \right ] \Delta _{0}

= -2.4 \Delta _{0}

 


Option 1)

0.6 \Delta_{o}

This answer is incorrect

Option 2)

-3.6 \Delta_{o}

This answer is incorrect

Option 3)

-2.4 \Delta_{o}

This answer is correct

Option 4)

-0.4 \Delta_{o}

This answer is incorrect

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