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  • I have a doubt, kindly clarify. - Coordination compounds: - NEET

Crystal field splitting energy for high spin d^{4} octahedral complex is:

  • Option 1)

    -1.6\Delta _{0}

  • Option 2)

    -1.2\Delta _{0}

  • Option 3)

    -0.6\Delta _{0}

  • Option 4)

    -0.8\Delta _{0}

 

Answers (1)

best_answer

As we learned 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

-

 

 CFSE (Octahedral complex) = [\frac-{2}{5} (e^{-} in t_{2}g) +\frac{3}{5}(e^{-} in lg)]

for high spin octahedral complex, e^{-} config is

\therefore CFSE= [-\frac{2}{5}\times (3) +\frac{3}{5}(1)] Do

=(-1.2+0.6) Do=-0.6 Do

 


Option 1)

-1.6\Delta _{0}

This is incorrect option

Option 2)

-1.2\Delta _{0}

This is incorrect option

Option 3)

-0.6\Delta _{0}

This is correct option

Option 4)

-0.8\Delta _{0}

This is incorrect option

Posted by

divya.saini

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