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  • Consider the following metal complexes : <img alt="\begin{aligned} &{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}} \\ &{\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\r

Consider the following metal complexes :

\begin{aligned} &{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}} \\ &{\left[\mathrm{CoCl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{2+}} \\ &{\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}} \\ &{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{3+}} \end{aligned}

The spin-only magnetic moment value of the complex that absorbes light with shortest wavelength is________ B.M. (Nearest integer)

Option: 1

0


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

The shortest wavelength mean largest \Delta _{\text{o}} (splitting) value.

them \text{CN}^{\ominus } is SFL which has maximum CFSE.

So, \mathrm{\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}} will be \mathrm{d^{2}sp^{3}} and \mathrm{3d^{6}4s^{o}}

Due to SFL all electron will be paired.

So, \mu =0 due to \text{n} =0

Answer = 0

Posted by

Sayak

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