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[Pd(F)(Cl)(Br)(I)]^{2-} has n number of geometrical isomers. Then, the spin- only magnetic moment and crystal field stabilisation energy [CFSE] of [Fe(CN)_{6}]^{n-6}, respectively are ;

[ Note : Ignore the pairing energy]

Option: 1

1.73 BM\ and -2.0\Delta _{0}


Option: 2

2.84\ BM\ and -1.6\Delta _{0}


Option: 3

0\ BM\ and -2.4\Delta _{0}


Option: 4

5.92\ BM\ and\ 0


Answers (1)

best_answer

As we have learnt,

Square planar complexes of type [Mabcd] have 3 geometrical isomers:

\mathrm{(a\ trans\ b); (a\ trans\ c); (b\ trans\ c)}

Therefore, number of Geometrical Isomers in square planar [PdFClBrI]^{2-} is 3
Hence, n = 3

Now, the given complex is [Fe(CN)_6]^{3-} which contains Fe^{3+} having a d^5 configuration. According to CFT, the configuration is  t_{2g} ^{5}e_g^{0}  and the central ion has only 1 unpaired electron.

\begin{array}{l}{\mu=\sqrt{n(n+2)}=1.73 \mathrm{B} . \mathrm{M}} \\\\ {\mathrm{CFSE}=-0.4 \Delta_{0} \times n t_{2 g}+0.6 \Delta_{0} \times n_{e g}} \\\\ {=-0.4 \Delta_{0} \times 5=-2.0 \Delta_{0}}\end{array}

Hence, the option number (1) is correct.

Posted by

Suraj Bhandari

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