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CBSE Class 12 Chemistry 2016 Outside Delhi Set 1 Solved Paper

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Question : 22 of 26
Marks: +1, -0
(a) For the complex [Fe(H2O)6]3+[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{3+}, write the hybridization, magnetic character and spin of the complex. (At. number: Fe=26\mathrm{Fe}=26 )
(b) Draw one of the geometrical isomers of the complex [Pt(en)2Cl2]2+[\mathrm{Pt}(\mathrm{en})_2\mathrm{Cl}_2]^{2+} which is optically inactive.
Solution:  
(a) [Fe(H2O)6]3+[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{3+}
Since H2O\mathrm{H}_2\mathrm{O} is a weak field ligand it cannot participate in pairing of electrons. Therefore, the number of unpaired electrons is 5 .
μ=n(n+2)=5(5+2)\mu=\sqrt{n(n+2)}=\sqrt{5(5+2)} =35=5.92BM.=\sqrt{35}=5.92\,\mathrm{BM}.
Thus, it is strongly paramagnetic (due to presence of unpaired electrons).
In [Fe(H2O)6]3+[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{3+}, outer d-orbitals are used in hybridization and it is high spin or spin free complex.
(b) Geometrical isomer of [Pt(en)2Cl2]2+[\mathrm{Pt}(\mathrm{en})_2\mathrm{Cl}_2]^{2+} which is optically active,
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