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NCERT Class XII Chemistry
Chapter - Coordination Compounds
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Question : 24 of 43
Marks: +1, -0
Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex :
(i) K[Cr(H2O)2(C2O4)2]â‹…3H2O\mathrm{K[Cr(H_2O)_2(C_2O_4)_2]}\cdot 3\mathrm{H_2O}
(ii)[Co(NH3)5Cl]Cl2\mathrm{[Co(NH_3)_5Cl]Cl_2}
(iii) CrCl3(py)3\mathrm{CrCl_3(py)_3}
(iv) Cs[FeCl4]\mathrm{Cs[FeCl_4]}
(v) K4[Mn(CN)6]\mathrm{K_4[Mn(CN)_6]}
Solution:  
(i) K[Cr(H2O)2(C2O4)2]â‹…3H2O\mathrm{K[Cr(H_2O)_2(C_2O_4)_2]}\cdot 3\mathrm{H_2O} : Potassiumdiaquadioxalatochromate(III) hydrate
Oxidation state = +3; C.N. = 6; configuration = t3{}^32g{}_2g
μ=n(n+2)\mu=\sqrt{n(n+2)}
=3×5=\sqrt{3 \times 5}
=15=\sqrt{15}
=3.87 B.M.=3.87\,\text{B.M.}
(ii)[Co(NH3)5Cl]Cl2\mathrm{[Co(NH_3)_5Cl]Cl_2} : Pentaamminechloridocobalt(III) chloride
Co(27):4s23d7;Co3+:4s03d6\mathrm{Co}(27) : 4s^{2} 3d^{7} ; \mathrm{Co}^{3+} : 4s^{0} 3d^{6}
The oxidation state of Co = +3
C.N. = 6; Configuration = t6{}^6 2g; μ{}_2g;\ \mu = 0 B.M.
(iv) Cs[FeCl4]\mathrm{Cs[FeCl_4]} : Trichloridotripyridinechromium (III) Coordination No. of Cr = 6; Oxidation state = +3; Configuration = t3{}^32g{}_2g;
Cr(24):4s13d5;Cr3+:4s03d3\mathrm{Cr}(24) : 4s^{1} 3d^{5} ; \mathrm{Cr}^{3+} : 4s^{0} 3d^{3}
μ=n(n+2)\mu=\sqrt{n(n+2)}
=15=\sqrt{15}
=3.87 B.M.=3.87\,\text{B.M.}
(iv) Cs[FeCl4]\mathrm{Cs[FeCl_4]}: Caesiumtetrachloridoferrate III)Oxidation state = +3; Coordination No. = 4; Configuration = t3{}^32g{}_2g e2{}^2g{}_g
Fe Fe(26):4s23d6;Fe3+:4s03d5\mathrm{Fe}(26) : 4s^{2} 3d^{6} ; \mathrm{Fe}^{3+} : 4s^{0} 3d^{5}
It is paramagnetic due to presence of 5 unpaired electrons.
μ=n(n+2)\mu=\sqrt{n(n+2)}
=5×7=\sqrt{5 \times 7}
=35=\sqrt{35}
=5.92 B.M.=5.92\,\text{B.M.};
(v) K4[Mn(CN)6]\mathrm{K_4[Mn(CN)_6]} : Potassium hexacyanomanganese(II)Oxidation state = +2; Coordination no. = 6; Configuration = t2g5t_{2g}^{5};
Mn(25) : [Ar]4s23d5;Mn2+: [Ar]4s03d5:\,[\mathrm{Ar}] 4s^{2} 3d^{5} ; \mathrm{Mn}^{2+} :\,[\mathrm{Ar}] 4s^{0} 3d^{5}
μ=n(n+2)\mu=\sqrt{n(n+2)}
=1×3=\sqrt{\sqrt{1 \times 3}}
=1.732 B.M.=1.732\,\text{B.M.}
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