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Chemistry Question on Coordination Compounds

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].3H2O(i) K[Cr(H_2O)_2(C_2O_4)_2].3H_2O
(ii)[Co(NH3)5Cl]Cl2(ii)[Co(NH_3)_5Cl]Cl_2
(iii)CrCl3(py)3(iii) CrCl_3(py)_3
(iv)Cs[FeCl4](iv) Cs[FeCl_4]
(v)K4[Mn(CN)6](v)K_4[Mn(CN)_6]

Answer

(i) Potassium diaquadioxalatochromate (III) trihydrate.
Oxidation state of chromium=3
Electronic configuration: 3d3:t2g33d ^{3} : t_{2g}^ 3
Coordination number = 6 Shape: octahedral
Stereochemistry:

Magnetic moment, A~ZˇA^14=n(n+2)ÃŽÂ\frac{1}{4}=\sqrt{n(n+2)}
=3(3+2)=\sqrt{3(3+2)}
=15=\sqrt{15}
A~¢E¨A^¼4BM∼ 4BM

(ii)[Co(NH3)5Cl]Cl2(ii) [Co(NH_3)_5Cl]Cl_2
IUPAC name: Pentaamminechloridocobalt(III) chloride
Oxidation state of Co=+3Co=+3
Coordination number=6 Shape: octahedral.
Electronic configuration: d6:t2g6.d ^{6} : t_{2g}^ 6 .
Stereochemistry:


Magnetic Moment=0

(iii)CrCl3(py)3(iii) CrCl_3(py)_3
IUPAC name: Trichloridotripyridinechromium (III)
Oxidation state of chromium = +3
Electronic configuration for d3=t2g3d^{ 3} = t_{2g}^ 3
Coordination number = 6
Shape: octahedral.
Stereochemistry:

Both isomers are optically active.
Therefore, a total of 4 isomers exist.
Magnetic moment,A~ZˇA^14=n(n+2), ÃŽÂ\frac{1}{4}=\sqrt{n(n+2)}
=3(3+2)=\sqrt{3(3+2)}
=15=\sqrt{15}
∼ 4BM

(iv)Cs[FeCl4](iv) Cs[FeCl_4]
IUPAC name: Caesium tetrachloroferrate (III)
Oxidation state ofFe=+3 Fe = +3
Electronic configuration of d6=eg2t2g3d ^{6} = eg^{ 2} t_{2g}^ 3
Coordination number = 4 Shape: tetrahedral
Stereochemistry: optically inactive Magnetic moment: A~ZˇA^14=n(n+2)ÃŽÂ\frac{1}{4}=\sqrt{n(n+2)}
=5(5+2)=\sqrt{5(5+2)}
=35=\sqrt{35}
~6BM

(v)K4[Mn(CN)6](v) K_4[Mn(CN)_6] Potassium hexacyanomanganate(II)
Oxidation state of manganese=+2
Electronic configuration: d5+:t2g5d ^{5+}: t_{2g}^ 5
Coordination number = 6 Shape: octahedral.
Streochemistry: optically inactive
Magnetic moment,A~ZˇA^14=n(n+2) ÃŽÂ\frac{1}{4}=\sqrt{n(n+2)}
=1(1+2)=\sqrt{1(1+2)}
=3=\sqrt{3}
=1.732BM=1.732BM