The total number of possible isomers of the complex compound $[Cu(NH_3)_4] [PtCl_4]$ is:
- A$3$
- B$6$
- C$5$
- D$4$
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Similar Questions
How many geometrical isomers and stereoisomers are possible for $[Pt(NO_2)(NH_3)(NH_2OH)(Py)]^+$ and $[Pt(Br)(Cl)(I)(NO_2)(NH_3)(Py)]$ respectively?
The spin-only magnetic moments of $[Mn(CN)_6]^{2-}$ and $[MnBr_4]^{2-}$ in Bohr magnetons,respectively are
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View SolutionWhich of the following complexes has a zero magnetic moment (spin only)?
MediumAIIMS 2009
View Solution$[Co(NH_3)_5SO_4]Br$ and $[Co(NH_3)_5Br]SO_4$ are a pair of isomers.
Match the following:
List-$I$ (Complex) List-$II$ (Spin only Magnetic moment) $A$. $[CoF_6]^{3-}$ $I$. $0$ $B$. $[Co(C_2O_4)_3]^{3-}$ $II$. $\sqrt{24}$ $C$. $[FeF_6]^{3-}$ $III$. $\sqrt{8}$ $D$. $[Mn(CN)_6]^{3-}$ $IV$. $\sqrt{35}$ $V$. $\sqrt{15}$
| List-$I$ (Complex) | List-$II$ (Spin only Magnetic moment) |
| $A$. $[CoF_6]^{3-}$ | $I$. $0$ |
| $B$. $[Co(C_2O_4)_3]^{3-}$ | $II$. $\sqrt{24}$ |
| $C$. $[FeF_6]^{3-}$ | $III$. $\sqrt{8}$ |
| $D$. $[Mn(CN)_6]^{3-}$ | $IV$. $\sqrt{35}$ |
| $V$. $\sqrt{15}$ |
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