A 3d series metal cation forms octahedral complexes with both -SCN as well as -CN ions as ligands- The complex -MCN6-3- has a spin only magnetic moment of -3 BM whereas the complex -MSCN6-3- has a spin only magnetic moment of -35 BM- Then out of the following- metal M can be-

As we know, μ =√(n(n+2)) BM M^3+ nbsp;with SCN^ nbsp;has 5 unpaired e^ M^3+ with CN^ nbsp;has 1 unpaired e^ So M^3+ nbsp;has d^5 nbsp;configuratio ...

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Standard XII

Chemistry

Shape and Geometry Using VSEPR

A 3d series m...

Question

A $3 d$ series metal cation forms octahedral complexes with both $^{-} S C N$ as well as $^{-} C N$ ions as ligands. The complex $[M (C N)_{6}]^{3 -}$ has a spin only magnetic moment of $\sqrt{3}$ BM whereas the complex $[M (S C N)_{6}]^{3 -}$ has a spin only magnetic moment of $\sqrt{35}$ BM. Then out of the following, metal $M$ can be:

C o

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N i

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F e

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M n

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Solution

The correct option is C $F e$
As we know,
$μ = \sqrt{n (n + 2)} B M$
$M^{3 +}$ with $S C N^{-}$ has $5$ unpaired $e^{-}$
$M^{3 +}$ with $C N^{-}$ has $1$ unpaired $e^{-}$
So $M^{3 +}$ has $d^{5}$ configuration
So $M$ has to be $s^{2} d^{6}$ configuration.

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Similar questions

Q. For the octahedral complexes of

F e^{3 +}

S C N^{-}

(thiocyanato-S) and in

C N^{-}

ligand environments, the difference between the spin-only magnetic moment in BM (when approximated to the nearest integer) is
[Atomic number of

F e = 26

]

Q. The spin only magnetic moment of

F e

in the complex

[F e (C N)_{6}]^{3 -}

in BM is:

Q. A complex

{[{M F}_{6}]}^{- x}

has magnetic moment of

4.91 B . M

whereas

{[M {(C N)}_{6}]}^{- x}

has zero magnetic moment, metal ion can be :

Q. A complex of certain metal has the magnetic moment of

4.89 B M

whereas another complex of the same metal with same oxidation state has zero magnetic moment. The metal ion could be :

Q. A complex

{[M F_{6}]}^{- x}

has magnetic moment of 4.91 B.M. whereas

{[M {(C N)}_{6}]}^{- x}

has zero magnetic moment, metal ion can be :

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A 3d series metal cation forms octahedral complexes with both −SCN as well as −CN ions as ligands. The complex [M(CN)6]3− has a spin only magnetic moment of √3 BM whereas the complex [M(SCN)6]3− has a spin only magnetic moment of √35 BM. Then out of the following, metal M can be:

The correct option is C FeAs we know, μ=√n(n+2) BM M3+ with SCN− has 5 unpaired e− M3+ with CN− has 1 unpaired e− So M3+ has d5 configuration So M has to be s2d6 configuration.

The correct option is C $F e$
As we know,
$μ = \sqrt{n (n + 2)} B M$
$M^{3 +}$ with $S C N^{-}$ has $5$ unpaired $e^{-}$
$M^{3 +}$ with $C N^{-}$ has $1$ unpaired $e^{-}$
So $M^{3 +}$ has $d^{5}$ configuration
So $M$ has to be $s^{2} d^{6}$ configuration.