Which of the following is paramagnetic octahedral complex?

{[C o F_{6}]}^{3 -}

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{[C o {({N H}_{3})}_{6}]}^{3 +}

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{[C u {({N H}_{3})}_{4}]}^{2 +}

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[N i {(C O)}_{4}]

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Solution

The correct option is A ${[C o F_{6}]}^{3 -}$
$[C o F_{6}]^{3 -} \to C o^{3 +} \to d^{6} \to t_{2 g}^{4} e g^{2} \to n = 4$ (Unpaired electrons)
Hence it is paramagnetic since $F^{-}$ is weaker field ligand than $N H_{3}$ ligand hence $(¯ ¯¯ ¯ F)$ not causes pairing.

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Similar questions

\begin{matrix} Column - I & Column - II i . & [N i (C O)_{4}] & Planar octahedral, diamagnetic i i . & [F e (N O) (H_{2} O)_{5}]^{2 +} & Planar, paramagnetic i i i . & [C o (N O_{2})_{6}]^{3 -} & Octahedral, diamagnetic i v . & [C u (N H_{3})_{4}]^{2 +} & Planar octahedral, paramagnetic Octahedral, paramagnetic \end{matrix}

Q. How many in each of the following complexes obey the rule of

18

(EAN rule)?

(a)

[C u (N H_{3})_{4}]^{2 +}, [C u (e n)_{3}]^{2 +}, [C u (C N)_{4}]^{3 -}

(b)

[F e (C N)_{6}]^{4 -}, [F e (C N)_{6}]^{3 -}, F e (C O)_{6}

(c)

[N i (N H_{3})_{6}]^{2 +}, [N i (C N)_{4}]^{2 -}, [N i (C O)_{4}]

(d)

[C r (N H_{3})_{6}]^{3 +}, [C r (C O)_{6}]

(e)

[C o (N H_{3})_{6}]^{3 +}, [C o C l_{4}]^{2 -}

Among $[N i (C o)_{4}], [N i C l_{4}]^{2 -}, [C o (N H_{3})_{4} C l_{2}] C l, N a_{2} [C o F_{6}]^{3 +}, N a_{2} O_{2} a n d C s O_{2},$ the total number of paramagnetic compounds is

Q. All the following complexes are found to be paramagnetic:

P : [F e F_{6}]^{3 -}; Q : [C o F_{6}]^{3 -}; R : [V (H_{2} O)_{6}]^{3 +}; S : [T i (H_{2} O)_{6}]^{3 +}

The correct order of their paramagnetic moment (spin only) is:

Q. Among

[N i (C O)_{4},, [N i C l_{4}]^{2 -}, [C o (N H_{2}), C l_{2}] C l, N a_{3} [C o F_{6}], N a_{2} O_{2} and C s O_{2} .]

the total number of paramagnetic compound is

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Which of the following is paramagnetic octahedral complex?

The correct option is A [CoF6]3−[CoF6]3−→Co3+→d6→t42geg2→n=4 (Unpaired electrons)Hence it is paramagnetic since F− is weaker field ligand than NH3 ligand hence (¯¯¯¯F) not causes pairing.

The correct option is A ${[C o F_{6}]}^{3 -}$
$[C o F_{6}]^{3 -} \to C o^{3 +} \to d^{6} \to t_{2 g}^{4} e g^{2} \to n = 4$ (Unpaired electrons)
Hence it is paramagnetic since $F^{-}$ is weaker field ligand than $N H_{3}$ ligand hence $(¯ ¯¯ ¯ F)$ not causes pairing.