If an iron (III) complex with the formula [Fe | vedclass

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(B) The complex is $[Fe(NH_3)_{x}(CN)_{y}]^{-}$.The oxidation state of $Fe$ is $+3$, so the electronic configuration of $Fe^{3+}$ is $3d^5$.For the $e

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$6$

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(B) The complex is $[Fe(NH_3)_{x}(CN)_{y}]^{-}$.The oxidation state of $Fe$ is $+3$, so the electronic configuration of $Fe^{3+}$ is $3d^5$.For the $e_g$ orbital to have no electrons, all $5$ electrons must occupy the $t_{2g}$ orbitals.This requires a strong field ligand environment to cause large crystal field splitting $(\Delta_o > P)$.$CN^-$ is a strong field ligand, while $NH_3$ is a moderate field ligand.For the complex to be low-spin with $t_{2g}^5 e_g^0$, the coordination number must be $6$ to satisfy the geometry of an octahedral complex.Thus, $x+y = 6$.For example, in $[Fe(CN)_6]^{3-}$, $x=0$ and $y=6$, which satisfies the condition.

List-$I$ (Hybridisation)	List-$II$ (Compound/ion)
$A. sp^3d$	$I. [PtCl_4]^{2-}$
$B. sp^3d^2$	$II. SF_6$
$C. dsp^2$	$III. BCl_3$
$D. dsp^3$	$IV. PCl_5$
	$V. ClF_3$

If an iron $(III)$ complex with the formula $[Fe(NH_3)_{x}(CN)_{y}]^{-}$ has no electron in its $e_g$ orbital, then the value of $x+y$ is

Similar Questions

The $[Co(H_2O)_6]^{2+}$ ion has three unpaired electrons. The hybridization of $Co$ in $[Co(H_2O)_6]^{2+}$ is

Match the following:
List-$I$ (Hybridisation) List-$II$ (Compound/ion)
$A. sp^3d$ $I. [PtCl_4]^{2-}$
$B. sp^3d^2$ $II. SF_6$
$C. dsp^2$ $III. BCl_3$
$D. dsp^3$ $IV. PCl_5$
$V. ClF_3$

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Which of the following statements is not true?

Nickel $(Z = 28)$ combines with a uninegative monodentate ligand $X^{-}$ to form a paramagnetic complex $[NiX_4]^{-2}.$ The number of unpaired electron$(s)$ in the nickel and geometry of this complex ion are,respectively $:$

Which of the following statements is correct?

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