Amongst TiF_6^{2-},CoF_6^{3-},Cu_2Cl_2 and Ni | vedclass

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(D) The colourless species are $TiF_6^{2-}$ and $Cu_2Cl_2$. In $TiF_6^{2-}$,the oxidation state of $Ti$ is $+4$,which corresponds to a $d^0$ configura

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$TiF_6^{2-}$ and $Cu_2Cl_2$

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(D) The colourless species are $TiF_6^{2-}$ and $Cu_2Cl_2$. In $TiF_6^{2-}$,the oxidation state of $Ti$ is $+4$,which corresponds to a $d^0$ configuration. Since there are no $d$-electrons,$d-d$ transitions are not possible,making it colourless. In $Cu_2Cl_2$,the oxidation state of $Cu$ is $+1$,which corresponds to a $d^{10}$ configuration. Since the $d$-subshell is completely filled,$d-d$ transitions are not possible,making it colourless.

Complex Ion	Label
$[Mn(CN)_6]^{3-}$	$A$
$[Fe(CN)_6]^{3-}$	$B$
$[CoF_6]^{3-}$	$C$
$[Co(C_2O_4)_3]^{3-}$	$D$

Amongst $TiF_6^{2-}$,$CoF_6^{3-}$,$Cu_2Cl_2$ and $NiCl_4^{2-}$,the colourless species are

Similar Questions

How many geometrical isomers will $[Pt(NH_3)(NH_2OH)(H_2O)(py)]^+$ form?

The compound$(s)$ that exhibit$(s)$ geometrical isomerism is(are):
$A$. $[Pt(en)Cl_2]$
$B$. $[Pt(en)_2]Cl_2$
$C$. $[Pt(en)_2Cl_2]Cl_2$
$D$. $[Pt(NH_3)_2Cl_2]$

Observe the following complex ions:
Complex Ion Label
$[Mn(CN)_6]^{3-}$ $A$
$[Fe(CN)_6]^{3-}$ $B$
$[CoF_6]^{3-}$ $C$
$[Co(C_2O_4)_3]^{3-}$ $D$

Identify the option in which the unpaired electrons in the complex ions are in correct increasing order.

Give an example of solvate isomerism.

The number of geometrical isomers for octahedral $[CoCl_4(NH_3)_2]^-$,square planar $[AuBr_2Cl_2]^-$ and $[PtCl_2(en)]$ are

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