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(D) In $[Cu(CN)_4]^{3-}$,$Cu$ is in $+1$ oxidation state ($3d^{10}$ configuration). It is colourless,tetrahedral,and diamagnetic. This is correct.In $

Vedclass · Accepted Answer

$[Ni(dmg)_2] \Rightarrow 	ext{Coloured, tetrahedral, } H^{-} 	ext{ bond present}$

Vedclass · Answer

(D) In $[Cu(CN)_4]^{3-}$,$Cu$ is in $+1$ oxidation state ($3d^{10}$ configuration). It is colourless,tetrahedral,and diamagnetic. This is correct.In $[Cu(NH_3)_4]^{2+}$,$Cu$ is in $+2$ oxidation state ($3d^9$ configuration). It is coloured due to $d-d$ transition and has a square planar geometry. This is correct.In $[Pt(NH_3)_2Cl_2]$,$Pt^{2+}$ is a $5d^8$ metal ion,which forms square planar complexes. It is diamagnetic and exhibits geometrical isomerism (cis-trans). This is correct.In $[Ni(dmg)_2]$,$Ni^{2+}$ is a $3d^8$ ion. It forms a square planar complex (not tetrahedral) and is diamagnetic. It contains intramolecular hydrogen bonding. Therefore,the statement 'tetrahedral' is incorrect.

List $-I$	List $-II$
$A$. $[PtCl_4]^{2-}$	$I$. $sp^3d$
$B$. $BrF_5$	$II$. $d^2sp^3$
$C$. $PCl_5$	$III$. $dsp^2$
$D$. $[Co(NH_3)_6]^{3+}$	$IV$. $sp^3d^2$

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$d^2sp^3$ hybridisation leads to

Match List $-I$ with List $-II$:

List $-I$ List $-II$

$A$. $[PtCl_4]^{2-}$ $I$. $sp^3d$

$B$. $BrF_5$ $II$. $d^2sp^3$

$C$. $PCl_5$ $III$. $dsp^2$

$D$. $[Co(NH_3)_6]^{3+}$ $IV$. $sp^3d^2$

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