The correct match of the complexes with their | vedclass

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(A) The correct match is:$i$. $NiCl_4^{2-}$: $Ni^{2+}$ is $d^8$. $Cl^-$ is a weak field ligand,so it forms a tetrahedral complex which is paramagnetic

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$i$ $\rightarrow q, ii$ $\rightarrow p, iii$ $\rightarrow r, iv$ $\rightarrow r$

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(A) The correct match is:$i$. $NiCl_4^{2-}$: $Ni^{2+}$ is $d^8$. $Cl^-$ is a weak field ligand,so it forms a tetrahedral complex which is paramagnetic due to two unpaired electrons. Match: $i \rightarrow q$.$ii$. $Ni(CO)_4$: $Ni$ is $d^{10}$. $CO$ is a strong field ligand,forming a tetrahedral complex which is diamagnetic. Match: $ii \rightarrow p$.$iii$. $PtCl_4^{2-}$: $Pt^{2+}$ is $5d^8$. $Cl^-$ is a weak field ligand,but for $5d$ series elements,the crystal field splitting energy is high,resulting in a square planar diamagnetic complex. Match: $iii \rightarrow r$.$iv$. $Ni(CN)_4^{2-}$: $Ni^{2+}$ is $d^8$. $CN^-$ is a strong field ligand,forcing pairing,resulting in a square planar diamagnetic complex. Match: $iv \rightarrow r$.Thus,the correct sequence is $i$ $\rightarrow q, ii$ $\rightarrow p, iii$ $\rightarrow r, iv$ $\rightarrow r$.

Complex	Structure and magnetic property
$i$. $NiCl_4^{2-}$	$p$. tetrahedral and diamagnetic
$ii$. $Ni(CO)_4$	$q$. tetrahedral and paramagnetic
$iii$. $PtCl_4^{2-}$	$r$. square planar and diamagnetic
$iv$. $Ni(CN)_4^{2-}$	$s$. square planar and paramagnetic

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