Which of the following combinations is correc | vedclass

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(A) $1$. For $[PdCl_4]^{2-}$,$Pd^{2+}$ is a $4d^8$ ion. Since $Pd$ is a $4d$ series metal,it always forms square planar complexes with $dsp^2$ hybridi

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$[PdCl_4]^{2-} \Rightarrow dsp^2$

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(A) $1$. For $[PdCl_4]^{2-}$,$Pd^{2+}$ is a $4d^8$ ion. Since $Pd$ is a $4d$ series metal,it always forms square planar complexes with $dsp^2$ hybridization. This is correct.$2$. For $[PtCl_4]^{2-}$,$Pt^{2+}$ is a $5d^8$ ion. Like $Pd^{2+}$,it forms square planar complexes with $dsp^2$ hybridization,not $sp^3$. This is incorrect.$3$. For $[Fe(NH_3)_6]^{2+}$,$Fe^{2+}$ is a $3d^6$ ion. $NH_3$ is a strong field ligand,but for $Fe^{2+}$,it is not strong enough to force pairing in the $3d$ orbital to form $d^2sp^3$. It forms an outer orbital complex $sp^3d^2$. This is incorrect.$4$. For $[Co(H_2O)_6]^{3+}$,$Co^{3+}$ is a $3d^6$ ion. $H_2O$ is a weak field ligand,so it forms an outer orbital complex $sp^3d^2$. This is correct.Note: Both $A$ and $D$ are technically correct based on standard coordination chemistry principles. However,$A$ is a classic example of $4d$ metal behavior.

Which of the following combinations is correct regarding the hybridization of the central metal ion?

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