The correct order of the calculated spin-only | vedclass

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(A) $1$. $Ni(CO)_{4}$: $Ni$ is in $0$ oxidation state $(3d^{8} 4s^{2})$. $CO$ is a strong field ligand,causing pairing of electrons. All electrons are

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$A \approx C \approx D < B$

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(A) $1$. $Ni(CO)_{4}$: $Ni$ is in $0$ oxidation state $(3d^{8} 4s^{2})$. $CO$ is a strong field ligand,causing pairing of electrons. All electrons are paired,so $\mu_{m} = 0 \ B.M.$$2$. $[Ni(H_{2}O)_{6}]Cl_{2}$: $Ni$ is in $+2$ oxidation state $(3d^{8})$. $H_{2}O$ is a weak field ligand. It has $2$ unpaired electrons. $\mu_{m} = \sqrt{n(n+2)} = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \ B.M.$$3$. $Na_{2}[Ni(CN)_{4}]$: $Ni$ is in $+2$ oxidation state $(3d^{8})$. $CN^{-}$ is a strong field ligand,causing pairing of electrons. All electrons are paired,so $\mu_{m} = 0 \ B.M.$$4$. $PdCl_{2}(PPh_{3})_{2}$: $Pd^{2+}$ is a $4d^{8}$ ion. Due to the large size of $4d$ orbitals,the crystal field splitting energy is high,leading to pairing of electrons. Thus,$\mu_{m} = 0 \ B.M.$Comparing the values: $A (0) \approx C (0) \approx D (0) Therefore,the correct order is $A \approx C \approx D < B$.

The correct order of the calculated spin-only magnetic moments of complexes $A$ to $D$ is:
$A$. $Ni(CO)_{4}$
$B$. $[Ni(H_{2}O)_{6}]Cl_{2}$
$C$. $Na_{2}[Ni(CN)_{4}]$
$D$. $PdCl_{2}(PPh_{3})_{2}$

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The correct order of the calculated spin-only magnetic moments of complexes $A$ to $D$ is:$A$. $Ni(CO)_{4}$$B$. $[Ni(H_{2}O)_{6}]Cl_{2}$$C$. $Na_{2}[Ni(CN)_{4}]$$D$. $PdCl_{2}(PPh_{3})_{2}$