The correct order of the increasing magnetic | vedclass

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(A) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ using the formula $\mu = \sqrt{n(n+2)} \ BM$.$1$. $Ni(CN)_4^{

Vedclass · Accepted Answer

$Ni(CN)_4^{2-} < Cu(H_2O)_6^{2+} < NiCl_4^{2-} < Fe(H_2O)_6^{2+}$

Vedclass · Answer

(A) To determine the magnetic moment,we calculate the number of unpaired electrons $(n)$ using the formula $\mu = \sqrt{n(n+2)} \ BM$.$1$. $Ni(CN)_4^{2-}$: $Ni^{2+}$ is $d^8$. $CN^-$ is a strong field ligand,causing pairing. $n = 0$,$\mu = 0 \ BM$.$2$. $Cu(H_2O)_6^{2+}$: $Cu^{2+}$ is $d^9$. $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$3$. $NiCl_4^{2-}$: $Ni^{2+}$ is $d^8$. $Cl^-$ is a weak field ligand,no pairing. $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.$4$. $Fe(H_2O)_6^{2+}$: $Fe^{2+}$ is $d^6$. $H_2O$ is a weak field ligand. $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.The increasing order of magnetic moments is $Ni(CN)_4^{2-} < Cu(H_2O)_6^{2+} < NiCl_4^{2-} < Fe(H_2O)_6^{2+}$.

The correct order of the increasing magnetic moments for the following ions is $NiCl_4^{2-}$,$Fe(H_2O)_6^{2+}$,$Ni(CN)_4^{2-}$,and $Cu(H_2O)_6^{2+}$.

Similar Questions

An aqueous solution of $[Ni(H_2O)_6]^{2+}$ contains $Ni^{2+}$ ions and has a magnetic moment of $2.83 \ BM$. What is the effect on the magnetic moment of the solution when ammonia is added?

$[Fe(CN)_6]^{4-}$; $[Fe(CN)_6]^{3-}$; $[Ti(CN)_6]^{3-}$; $[Ni(CN)_4]^{2-}$; $[Co(CN)_6]^{3-}$
Among the given complexes,the number of paramagnetic complexes is .... .

Which one of the following complexes is not expected to exhibit isomerism?

The two complexes given below are $............$.

Number of complexes from the following with an even number of unpaired $d$ electrons is . . . . . . . . . .
$[V(H_2O)_6]^{3+}, \quad [Cr(H_2O)_6]^{2+}, \quad [Fe(H_2O)_6]^{3+}, \quad [Ni(H_2O)_6]^{3+}, \quad [Cu(H_2O)_6]^{2+}$
[Given atomic numbers: $V=23, Cr=24, Fe=26, Ni=28, Cu=29$]

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