Magnetic moment (spin only) of an octahedral | vedclass

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(C) For an octahedral complex,$CFSE = (-0.4n_{t_{2g}} + 0.6n_{e_g}) \Delta_0$.Given $CFSE = -0.8 \Delta_0$,we test configurations for $d^n$ ions with

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Both $(A)$ and $(B)$

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(C) For an octahedral complex,$CFSE = (-0.4n_{t_{2g}} + 0.6n_{e_g}) \Delta_0$.Given $CFSE = -0.8 \Delta_0$,we test configurations for $d^n$ ions with weak field ligands.For a $d^7$ ion $(t_{2g}^5 e_g^2)$: $CFSE = (-0.4 	imes 5 + 0.6 	imes 2) \Delta_0 = (-2.0 + 1.2) \Delta_0 = -0.8 \Delta_0$. This has $n = 3$ unpaired electrons,so $\mu = \sqrt{3(3+2)} = \sqrt{15} \, BM$.For a $d^2$ ion $(t_{2g}^2 e_g^0)$: $CFSE = (-0.4 	imes 2 + 0.6 	imes 0) \Delta_0 = -0.8 \Delta_0$. This has $n = 2$ unpaired electrons,so $\mu = \sqrt{2(2+2)} = \sqrt{8} \, BM$.Since both configurations are possible with weak field ligands,both $(A)$ and $(B)$ are correct.

List-$I$	List-$II$
$A$. Ziegler-Natta	$i$. $Fe_4[Fe(CN)_6]_3$
$B$. Brown ring complex	$ii$. $[Fe(H_2O)_5NO]SO_4$
$C$. Prussian Blue	$iii$. $Al(C_2H_5)_3 + TiCl_4$
$D$. Turnbull Blue	$iv$. $Fe_3[Fe(CN)_6]_2$

Magnetic moment (spin only) of an octahedral complex having $CFSE = -0.8 \Delta_0$ and surrounded by weak field ligands can be

Similar Questions

Identify the correct set from the following table:

Match List-$I$ with List-$II$ and select the correct code:

List-$I$ List-$II$

$A$. Ziegler-Natta $i$. $Fe_4[Fe(CN)_6]_3$

$B$. Brown ring complex $ii$. $[Fe(H_2O)_5NO]SO_4$

$C$. Prussian Blue $iii$. $Al(C_2H_5)_3 + TiCl_4$

$D$. Turnbull Blue $iv$. $Fe_3[Fe(CN)_6]_2$

Correct code is:

The sum of the total number of $Cr-O$ bonds in chromate $(CrO_4^{2-})$ and dichromate $(Cr_2O_7^{2-})$ ions is:

Which will give $Fe^{3+}$ ions in solution?

Which of the following has a yellow colour?

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