JEE MAIN - Chemistry (2024 - 27th January Morning Shift - No. 8)
Consider the following complex ions
$$\begin{aligned} & \mathrm{P}=\left[\mathrm{FeF}_6\right]^{3-} \\ & \mathrm{Q}=\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \\ & \mathrm{R}=\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \end{aligned}$$
The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :
Explanation
$$\left[\mathrm{FeF}_6\right]^{3-}: \mathrm{Fe}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^5$$
F : Weak field Ligand
_27th_January_Morning_Shift_en_8_1.png)
No. of unpaired electron's $$=5$$
$$\begin{aligned} & \mu=\sqrt{5(5+2)} \\ & \mu=\sqrt{35} \mathrm{~BM} \\ & {\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{+2}: \mathrm{V}^{+2}: 3 \mathrm{~d}^3} \end{aligned}$$
_27th_January_Morning_Shift_en_8_2.png)
No. of unpaired electron's $$=3$$
$$\begin{aligned} & \mu=\sqrt{3(3+2)} \\ & \mu=\sqrt{15} \mathrm{~BM} \\ & {\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{+2}: \mathrm{Fe}^{+2}: 3 \mathrm{~d}^6} \end{aligned}$$
H$$_2$$O : Weak field Ligand
_27th_January_Morning_Shift_en_8_3.png)
No. of unpaired electron's $$=4$$
$$\begin{aligned} & \mu=\sqrt{4(4+2)} \\ & \mu=\sqrt{24} \mathrm{~BM} \end{aligned}$$
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