નીચે આપેલા સંકીર્ણોમાંથી ક્યું એક સૌથી ઓછી અનુચુંબકીય વર્તણુંક પ્રદશિત કરશે? [આપેલ : પ૨માણુક્રમાંક $\mathrm{Cr}=24, \mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27$ ]

Question

A$\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
B$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
C$\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
D$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

JEE MAIN 2024, Medium

Download our app for free and get started

Solution

a

	Number of unpaired $\mathrm{e}^{-}$	$\mu=\sqrt{n(n+2)}$ $B.M.$
$\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$	$3$	$3.87$
$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$	$4$	$4.89$
$\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$	$5$	$5.92$
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$	$4$	$4.89$

Least paramagnetic behaviour $=\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

	Number of unpaired \(\mathrm{e}^{-}\)	\(\mu=\sqrt{n(n+2)}\) \(B.M.\)
\(\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)	\(3\)	\(3.87\)
\(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)	\(4\)	\(4.89\)
\(\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)	\(5\)	\(5.92\)
\(\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)	\(4\)	\(4.89\)

Download our app
and get started for free

Similar Questions

Download our appand get started for free

Similar Questions

Download our app
and get started for free