Question

The value of 'spin only' magnetic moment, follows the correct order:

• A. $\left[Fe\right(CN{)}_6{]}^{4-}>[CoC{l}_4{]}^{2-}>[MnC{l}_4{]}^{2-}$
• B. $[MnC{l}_4{]}^{2-}>[Fe(CN{)}_6{]}^{4-}>[CoC{l}_4{]}^{2-}$
• C. $\left[Fe\right(CN{)}_6{]}^{4-}>[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}$
• D. $[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}>\left[Fe\right(CN{)}_6{]}^{4-}$

Answer 1

The correct option is D $[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}>\left[Fe\right(CN{)}_6{]}^{4-}$

$\left[Fe\right(CN{)}_6{]}^{4-}$

$F{e}^{2+}⟶4s^{0}3d^6$

(Refer to Image 1)

Unpaired $e^{-}=0$

$\therefore M=0$

$[CoC{l}_4{]}^{2-}$

$C{o}^{2+}⟶4s^{0}3d^7$

(Refer to Image 2)

$3$ unpaired $e^{-}$

$\therefore M=\sqrt{3(3+2)}$

$=\sqrt{15}$

$=3.87$

$[MnC{l}_4{]}^{2-}$

$M{n}^{2+}⟶4s^{0}3d^5$

(Refer to Image 3)

No. of unpaired $e^{-}=5$

$\therefore M=\sqrt{5(5+2)}$

$=\sqrt{35}$

$=5.91$

$\therefore$ Order: $[MnC{l}_4{]}^{2-}>[CoC{l}_4{]}^{2-}>\left[Fe\right(CN{)}_6{]}^{4-}$

Answer is option D.