Question

Calculate the spin only magnetic moment for k4[Mn(NCS)6]

Answer 1

Answer:

2.2 is the spin only magnetic moment for K4[Mn(NCS)6]

Answer 2

To find:

The spin only magnetic moment for k4[Mn(NCS)6]

Calculation:

General formula for spin only- magnetic moment in transition metals is:

$\boxed{ \sf{ \mu = \sqrt{n(n + 2)} \: BM}}$

NCS- is a strong field lig-and and will try to pair up the valence orbital electrons in the transition metal (Mn in this case)

So, 3d orbital of Manganese in +2 state in presence of NCS:

$\bf{ {Mn}^{ + 2} \implies \boxed{ \uparrow \downarrow | \uparrow \downarrow | \uparrow | - | - }}$

So, number of unpaired electrons = 1.

$\therefore \: \sf{ \mu = \sqrt{n(n + 2)} \: BM}$

$= > \: \sf{ \mu = \sqrt{1(1+ 2)} \: BM}$

$= > \: \sf{ \mu = \sqrt{3} \: BM}$

$= > \: \sf{ \mu = 1.73 \: BM}$

So, final answer is:

$\boxed{\: \bf{ \mu = 1.73 \: BM}}$