Question

The spin only magnetic moments of [Mn(CN)6]4 - & [MnBr4]2 - in bhor magnetons, respectively,are (1.73 and 5.92) . Sir/Madam can you please explain how this result was obtained.

Answer 1

Hey bae,

● Explaination-
(a) [Mn(CN)6]4-
Unpaired electrons are less as CN is strong ligand.
Unpaired eletrons = 1

Spin magnetic moment = √(n)(n+2)
Spin magnetic moment = √(1×3)
Spin magnetic moment = √(3)
Spin magnetic moment = 1.73 B

(a) [MnBr4]2-
Unpaired electrons are more as Br is weak ligand.
Unpaired eletrons = 5

Spin magnetic moment = √(n)(n+2)
Spin magnetic moment = √(5×7)
Spin magnetic moment = √(35)
Spin magnetic moment = 5.92 B

Hope this is useful...

Answer 2

Answer: The results are proved below.

Explanation:

To calculate the spin only magnetic moment, we use the equation given as:

$\sqrt{n(n+2)}$

where, n = number of unpaired electrons of metal ion

• For complex $[Mn(CN)_6]^{4-}$

Electronic configuration of manganese element = $s^2d^5$

Electronic configuration of $Mn^{2+}$ ion = $d^5$

Metal ion is present in +2 oxidation state. The ligand (cyanide ion) attached to the metal ion has a strong field and makes the pairing of electrons of metal ion. Thus, the 5 unpaired electrons of $Mn^{2+}$ ion gets paired up and only 1 electron remains unpaired.

Putting this value in equation 1, we get:

$(n=1):\sqrt{1(1+2)}=\sqrt{3}=1.73BM$

• For complex $[Mn(Br)_4]^{2-}$

Electronic configuration of manganese element = $s^2d^5$

Electronic configuration of $Mn^{2+}$ ion = $d^5$

Metal ion is present in + 2 oxidation state. The ligand (bromide ion) attached to the metal ion has a weak field and does not pair up the electrons of metal ion. Thus, the 5 unpaired electrons of $Mn^{2+}$ ion remains as such.

Putting this value in equation 1, we get:

$(n=5):\sqrt{5(5+2)}=\sqrt{35}=5.92BM$

Hence, the results are proved above.