Question

The correct order of magnetic moments (spin only values in
B.M.) among is
(a) [Fe(CN)₆ ]⁴⁻ [MnCl₄]²⁻ [CoCl₄]²⁻
(b) [MnCl₄ ]²⁻ [Fe(CN)₆ ]⁴⁻ [CoCl₄]²⁻
(c) [MnCl₄ ]²⁻ [CoCl₄]²⁻ [Fe(CN)₆ ]⁴⁻
(d) [Fe(CN)₆ ]⁴⁻ [CoCl₄]²⁻ [MnCl₄]²⁻
(Atomic nos. : Mn = 25, Fe = 26, Co = 27)

Answer 1

Answer:

(c) [MnCl₄ ]²⁻ [CoCl₄]²⁻ [Fe(CN)₆ ]⁴⁻

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Answer 2

[MnCl4]2– > [CoCl4]2– > [Fe(CN)6]4– is the correct order.

Explanation:

• The magnetic moment is defined as the paramagnetism shown by a compound or atom.

• It is dependent on the number of unpaired electrons in the structure.

• In case of [MnCl4]2–, charge of manganese is Mn2+.

• So the electronic configuration shows that it has 5 unpaired electrons in its 3d orbital.

• In case of [CoCl4]2–, the charge of Cobalt is Co2+.

• So the electronic configuration shows that it has 3 unpaired electrons in its 3d orbital.

• Similarly in case of [Fe(CN)6]4–, charge of iron is Fe2+.

• So the electronic configuration shows that it has 0 unpaired electrons in its 3d orbital.

• As the magnetic moment is directly proportional to number of unpaired electrons, so the trend of magnetic moment among these complexes is

• [MnCl4]2– > [CoCl4]2– > [Fe(CN)6]4–

For more information about magnetic moment,

https://brainly.in/question/5952649

The ratio of spin only magnetic moments of Fe3+and Co2+ is

a)241/2:151/2

b)7:3

c)351/2:151/2

d)51/2:71/2

https://brainly.in/question/8798649

The ratio of magnetic moments of Fe(III) and Co(II) is

(a) 7 : 3

(b) 3 : 7

(c) \sqrt{7} : \sqrt{3}

(d) \sqrt{3} : \sqrt{7}