The pair in which both the species have

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Target Exam

CUET

Subject

Chemistry

Chapter

Inorganic: Coordination Compounds

Question:

The pair in which both the species have the same magnetic moment (spin only value) is:

Options:

$[Cr(H_2O)_6]^{2+}, [CoCl_4]^{2-}$

$[Cr(H_2O)_6]^{2+}, [Fe(H_2O)_6]^{2+}$

$[Mn(H_2O)_6]^{2+}, [Cr(H_2O)_6]^{2+}$

$[CoCl_4]^{2-}, [Fe(H_2O)_6]^{2+}$

Correct Answer:

$[Cr(H_2O)_6]^{2+}, [Fe(H_2O)_6]^{2+}$

Explanation:

The correct answer is option (2) $[Cr(H_2O)_6]^{2+}, [Fe(H_2O)_6]^{2+}$ .

The magnetic moment ( $\mu$ ) of a species is related to the number of unpaired electrons. The formula for the magnetic moment in terms of the number of unpaired electrons ( $n$ ) is given by:

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

where

$n$ is the number of unpaired electrons.

Now, let's analyze each species in the given pairs:

1.

$[Cr(H_2O)_6]^{2+}, [CoCl_4]^{2-}$ :
-

$Cr^{2+}$ has

$4$ unpaired electrons (

$n = 4$ ).
-

$Co^{2-}$ has

$4$ unpaired electrons (

$n = 4$ ).

2.

$[Cr(H_2O)_6]^{2+}, [Fe(H_2O)_6]^{2+}$ :
-

$Cr^{2+}$ has

$4$ unpaired electrons (

$n = 4$ ).
-

$Fe^{2+}$ has

$4$ unpaired electrons (

$n = 4$ ).

3.

$[Mn(H_2O)_6]^{2+}, [Cr(H_2O)_6]^{2+}$ :
-

$Mn^{2+}$ has

$3$ unpaired electrons (

$n = 3$ ).
-

$Cr^{2+}$ has

$4$ unpaired electrons (

$n = 4$ ).

4.

$[CoCl_4]^{2-}, [Fe(H_2O)_6]^{2+}$ :
-

$Co^{2-}$ has

$7$ unpaired electrons (

$n = 7$ ).
-

$Fe^{2+}$ has

$4$ unpaired electrons (

$n = 4$ ).

Now, let's compare the pairs to see which one has the same magnetic moment:

The pair in which both species have the same magnetic moment is: 2.

$[Cr(H_2O)_6]^{2+}, [Fe(H_2O)_6]^{2+}$

Both

$Cr^{2+}$ and

$Fe^{2+}$ have

$4$ unpaired electrons (

$n = 4$ ), so they will have the same magnetic moment according to the given formula.

Therefore, the correct answer is option 2.

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