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(C) The magnetic moment (spin only) is given by $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.$1. [CoCl_4]^{2-}: Co^{

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

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(C) The magnetic moment (spin only) is given by $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.$1. [CoCl_4]^{2-}: Co^{2+} (d^7)$ is a tetrahedral complex. The configuration is $e^4 t_2^3$,so $n = 3$.$2. [Fe(H_2O)_6]^{2+}: Fe^{2+} (d^6)$ is an octahedral complex. The configuration is $t_{2g}^4 e_g^2$,so $n = 4$.$3. [Mn(H_2O)_6]^{2+}: Mn^{2+} (d^5)$ is an octahedral complex. The configuration is $t_{2g}^3 e_g^2$,so $n = 5$.$4. [Cr(H_2O)_6]^{2+}: Cr^{2+} (d^4)$ is an octahedral complex. The configuration is $t_{2g}^3 e_g^1$,so $n = 4$.Comparing the number of unpaired electrons,both $[Fe(H_2O)_6]^{2+}$ and $[Cr(H_2O)_6]^{2+}$ have $n = 4$,thus they have the same magnetic moment.

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

Similar Questions

$A$ complex of a certain metal has a magnetic moment of $4.91 \ BM$,whereas another complex of the same metal with the same oxidation state has zero magnetic moment. The metal ion could be:

The complex compound $[Co(SCN)_2(NH_3)_4]Cl$ does not exhibit which of the following types of isomerism?

Given below are two statements:
Statement $I$: $A$ solution of $[Ni(H_2O)_6]^{2+}$ is green in colour.
Statement $II$: $A$ solution of $[Ni(CN)_4]^{2-}$ is colourless.
In the light of the above statements,choose the most appropriate answer from the options given below:

The magnetic moment (spin only) of $[NiCl_{4}]^{2-}$ is ....... $B.M.$

If $n$ is the number of unpaired electrons,then the magnetic moment $(B.M.)$ of a transition metal/ion is given by:

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