Although Cr^{3+} and Co^{2+} ions have the sa | vedclass

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(B) Both $Cr^{3+}$ $(3d^3)$ and $Co^{2+}$ $(3d^7)$ have $3$ unpaired electrons.According to the spin-only formula,$\mu = \sqrt{n(n+2)} \ B.M.$,both sh

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Due to orbital contribution in $Co^{2+}$ ion.

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(B) Both $Cr^{3+}$ $(3d^3)$ and $Co^{2+}$ $(3d^7)$ have $3$ unpaired electrons.According to the spin-only formula,$\mu = \sqrt{n(n+2)} \ B.M.$,both should have a magnetic moment of $\sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ B.M.$However,$Cr^{3+}$ has a symmetrical $t_{2g}^3$ configuration,resulting in no orbital contribution to the magnetic moment.In contrast,$Co^{2+}$ has an unsymmetrical configuration,which leads to a significant orbital contribution,increasing the observed magnetic moment to approximately $4.87 \ B.M.$

Column-$I$	Column-$II$
$(a) [Pt(NH_3)_3Cl]Br$	$(P) \text{Coordination isomerism}$
$(b) [Cr(NH_3)_5(NCS)]^{+2}$	$(Q) \text{Ionisation isomerism}$
$(c) [Co(H_2O)_6]Cl_3$	$(R) \text{Linkage isomerism}$
$(d) [PtCl_4][Pt(NH_3)_4]$	$(S) \text{Hydrate isomerism}$

Although $Cr^{3+}$ and $Co^{2+}$ ions have the same number of unpaired electrons,why is the magnetic moment of $Cr^{3+}$ $3.87 \ B.M.$ and that of $Co^{2+}$ $4.87 \ B.M.?$

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Total number of geometrical isomers possible for the complexes $[NiCl_4]^{2-}$,$[CoCl_2(NH_3)_4]^{+}$,$[Co(NH_3)_3(NO_2)_3]$ and $[Co(NH_3)_5Cl]^{2+}$ is

The correct order of magnetic moments (spin values in $BM$) among the following is:

Match the column:
Column-$I$ Column-$II$
$(a) [Pt(NH_3)_3Cl]Br$ $(P) \text{Coordination isomerism}$
$(b) [Cr(NH_3)_5(NCS)]^{+2}$ $(Q) \text{Ionisation isomerism}$
$(c) [Co(H_2O)_6]Cl_3$ $(R) \text{Linkage isomerism}$
$(d) [PtCl_4][Pt(NH_3)_4]$ $(S) \text{Hydrate isomerism}$

The magnitude of magnetic moment (spin only) of $[NiCl_4]^{2-}$ will be ........ $B.M.$

Which of the following is diamagnetic?

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