If the spin-only magnetic moment of VCl_x is | vedclass

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(C) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 1.73 \

Vedclass · Accepted Answer

$VCl_4$

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(C) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 1.73 \ BM$,we have $\sqrt{n(n+2)} = 1.73$,which implies $n(n+2) = 3$,so $n = 1$.Vanadium $(V)$ has the atomic number $23$ and electronic configuration $[Ar] 3d^3 4s^2$.For $VCl_4$,the oxidation state of $V$ is $+4$,resulting in the configuration $[Ar] 3d^1$.Since there is $1$ unpaired electron $(n=1)$,the magnetic moment is $\sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.Therefore,the correct formula is $VCl_4$.

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

If the spin-only magnetic moment of $VCl_x$ is $1.73 \ BM$,then the correct formula is:

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

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