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Question

(C) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.For $\mu = 3.87 \ B.M

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$V^{2+}$

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(C) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.For $\mu = 3.87 \ B.M.$,we have $\sqrt{n(n+2)} = 3.87$,which implies $n(n+2) \approx 15$,so $n = 3$.$Cr^{2+}: [Ar] 3d^4, n=4, \mu = \sqrt{4(6)} = 4.89 \ B.M.$$Mn^{2+}: [Ar] 3d^5, n=5, \mu = \sqrt{5(7)} = 5.91 \ B.M.$$V^{2+}: [Ar] 3d^3, n=3, \mu = \sqrt{3(5)} = 3.87 \ B.M.$$Ti^{2+}: [Ar] 3d^2, n=2, \mu = \sqrt{2(4)} = 2.82 \ B.M.$Thus,the metal ion is $V^{2+}$.

List-$I$	List-$II$
$(1)$ $Ti^+$	$(A)$ $(n-1)d^4ns^2$
$(2)$ $_{30}Zn$	$(B)$ $(n-1)d^{10}ns^1$
$(3)$ $_{24}Cr$	$(C)$ $(n-1)d^2ns^0$
$(4)$ $_{29}Cu$	$(D)$ $(n-1)d^5ns^1$
	$(E)$ $(n-1)d^0ns^2$
	$(F)$ $(n-1)d^{10}ns^2$

The magnetic moment of a transition metal compound has been calculated to be $3.87 \ B.M.$ The metal ion is

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The electronic configuration of the last element of the volatile metal group is:

Match the elements in List-$I$ with their electronic configurations in List-$II$.

List-$I$ List-$II$

$(1)$ $Ti^+$ $(A)$ $(n-1)d^4ns^2$

$(2)$ $_{30}Zn$ $(B)$ $(n-1)d^{10}ns^1$

$(3)$ $_{24}Cr$ $(C)$ $(n-1)d^2ns^0$

$(4)$ $_{29}Cu$ $(D)$ $(n-1)d^5ns^1$

$(E)$ $(n-1)d^0ns^2$

$(F)$ $(n-1)d^{10}ns^2$

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The number of unpaired electrons in $Zn^{2+}$ is

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