mu = sqrt{15} is true for the pair :- | vedclass

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Question

(A) 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 $\mu = \sqrt{15}$

Vedclass · Accepted Answer

$Co^{+2}, Cr^{+3}$

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(A) 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 $\mu = \sqrt{15}$,we have $n(n+2) = 15$,which implies $n^2 + 2n - 15 = 0$. Solving this quadratic equation,we get $(n+5)(n-3) = 0$,so $n = 3$.Thus,the pair must have $3$ unpaired electrons each.For $Co^{+2}$ $(Z=27)$: $[Ar] 3d^7$,which has $3$ unpaired electrons.For $Cr^{+3}$ $(Z=24)$: $[Ar] 3d^3$,which has $3$ unpaired electrons.Therefore,the pair $Co^{+2}, Cr^{+3}$ satisfies the condition.

List $I$ (Complex)	List $II$ (Type of isomerism)
$A$. $[Pt(NH_3)_2Cl_2]$	$I$. Optical
$B$. $[Co(en)_3]^{3+}$	$II$. Solvate
$C$. $[Co(NH_3)_5NO_2]Cl_2$	$III$. Geometrical
$D$. $[Cr(H_2O)_6]Cl_3$	$IV$. Linkage

$\mu = \sqrt{15}$ is true for the pair :-

Similar Questions

Which of the following statements is true?

The complex(es),which can exhibit the type of isomerism shown by $[Pt(NH_3)_2 Br_2]$,is(are) [en $= H_2 NCH_2 CH_2 NH_2$ ]
$(A)$ $[Pt(en)(SCN)_2]$
$(B)$ $[Zn(NH_3)_2 Cl_2]$
$(C)$ $[Pt(NH_3)_2 Cl_4]$
$(D)$ $[Cr(en)_2(H_2 O)(SO_4)]^{+}$

$A$ compound absorbs light in the wavelength region $490-500 \,nm$. Its complementary colour is

Match List $I$ with List $II$ :
List $I$ (Complex) List $II$ (Type of isomerism)
$A$. $[Pt(NH_3)_2Cl_2]$ $I$. Optical
$B$. $[Co(en)_3]^{3+}$ $II$. Solvate
$C$. $[Co(NH_3)_5NO_2]Cl_2$ $III$. Geometrical
$D$. $[Cr(H_2O)_6]Cl_3$ $IV$. Linkage

Choose the correct answer from the options given below :

The spin-only magnetic moment of $[ZCl_{4}]^{2-}$ is $3.87 \, BM,$ where $Z$ is $.....$

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