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(B) The magnetic moment is given as $0 \, B.M.$,which implies that there are no unpaired electrons in the titanium ion. For an octahedral complex,the

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$[Ti(H_2O)_6]Cl_4$

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(B) The magnetic moment is given as $0 \, B.M.$,which implies that there are no unpaired electrons in the titanium ion. For an octahedral complex,the coordination number is $6$. In the complex $[Ti(H_2O)_6]Cl_4$,the oxidation state of $Ti$ is calculated as $x + 6(0) = +4$,so $Ti$ is in the $+4$ oxidation state. The electronic configuration of $Ti$ $(Z=22)$ is $[Ar] 3d^2 4s^2$. For $Ti^{4+}$,the configuration is $[Ar] 3d^0$,which has $0$ unpaired electrons. Therefore,the magnetic moment is $0 \, B.M.$

An aqueous solution of titanium chloride,when subjected to magnetic measurement,measured zero magnetic moment. Assuming the octahedral complex in aqueous solution,the formula of the complex is

Similar Questions

Which of the following will have two stereoisomeric forms?
$I. [Cr(NO_3)_3(NH_3)_3]$
$II. K_3[Fe(C_2O_4)_3]$
$III. [CoCl_2(en)_2]^+$
$IV. [CoBrCl(ox)_2]^{3-}$

Among the following complex ions,the one which shows geometrical isomerism will be

$A$ compound of a metal ion $M^{x+}$ $(Z = 24)$ has a spin-only magnetic moment of $\sqrt{15} \ BM$. The number of unpaired electrons in the compound is:

Among the following complexes,the total number of diamagnetic species is. . . . $[Mn(NH_3)_6]^{3+}, [MnCl_6]^{3-}, [FeF_6]^{3-}, [CoF_6]^{3-}, [Fe(NH_3)_6]^{3+}$ and $[Co(en)_3]^{3+}$
[Given,atomic number: $Mn=25, Fe=26, Co=27$; $en=H_2NCH_2CH_2NH_2$ ]

The correct order of the number of unpaired electrons in the given complexes is
$A$. $[Fe(CN)_6]^{3-}$; $B$. $[FeF_6]^{3-}$; $C$. $[CoF_6]^{3-}$; $D$. $[Cr(oxalate)_3]^{3-}$; $E$. $[Ni(CO)_4]$
Choose the correct answer from the options given below:

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