The complex $[Ti(H_2O)_6]^{3+}$ is paramagnetic in nature because...
- AIt contains one unpaired electron.
- BIt contains two paired electrons.
- CIt contains three unpaired electrons.
- DIt contains no unpaired electrons.
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Similar Questions
Correct increasing order of spin magnetic moment $($in $B.M.)$ for the following complexes is:
$(I) \ [FeF_6]^{3-} \ (II) \ [V(H_2O)_6]^{2+} \ (III) \ [Fe(H_2O)_6]^{2+}$
$(I) \ [FeF_6]^{3-} \ (II) \ [V(H_2O)_6]^{2+} \ (III) \ [Fe(H_2O)_6]^{2+}$
Medium
View SolutionWhich one of the following complexes will have four isomers?
Difficult
View SolutionThe compound$(s)$ that exhibit$(s)$ geometrical isomerism is(are):
$A$. $[Pt(en)Cl_2]$
$B$. $[Pt(en)_2]Cl_2$
$C$. $[Pt(en)_2Cl_2]Cl_2$
$D$. $[Pt(NH_3)_2Cl_2]$
$A$. $[Pt(en)Cl_2]$
$B$. $[Pt(en)_2]Cl_2$
$C$. $[Pt(en)_2Cl_2]Cl_2$
$D$. $[Pt(NH_3)_2Cl_2]$
DifficultIIT 2009
View Solution$[PdCl_2(PMe_3)_2]$ is a diamagnetic complex of $Pd(II)$. How many total isomers are possible for the analogous paramagnetic complex of $Ni(II)$?
Match the column:
Column-$I$ Column-$II$ $(a) [Pt(NH_3)_3Cl]Br$ $(P) \text{Coordination isomerism}$ $(b) [Cr(NH_3)_5(NCS)]^{+2}$ $(Q) \text{Ionisation isomerism}$ $(c) [Co(H_2O)_6]Cl_3$ $(R) \text{Linkage isomerism}$ $(d) [PtCl_4][Pt(NH_3)_4]$ $(S) \text{Hydrate isomerism}$
| Column-$I$ | Column-$II$ |
|---|---|
| $(a) [Pt(NH_3)_3Cl]Br$ | $(P) \text{Coordination isomerism}$ |
| $(b) [Cr(NH_3)_5(NCS)]^{+2}$ | $(Q) \text{Ionisation isomerism}$ |
| $(c) [Co(H_2O)_6]Cl_3$ | $(R) \text{Linkage isomerism}$ |
| $(d) [PtCl_4][Pt(NH_3)_4]$ | $(S) \text{Hydrate isomerism}$ |
Medium
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