The metal $Ni$ $(Z = 28)$ combines with a unidentate ligand $X^-$ to form a paramagnetic complex $[NiX_4]^{2-}$. What are its geometry and the number of unpaired electrons in $Ni$?
- A$2$; Square planar
- B$1$; Tetrahedral
- C$2$; Tetrahedral
- D$1$; Square planar
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Similar Questions
What type of hybridization is present in $[Co(NH_3)_6]^{3+}$ complex?
In $[NiCl_4]^{2-}$,the number of unpaired electrons is
Medium
View SolutionMatch List-$I$ with List-$II$:
List-$I$ List-$II$ $A. K_2[Ni(CN)_4]$ $I. sp^3$ $B. [Ni(CO)_4]$ $II. sp^3d^2$ $C. [Co(NH_3)_6]Cl_3$ $III. dsp^2$ $D. Na_3[CoF_6]$ $IV. d^2sp^3$
Choose the correct answer from the options given below:
| List-$I$ | List-$II$ |
| $A. K_2[Ni(CN)_4]$ | $I. sp^3$ |
| $B. [Ni(CO)_4]$ | $II. sp^3d^2$ |
| $C. [Co(NH_3)_6]Cl_3$ | $III. dsp^2$ |
| $D. Na_3[CoF_6]$ | $IV. d^2sp^3$ |
Choose the correct answer from the options given below:
Match the following:
List-$I$ (Hybridisation) List-$II$ (Compound/ion) $A. sp^3d$ $I. [PtCl_4]^{2-}$ $B. sp^3d^2$ $II. SF_6$ $C. dsp^2$ $III. BCl_3$ $D. dsp^3$ $IV. PCl_5$ $V. ClF_3$
The correct match is:
| List-$I$ (Hybridisation) | List-$II$ (Compound/ion) |
|---|---|
| $A. sp^3d$ | $I. [PtCl_4]^{2-}$ |
| $B. sp^3d^2$ | $II. SF_6$ |
| $C. dsp^2$ | $III. BCl_3$ |
| $D. dsp^3$ | $IV. PCl_5$ |
| $V. ClF_3$ |
The correct match is:
If $Hund's$ rule is violated,then select the $CORRECT$ statement regarding $[Ni(NH_3)_6]^{2+}$:
Difficult
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