Dimethyl glyoxime forms a square planar complex with $Ni^{2+}$. This complex should be

  • A
    diamagnetic
  • B
    paramagnetic having $1$ unpaired electron
  • C
    paramagnetic having $2$ unpaired electrons
  • D
    ferromagnetic

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Similar Questions

Both $[Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$ are diamagnetic. The hybridization of nickel in these complexes,respectively,are

Match the following hybridization in Column $I$ with the corresponding coordination complexes in Column $II$.
$A. sp^3$$(i). [Co(NH_3)_6]^{3+}$
$B. dsp^2$$(ii). [Ni(CO)_4]$
$C. sp^3d^2$$(iii). [Pt(NH_3)_2Cl_2]$
$D. d^2sp^3$$(iv). [CoF_6]^{3-}$
$(v). [Fe(CO)_5]$

The geometries of $Ni(CO)_4$ and $Ni(PPh_3)_2Cl_2$ are:

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Give the correct sequence of initials $T$ or $F$ for the following statements. Use $T$ if the statement is true and $F$ if it is false.
$(I) \ Co(III)$ is stabilized in the presence of weak field ligands,while $Co(II)$ is stabilized in the presence of strong field ligands.
$(II)$ Four-coordinated complexes of $Pd(II)$ and $Pt(II)$ are diamagnetic and square planar.
$(III) \ [Ni(CN)_4]^{4-}$ ion and $[Ni(CO)_4]$ are diamagnetic tetrahedral and square planar respectively.
$(IV) \ Ni^{2+}$ ion does not form inner orbital octahedral complexes.

Which of the following statements is incorrect regarding $[Ni(dmg)_2]$?

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