Which of the following sets of atomic orbitals results in a square planar complex?
- A$s, p_x, p_y, p_z$
- B$s, p_x, p_y, d_{x^2-y^2}$
- C$s, p_x, p_y, d_{z^2}$
- D$s, p_x, p_z, d_{zy}$
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Which of the following statements are correct?
$i$. $[Mn(CN)_6]^{3-}, [MnCl_6]^{3-}$ both are paramagnetic
$ii$. Both $[NiCl_4]^{2-}, [Ni(CO)_4]$ are tetrahedral and diamagnetic
$iii$. $[Co(NH_3)_6]^{3+}$ is an inner orbital complex,but $[Ni(NH_3)_6]^{2+}$ is an outer orbital complex
$i$. $[Mn(CN)_6]^{3-}, [MnCl_6]^{3-}$ both are paramagnetic
$ii$. Both $[NiCl_4]^{2-}, [Ni(CO)_4]$ are tetrahedral and diamagnetic
$iii$. $[Co(NH_3)_6]^{3+}$ is an inner orbital complex,but $[Ni(NH_3)_6]^{2+}$ is an outer orbital complex
Which of the following has a square planar geometry?
(At. nos.: $Fe=26, Co=27, Ni=28, Pt=78$)
(At. nos.: $Fe=26, Co=27, Ni=28, Pt=78$)
Which of the following statements about $Fe(CO)_5$ is correct?
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View SolutionIn $[Pt(NH_3)_2Cl_2]$,the $Pt-Cl$ bond length is $2 \ \mathring{A}$ and the $Cl-Cl$ distance is $2.88 \ \mathring{A}$. What is the geometry of the compound?
$[Co(NH_3)_6]^{+2}$ is paramagnetic in nature due to the presence of unpaired $e^-$ in
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