Square planar complex among the following is

[N i C l_{4}]^{2 -}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[P d C l_{4}]^{2 -}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

[N i B r_{4}]^{2 -}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of the above

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $[P d C l_{4}]^{2 -}$
Both $C l^{-}$ and $B r^{-}$ are weak field ligands . However as we move from first transition series to second and third, $Δ_{0}$ increases accordingly and as a result it gives rise to square planar geometry. Thus here $[N i C l_{4}]^{2 -}$ and $[N i B r_{4}]^{2 -}$ have tetrahedral geometry but $[P d C l_{4}]^{2 -}$ has square planar shape.

Suggest Corrections

Similar questions

X e F_{4}

S F_{4}

[N i C l_{4}]^{2 -}

[P d C l_{4}]^{2 -}

[P t (N H_{3})_{4}]^{2 +} [C o (S C N)_{4}]^{2 -} [C u (e n)_{2}]^{2 +} [H g I_{4}]^{2 -}

[N i (C O)_{4}]

[N i C l_{4}]^{2 -}

Which of the following complexes is/are square planar in shape ?
(I) [Ni(CN)4]2− (II) [MnCl4]2− (III) [ZnCl4]2− (IV) [Ni(NH3)4]2+

N i^{2 +}, P t^{2 +}

Z n^{2 +},

Join BYJU'S Learning Program