The shape of $[C u (N H_{3})_{4}] S O_{4}$ is:

square planar

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

pyramidal

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

octahedral

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

tetrahedral

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

Open in App

Solution

The correct option is A square planar
Since $- N H_{3}$ is a strong field ligand, it will form square planar complex with $C u^{2 +}$ with $d s p^{2}$ hybridization.

Suggest Corrections

Similar questions

IUPAC name for the complex $[C u (N H_{3})_{4}] S O_{4}$ is:

Q. The solution of the complex

[C u (N H_{3})_{4}] S O_{4}

in water will give

Q. Using IUPAC norms, write the systematic names of the following.
(i)

K_{4} [F e (C N)_{6}]

(ii)

[C u (N H_{3})_{4}] S O_{4}

The effective atomic number of central Cu (At. No. 29) metal in $[C u (N H_{3})_{4}] S O_{4}$ is

Write balanced equations for the following conversions
(a) $Z n S O_{4} A ⟶ Z n (O H)_{2} B ⟶$
(b) $C u S O_{4} A ⟶ C u (O H)_{2} B ⟶ [C u (N H_{3}) 4] S O_{4}$

Join BYJU'S Learning Program

The shape of [Cu(NH3)4]SO4 is:

The correct option is A square planarSince −NH3 is a strong field ligand, it will form square planar complex with Cu2+ with dsp2 hybridization.

The shape of $[C u (N H_{3})_{4}] S O_{4}$ is:

The correct option is A square planar
Since $- N H_{3}$ is a strong field ligand, it will form square planar complex with $C u^{2 +}$ with $d s p^{2}$ hybridization.