When an aqueous solution of copper (II) sulphate is saturated with ammonia, the blue compound crystallines on evaporation. The formula of this blue compound is:

[C u {({N H}_{3})}_{4}] {S O}_{4} \cdot H_{2} O

(square planar)

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[C u {({N H}_{3})}_{4}] {S O}_{4}

(Tetrahedral)

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[C u {({N H}_{3})}_{6}] {S O}_{4}

(Octahedral)

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[C u {({S O}_{4}) ({N H}_{3})}_{5}]

(Octahedral)

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Solution

The correct option is A $[C u {({N H}_{3})}_{4}] {S O}_{4} \cdot H_{2} O$ (square planar)
The aqueous solution of copper(II) sulphate reacts with ammonia to form blue colored crystalline complex $[C u {({N H}_{3})}_{4}] {S O}_{4} \cdot H_{2} O$

$C u S O_{4} + 4 N H_{3} + H_{2} O \to [C u {({N H}_{3})}_{4}] {S O}_{4} \cdot H_{2} O$

The $C u^{2 +}$ ion undergoes $d s p^{2}$ hybridization which results in square planar geometry.

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\begin{matrix} Column - I & Column - II i . & [N i (C O)_{4}] & Planar octahedral, diamagnetic i i . & [F e (N O) (H_{2} O)_{5}]^{2 +} & Planar, paramagnetic i i i . & [C o (N O_{2})_{6}]^{3 -} & Octahedral, diamagnetic i v . & [C u (N H_{3})_{4}]^{2 +} & Planar octahedral, paramagnetic Octahedral, paramagnetic \end{matrix}

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