When $C u S O_{4}$ is treated with excess KCN solution, what compound(s) is/are formed?

K [C u (C N)_{2}]

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

K_{3} [C u (C N)_{4}]

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

K_{2} [C u (C N)_{4}]

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

N \equiv C - C \equiv N

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

Open in App

Solution

The correct option is D $N \equiv C - C \equiv N$
$(C u S O_{4} + 2 K C N ⟶ C u (C N)_{2} + K_{2} S O_{4}) \times 2$
$2 C u (C N)_{2} ⟶ C u_{2} (C N)_{2} + (C N)_{2}$
$C u_{2} (C N)_{2} + 6 K C N ⟶ 2 K_{3} [C u (C N)_{4}]$
So,adding these reactions we get,
$2 C u S O_{4} + 10 K C N ⟶ 2 K_{3} [C u (C N)_{4}] + 2 K_{2} S O_{4} + (C N)_{2}$

Suggest Corrections

Similar questions

C u S O_{4}

K C N

Blue solution of $C u S O_{4}$ when reacts with excess of $K C N$ gives colourless solution due to the :

K C N

{M C l}_{2}

K C N

H_{2} S

M S

M

Join BYJU'S Learning Program