Question

The coordination number of $N{i}^{2+}$ is $4$.
$NiC{l}_2+KCN\left(excess\right)\to A\left(cyanocomplex\right)$
$NiC{l}_2+conc.HCl\left(excess\right)\to B\left(Chlorocomplex\right)$

The IUPAC name of A and B are

• A. Potassium tetracyanonickelate (II),
  Potassium tetrachloronickelate (II).
• B. Tetracyanopotassiumnickelate (II),
  Tetrachloropotassiumnickelate (II).
• C. Tetracyanonickel (II), Tetrachloronickel (II)
• D. Potassium tetracyanonickel (II), Tetracholoronickel (II).

Answer 1

The correct option is A Potassium tetracyanonickelate (II),

Potassium tetrachloronickelate (II).
Here are the reactions and the products:

$NiC{l}_2+4KCN\to K_2\left[Ni\right(CN{)}_4]+2KCl$
A is $K_2\left[Ni\right(CN{)}_4]$
This is clearly going to be potassium tetracyanonickelate(II).

$K_2\left[Ni\right(CN{)}_4]+4HCl(conc.)\to ,K_2[Ni\left(CN{)}_4\right]+4HCN$
$Bis{K}_2\left[NiC{l}_4\right]$
We can name this potassium tetrachloridonickelate(II).