The hybridization of compound $\mathbf{\text{'}}\mathbf{V}\mathbf{\text{'}}$ is:

When boron reacts with air, two products formed. One product formed when oxygen gas reacts with boron and another product formed when nitrogen gas reacts with boron. As quantity of nitrogen is high in air, this implies that the product formed when nitrogen reacts with boron is the major product.

$\begin{array}{ccc}4\mathrm{B}+3{\mathrm{O}}_2& \to & \underset{\mathrm{Minor} \mathrm{product}}{2{\mathrm{B}}_2{\mathrm{O}}_3}\\ 2\mathrm{B}+{\mathrm{N}}_2& \to & \underset{\mathrm{Major} \mathrm{product} \left(\mathrm{X}\right)}{2\mathrm{BN}}\end{array}$

BN reacts with water to form H₃BO₃ and ammonia gas.

$\mathrm{BN}+3{\mathrm{H}}_2\mathrm{O}\to \underset{\mathrm{Y}}{{\mathrm{H}}_3{\mathrm{BO}}_3}+\underset{\mathrm{Z}}{{\mathrm{NH}}_3\uparrow }$

Reaction of NH₃ with CuSO₄ forms a blue color compound, [Cu(NH₃)₄]SO₄.

$\underset{\mathrm{Y}}{{\mathrm{NH}}_3}+{\mathrm{CuSO}}_4\to \underset{\mathrm{V} (\mathrm{blue} \mathrm{color} \mathrm{product})}{\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathrm{SO}}_4}$

[Cu(NH₃)₄]SO₄ exists as $[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4{]}^{2+}$ and ${\mathrm{SO}}_4^{2-}.$ Ammonia is the strong field ligand, results in the pairing of Cu electrons. Hence, hybridization of Cu in $[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4{]}^{2+}$ is dsp².