Question

in $sp^{2}$ -hybridized orbitals, how many p - orbitals remain to form multiple bonds?
answer:

Explanation:

In \(sp^{2}\) hybridization, one \(s\) orbital and two \(p\) orbitals hybridize. Since there are three \(p\) orbitals (\(p_{x}\), \(p_{y}\), \(p_{z}\)) in total for a given energy level, if two \(p\) - orbitals are used in \(sp^{2}\) hybridization, then the number of un - hybridized \(p\) - orbitals is \(3 - 2=1\). This un - hybridized \(p\) - orbital is used to form \(\pi\) bonds (a type of multiple bond).

Answer:

One.