Question

using the
as\ idea, how many electrons are \available\ for the compound pf3?
n = 32
32 26 12

Explanation:

Step1: Recall the formula for available electrons (A)

In the NAS (N = A + S) idea, where N is the total number of electrons in the octet - like structure of the molecule, A is the number of available electrons (valence electrons of atoms), and S is the number of shared electrons. For \(PF_3\), we first find the valence electrons of each atom. Phosphorus (\(P\)) has 5 valence electrons (\(n = 5\)) and each fluorine (\(F\)) has 7 valence electrons (\(m = 7\)). The formula for \(A\) is \(A=n + 3m\).

Step2: Calculate the value of \(A\)

Substitute \(n = 5\) and \(m = 7\) into the formula \(A=n + 3m\). So, \(A=5+3\times7\). First, calculate \(3\times7 = 21\), then \(A=5 + 21\).

Answer:

\(26\)