Question

f25
consider the structure below for the $eo_{2}f^{2 - }$ ion. note: e represents an unidentified element.

(a) what is the formal charge on e?
(b) which element could e represent? (although there is more than one possibility, you must enter only one answer. enter the correct atomic symbol for your choice.)

Explanation:

Step1: Calculate formal charge

The formula for formal charge is \(FC = V - N - \frac{B}{2}\), where \(V\) is the number of valence electrons in the free atom, \(N\) is the number of non - bonding electrons, and \(B\) is the number of bonding electrons.
For \(E\), assume \(E\) has \(V\) valence electrons. From the structure, \(N = 2\) (lone pair on \(E\)), and \(B= 6\) (3 bonds: one single bond to \(F\) and two double bonds to \(O\)).

$$FC=V - 2-\frac{6}{2}$$

Since the ion is \([EO_{2}F]^{2 -}\), and \(O\) has \(6\) valence electrons (\(2\) non - bonding pairs and \(2\) bonding electrons in a double bond, formal charge on each \(O\) in double bond: \(6-(4 + \frac{4}{2})=0\), \(F\) has \(7\) valence electrons (\(6\) non - bonding and \(2\) bonding in a single bond, formal charge on \(F\): \(7-(6+\frac{2}{2}) = 0\)). Let the formal charge on \(E\) be \(x\). The sum of formal charges in the ion \([EO_{2}F]^{2-}\) is \(- 2\). So \(x+0 + 0+0=-2\). But using the formula \(FC = V - N-\frac{B}{2}\), if we assume \(E\) is in group \(6\) (e.g., \(S\), \(V = 6\)):

$$FC=6 - 2-\frac{6}{2}=6 - 2 - 3=1$$

(wrong). If \(E\) is in group \(4\) (e.g., \(C\), \(V = 4\)):

$$FC=4 - 2-\frac{6}{2}=4 - 2 - 3=-1$$

(wrong). If \(E\) is in group \(3\) (e.g., \(B\), \(V = 3\)):

$$FC=3 - 2-\frac{6}{2}=3 - 2 - 3=-2$$

Step2: Identify element

For part (b), based on the formal charge calculation and valence electron requirements. Boron (\(B\)) has \(3\) valence electrons.

Answer:

(a) \(-2\)
(b) \(B\)