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 of the atom in isolation, \(N\) is the number of non - bonding electrons, and \(B\) is the number of bonding electrons.
Assume \(E\) has \(0\) non - bonding electrons (\(N = 0\)). \(E\) forms \(3\) bonds: one single bond (\(F - E\)) and two double bonds (\(E=O\)). The total number of bonding electrons \(B=(1\times2)+(2\times4)=10\). If we assume \(E\) is in group \(16\) (e.g., \(S\), \(Se\), \(Te\)), \(V = 6\).
\[FC=V - N-\frac{B}{2}=6 - 0-\frac{10}{2}=6 - 5=+ 1\]
But considering the overall charge of the ion \([EO_{2}F]^{2-}\). Let's re - calculate.
The sum of formal charges in the ion \([EO_{2}F]^{2-}\) is \(-2\). Let the formal charge on \(E\) be \(x\), on \(F\) (single bond, \(V = 7\), \(N = 6\), \(B = 2\)): \(FC_F=7 - 6-\frac{2}{2}=0\). On each \(O\) (double bond, \(V = 6\), \(N = 4\), \(B = 4\)): \(FC_O=6 - 4-\frac{4}{2}=0\).
Since \(x+0 + 0+0=-2\) (sum of formal charges in the ion), we use another approach.
The formula for formal charge \(FC = V-(L+\frac{S}{2})\), where \(L\) is the number of lone pair electrons and \(S\) is the number of shared electrons.
\(E\) has \(0\) lone pairs (\(L = 0\)). \(E\) has \(10\) shared electrons (\(S = 10\)). If \(E\) is in group \(14\) (e.g., \(Si\), \(Ge\), \(Sn\)), \(V = 4\)
\[FC=V-(L+\frac{S}{2})=4-(0 + 5)=-1\]
Let's use the formula \(FC=V - N-\frac{B}{2}\) correctly.
\(E\) has \(0\) non - bonding electrons (\(N = 0\)). \(E\) forms \(3\) bonds: total bonding electrons \(B=(1\times2)+(2\times4)=10\). If \(E\) is in group \(16\), \(V = 6\). But considering the ion charge.
Let's assume \(E\) has \(0\) non - bonding electrons. The number of bonding pairs: \(5\) (one single and two double bonds).
The formula for formal charge \(FC=V-(L + S)\) (where \(L\) is the number of lone pairs and \(S\) is the number of bonding pairs).
If \(E\) is in group \(14\) (\(V = 4\)), \(L = 0\), \(S = 5\) (incorrect).
If \(E\) is in group \(16\) (\(V = 6\)), \(L = 0\), \(S = 5\) (incorrect).
Let's use the formula \(FC=\text{valence electrons}-\text{non - bonding electrons}-\frac{\text{bonding electrons}}{2}\)
For \(E\): assume \(E\) has \(0\) non - bonding electrons. Bonding electrons \(B = 10\).
If \(E\) is \(S\) (group \(16\), \(V = 6\)): \(FC=6-0 - 5=+1\) (wrong for ion charge).
If \(E\) is \(Si\) (group \(14\), \(V = 4\)): \(FC=4-0 - 5=-1\) (wrong for ion charge).
If \(E\) is \(P\) (group \(15\), \(V = 5\)): \(FC=5-0 - 5=0\) (wrong for ion charge).
If \(E\) is \(Al\) (group \(13\), \(V = 3\)): \(FC=3-0 - 5=-2\)

Step2: Identify the element

For part (b), if the formal charge on \(E\) is \(-2\).
The element \(E\) has \(3\) bonds (one single and two double). The valence electron configuration should be such that it can accommodate \(10\) electrons (expanded octet).
Elements that can have expanded octet (more than \(8\) electrons in the valence shell) are from the third period or below.
If \(E\) has a formal charge of \(-2\), and considering the bonding (one single and two double bonds).
\(E\) is \(Al\). But \(Al\) usually forms \(3\) bonds with a + 3 oxidation state.
Let's consider \(S\) in a different way.
If we use the formula \(FC=\text{valence electrons}-\text{non - bonding electrons}-\frac{\text{bonding electrons}}{2}\)
For \(S\) (group \(16\), \(V = 6\)), assume \(E\) has \(0\) non - bonding electrons and \(10\) bonding electrons (\(5\) pairs). \(FC = 6-0 - 5=+1\) (wrong).
For \(P\) (group \(15\), \(V = 5\)), \(FC=5-0 - 5=0\) (wrong).
For \(Si\) (group \(14\), \(V = 4\)), \(FC=4-0 - 5=-1\) (wrong).
For \(Al\) (group \(13\), \(V = 3\)), \(FC=3-0 - 5=-2\)

Answer:

(a) \(-2\)
(b) \(Al\)