Question

Consider the following Lewis structure, where E is an unknown element:

What are some possible identities for element E? Predict the molecular structure (including bond angles) for this ion. (See Exercises 97 and 98.)

Short answer

Some possible identities for element E are

1. As we see the complex contains a negative charge. Hence the central atom must contain a positive charge.
2. With the help of valence electrons we can predict that E must have 6 electrons missing.
3. After considering Lewis's structure we can predict it must be group 16 element.The possible identities of element E are O, S, Se, and Te.

The geometry of the electron pair is trigonal bipyramidal due to the presence of two lone pairs in the central atom and has a T-shape.

Step-by-step solution

Step-1 Finding total number of valence electron of the compound

First, we need to count the total number of valence electrons present.

The total number of valence electrons is defined as:

Total number of valence electrons = (No. of lone pairs + No. of bond pairs)(2)

The number of lone pairs in the figure is 11 and the number of bond pairs is 3.

Total number of valence electrons = (11+3)(2)

Total number of valence electrons = 28

Step-2 Finding missing  electrons of central storm E

After finding valence electrons we can find the missing electron of central atom E as Oxygen contributes 6 valence electrons and both oxygen contribute 7 valence electrons. Hence, it proves that central atom E has 6 electrons missing.

Step-3 Compare the previous result with elements of periodic table

When we compare it with the periodic table we will get that the missing element must come from Group 16 or VI-A. The possible identities of element E are O, S, Se, and Te.