Which statements are NOT TRUE about $$XeO_2 F_2$$?

A. It has a see-saw shape.
B. Xe has 5 electron pairs in its valence sheU in XeO 2 F 2.
C. The $$O - Xe- O$$ bond angle is close to $$180^{o}$$.
D. The $$F- Xe -F$$ bond angle is close to $$180^{o}$$.
E. $$Xe$$ has 16 valence electrons in $$XeO_2 F_2$$.

Choose the correct answer from the options given below:

To determine which statements are not true about $$XeO_2F_2$$, we analyze its molecular structure using VSEPR theory and Lewis structure concepts.

First, recall that xenon (Xe) has 8 valence electrons. In $$XeO_2F_2$$, the central Xe atom is bonded to two oxygen (O) atoms and two fluorine (F) atoms. The Lewis structure shows:

• Two Xe-F single bonds (each bond is a single pair of electrons).
• Two Xe-O double bonds (each double bond consists of two pairs of electrons).
• One lone pair on Xe (since Xe contributes 6 electrons to bonding and has 2 electrons left).

Total valence electrons in $$XeO_2F_2$$: Xe contributes 8, each O contributes 6 (total 12 for two O), each F contributes 7 (total 14 for two F), summing to $$8 + 12 + 14 = 34$$ electrons. After accounting for bonding and lone pairs on terminal atoms, Xe has one lone pair.

Now, evaluate each statement:

Statement A: It has a see-saw shape.
Using VSEPR theory, the steric number (SN) is the number of electron domains around Xe. Each bond (single or double) counts as one domain, and the lone pair counts as one domain. Thus, SN = 4 bonding domains (two from Xe-F and two from Xe-O) + 1 lone pair domain = 5. The electron geometry is trigonal bipyramidal. With one lone pair in an equatorial position, the molecular shape is see-saw. Thus, this statement is TRUE.

Statement B: Xe has 5 electron pairs in its valence shell in $$XeO_2F_2$$.
"Electron pairs" refer to both bonding pairs and lone pairs. In the Lewis structure: - Lone pair on Xe: 1 pair. - Bonding pairs: Two Xe-F single bonds contribute 2 bonding pairs, and two Xe-O double bonds contribute 4 bonding pairs (since each double bond has two pairs). Total = $$1 + 2 + 4 = 7$$ electron pairs. The statement says 5, which is incorrect. Thus, this statement is NOT TRUE.

Statement C: The $$O - Xe - O$$ bond angle is close to $$180^\circ$$.
In the see-saw shape (trigonal bipyramidal with one lone pair), the lone pair occupies an equatorial position. The two O atoms are typically in equatorial positions (due to bond length and electronegativity considerations). The ideal equatorial bond angle is $$120^\circ$$, but the lone pair repulsion reduces it to less than $$120^\circ$$. It is not close to $$180^\circ$$. Thus, this statement is NOT TRUE.

Statement D: The $$F - Xe - F$$ bond angle is close to $$180^\circ$$.
In the trigonal bipyramidal arrangement, the two F atoms are in axial positions. The axial bond angle is $$180^\circ$$. Thus, this statement is TRUE.

Statement E: Xe has 16 valence electrons in $$XeO_2F_2$$.
The number of valence electrons assigned to Xe in the Lewis structure is calculated as: - Lone pair electrons: 2 electrons. - Half the bonding electrons: Each Xe-F bond (2 electrons) contributes 1 electron to Xe, so two bonds give 2 electrons. Each Xe-O double bond (4 electrons) contributes 2 electrons to Xe, so two double bonds give 4 electrons. Total = $$2 + 2 + 4 = 8$$ valence electrons. The statement says 16, which is incorrect. Thus, this statement is NOT TRUE.

The NOT TRUE statements are B, C, and E. The correct option is A. B, C and E Only.