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Calculate the formal charge on the indicated atom in each of the following molecules or ions: (a) the central oxygen atom in \(\mathrm{O}_{3}\), (b) phosphorus in \(\mathrm{PF}_{6}^{-}\), (c) nitrogen in \(\mathrm{NO}_{2}\), (d) iodine in \(\mathrm{ICl}_{3}\), (e) chlorine in \(\mathrm{HClO}_{4}\) (hydrogen is bonded to \(\mathrm{O}\) ).

Short Answer

Expert verified
The formal charges on the indicated atoms in the molecules or ions are: (a) central oxygen atom in O3 = +2, (b) phosphorus in PF6^- = -1, (c) nitrogen in NO2 = +2, (d) iodine in ICl3 = 0, (e) chlorine in HClO4 = +3.

Step by step solution

01

Calculate Formal Charge on the central Oxygen in O3

1) Draw the Lewis structure of O3 Oxygen has 6 valence electrons, so in O3, we have 18 valence electrons in total. The Lewis structure will be: O=O-O with a lone pair on the terminal oxygen, and three lone pairs on each of the other oxygens. 2) Calculate Formal Charge on the central oxygen atom Valence Electrons of Central Oxygen: 6 Non-bonding electrons: 2 (from the lone pair) Bonding electrons: 4 (2 single bonds) Formal Charge = 6 - 2 - (1/2 x 4) = 6 - 2 - 2 = +2
02

Calculate Formal Charge on Phosphorus in PF6^-

1) Draw the Lewis structure of PF6^- Phosphorus has 5 valence electrons, and fluorine has 7 valence electrons. In PF6^-, we have a total of 48 valence electrons [(5 + 6 * 7) +1 = 48]. The Lewis structure will be a central phosphorus atom bonded to six fluorine atoms, with three lone pairs on each fluorine. 2) Calculate Formal Charge on Phosphorus Valence Electrons of Phosphorus: 5 Non-bonding electrons: 0 Bonding electrons: 12 (6 single bonds) Formal Charge = 5 - 0 - (1/2 x 12) = 5 - 6 = -1
03

Calculate Formal Charge on Nitrogen in NO2

1) Draw the Lewis structure of NO2 Nitrogen has 5 valence electrons and oxygen has 6 valence electrons, so in NO2 we have a total of 17 valence electrons (5 + 2 * 6 = 17). The Lewis structure will be: O=N-O with one lone pair on the singly bonded oxygen and two lone pairs on the other oxygen. 2) Calculate Formal Charge on Nitrogen Valence Electrons of Nitrogen: 5 Non-bonding electrons: 0 Bonding electrons: 6 (1 double bond and 1 single bond) Formal Charge = 5 - 0 - (1/2 x 6) = 5 - 3 = +2
04

Calculate Formal Charge on Iodine in ICl3

1) Draw the Lewis structure of ICl3 Iodine has 7 valence electrons and chlorine has 7 valence electrons, so in ICl3 we have a total of 28 valence electrons (7 + 3 * 7 = 28). The Lewis structure will be a central iodine atom bonded to three chlorine atoms and two lone pairs on the iodine. Each chlorine atom has three lone pairs. 2) Calculate Formal Charge on Iodine Valence Electrons of Iodine: 7 Non-bonding electrons: 4 (from the two lone pairs) Bonding electrons: 6 (3 single bonds) Formal Charge = 7 - 4 - (1/2 x 6) = 3 - 3 = 0
05

Calculate Formal Charge on Chlorine in HClO4

1) Draw the Lewis structure of HClO4 Chlorine has 7 valence electrons, oxygen has 6 valence electrons, and hydrogen has 1 valence electron, so in HClO4 we have a total of 32 valence electrons (7 + 4 * 6 + 1 = 32). The Lewis structure will consist of a central chlorine atom bonded to four oxygen atoms, with one of the oxygen atoms bonded to the hydrogen. Each oxygen will have two lone pairs and the hydrogen will have no lone pairs. 2) Calculate Formal Charge on Chlorine Valence Electrons of Chlorine: 7 Non-bonding electrons: 0 Bonding electrons: 8 (4 single bonds) Formal Charge = 7 - 0 - (1/2 x 8) = 7 - 4 = +3

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Most popular questions from this chapter

One scale for electronegativity is based on the concept that the electronegativity of any atom is proportional to the ionization energy of the atom minus its electron affinity: electronegativity \(=k(I-E A)\), where \(k\) is a proportionality constant. (a) How does this definition explain why the electronegativity of \(\mathrm{F}\) is greater than that of \(\mathrm{Cl}\) even though \(\mathrm{Cl}\) has the greater electron affinity? (b) Why are both ionization energy and electron affinity relevant to the notion of electronegativity? (c) By using data in Chapter 7 , determine the value of \(k\) that would lead to an electronegativity of \(4.0\) for \(F\) under this definition. (d) Use your result from part (c) to determine the electronegativities of \(\mathrm{Cl}\) and \(\mathrm{O}\) using this scale. (e) Another scale for electronegativity defines electronegativity as the average of an atom's first ionization energy and its electron affinity. Using this scale, calculate the electronegativities for the halogens, and scale them so fluorine has an electronegativity of 4.0. On this scale, what is Br's electronegativity?

Which of the following statements about electronegativity is false? (a) Electronegativity is the ability of an atom in a molecule to attract electron density toward itself. (b) Electronegativity is the same thing as electron affinity. (c) The numerical values for electronegativity have no units. (d) Fluorine is the most electronegative element. (e) Cesium is the least electronegative element.

Predict the chemical formula of the ionic compound formed between the following pairs of elements: (a) \(\mathrm{Al}\) and \(\mathrm{F}\), (b) \(\mathrm{K}\) and \(\mathrm{S}\), (c) \(\mathrm{Y}\) and \(\mathrm{O}\), (d) \(\mathrm{Mg}\) and \(\mathrm{N}\).

Consider the collection of nonmetallic elements \(\mathrm{O}, \mathrm{P}, \mathrm{Te}\), I and B. (a) Which two would form the most polar single bond? (b) Which two would form the longest single bond? (c) Which two would be likely to form a compound of formula \(\mathrm{XY}_{2}\) ? (d) Which combinations of elements would likely yield a compound of empirical formula \(\mathrm{X}_{2} \mathrm{Y}_{3}\) ?

(a) True or false: The hydrogen atom is most stable when it has a full octet of electrons. (b) How many electrons must a sulfur atom gain to achieve an octet in its valence shell? (c) If an atom has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{3}\), how many electrons must it gain to achieve an octet?

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