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Chapter 7: Ques 53 E (page 398)

Calculate the formal charge of chlorine in the molecules \({\rm{C}}{{\rm{l}}_{\rm{2}}}\), \({\rm{BeC}}{{\rm{l}}_{\rm{2}}}\), and \({\rm{Cl}}{{\rm{F}}_{\rm{5}}}\).

Short Answer

Expert verified

The formal charge of \({\rm{Cl}}\) in the compound \({\rm{C}}{{\rm{l}}_{\rm{2}}}\) is \(0\), \({\rm{Be}}\) and \({\rm{Cl}}\) in the compound \({\rm{BeC}}{{\rm{l}}_{\rm{2}}}\) is \(0\)and, \({\rm{Cl}}\) and \({\rm{F}}\) in the compound \({\rm{Cl}}{{\rm{F}}_{\rm{5}}}\) is \(0\).

Step by step solution

01

Concept Introduction

A formal charge (\({\rm{F}}{\rm{.C}}{\rm{.}}\)or\({\rm{q}}\)) is the charge on an atom/molecule in covalent bonding, assuming electrons are shared equally in bonds made.

02

Formal Charge of Chlorine

To find the Formal charge use the formula –

Formal Charge \({\rm{ = }}\) \({\rm{(}}\)Number of valence electrons on atom) \({\rm{ - }}\) (non-bonded electrons\({\rm{ + }}\) Bonding electrons\({\rm{)}}\).

Formal Charge for \({\rm{Cl}}\) is:

\(\begin{aligned}&= \left( 7 \right){\rm{ - }}\left( {{\rm{6 + 1}}} \right)\\&= 0\end{aligned}\)

The formal charge for\({\rm{C}}{{\rm{l}}_{\rm{2}}}\) is found below –

Element

Bonding Electrons

Non-bonded Electrons

Valence Electrons

Formal Charge

\({\rm{Cl}}\)

\(1\)

\(6\)

\(7\)

\(0\)

Therefore, the formal charge for \({\rm{Cl}}\)is \(0\).

03

Formal Charge of Beryllium Chloride

To find the Formal charge use the formula –

Formal Charge\({\rm{ = }}\) \({\rm{(}}\)Number of valence electrons on atom) \({\rm{ - }}\) (non-bonded electrons \({\rm{ + }}\) Bonding electrons\({\rm{)}}\).

Formal Charge for \({\rm{Be}}\) is:

\(\begin{aligned}&= \left( {\rm{2}} \right){\rm{ - }}\left( {{\rm{0 + 2}}} \right)\\&= \end{aligned}\)

Formal Charge for \({\rm{Cl}}\) is:

\(\begin{aligned}&= \left( 7 \right){\rm{ - }}\left( {{\rm{6 + 1}}} \right)\\&= 0\end{aligned}\)

The formal charge for \({\rm{BeC}}{{\rm{l}}_{\rm{2}}}\)is found below –

Element

Bonding Electrons

Non-bonded Electrons

Valence Electrons

Formal Charge

\({\rm{Be}}\)

\(2\)

\(0\)

\(2\)

\(0\)

\({\rm{Cl}}\)

\(1\)

\(6\)

\(7\)

\(0\)

Therefore, the formal charge for\({\rm{Be}}\) and\({\rm{Cl}}\)is\(0\).

04

Formal Charge of Chlorine Pentafluoride

To find the Formal charge use the formula –

Formal Charge \({\rm{ = }}\) \({\rm{(}}\)Number of valence electrons on atom)\({\rm{ - }}\) (non-bonded electrons \({\rm{ + }}\) Bonding electrons\({\rm{)}}\).

Formal Charge for\({\rm{Cl}}\) is:

\(\begin{aligned} &= \left( 7 \right){\rm{ - }}\left( {{\rm{2 + 5}}} \right)\\&= 0\end{aligned}\)

Formal Charge for \({\rm{F}}\) is:

\(\begin{aligned}&= \left( 7 \right){\rm{ - }}\left( {{\rm{6 + 1}}} \right)\\&= 0\end{aligned}\)

The formal charge for\({\rm{Cl}}{{\rm{F}}_{\rm{5}}}\) is found below –

Element

Bonding Electrons

Non-bonded Electrons

Valence Electrons

Formal Charge

\({\rm{Cl}}\)

\(5\)

\(2\)

\(7\)

\(0\)

\({\rm{F}}\)

\(1\)

\(6\)

\(7\)

\(0\)

Therefore, the formal charge for \({\rm{Cl}}\) and\({\rm{F}}\)is \(0\).

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Most popular questions from this chapter

Use the Molecule Shape simulator (http://openstaxcollege.org/l/16MolecShape) to explore real molecules. On the Real Molecules tab, select H2O. Switch between the “real” and “model” modes. Explain the difference observed.

The lattice energy of \({\rm{LiF}}\) is \({\rm{1023 kJ/mol}}\), and the \({\rm{Li - F}}\) distance is \({\rm{201 pm}}\). \({\rm{MgO}}\) crystallizes in the same structure as \({\rm{LiF}}\) but with a \({\rm{Mg - O}}\) distance of \({\rm{205 pm}}\). Which of the following values most closely approximates the lattice energy of \({\rm{MgO}}\): \({\rm{256 kJ/mol, 512 kJ/mol, 1023 kJ/mol, 2046 kJ/mol,}}\) or \({\rm{4008 kJ/mol}}\)? Explain your choice.

Which compound in each of the following pairs has the larger lattice energy? Note: \({\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}\) and \({\rm{L}}{{\rm{i}}^{\rm{ + }}}\) have similar radii; \({{\rm{O}}^{{\rm{2 - }}}}\) and \({{\rm{F}}^{\rm{ - }}}\) have similar radii. Explain your choices.

(a) \({\rm{MgO}}\) or \({\rm{MgSe}}\)

(b) \({\rm{LiF}}\) or \({\rm{MgO}}\)

(c) \({\rm{L}}{{\rm{i}}_{\rm{2}}}{\rm{O}}\) or \({\rm{LiCl}}\)

(d) \({\rm{L}}{{\rm{i}}_{\rm{2}}}{\rm{Se}}\) or \({\rm{MgO}}\)

Which compound in each of the following pairs has the larger lattice energy? Note: \({\rm{B}}{{\rm{a}}^{{\rm{2 + }}}}\) and \({{\rm{K}}^{\rm{ + }}}\) have similar radii; \({{\rm{S}}^{{\rm{2 - }}}}\) and \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\) have similar radii. Explain your choices.

(a) \({{\rm{K}}_{\rm{2}}}{\rm{O}}\) or \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{O}}\)

(b) \({{\rm{K}}_{\rm{2}}}{\rm{S}}\) or \({\rm{BaS}}\)

(c) \({\rm{KCl}}\) or \({\rm{BaS}}\)

(d) \({\rm{BaS}}\) or \({\rm{BaC}}{{\rm{l}}_{\rm{2}}}\)

Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions:\({\rm{P, I, Mg, Cl, In, Cs, O, Pb, Co}}\)?
See all solutions

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