The molecular geometries of the given ions.

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Answer 1

Question

Chapter 5, Problem 5.26QA

Interpretation Introduction

To find:

The molecular geometries of the given ions.

Expert Solution & Answer

Answer to Problem 5.26QA

Solution:

The molecular geometries of the ions are (a) linear, (b) tetrahedral, (c) linear, and (d) trigonal pyramidal.

Explanation of Solution

a) SCN^-:

i. The Lewis structure:

The total number of valence electrons in a molecule of SCN^- is 16.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
S	1	6	1 x 6 = 6
C	1	4	1 x 4 = 4
N	1	5	1 x 5 = 5
Negative charge			1
Valence electrons in SCN^-			16

C is the central atom because C is least electronegative atom. The Lewis structure of SCN^- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 1

Formula to calculate steric number SN =number of atoms bonded to central atom+Number of lone pairs on central atom

ii. In the Lewis structure there is one S and one N atom attached to central C atom having no lone pair of electrons. So SN = 2. There is a -1 formal charge on S atom

iii. For SN = 2 and as the bond angle is 180^o, molecular geometry is linear.

b) CH₃PCl₃⁺:

i. The Lewis structure:

The total number of valence electrons in a molecule of CH₃PCl₃⁺ is 32.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
C	1	4	1 x 4 = 4
H	3	1	3 x 1 = 3
P	1	5	1 x 5 = 5
Cl	3	7	3 x 7 = 21
Positive charge			-1
Valence electrons in CH₃PCl₃⁺			32

P is the central atom. The Lewis structure of CH₃PCl₃⁺ is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 2

ii. In the Lewis structure there are three chlorine atoms and one CH3 group via C atom bonded to central P atom. So SN = 4. There is +1 formal charge on three O atoms. So SN = 4.

iii. For SN = 4, the molecular geometry is tetrahedral and bond angle is 109.5^o.

c) ICl₂^-:

i. The Lewis structure:

The total number of valence electrons in a molecule of ICl₂^- is 22.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
I	1	7	1 x 7 = 7
Cl	2	7	2 x 7 = 14
Negative charge			1
Valence electrons in ICl₂^-			22

I is the central atom because I is least electronegative atom. The Lewis structure of ICl₂^- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 3

ii. In the Lewis structure there are two oxygen atoms bonded to central I atom. There are three lone pair on I atom with -1 formal charge. So SN = 5.

iii. For SN = 5, the electron geometry is trigonal bipyramidal. As all lone pairs are at equatorial position to minimize the repulsion the molecular geometry is linear.

d) PO₃^3-:

i. The Lewis structure:

The total number of valence electrons in a molecule of PO₃^3- is 26.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
P	1	5	1 x 5 = 5
O	3	6	3 x 6 = 18
Negative charge			3
Valence electrons in PO₃^3-			26

P is the central atom because P is least electronegative atom. The Lewis structure of PO₃^3- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 4

ii. In the Lewis structure there are three O atoms bonded to central P atom. There is one lone pair on P atom. So SN = 4.

iii. If SN = 4, the electron geometry is tetrahedral but as there is one lone pair on P atom and it is bonded to three O atoms the geometry is trigonal pyramidal.

Conclusion:

The molecular geometries of the ions are (a) linear, (b) tetrahedral, (c) linear, and (d) trigonal pyramidal.

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For the reaction 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed: N2O5 →> NO₂+ NO3_(K1) NO2 + NO3 → N2O5 (k-1) NO2 + NO3 → → NO2 + O2 + NO (K2) NO + N2O5- NO2 + NO2 + NO2 (K3) d[N₂O5] __2k‚k₂[N2O5] Indicate whether the following rate expression is acceptable: dt k₁₁+ k₂

Consider the following decomposition reaction of N2O5(g): For the reaction 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed: N2O5 → NO2 + NO3 (K1) NO2 + NO3 → N2O5 (k-1) NO2 + NO3 → NO2 + O2 + NO (K2) NO + N2O5 → NO2 + NO2 + NO2 (K3) Indicate whether the following rate expression is acceptable: d[N2O5] = -k₁[N₂O₂] + K¸₁[NO₂][NO3] - K¸[NO₂]³ dt

In a reaction of A + B to give C, another compound other than A, B or C may appear in the kinetic equation.

Answer 2

Question

Chapter 5, Problem 5.26QA

Interpretation Introduction

To find:

The molecular geometries of the given ions.

Expert Solution & Answer

Answer to Problem 5.26QA

Solution:

The molecular geometries of the ions are (a) linear, (b) tetrahedral, (c) linear, and (d) trigonal pyramidal.

Explanation of Solution

a) SCN^-:

i. The Lewis structure:

The total number of valence electrons in a molecule of SCN^- is 16.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
S	1	6	1 x 6 = 6
C	1	4	1 x 4 = 4
N	1	5	1 x 5 = 5
Negative charge			1
Valence electrons in SCN^-			16

C is the central atom because C is least electronegative atom. The Lewis structure of SCN^- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 1

Formula to calculate steric number SN =number of atoms bonded to central atom+Number of lone pairs on central atom

ii. In the Lewis structure there is one S and one N atom attached to central C atom having no lone pair of electrons. So SN = 2. There is a -1 formal charge on S atom

iii. For SN = 2 and as the bond angle is 180^o, molecular geometry is linear.

b) CH₃PCl₃⁺:

i. The Lewis structure:

The total number of valence electrons in a molecule of CH₃PCl₃⁺ is 32.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
C	1	4	1 x 4 = 4
H	3	1	3 x 1 = 3
P	1	5	1 x 5 = 5
Cl	3	7	3 x 7 = 21
Positive charge			-1
Valence electrons in CH₃PCl₃⁺			32

P is the central atom. The Lewis structure of CH₃PCl₃⁺ is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 2

ii. In the Lewis structure there are three chlorine atoms and one CH3 group via C atom bonded to central P atom. So SN = 4. There is +1 formal charge on three O atoms. So SN = 4.

iii. For SN = 4, the molecular geometry is tetrahedral and bond angle is 109.5^o.

c) ICl₂^-:

i. The Lewis structure:

The total number of valence electrons in a molecule of ICl₂^- is 22.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
I	1	7	1 x 7 = 7
Cl	2	7	2 x 7 = 14
Negative charge			1
Valence electrons in ICl₂^-			22

I is the central atom because I is least electronegative atom. The Lewis structure of ICl₂^- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 3

ii. In the Lewis structure there are two oxygen atoms bonded to central I atom. There are three lone pair on I atom with -1 formal charge. So SN = 5.

iii. For SN = 5, the electron geometry is trigonal bipyramidal. As all lone pairs are at equatorial position to minimize the repulsion the molecular geometry is linear.

d) PO₃^3-:

i. The Lewis structure:

The total number of valence electrons in a molecule of PO₃^3- is 26.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
P	1	5	1 x 5 = 5
O	3	6	3 x 6 = 18
Negative charge			3
Valence electrons in PO₃^3-			26

P is the central atom because P is least electronegative atom. The Lewis structure of PO₃^3- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 4

ii. In the Lewis structure there are three O atoms bonded to central P atom. There is one lone pair on P atom. So SN = 4.

iii. If SN = 4, the electron geometry is tetrahedral but as there is one lone pair on P atom and it is bonded to three O atoms the geometry is trigonal pyramidal.

Conclusion:

The molecular geometries of the ions are (a) linear, (b) tetrahedral, (c) linear, and (d) trigonal pyramidal.

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Answer 3

Question

Chapter 5, Problem 5.26QA

Interpretation Introduction

To find:

The molecular geometries of the given ions.

Expert Solution & Answer

Answer to Problem 5.26QA

Solution:

The molecular geometries of the ions are (a) linear, (b) tetrahedral, (c) linear, and (d) trigonal pyramidal.

Explanation of Solution

a) SCN^-:

i. The Lewis structure:

The total number of valence electrons in a molecule of SCN^- is 16.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
S	1	6	1 x 6 = 6
C	1	4	1 x 4 = 4
N	1	5	1 x 5 = 5
Negative charge			1
Valence electrons in SCN^-			16

C is the central atom because C is least electronegative atom. The Lewis structure of SCN^- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 1

Formula to calculate steric number SN =number of atoms bonded to central atom+Number of lone pairs on central atom

ii. In the Lewis structure there is one S and one N atom attached to central C atom having no lone pair of electrons. So SN = 2. There is a -1 formal charge on S atom

iii. For SN = 2 and as the bond angle is 180^o, molecular geometry is linear.

b) CH₃PCl₃⁺:

i. The Lewis structure:

The total number of valence electrons in a molecule of CH₃PCl₃⁺ is 32.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
C	1	4	1 x 4 = 4
H	3	1	3 x 1 = 3
P	1	5	1 x 5 = 5
Cl	3	7	3 x 7 = 21
Positive charge			-1
Valence electrons in CH₃PCl₃⁺			32

P is the central atom. The Lewis structure of CH₃PCl₃⁺ is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 2

ii. In the Lewis structure there are three chlorine atoms and one CH3 group via C atom bonded to central P atom. So SN = 4. There is +1 formal charge on three O atoms. So SN = 4.

iii. For SN = 4, the molecular geometry is tetrahedral and bond angle is 109.5^o.

c) ICl₂^-:

i. The Lewis structure:

The total number of valence electrons in a molecule of ICl₂^- is 22.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
I	1	7	1 x 7 = 7
Cl	2	7	2 x 7 = 14
Negative charge			1
Valence electrons in ICl₂^-			22

I is the central atom because I is least electronegative atom. The Lewis structure of ICl₂^- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 3

ii. In the Lewis structure there are two oxygen atoms bonded to central I atom. There are three lone pair on I atom with -1 formal charge. So SN = 5.

iii. For SN = 5, the electron geometry is trigonal bipyramidal. As all lone pairs are at equatorial position to minimize the repulsion the molecular geometry is linear.

d) PO₃^3-:

i. The Lewis structure:

The total number of valence electrons in a molecule of PO₃^3- is 26.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
P	1	5	1 x 5 = 5
O	3	6	3 x 6 = 18
Negative charge			3
Valence electrons in PO₃^3-			26

P is the central atom because P is least electronegative atom. The Lewis structure of PO₃^3- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 4

ii. In the Lewis structure there are three O atoms bonded to central P atom. There is one lone pair on P atom. So SN = 4.

iii. If SN = 4, the electron geometry is tetrahedral but as there is one lone pair on P atom and it is bonded to three O atoms the geometry is trigonal pyramidal.

Conclusion:

The molecular geometries of the ions are (a) linear, (b) tetrahedral, (c) linear, and (d) trigonal pyramidal.

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Answer 4

Question

Chapter 5, Problem 5.26QA

Interpretation Introduction

To find:

The molecular geometries of the given ions.

Expert Solution & Answer

Answer to Problem 5.26QA

Solution:

The molecular geometries of the ions are (a) linear, (b) tetrahedral, (c) linear, and (d) trigonal pyramidal.

Explanation of Solution

a) SCN^-:

i. The Lewis structure:

The total number of valence electrons in a molecule of SCN^- is 16.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
S	1	6	1 x 6 = 6
C	1	4	1 x 4 = 4
N	1	5	1 x 5 = 5
Negative charge			1
Valence electrons in SCN^-			16

C is the central atom because C is least electronegative atom. The Lewis structure of SCN^- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 1

Formula to calculate steric number SN =number of atoms bonded to central atom+Number of lone pairs on central atom

ii. In the Lewis structure there is one S and one N atom attached to central C atom having no lone pair of electrons. So SN = 2. There is a -1 formal charge on S atom

iii. For SN = 2 and as the bond angle is 180^o, molecular geometry is linear.

b) CH₃PCl₃⁺:

i. The Lewis structure:

The total number of valence electrons in a molecule of CH₃PCl₃⁺ is 32.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
C	1	4	1 x 4 = 4
H	3	1	3 x 1 = 3
P	1	5	1 x 5 = 5
Cl	3	7	3 x 7 = 21
Positive charge			-1
Valence electrons in CH₃PCl₃⁺			32

P is the central atom. The Lewis structure of CH₃PCl₃⁺ is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 2

ii. In the Lewis structure there are three chlorine atoms and one CH3 group via C atom bonded to central P atom. So SN = 4. There is +1 formal charge on three O atoms. So SN = 4.

iii. For SN = 4, the molecular geometry is tetrahedral and bond angle is 109.5^o.

c) ICl₂^-:

i. The Lewis structure:

The total number of valence electrons in a molecule of ICl₂^- is 22.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
I	1	7	1 x 7 = 7
Cl	2	7	2 x 7 = 14
Negative charge			1
Valence electrons in ICl₂^-			22

I is the central atom because I is least electronegative atom. The Lewis structure of ICl₂^- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 3

ii. In the Lewis structure there are two oxygen atoms bonded to central I atom. There are three lone pair on I atom with -1 formal charge. So SN = 5.

iii. For SN = 5, the electron geometry is trigonal bipyramidal. As all lone pairs are at equatorial position to minimize the repulsion the molecular geometry is linear.

d) PO₃^3-:

i. The Lewis structure:

The total number of valence electrons in a molecule of PO₃^3- is 26.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
P	1	5	1 x 5 = 5
O	3	6	3 x 6 = 18
Negative charge			3
Valence electrons in PO₃^3-			26

P is the central atom because P is least electronegative atom. The Lewis structure of PO₃^3- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 4

ii. In the Lewis structure there are three O atoms bonded to central P atom. There is one lone pair on P atom. So SN = 4.

iii. If SN = 4, the electron geometry is tetrahedral but as there is one lone pair on P atom and it is bonded to three O atoms the geometry is trigonal pyramidal.

Conclusion:

The molecular geometries of the ions are (a) linear, (b) tetrahedral, (c) linear, and (d) trigonal pyramidal.

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Answer 5

Chapter 5, Problem 5.26QA

Interpretation Introduction

To find:

The molecular geometries of the given ions.

Expert Solution & Answer

Answer to Problem 5.26QA

Solution:

The molecular geometries of the ions are (a) linear, (b) tetrahedral, (c) linear, and (d) trigonal pyramidal.

Explanation of Solution

a) SCN^-:

i. The Lewis structure:

The total number of valence electrons in a molecule of SCN^- is 16.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
S	1	6	1 x 6 = 6
C	1	4	1 x 4 = 4
N	1	5	1 x 5 = 5
Negative charge			1
Valence electrons in SCN^-			16

C is the central atom because C is least electronegative atom. The Lewis structure of SCN^- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 1

Formula to calculate steric number SN =number of atoms bonded to central atom+Number of lone pairs on central atom

ii. In the Lewis structure there is one S and one N atom attached to central C atom having no lone pair of electrons. So SN = 2. There is a -1 formal charge on S atom

iii. For SN = 2 and as the bond angle is 180^o, molecular geometry is linear.

b) CH₃PCl₃⁺:

i. The Lewis structure:

The total number of valence electrons in a molecule of CH₃PCl₃⁺ is 32.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
C	1	4	1 x 4 = 4
H	3	1	3 x 1 = 3
P	1	5	1 x 5 = 5
Cl	3	7	3 x 7 = 21
Positive charge			-1
Valence electrons in CH₃PCl₃⁺			32

P is the central atom. The Lewis structure of CH₃PCl₃⁺ is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 2

ii. In the Lewis structure there are three chlorine atoms and one CH3 group via C atom bonded to central P atom. So SN = 4. There is +1 formal charge on three O atoms. So SN = 4.

iii. For SN = 4, the molecular geometry is tetrahedral and bond angle is 109.5^o.

c) ICl₂^-:

i. The Lewis structure:

The total number of valence electrons in a molecule of ICl₂^- is 22.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
I	1	7	1 x 7 = 7
Cl	2	7	2 x 7 = 14
Negative charge			1
Valence electrons in ICl₂^-			22

I is the central atom because I is least electronegative atom. The Lewis structure of ICl₂^- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 3

ii. In the Lewis structure there are two oxygen atoms bonded to central I atom. There are three lone pair on I atom with -1 formal charge. So SN = 5.

iii. For SN = 5, the electron geometry is trigonal bipyramidal. As all lone pairs are at equatorial position to minimize the repulsion the molecular geometry is linear.

d) PO₃^3-:

i. The Lewis structure:

The total number of valence electrons in a molecule of PO₃^3- is 26.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
P	1	5	1 x 5 = 5
O	3	6	3 x 6 = 18
Negative charge			3
Valence electrons in PO₃^3-			26

P is the central atom because P is least electronegative atom. The Lewis structure of PO₃^3- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 4

ii. In the Lewis structure there are three O atoms bonded to central P atom. There is one lone pair on P atom. So SN = 4.

iii. If SN = 4, the electron geometry is tetrahedral but as there is one lone pair on P atom and it is bonded to three O atoms the geometry is trigonal pyramidal.

Conclusion:

The molecular geometries of the ions are (a) linear, (b) tetrahedral, (c) linear, and (d) trigonal pyramidal.

Answer 6

Chapter 5, Problem 5.26QA

Interpretation Introduction

To find:

The molecular geometries of the given ions.

Expert Solution & Answer

Answer to Problem 5.26QA

Solution:

The molecular geometries of the ions are (a) linear, (b) tetrahedral, (c) linear, and (d) trigonal pyramidal.

Explanation of Solution

a) SCN^-:

i. The Lewis structure:

The total number of valence electrons in a molecule of SCN^- is 16.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
S	1	6	1 x 6 = 6
C	1	4	1 x 4 = 4
N	1	5	1 x 5 = 5
Negative charge			1
Valence electrons in SCN^-			16

C is the central atom because C is least electronegative atom. The Lewis structure of SCN^- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 1

Formula to calculate steric number SN =number of atoms bonded to central atom+Number of lone pairs on central atom

ii. In the Lewis structure there is one S and one N atom attached to central C atom having no lone pair of electrons. So SN = 2. There is a -1 formal charge on S atom

iii. For SN = 2 and as the bond angle is 180^o, molecular geometry is linear.

b) CH₃PCl₃⁺:

i. The Lewis structure:

The total number of valence electrons in a molecule of CH₃PCl₃⁺ is 32.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
C	1	4	1 x 4 = 4
H	3	1	3 x 1 = 3
P	1	5	1 x 5 = 5
Cl	3	7	3 x 7 = 21
Positive charge			-1
Valence electrons in CH₃PCl₃⁺			32

P is the central atom. The Lewis structure of CH₃PCl₃⁺ is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 2

ii. In the Lewis structure there are three chlorine atoms and one CH3 group via C atom bonded to central P atom. So SN = 4. There is +1 formal charge on three O atoms. So SN = 4.

iii. For SN = 4, the molecular geometry is tetrahedral and bond angle is 109.5^o.

c) ICl₂^-:

i. The Lewis structure:

The total number of valence electrons in a molecule of ICl₂^- is 22.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
I	1	7	1 x 7 = 7
Cl	2	7	2 x 7 = 14
Negative charge			1
Valence electrons in ICl₂^-			22

I is the central atom because I is least electronegative atom. The Lewis structure of ICl₂^- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 3

ii. In the Lewis structure there are two oxygen atoms bonded to central I atom. There are three lone pair on I atom with -1 formal charge. So SN = 5.

iii. For SN = 5, the electron geometry is trigonal bipyramidal. As all lone pairs are at equatorial position to minimize the repulsion the molecular geometry is linear.

d) PO₃^3-:

i. The Lewis structure:

The total number of valence electrons in a molecule of PO₃^3- is 26.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
P	1	5	1 x 5 = 5
O	3	6	3 x 6 = 18
Negative charge			3
Valence electrons in PO₃^3-			26

P is the central atom because P is least electronegative atom. The Lewis structure of PO₃^3- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 4

ii. In the Lewis structure there are three O atoms bonded to central P atom. There is one lone pair on P atom. So SN = 4.

iii. If SN = 4, the electron geometry is tetrahedral but as there is one lone pair on P atom and it is bonded to three O atoms the geometry is trigonal pyramidal.

Conclusion:

The molecular geometries of the ions are (a) linear, (b) tetrahedral, (c) linear, and (d) trigonal pyramidal.

Answer 7

Chapter 5, Problem 5.26QA

Interpretation Introduction

To find:

The molecular geometries of the given ions.

Answer 8

Expert Solution & Answer

Answer to Problem 5.26QA

Solution:

The molecular geometries of the ions are (a) linear, (b) tetrahedral, (c) linear, and (d) trigonal pyramidal.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

a) SCN^-:

i. The Lewis structure:

The total number of valence electrons in a molecule of SCN^- is 16.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
S	1	6	1 x 6 = 6
C	1	4	1 x 4 = 4
N	1	5	1 x 5 = 5
Negative charge			1
Valence electrons in SCN^-			16

C is the central atom because C is least electronegative atom. The Lewis structure of SCN^- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 1

Formula to calculate steric number SN =number of atoms bonded to central atom+Number of lone pairs on central atom

ii. In the Lewis structure there is one S and one N atom attached to central C atom having no lone pair of electrons. So SN = 2. There is a -1 formal charge on S atom

iii. For SN = 2 and as the bond angle is 180^o, molecular geometry is linear.

b) CH₃PCl₃⁺:

i. The Lewis structure:

The total number of valence electrons in a molecule of CH₃PCl₃⁺ is 32.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
C	1	4	1 x 4 = 4
H	3	1	3 x 1 = 3
P	1	5	1 x 5 = 5
Cl	3	7	3 x 7 = 21
Positive charge			-1
Valence electrons in CH₃PCl₃⁺			32

P is the central atom. The Lewis structure of CH₃PCl₃⁺ is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 2

ii. In the Lewis structure there are three chlorine atoms and one CH3 group via C atom bonded to central P atom. So SN = 4. There is +1 formal charge on three O atoms. So SN = 4.

iii. For SN = 4, the molecular geometry is tetrahedral and bond angle is 109.5^o.

c) ICl₂^-:

i. The Lewis structure:

The total number of valence electrons in a molecule of ICl₂^- is 22.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
I	1	7	1 x 7 = 7
Cl	2	7	2 x 7 = 14
Negative charge			1
Valence electrons in ICl₂^-			22

I is the central atom because I is least electronegative atom. The Lewis structure of ICl₂^- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 3

ii. In the Lewis structure there are two oxygen atoms bonded to central I atom. There are three lone pair on I atom with -1 formal charge. So SN = 5.

iii. For SN = 5, the electron geometry is trigonal bipyramidal. As all lone pairs are at equatorial position to minimize the repulsion the molecular geometry is linear.

d) PO₃^3-:

i. The Lewis structure:

The total number of valence electrons in a molecule of PO₃^3- is 26.

Element		Valence electrons
Symbol	Number of atoms	In one atom	Total
P	1	5	1 x 5 = 5
O	3	6	3 x 6 = 18
Negative charge			3
Valence electrons in PO₃^3-			26

P is the central atom because P is least electronegative atom. The Lewis structure of PO₃^3- is drawn by completing the octet of each element present in the molecule by adding lone pairs on it.

Chemistry: An Atoms-Focused Approach, Chapter 5, Problem 5.26QA , additional homework tip 4

ii. In the Lewis structure there are three O atoms bonded to central P atom. There is one lone pair on P atom. So SN = 4.

iii. If SN = 4, the electron geometry is tetrahedral but as there is one lone pair on P atom and it is bonded to three O atoms the geometry is trigonal pyramidal.

Conclusion:

The molecular geometries of the ions are (a) linear, (b) tetrahedral, (c) linear, and (d) trigonal pyramidal.