Chapter 5, Problem 5.28QA

Interpretation Introduction

To find:

The molecular geometries of the given ions.

Answer to Problem 5.28QA

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.

Formula to calculate steric number SN $=\left(\begin{array}{c}number of atoms\\ bonded to central \\ atom\end{array}\right)+\left(\begin{array}{c}Number of lone pairs \\ on central atom\end{array}\right)$

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.

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.

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 PO33-			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.

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.