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SCCH 161 Homework 3, Exercises of Geometry

Geometry

Give the number of lone pairs around the central atom and the molecular geometry of CBr4. Answer: Carbon has 4 valence electrons and bonds to four bromine ...

Typology: Exercises

2022/2023

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SCCH 161 Homework 3
1. Give the number of lone pairs around the central atom and the molecular geometry of CBr4.
Answer: Carbon has 4 valence electrons and bonds to four bromine atoms (each has 7
VE’s). As the carbon now has a full octet, it has no lone pairs. This type of molecule is an
AB4 system where the bromine atoms are at the corners of a tetrahedron. The geometry of
this molecule is tetrahedral.
2. Give the number of lone pairs around the central atom and the geometry of the ion IBr2–
(hint-halogens do not normally form more than 1 bond).
A. 0 lone pairs, linear
B. 1 lone pair, bent
C. 2 lone pairs, bent
D. 3 lone pairs, bent
E. 3 lone pairs, linear
Answer: E. I and Br both have 7 VE, so total number of VE is 21 + 1 extra electron = 22 VE.
The I is less electronegative so this is in the center of the Lewis structure. The Lewis
structure below satisfies the arrangement:
This leaves 3 bonded pairs around the central iodine atom, and two bonded pairs. As non
bonded pairs experience more repulsion than bonded-non bonded pairs, they have an
angle of 120o between each other, putting the Br atoms at an angle of 180o from each
other. This system is thus an AB2E3 system, with a trigonal bipyramidal arrangement of
electron pairs. However, the molecule is linear as there are only 2 atoms attached to the
central iodine atom.
3. According to the VSEPR theory, which one of the following species is linear?
A. H2S
B. HCN
C. BF3
D. H2CO
E. SO2
Answer: B. From the Lewis structures, B is an AB2 system (linear) with no non bonded pairs
on the central carbon atom.
4. Which of the following substance is/are planar?
(i) SO3 (ii) SO32– (iii) NO3– (iv) PF3 (v) BF3
A. (i) and (ii)
pf3
pf4
pf5

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SCCH 161 Homework 3

  1. Give the number of lone pairs around the central atom and the molecular geometry of CBr 4. Answer: Carbon has 4 valence electrons and bonds to four bromine atoms (each has 7 VE’s). As the carbon now has a full octet, it has no lone pairs. This type of molecule is an AB 4 system where the bromine atoms are at the corners of a tetrahedron. The geometry of this molecule is tetrahedral.
  2. Give the number of lone pairs around the central atom and the geometry of the ion IBr 2 – (hint-halogens do not normally form more than 1 bond). A. 0 lone pairs, linear B. 1 lone pair, bent C. 2 lone pairs, bent D. 3 lone pairs, bent E. 3 lone pairs, linear Answer: E. I and Br both have 7 VE, so total number of VE is 21 + 1 extra electron = 22 VE. The I is less electronegative so this is in the center of the Lewis structure. The Lewis structure below satisfies the arrangement:

This leaves 3 bonded pairs around the central iodine atom, and two bonded pairs. As non bonded pairs experience more repulsion than bonded-non bonded pairs, they have an angle of 120o^ between each other, putting the Br atoms at an angle of 180o^ from each other. This system is thus an AB 2 E 3 system, with a trigonal bipyramidal arrangement of electron pairs. However, the molecule is linear as there are only 2 atoms attached to the central iodine atom.

  1. According to the VSEPR theory, which one of the following species is linear? A. H 2 S B. HCN C. BF 3 D. H 2 CO E. SO 2 Answer: B. From the Lewis structures, B is an AB 2 system (linear) with no non bonded pairs on the central carbon atom.
  2. Which of the following substance is/are planar? (i) SO 3 (ii) SO 32 –^ (iii) NO 3 –^ (iv) PF 3 (v) BF 3 A. (i) and (ii)

B. (i), (iii), and (v) C. (iv) D. (i), (ii), (iii), and (v) E. (i), (iii), (iv), and (v) Answer: B. SO 3 , and NO 3 -^ both have 24 VE and have Lewis structures having no non bonded pairs on the central atm. BF 3 also has no non bonded pairs on the central atom. These are all trigonal planar AB 3 systems. SO 3 2-^ has 3 groups surrounding the sulfur, which has a lone pair (26 VE). It is an AB 3 E system (trigonal pyramidal). PF 3 also has a non bonded pair on the central atom (26 VE) and is trigonal pyramidal.

  1. The bond angles in SF 5 +^ are expected to be A. 90. B. 120. C. 90 and 120. D. 90 and 180. E. 90, 120, and 180. Answer: E. SF 5 +^ has 40 VE (6 from S, 7 each from F, and -1 to get positive charge). All are used to form stable octets for all F atoms, with 5 single bonds attaching these to the central sulfur. The sulfur has no non bonded pairs, and is an AB 5 system. This is a trigonal bipyramidal geometry, with angles of 90o, 120o^ and 180o^ between atoms (see below).

Answer: This center has no non bonded pairs, and three bonded atoms. We predict that it will be sp^2 hybridised.

  1. Indicate the type of hybrid orbitals used by the central atom in CCl 4. A. sp B. sp^2 C. sp^3 D. sp^3 d E. sp^3 d^2 Answer: C. This molecule has no lone pairs on the carbon and 4 bonded atoms. The carbon is predicted to be sp^3 hybridised.
  2. In which one of the following molecules is the central atom sp^2 hybridized? A. SO 2 B. N 2 O C. BeCl 2 D. NF 3 E. PF 5 Answer: A. Draw all Lewis structures. Of these, only SO 2 has 3 groups (lone pair + 2 bonded atoms) around the sulfur atom predicting that it should be sp^2 hybridised.
  3. According to Valence Bond Theory, which orbital is left vacant in the molecule BH 3? A. sp^3 B. sp^2 C. sp D. p E. s

Answer: D. The ground state of boron is 1s^2 2s^2 2p^1 , meaning there are 3 VE. Boron needs to form 3 bonds of equal energy, so needs 3 hybrid orbitals- sp^2 hybrid orbitals. This leaves one p orbital unused and vacant

  1. Which of the following species have the same geometries? A. NH 2 –^ and H 2 O B. NH 2 –^ and BeH 2 C. H 2 O and BeH 2 D. NH 2 – , H 2 O, and BeH 2 Answer: A. BeH 2 is a linear molecule (AB 2 ). H 2 O and NH 2 -^ both have 8 VE and two non bonded pairs on the central atoms, so they cannot be linear due to electron repulsion. These are AB 2 E 2 systems which are bent.
  2. The number of pi bonds in the molecule below is

A. 1

B. 2

C. 3

D. 5

E. 9

Answer: C. Three pi bonds are in this molecule- one for the double bond and two for the triple bond.

  1. According to Molecular Orbital Theory, two separate px orbitals interact about the x-axis to form what molecular orbitals? A.  and * B.  and * C. , * and  D. , , and  E. , , , and * Answer: A. p-p overlap here is along the x-axis, so is end-to end which is aand  ***** situation
  2. Consider the species O 2 – , O 2 , and O 2 +. Which of these species will be paramagnetic? A. O 2 and O 2 – B. O 2 +^ and O 2 C. O 2 +^ and O 2 –

Answer: False. The Lewis structure is:

There is 1 lone pair around the central atom. Bonding/non bonding pair repulsion is greater than repulsion between two bonded pairs so the magnitudes of these are not equal.

  1. Which of the following is not true of molecular orbitals and MO theory? A. The number of molecular orbitals formed is always equal to the number of atomic orbitals combined. B. A molecular orbital can accommodate up to two electrons. C. When electrons are added to orbitals of the same energy, the most stable arrangement is predicted by Hund's rule. D. Low-energy molecular orbitals fill before high-energy molecular orbitals fill. E. Molecular orbital theory cannot be used to predict bond lengths Answer: E. MO theory can be used to predict bond lengths through calculation of bond orders. The others are all true