Question

warm up molecular geometry

1. count the total valence e - of the compound.

b = 3a group
i = 7a group
(1)(3)+(3)(7)=24e

1. count the number of electrons used.

eused = 30e
b = boron
i = iodine
b. c₂h₂

1. count the total valence e - of the compound.

c = 4a group
h = 1a group
(2)(4)+(2)(1)=10e

1. count the number of electrons used.

eused = 13e

1. form the skeleton of the compound. note: each line in the skeleton contains two electrons.
2. complete the octet (8 e - ) of the outer atoms. note: for h, duet rule (2 e - ) only applies.
3. calculate the no. of electrons left and add them to the central atom. if necessary, form double or triple bonds.

no. of e - left = total valence e - - no. of electrons used

1. determine and draw the molecular geometry.

molecular geometry: trigonal planar
polar or non - polar: nonpolar
fc = 0
3 - 3=0
b. c₂h₂

1. form the skeleton of the compound. note: each line in the skeleton contains two electrons.
2. complete the octet (8 e - ) of the outer atoms. note: for h, duet rule (2 e - ) only applies.
3. calculate the no. of electrons left and add them to the central atom. if necessary, form double or triple bonds.

no. of e - left = total valence e - - no. of electrons used

1. determine and draw the molecular geometry.

molecular geometry: linear
polar or non - polar: nonpolar
x boron always has 3 shared electrons
x we want 1 h₂ atom to h₂→2e - octet
x we want 1 c₂ atom to c→8e - octet

Explanation:

Step1: Identify the problem type

This is about determining molecular - geometry and related properties of compounds.

Step2: Analyze \(BI_3\)

• Valence - electron count: Boron (B) is in group 13 with 3 valence electrons and iodine (I) is in group 17 with 7 valence electrons. For \(BI_3\), the total number of valence electrons is \(3+(3\times7)=24\) (not 25 as in the hand - written work).
• Skeleton formation: The skeleton has B as the central atom bonded to 3 I atoms. Each bond uses 2 electrons, so 6 electrons are used in bonding initially.
• Octet completion: After forming the skeleton, we need to complete the octets of the outer atoms (Iodine). Each Iodine needs 6 more electrons to complete its octet. So, \(3\times6 = 18\) more electrons are added as lone - pairs on Iodine atoms.
• Molecular geometry: With 3 bonding pairs and 0 lone pairs on the central B atom, the molecular geometry is trigonal planar. Since the electronegativity difference between B and I is not zero, the molecule is polar.

Step3: Analyze \(C_2H_2\)

• Valence - electron count: Carbon (C) is in group 14 with 4 valence electrons and hydrogen (H) is in group 1 with 1 valence electron. For \(C_2H_2\), the total number of valence electrons is \((2\times4)+(2\times1)=10\).
• Skeleton formation: The skeleton is \(H - C\equiv C - H\). The triple bond between the two C atoms uses 6 electrons and the two \(C - H\) bonds use 2 electrons each, so a total of 10 electrons are used in bonding.
• Octet completion: The octets of carbon atoms are completed by the triple bond and single bonds. Hydrogen has a duet.
• Molecular geometry: With 2 bonding pairs and 0 lone pairs on each carbon atom, the molecular geometry is linear. The molecule is non - polar because of the symmetrical arrangement of the bonds and the equal sharing of electrons in the triple bond and \(C - H\) bonds.

Answer:

For \(BI_3\): Molecular geometry is trigonal planar, and it is polar.
For \(C_2H_2\): Molecular geometry is linear, and it is non - polar.