Question

question 17 (1 point)
the number of electron groups exists in the valence shell of the boron atom in the bcl₃ molecule is?
1
2
3
4
5
question 18 (1 point)
resonance structures are needed to describe the bonding in which of the following?
co₂
h₂o
clf₃
hno₃
ch₄

Explanation:

Question 17

Step1: Determine the electron - group formula

The formula for the number of electron groups around an atom is \( \text{Number of electron groups}=\text{number of bonding pairs}+\text{number of lone pairs}\).
For \(BCl_{3}\), boron (\(B\)) has \(3\) valence electrons. Each \(Cl\) atom forms a single bond with \(B\). So the number of bonding pairs is \(3\), and the number of lone pairs on \(B\) is \(0\) (since \(B\) shares all its \(3\) valence electrons with \(Cl\) atoms).

Step2: Calculate the number of electron groups

Using the formula \( \text{Number of electron groups}=3 + 0=3\)

• For \(CO_{2}\), the Lewis structure is \(O = C=O\), and there is no resonance as the double - bond positions are fixed.
• For \(H_{2}O\), the Lewis structure is \(H - O - H\) with two lone pairs on \(O\), no resonance.
• For \(ClF_{3}\), the Lewis structure has a central \(Cl\) atom with three \(Cl - F\) bonds and two lone pairs, no resonance.
• For \(HNO_{3}\), the Lewis structure has resonance. The nitrate group (\(NO_{3}^{-}\) in \(HNO_{3}\)) has resonance structures because the double bond can be delocalized among the \(N - O\) bonds.
• For \(CH_{4}\), the Lewis structure is \(H - C - H\) with four single \(C - H\) bonds, no resonance.

Answer:

3

Question 18