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 a central atom is \( \text{Number of electron groups}=\text{Number of bonding pairs}+\text{Number of lone pairs}\).
For \(BCl_3\), the central atom is \(B\). Boron has 3 valence electrons (\(B:1s^{2}2s^{2}2p^{1}\)). Chlorine has 7 valence electrons (\(Cl:1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}\)). In \(BCl_3\), \(B\) forms 3 single bonds with \(Cl\) atoms (\(B - Cl\)). There are no lone pairs on the \(B\) atom because \(3\) valence electrons of \(B\) are used in bonding (\(3\) single bonds).

Step2: Calculate the number of electron groups

Using the formula \(\text{Number of electron groups}=\text{Number of bonding pairs}+\text{Number of lone pairs}\), with \(\text{Number of bonding pairs} = 3\) and \(\text{Number of lone pairs}=0\), we get \(\text{Number of electron groups}=3 + 0=3\).

• For \(CO_2\), the Lewis structure is \(O = C=O\), and there is no resonance as the double - bond positions are fixed.
• For \(H_2O\), the Lewis structure is \(H - O - H\) with two lone pairs on \(O\), no resonance.
• For \(ClF_3\), the Lewis structure has \(Cl\) as the central atom with 3 \(Cl - F\) bonds and 2 lone pairs, no resonance.
• For \(HNO_3\), the Lewis structure has resonance. The \(N\) atom is bonded to two \(O\) atoms via double bonds (in resonance forms) and one \(O\) atom via a single bond (with the \(H\) attached to that \(O\)).
• For \(CH_4\), the Lewis structure is \(H - C - H\) (tetrahedral), no resonance.

Answer:

3

Question 18