Quiz On Chemical Bonding And Molecular Structure
English
Intermediate
Hi There Good People , Test Your Skills On Chemical Bonding And Molecular Structure By Taking On This Quiz Of 30 Questions.A Must For Science Students Who Are Planning To Face Borad ,JEE , SAT Exams.All The Best...😊
Description
Chemistry, Bonding, Molecules, Trending, One Min, SAT, JEE, Boards, Helpful, Intermediate, Molecular Bonding, Must Do, NCERT
Tags
30
Questions
1 min
Per question
3:34
Average time
4.0
Contest Score
8
Participants
Educational explanations. 👍 Explanations cover why the correct answers are correct and why the wrong answers are wrong (when appropriate). E.g.: #q30, #q24, #q1
Question 12:
As you said, stability is directly proportional to the bond order: the higher the bond order, the higher the stability of the molecule/ion.
And, in fact, the right answer is the second one (O₂²⁻ < O₂⁻ < O₂ < O₂⁺).
But in the explanation you wrote the wrong values. The correct values (that reflect the stability order) are:
1) Bond order of O₂⁺ = 2.5
2) Bond order of O₂ = 2
3) Bond order of O₂⁻ = 1.5
4) Bond order of O₂²⁻ = 1
Question 20: I think the right answer is the second one (i and iii):
i) in BF₃ the central atom (boron) has 3 single bonds and no lone pairs: so the geometry is trigonal planar;
ii) in CH₃⁻ the central atom (carbon) has 3 single bonds and a lone pair (the negative charge): so the geometry is tetrahedral;
iii) in CH₃⁺ the central atom (carbon) has 3 single bonds and no lone pairs (it has a positive charge): so the geometry is trigonal planar;
iv) in CO₂ the central atom (carbon) has 2 double bonds and no lone pairs: so the geometry is linear.
After that, we can say that the two molecoles having the same geometry are BF₃ and CH₃⁺ (i and iii) since they are both trigonal planar.
But in question 20 you selected "i and ii" as right answer and then in the explanation you gave information to support that "i and ii" is the right question.
This can not be considered a typos. This is a mistake.