Easy MCQ +4 / -1 PYQ · JEE Mains 2022

Match List - I with List - II.

List - I List - II
(A) $\psi_{\mathrm{MO}}=\psi_{\mathrm{A}}-\psi_{\mathrm{B}}$ (I) Dipole moment
(B) $\mu=Q \times r$ (II) Bonding molecular orbital
(C) $\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}$ (III) Anti-bonding molecular orbital
(D) $\psi_{\mathrm{MO}}=\psi_{\mathrm{A}}+\psi_{\mathrm{B}}$ (IV) Bond order

Choose the correct answer from the options given below :

  1. A $$(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{IV}),(\mathrm{D})-(\mathrm{III})$$
  2. B $$(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{II})$$
  3. C $$(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{IV}),(\mathrm{D})-(\mathrm{II})$$ Correct answer
  4. D $$(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{I})$$

Solution

$\Psi_{A}-\Psi_{B}=\Psi_{M O}$ is anti-boding molecular orbital <br/><br/> $$ \begin{aligned} &\mu=Q \times r \text { is dipole moment } \\\\ &\frac{N_{b}-N_{a}}{2}=\text { bond order } \\\\ &\Psi_{\mathrm{A}}+\Psi_{\mathrm{B}}=\Psi_{\mathrm{MO}} \text { is boding molecular orbital. } \end{aligned} $$

About this question

Subject: Chemistry · Chapter: Chemical Bonding and Molecular Structure · Topic: Ionic and Covalent Bonding

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