Question

$\text{Match List-I and List-II}$

$\begin{array}{cccc}& \text{List-I}& & \text{List-II}\\ \text{(A)}& {\psi }_{MO}={\psi }_A-{\psi }_B& \text{(I)}& \text{Dipole moment}\\ \text{(B)}& \mu =Q\times r& \text{(II)}& \text{Bonding molecular orbital}\\ \text{(C)}& \frac{N_b-N_a}{2}& \text{(III)}& \text{Anti-bonding molecular orbital}\\ \text{(D)}& {\psi }_{MO}={\psi }_A+{\psi }_B& \text{(IV)}& \text{Bond order}\end{array}$

Choose the correct answer from options given below:

• A. $\text{(A)-(III), (B)-(IV), (C)-(I), (D)-(II)}$
• B. $\text{(A)-(III), (B)-(I), (C)-(IV), (D)-(II)}$
• C. $\text{(A)-(II), (B)-(I), (C)-(IV), (D)-(III)}$
• D. $\text{(A)-(III), (B)-(IV), (C)-(II), (D)-(I)}$

Answer 1

The correct option is B $\text{(A)-(III), (B)-(I), (C)-(IV), (D)-(II)}$

${\psi }_A-{\psi }_B={\psi }_{MO}$ is anti-boding molecular orbital

$\mu =Q\times r$ is dipole moment

$\frac{N_b-N_a}{2}=\text{bond order}$

${\psi }_A+{\psi }_B={\psi }_{MO}$ is boding molecular orbital