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Match List - I with List - II.

List I List II
\mathrm{(A) \: \psi_{\mathrm{MO}}=\psi_{\mathrm{A}}-\psi_{\mathrm{B}}}

(I) Dipole moment
\mathrm{(B) \: \mu=\mathrm{Q} \times \mathrm{r}}

(II) Bonding molecular orbital
\mathrm{(C) \: \: \frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}}

(III) Anti-bonding molecular orbital
\mathrm{(D) \: \psi_{\mathrm{MO}}=\psi_{\mathrm{A}}+\psi_{\mathrm{B}}}

(IV) Bond order

Choose the correct answer from the options given below :
 

Option: 1

(A) - (II), (B) - (I), (C) - (IV), (D) - (III)
 


Option: 2

(A) - (III), (B) - (IV) ,(C) - (I) ,(D) - (II)


Option: 3

(A) - (III), (B) - (I), (C) - (IV), (D) - (II)


Option: 4

(A) - (III), (B) - (IV), (C) - (II), (D) - (I)


Answers (1)

best_answer

\mathrm{(A) \: \Psi_{M 0}=\Psi_{A}-\Psi_{B} \rightarrow (III)\: \text{Anti-Bonding molecular orbital} }
\mathrm{(B) \: \mu=Q \times r \quad \longrightarrow (I) \: Dipole \: moment }
\mathrm{(C) \: \frac{\mathrm{N}_{b}-\mathrm{Na}}{2} \longrightarrow (IV) \: Bond \: order }
\mathrm{(D) \: \Psi_{M_{0}}=\Psi_{A}+\Psi_{B} \rightarrow (II)\text{ Bonding molecular orbital} }

\mathrm{So, (A)-(III) ,(B)-(I),(C)-( IV ),(D)-(I I) }

Option (3) is correct.

Posted by

Sayak

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