Q97 · CSIR-NET Chemistry, December 2012

Paper: CSIR-NET December 2012 · Subject: Physical Chemistry · Chapter: Quantum Chemistry · Topic: Huckel MO Theory · Marks: 2 · Difficulty: Medium

One molecular orbital of a polar molecule $A B$ has the form $c_{A} \psi_{A}+c_{B} \psi_{B}$, where $\psi_{A}$ and $\psi_{B}$ are normalized atomic orbital's centred on A and B , respectively. The electron in this orbital is found on atom B with a probability of 90\% neglecting the overlap between $\psi_{\mathrm{A}}$ and $\psi_{\mathrm{B}}$ a possible set of $\mathrm{c}_{\mathrm{A}}$ and $\mathrm{c}_{\mathrm{B}}$ is:

(a)$\mathrm{c}_{\mathrm{A}}=0.95, \mathrm{c}_{\mathrm{B}}=0.32$

(b)$\mathrm{c}_{\mathrm{A}}=0.10, \mathrm{c}_{\mathrm{B}}=0.90$

(c)$\mathrm{c}_{\mathrm{A}}=-0.95, \mathrm{c}_{\mathrm{B}}=0.32$

(d)$\mathrm{c}_{\mathrm{A}}=0.32, \mathrm{c}_{\mathrm{B}}=0.95$

Answer

Answer: D ✓ checked by 4AB · confidence high

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation

With zero overlap the probability on B is c_B² = 0.90 ⇒ c_B = 0.95, and normalisation gives c_A = √0.10 = 0.32.

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