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BF3 has 6 normal modes of vibration (A1', A2'', 2E'), which are made up of SALCs of 3 stretches and 3 bends. The observed bands around 1450-1500 cm^-1 in the spectrum of BF3 are assigned to asymmetric B-F stretches (E') shows marked splitting because of 10-B and 11-B abundance in natural boron. The Raman band at 888 cm^-1 assigned to the A1' symmetric B-F stretch shows no isotopic effects. Explain why only the asymmetric stretches show isotopic effects. 

I can't figure out why only the asymmetric stretches show isotopic effects. I think that since the frequency of symmetric stretch is the same it wouldn't show splitting. During asymmetric stretch there are two different frequencies. Thus, showing splitting. But I don't know if the answer is logical and how do I relate it to isotopic effect.

Thanks. 

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