a) Which process is effecting analyte ion formation in CT-CI?
CT stands for charge transfer (formerly charge exchange, CE). The ionic charge of the reagent ion is transferred to the analyte by electron transfer.
b) Are analyte ions from CT even- or odd-electron species?
Due to the loss of a single electron which is abstracted by the reagent ion, the resulting analyte ions are odd-electron ions, i.e., molecular ions, M+., as in EI. Nonetheless, their internal energy is significantly lower.
c) What is the driving force for the CT process?
The difference between recombination energy (RE) of the reagent ion and ionization energy of analyte defines whether charge exchange can proceed or not. CT will occur only if the analyte has a lower ionization energy than the reagent gas.
d) Assume xenon as the reagent gas in CT-CI. Which analyte should yield the lower abundance of fragment ions, n-decane or n-propane?
Xenon forms two relevant electronic states corresponding to REs of 12.1 and 13.4 eV. The ionization energy of n-propane is 10.9 eV while that of n-decane is just 9.7 eV (Table 2.1). Thus, the difference RE(X+.) – IE(M) is larger in case of n-decane by 1.2 eV causing a higher degree of fragmentation. (However, being the larger molecule it possesses more degrees of freedom to distribute the internal energy received upon CT).