Q54 / 160 · JIPMER 2012 · Chemistry
The correct order of decreasing first ionisation energy is
- A$\mathrm{C}\gt\mathrm{B}\gt\mathrm{Be}\gt\mathrm{Li}$
- Correct answerB$\mathrm{C}\gt\mathrm{Be}\gt\mathrm{B}\gt\mathrm{Li}$
- C$\mathrm{B}\gt\mathrm{C}\gt\mathrm{Be}\gt\mathrm{Li}$
- D$\mathrm{Be}\gt\mathrm{Li}\gt\mathrm{B}\gt\mathrm{C}$
Solution
On moving from left to right in a period, as the size decreases, ionisation energy increases. But first IE of elements of II A group is higher than that of the elements of III A group because of the stable $n s^2$ configuration (of II A group). Hence, the order of first ionisation energy is
\(\begin{array}{c} & \mathrm{C} & > & \mathrm{Be} & > & \mathrm{B} & > & \mathrm{Li} \\ \text{Group no.} & \text{ (IV) } & & \text { (II) } & & \text { (III) } & & \text { (I) } \end{array}\)
Hence, the correct answer is option: 'B'.