The `IE_(1)` of `C` atom is greater than that of boron `(B)` atom, whereas the reverse is true for `IE_(2)`. Explain?

Explanation

The electronic configuration of `C` and `B` are as follows.
`C: 1s^(2) 2s^(2) 2p^(1)`
B: `1s^(2) 2s^(2) 2p_(x)^(1)`
Due to the higher nuclear charge `(Z)` in `C`, the force of attraction towards valency electrons is more in `C` atom and hence `IE_(1)` of `C gt IE_(1)` of `B`.
After loss of one electron, the configuration of `C^(o+)` and `B^(o+)` is as follows:
`C^(o+) : 1s^(2) 2s^(2) 2p_(x)^(1) overset(Easy)rarr C^(2+) + e^(-)`
`B^(o+) : 1s^(2) 2s^(2)`
(Stable) `overset("Difficult")rarr B^(2+) + e^(-)`
The `B^(o+)` configuration is stable hence the removal of the `2nd` electron is difficult in comparison to `C^(o+)`. Hence, `IE_(2)` of `B gt IE_(2)` of `C`.