The electronic configuration of the ions obtained after removal of first electron from the elements of 2^nd period from left to right are: Li⁺ (1s²), Be⁺ (1s² 2s¹), B⁺ (1s² 2s²), O (1s² 2s² 2p¹), N⁺ (1s² 2s² 2p²), O^- (1s² 2s² 2p³), F (1s² 2s² 2p⁴), Ne⁺ (1s² 2s² 2p³). 

The following conclusions can be drawn from the above configurations: 

(i) Li⁺ has noble gas, i.e., He gas configuration, therefore, AlH₂ of Li is the highest in the second period. 

(ii) Since in B⁺, the electron has to be removed from a more stable fully filled 28- orbital while in Be⁺, it has to be lost from the less stable half-filled 2s-orbital and furthermore, the loss of an electron from Be+ gives more stable Be²⁺ ion with noble gas configuration, therefore, AjH₂ of Be is lower than that of B.

(iii) Since more energy is required to remove an s-electron than a p-electron of the same energy level, therefore, more energy is required to remove a 2s-electron from B⁺ (1s² 2s²) than a 2p-electron from C⁺ (1s² 2s² 2P¹). In other words, Δ_iH₂ of C is lower than that of B. 

(iv) As we move from C to N to O, the nuclear charge increases by one unit at a time, therefore, their Δ_iH₂ also increase accordingly. In other words, Δ_iH₂ of O is higher than that of N which, in turn, is higher than that of C. 

(v) In case of O⁺ (1s² 2s² 2p³) an electron is to be lost from an exactly half-filled 2p-orbital but in case of F⁺( 1s² 2s² 2p⁴) this is not so. However, loss of an electron from F⁺ gives an exactly half-filled 2p-orbital (i.e., F²⁺ ( 1s² 2s² 2p³), therefore, Δ_iH₂ of F should be lower than that of O. 

(vi) Like O⁺ (1s² 2s² 2p³) and F⁺ (1s² 2s² 2p⁴), in case of Ne⁺ (1s² 2s² 2p⁵) also an electron is to be removed from a 2p-orbital. Since Ne has the highest nuclear charge in 2nd period, Δ_iH₂ of Ne is expected to be much higher than that of O or F. From the above discussion, it follows that Δ_iH₂ of the elements of 2nd period increase in the order: Be < C < B < N < F < O < Ne < Li.