Position in Periodic Table

Hydrogen's electronic configuration is $1s^1$ -- one electron in the first shell. This matches the general valence-shell pattern $ns^1$ of the alkali metals (Li: $2s^1$, Na: $3s^1$, K: $4s^1$, ...), and hydrogen indeed behaves like an alkali metal in several ways:

1. It forms a unipositive ion, H⁺, just as alkali metals form Na⁺, K⁺, Cs⁺.
2. It forms analogous compounds -- halides (HX), oxides (H₂O), peroxides (H₂O₂) and sulphides (H₂S) -- mirroring the alkali metals' halides (NaX), oxides (Na₂O), peroxides (Na₂O₂) and sulphides (Na₂S).
3. It acts as a reducing agent, the same role alkali metals play in many reactions.

But the resemblance is only partial. Alkali metals have low first ionisation energies, 377-520 kJ mol⁻¹; hydrogen's is far higher, 1,314 kJ mol⁻¹ -- it is much harder to strip the electron off a hydrogen atom than off an alkali-metal atom.

Hydrogen also shows a second face, resembling the halogens. Just as a halogen atom can gain one electron to complete an octet and form a halide ion X⁻, a hydrogen atom can gain one electron to complete a duet (matching helium's configuration) and form the hydride ion H⁻. But hydrogen's electron affinity is much smaller than a halogen's, so its drive to form H⁻ is comparatively weak. This is quantified by the enthalpy changes for the two processes:

Forming H⁻ from H₂ actually costs energy (positive ΔH), while forming Br⁻ from Br₂ releases energy (negative ΔH) -- confirming halogens gain electrons far more readily than hydrogen does. …