Q.Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?

A single hydrogen atom has the configuration $1s^1$ — only one electron, which is far from the stable, filled duplet configuration of helium ($1s^2$). An isolated H atom is therefore highly reactive and thermodynamically unstable.

When two hydrogen atoms approach each other, their unpaired 1s electrons pair up to form a covalent bond, producing the diatomic molecule H2:

This process releases a large amount of energy — the H–H bond has a bond dissociation enthalpy of 435.88 kJ mol⁻¹, one of the strongest single bonds known — because each hydrogen atom in H2 now effectively shares a pair of electrons and attains the stable duplet (closed-shell, He-like) arrangement.

Because forming this bond is strongly energetically favourable, under normal conditions dihydrogen exists overwhelmingly as the diatomic molecule H2 rather than as free monoatomic hydrogen; free H atoms are found only under extreme conditions (very high temperature, electric discharge, or as very short-lived intermediates, e.g., in the atomic hydrogen torch).

Compare the electron configuration of atomic H (1s¹, unstable) with the stable duplet achieved by pairing in H2; cite the high H–H bond dissociation enthalpy as the thermodynamic driving force.

• Confusing "duplet" (2-electron stability, relevant to H/He) with "octet" (8-electron stability) — hydrogen's stable configuration is the 2-electron duplet, not an octet.
• Thinking atomic hydrogen is the normal/ground state and H2 is some special/excited form — it is the reverse.