Q.Arrange the following

(i) Electrical conductance — CaH2, BeH2, TiH2:

• BeH2 is covalent/polymeric (Be being small, highly polarising) — essentially a non-conductor.
• CaH2 is an ionic (saline) hydride — poor conductor as a solid (ions not mobile in the lattice), though it conducts when molten.
• TiH2 is a metallic (interstitial) hydride — retains metallic bonding/character and so conducts electricity well, like the parent metal.

(ii) Ionic character — LiH, NaH, CsH:

Ionic character of the alkali-metal hydride increases as the metal becomes more electropositive going down Group 1 (Li → Na → Cs), since the metal more readily and completely transfers its electron to hydrogen.

(iii) Bond dissociation enthalpy — H–H, D–D, F–F:

• F–F is unusually weak (~155–159 kJ/mol) due to strong repulsion between the small fluorine atoms' lone pairs of electrons.
• H–H is much stronger (435.88 kJ/mol), a typical strong single covalent bond.
• D–D is slightly stronger still than H–H, because the heavier deuterium nucleus gives a lower zero-point vibrational energy, making the bond marginally harder to break (the isotope effect).

(iv) Reducing property — NaH, MgH2, H2O:

Reducing power here tracks with the "hydridic"/ionic character of the H in the compound (H⁻ is a strong reductant; H in H2O, +1, is not reducing). …

For each series, identify the underlying structural/electronic trend (bond type for conductance, electropositivity for ionic character, lone-pair repulsion/isotope effect for …

• Ranking F–F above H–H by assuming "more electronegative = stronger bond" — F–F is actually anomalously WEAK due to lone-pair repulsion.
• Forgetting the isotope effect that makes D–D marginally stronger than H–H. …