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Q.Discuss the three types of covalent hydrides.

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Step 1. Electron-precise hydrides -- the central atom's valence electrons exactly match what is needed for bonds to every hydrogen attached, with nothing left as a lone pair and no shortfall either. Example: carbon (4 valence e⁻) forms exactly 4 bonds in CH4CH_4, using all 4 electrons.

Step 2. Electron-deficient hydrides -- the central atom does not have enough valence electrons to form ordinary two-centre, two-electron bonds to every hydrogen present. Example: boron (3 valence e⁻) in B2H6B_2H_6 (diborane) cannot form four normal B-H bonds per boron, so the molecule instead uses unusual three-centre, two-electron "banana" bridge bonds to hold the bridging hydrogens in place.

Step 3. Electron-rich hydrides -- the central atom has one or more lone pairs of electrons remaining after bonding to all its hydrogens. Example: nitrogen (5 valence e⁻) uses only 3 in bonding to three H atoms in NH3NH_3, leaving one lone pair; oxygen (6 valence e⁻) uses only 2 in bonding to two H atoms in H2OH_2O, leaving two lone pairs.

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Electron-precise hydrides (e.g. CH₄, C₂H₆, SiH₄, GeH₄) use exactly enough valence electrons for all their H bonds with none left over; electron-deficient hydrides (e.g. B₂H₆) lack enough electrons for conventional bonds to every H and use bridge bonding instead; electron-rich hydrides (e.g. NH₃, H₂O) have one or more lone pairs left over after bonding to all their H atoms.

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