Match List I with List II

	List I Type of Hydride		List II Example
A	Electron deficient hydride	I	MgH$$_2$$
B	Electron rich hydride	II	HF
C	Electron precise hydride	III	B$$_2$$H$$_6$$
D	Saline hydride	IV	CH$$_4$$

Choose the correct answer from the options given below :

Reason: Group 13 elements form electron-deficient hydrides. Diborane ($$B_{2}H_6$$) lacks sufficient valence electrons to satisfy a conventional octet network via normal 2-electron covalent bonds, necessitating unique 3-center-2-electron bridge bonds.
$$\text{}$$

Reason: Group 15, 16, and 17 elements form electron-rich compounds because they hold extra lone pairs of electrons not involved in bonding. Hydrogen fluoride ($$HF$$) features three lone pairs on its fluorine atom.
$$\text{}$$

Reason: Group 14 elements form electron-precise systems. Methane ($$CH_4$$) possesses exactly the right number of valence electrons (8 electrons) needed to form its 4 symmetric covalent bonds without remaining lone pairs.
$$\text{}$$

Reason: Saline (or ionic) hydrides are formed when hydrogen bonds with highly electropositive s-block elements (alkali and alkaline earth metals). Magnesium hydride ($$MgH_2$$) belongs cleanly to this ionic chemical class.