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Which of the following hydride is/ are "electron-precise" type?

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$HF$

$H_{2} O$

${SiH}_{4}$

${PH}_{3}$

answer is C.

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Detailed Solution

The hydrides that possess the necessary number of electrons to form covalent bonds are referred to as electron precise hydrides.

This hydride contains the exact number of electrons needed to form normal covalent bonds, with neither excess or deficit. These are formed by group 14 elements having four valence electrons, such as Si and C, e.g. ${SiH}_{4}$ , ${CH}_{4}$

Hence, ${SiH}_{4}$ is "electron-precise" type molecule

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