$A l H_{3}$ is an example for :

interstitial hydride

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electron rich hydride

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ionic hydride

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electron deficient hydride

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Solution

The correct option is D electron deficient hydride
According to the Bohr's octet rule, each element tends to completely fill its outermost orbit with 8 electrons in it. Electronic configuration of aluminium is 2,8,3 that means it still needs 5 more electrons to complete its octet. That's why aluminium is electron deficient.

Hence, option $D$ is correct.

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