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BeCI 2 reacts with LiAIH 4 to give:

1. Be + Li[AICI 4 ] + H 2
2. Be + AIH 3 + LiCI + HCI
3. BeH 2 + LiCl + AlCl 3
4. BeH 2 + Li[AlCl 4 ]

Correct answer: BeH 2 + LiCl + AlCl 3

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