
A hydride is a compound containing a hydride ion, \[{{H}^{-}}\]. Predict two elements whose hydrides would contain incomplete octets.
Answer
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Hint: Octet means electronic configuration of valence orbital having eight electrons. The elements that have lower atomic number cannot complete their octet while forming hydrides.
Complete step by step answer:
Octet rule states that the atoms combine to form compound in such a way that both contain 8 electrons in their valence shell- So, many elements like Sodium, Lithium, Beryllium makes hydrides and that are NaH, LiH and \[Be{{H}_{2}}\] respectively.
- We can see that NaH, sodium loses one electron and gains the complete octet state while hydrogen also completes its valence orbital.
- In LiH, lithium loses one electron and hydrogen gains one electron to complete their valence orbit which has 2 electrons each but Li ion here cannot fulfill its octet here because it only has two electrons. Actually the Lithium atom here has a filled valence shell but it consists of only two electrons. So, we can say that it does not have octet structure.
- In \[Be{{H}_{2}}\], Beryllium loses two electrons and gains a stable state containing a fully filled valence shell, the same case is here with hydrogen atoms also. Here also Beryllium ion has only two electrons in its valence shell. So we can say that it does not have octet structure.
So, we can conclude that both Lithium and Beryllium will form hydrides in which they will have incomplete octets .
Note: Note that octet rule does not apply to the elements that are of d-block or f-block because they have or need more than 8 electrons in order to complete their octet. In octet, only electrons of valence shells are counted.
Complete step by step answer:
Octet rule states that the atoms combine to form compound in such a way that both contain 8 electrons in their valence shell- So, many elements like Sodium, Lithium, Beryllium makes hydrides and that are NaH, LiH and \[Be{{H}_{2}}\] respectively.
- We can see that NaH, sodium loses one electron and gains the complete octet state while hydrogen also completes its valence orbital.
- In LiH, lithium loses one electron and hydrogen gains one electron to complete their valence orbit which has 2 electrons each but Li ion here cannot fulfill its octet here because it only has two electrons. Actually the Lithium atom here has a filled valence shell but it consists of only two electrons. So, we can say that it does not have octet structure.
- In \[Be{{H}_{2}}\], Beryllium loses two electrons and gains a stable state containing a fully filled valence shell, the same case is here with hydrogen atoms also. Here also Beryllium ion has only two electrons in its valence shell. So we can say that it does not have octet structure.
So, we can conclude that both Lithium and Beryllium will form hydrides in which they will have incomplete octets .
Note: Note that octet rule does not apply to the elements that are of d-block or f-block because they have or need more than 8 electrons in order to complete their octet. In octet, only electrons of valence shells are counted.
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