Question

Oxidation number H2SiOF6(Si)

Answer 1

Answer:

+4

Explanation:

You can calculate it by putting the most electronegative elements at their lowest oxidation states, F at -1, O at -2, and H at +1.

2(+1)+X+1(-2)+6(-1)=0 X=6

Which can't be right because if Si would lose all of its outer electrons it would only be a +4 — it's highest education state. The problem is that I assumed O would be -2, it normally is unless you have O-O bonds (like H2O2 where O is -1) or O-F bonds (OF2 where O is +2). If O is bonded to F and Si it would be “losing” an electron to F (+1 to O charge) and “taking” an electron from Si(-1 to O charge) leaving O at +0. This changes you calculation:

2(+1)+X+1(+0)+6(-1)=0 X=4

Si would be at its highest oxidation state +4.

Answer 2

Answer:

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Explanation:

l am still class 5