In H2SiF6 oxidation number of silicon is
Answers
Answer:
The oxidation number of silicon is +4
Explanation:
Let the oxidation number of Silicon be 'x'
From the given compound,H2SiF6,
The expression to find the oxidation number of Silicon is 2(H)+1(Si)+6(F)=0.
(H) is the oxidation number of Hydrogen,
(Si) is the oxidation number of Silicon,
(F) is the oxidation number of Fluorine.
2(H)+1(Si)+6(F)=0.
2(+1)+x+6(-1)=0
2+x-6=0
x-4=0
x=+4
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The oxidation state of Si in H2SiF6 is +4.
Given:
You can calculate it by swing the foremost negative components at their lowest oxidisation states, F at -1, O at -2, and H at +1.
2(+1)+X+1(-2)+6(-1)=0 X=6
Solution:
Which cannot be right as a result of if Si would lose all of its outer electrons it'd solely be a +4 — it's highest education state.
The matter is that I assumed O would be -2, it unremarkably is unless you've got O-O bonds (like H2O2 wherever O is -1) or O-F bonds (OF2 wherever O is +2). If O is secure to F Associate in Nursingd Si it'd be “losing” Associate in Nursing lepton to F (+1 to O charge) and “taking” an lepton from Si(-1 to O charge) feat O at +0.
This changes you calculation:
2(+1)+X+1(+0)+6(-1)=0 X=4
Si would be at its highest number +4.
hence,
The oxidation state of H in H2SiF6 is +1.
The oxidation state of H in H2SiF6 is +1. The oxidation state of F in H2SiF6 is -1.
The oxidation state of H in H2SiF6 is +1. The oxidation state of F in H2SiF6 is -1. The oxidation state of Si in H2SiF6 is +4.
The oxidation state of H in H2SiF6 is +1. The oxidation state of F in H2SiF6 is -1. The oxidation state of Si in H2SiF6 is +4.
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