find the oxidation number of the underline species in the following compounds
Answers
Answer:
Oxidation no of Na is +1 (since oxidation no of alkali metal is +1) oxidation no of H is -1 (hydrogen in hydrides will have -1) sum of oxidation numbers in a compound is 0, so let oxidation no of boron be x×(+1)+x+4×(−1)=0
On solving it we get x=+3
So, oxidation no of B is +3
2)
It is 4+. because H
+
has two molecules and Cl
−
has six, and the charge of the compound must be zero(0), so the charge of hydrogen (+1×2=2) plus charge of chloride (−1×6=6) and adding the Platinum charge must be positive four :
H
+
×2=2
Cl
−
6=−6
Pt
+
×4=4
total charge of compound= 0
So charge on H
2
PtCl
6
6Cl
−
is -6
2H
+
2
Pt is 4+,becausethere is only one atom of it
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Answer:
Oxidation no of Na is +1 (since oxidation no of alkali metal is +1) oxidation no of H is -1 (hydrogen in hydrides will have -1) sum of oxidation numbers in a compound is 0, so let oxidation no of boron be x×(+1)+x+4×(−1)=0
On solving it we get x=+3
So, oxidation no of B is +3
2)
It is 4+. because H
+
has two molecules and Cl
−
has six, and the charge of the compound must be zero(0), so the charge of hydrogen (+1×2=2) plus charge of chloride (−1×6=6) and adding the Platinum charge must be positive four :
H
+
×2=2
Cl
−
6=−6
Pt
+
×4=4
total charge of compound= 0
So charge on H
2
PtCl
6
6Cl
−
is -6
2H
+
2
Pt is 4+,because there is only one atom of it