Explanation
Oxidation number is the apparent charge on an atom of an element present in a molecule.In $$NCl_3$$, Electronegativity of Nitrogen is 3 while that of Chlorine is 3.2. Thus Chlorine being more electronegative has oxidation state=-1.
Sum of oxidation numbers of all elements present in a neutral compound is zero.
So, O.N. of Nitrogen+ 3(O.N. of $$Cl$$)=0
O.N. of $$N^{-3}=0$$
O.N. of $$N=+3$$
1)Structure: $$S$$ is double-bonded to$$ O$$ ($$— 2$$ electrons) and is bonded to $$2 C$$ ($$+2$$ electrons).
2)$$ +2 + -2 = 0. S$$has a$$ 0 $$Oxidation number in $$(CH_3)_2SO$$.
$$S$$ is double-bonded to$$ O (— 2$$ electrons$$) $$and is bonded to $$2 C (+2$$ electrons).
$$+2 + -2 = 0$$.
$$S$$ has a$$ 0 $$Oxidation number in $$(CH_3)_2SO$$.
Osmium exhibits oxidation states from $$0 - +8 $$ in its compounds, with the exception of $$+1$$.
well-characterized and stable compounds contain the element in$$ +2, +3, +4, +6,$$ and $$+8$$ states.
There are also carbonyl and organometallic compounds in the low oxidation states $$−2, 0, +1$$.
$$A.$$ $$Mn_2{(CO)}_{10}\Longrightarrow 2x+(10\times 0)=0\Longrightarrow x=0,$$ ($$x=$$oxidation state of metal.)
$$B.$$ $$[Ni{(CO)}_{4}]\Longrightarrow x+(4\times 0)=0\Longrightarrow x=0,$$ ($$x=$$oxidation state of metal.)
$$C.$$ $$[Cr{(C_6H_6)}_{2}]\Longrightarrow x+(2\times 0)=0\Longrightarrow x=0,$$ ($$x=$$oxidation state of metal.)
$$D.$$ $$K[PtCl_3{(C_2H_4)}]\Longrightarrow +1+x+3\times (-1)+0=0\Longrightarrow x=+2,$$ ($$x=$$oxidation state of metal.)
$$\therefore$$ $$Pt$$ is in $$+2$$ Oxidation state.
$$\therefore$$ Option $$D.$$ is the correct answer.
CrO5, also called as per chromate, has a butterfly shape and neutral charge.
Chromium lies in +6 oxidation state.
So, four oxygen will have -1 charge and one will have -2 charge, thereby satisfying the overall neutral charge.
Hence, option C is correct.
Here $$BaO_2+H_2SO_4\longrightarrow BaSO_4+H_2O_2$$ is a double displacement reaction where the cations and anions of the $$2$$ reactants switch places forming $$2$$ new compounds.
All other options involve change in oxidation status. Hence they are redox reactions.
The reaction of sulphur dioxide with acidified $$K_2Cr_2O_7$$ in acidic medium is
$$3SO_2 + Cr_2O_7^{2-} + 2 H^{+} \rightarrow 3 SO_4^{2-} + 2 Cr^{3+} + H_2O$$
So Oxidation state of Sulphur in $$SO_2$$ is $$+4$$ which is changed to $$+6$$ in $$SO_4^{2-}$$ .
Hence option $$C$$ is correct.
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