Maximum oxidation state of Os is :

+ 6

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+ 7

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+ 8

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+ 5

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Solution

The correct option is C $+ 8$
$O s - [X e] 4^{f} 145 d^{6} 6 s^{2}$
Maximum oxidation state of Os is +8.

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	Column I		Column II
(A)	An element which can show +8 oxidation state	(i)	$C e$
(B)	An element with 7+ as the most stable oxidation state in its oxides	(ii)	$P m$
(C)	Radioactive lanthanoid	(iii)	$O s$
(D)	Lanthanoid which shows +4 oxidation state	(iv)	$M n$

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