Compounds O.N.
(A) $K M n^{} O_{4}$ (1) +4
(B) $N i^{} (C O)_{4}$ (2) +7
(C) $[P t^{} (N H_{3}) C l_{2}] C l_{2}$ (3) 0
(D) $N a_{2} O_{2}^{}$ (4) -1
The correct code for the O.N. of asterisked atom would be:

A -1, B - 2, C - 3, D - 4

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A -4, B - 3, C - 2, D - 1

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A -2, B - 3, C - 1, D - 4

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A -4, B - 1, C - 2, D - 3

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Solution

The correct option is C A -2, B - 3, C - 1, D - 4
(c) $[p t (N H_{3}) C l_{2}] C l_{2} \to$ Let, $0 . N$ of $P_{t}$ is $x$ .
$\begin{matrix} x + 0 - 2 = 2 ∴ [x = 4] \end{matrix}$
(D) ${N a}_{2} O_{2} \to$ Sodium forms peroxide $(O_{2}^{2 -})$ .
$In O_{2}^{2 -},$ oxidation state of each $O$ is (-1) .
So, correct option is C

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Identify the correct match between the codons and coding functions.

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	Compounds		O.N.
(A)	$K M n^{*} O_{4}$	(1)	+4
(B)	$N i^{*} (C O)_{4}$	(2)	+7
(C)	$[P t^{*} (N H_{3}) C l_{2}] C l_{2}$	(3)	0
(D)	$N a_{2} O_{2}^{*}$	(4)	-1

CompoundsO.N.(A)KMn∗O4(1)+4(B)Ni∗(CO)4(2)+7(C)[Pt∗(NH3)Cl2]Cl2(3)0(D)Na2O∗2(4)-1The correct code for the O.N. of asterisked atom would be:

The correct option is C A -2, B - 3, C - 1, D - 4(c) [pt(NH3)Cl2]Cl2→ Let, 0.N of Pt is x.x+0−2=2∴[x=4](D) Na2O2→ Sodium forms peroxide (O2−2).InO2−2, oxidation state of each O is (-1) .So, correct option is C

Compounds O.N.
(A) $K M n^{} O_{4}$ (1) +4
(B) $N i^{} (C O)_{4}$ (2) +7
(C) $[P t^{} (N H_{3}) C l_{2}] C l_{2}$ (3) 0
(D) $N a_{2} O_{2}^{}$ (4) -1
The correct code for the O.N. of asterisked atom would be: