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Refer the below Table :

No Compound Oxydation state of sulphur
1 $H_2S$ -2(a)
2 $S_2Cl_2$ +2(b)
3 $SO_2$ +4(c)
4 $SO_3$ +6(d)

Choose correct one

 

$\begin{array}{1 1}(a)\;1-(a),2-(b),3-(c),4-(d)&(b)\;1-(c),2-(b),3-(d),4-(a)\\(c)\;1-(b),2-(a),3-(c),4-(d)&(d)\;1-(a),2-(b),3-(d),4-(c)\end{array}$

Can you answer this question?
 
 

1 Answer

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Sulphur exibits different oxidation states because of d orbitals.
Hence (a) is the correct answer.
answered Nov 27, 2013 by sreemathi.v
 
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