1.

Assign oxidation number to the underlined elements in each of the following species : (a) NaH_(2)underlinePO_(4) (b) NaHunderlineSO_(4) ( c) H_(4)underlineP_(2)O_(7) (d) K_(2)underline(Mn)O_(4) (e) CaunderlineO_(2) (f) NaunderlineBH_(4) (g) H_(2)underlineS_(2)O_(7) (h) KAl(underlineSO_(4))_(2)*12H_(2)O

Answer»

Solution :(a) `NaH_(2)underlinePO_(4)to1(Na)+2(H)+(P)+4(O)=0`
`therefore1(+1)+2(+1)+(P)+4(-2)=0`
`therefore1+2+P-8=0`
`thereforeP=+5`
(b) `NaHunderlineSO_(4)to(+1)+(+1)+(S)+4(-2)=0`
`therefore+2+S-8=0`
`thereforeS=+6`
( C) `H_(4)underlineP_(2)O_(7)to4(+1)+2(P)+7(-2)=0`
`therefore4+2P-14=0`
`therefore2P-10=0`
`thereforeP=+5`
(d) `K_(2)UNDERLINE(Mn)O_(4)to2(+1)+Mn+4(-2)=0`
`therefore+2+Mn-8=0`
`thereforeMn=+6`
(e) `CaunderlineO_(2)toCa+2(-1)=0`
`thereforeCa-2=0`
`thereforeCa=+2`
(F) `NaunderlineBH_(4)to1(+1)+B+4(-1)=0`
`thereforeB=+3`
(g) `H_(2)underlineS_(2)O_(7)to2(+1)+2(S)+7(-2)=0`
`thereforex=+6`
(h) `KAl(underlineSO_(4))_(2)*12H_(2)Oto1+3+2(S)+8(-2)+12(2xx1-2)=0`
`therefore2(S)+4-16`
`thereforeS=+6`
OR
As `H_(2)O` is NEUTRAL therefore total charge remain zero.
`therefore1+3+2(S)+8(-2)=0`
`thereforeS=+6`


Discussion

No Comment Found

Related InterviewSolutions