Answer:
(i)
\[Si{{H}_{4}}\]
: \[\overset{x}{\mathop{S}}\,i{{\overset{+1}{\mathop{H}}\,}_{4}}\]
\[x+4(+1)=0\]
or\[x=-4\]
(ii)
\[B{{H}_{3}}\]
\[:\overset{x}{\mathop{B}}\,{{\overset{+1}{\mathop{H}}\,}_{3}}\]
\[x+3(+1)=0\]
or \[x=-3\]
(iii)
\[B{{F}_{3}}\]
:\[\overset{x}{\mathop{B}}\,{{\overset{-1}{\mathop{F}}\,}_{3}}\]
\[x+3(-1)=0\]
or \[x=+3\]
(iv)
\[{{S}_{2}}{{O}_{3}}^{2-}\]
\[:{{\overset{x}{\mathop{S}}\,}_{2}}{{\overset{-2}{\mathop{O}}\,}_{3}}\]
\[2(x)+3(-2)=-2\]
or \[x=+2\]
(v)
\[BrO{{_{4}^{{}}}^{-}}\]
\[:\overset{x}{\mathop{B}}\,r\,{{\overset{-2}{\mathop{O}}\,}_{4}}\]
\[x+4(-2)=-1\]
or \[x=+7\]
(vi)
\[HP{{O}_{4}}^{2-}\]
: \[\overset{+1}{\mathop{H}}\,\overset{x}{\mathop{P}}\,\,{{\overset{-2}{\mathop{O}}\,}_{4}}\]
\[(+1)+x+4(-2)=-2\]
or \[x=+5\]
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