Question

Assign oxidation numbers to the underlined elements in each of the following species:
$Na{H}_{2}P{O}_4$

$NaH\underset{–}{S}{O}_4$

$H_4{\underset{–}{P}}_2{O}_7$

$K_2\underset{–}{Mn}{O}_4$

$Ca{\underset{–}{O}}_2$

$Na\underset{–}{B}{H}_4$

$H_2{\underset{–}{S}}_2{O}_7$

$KAl\underset{–}{S}{O}_4{)}_2.12H_{2}O$

Answer 1

$Na{H}_{2}P{O}_4$
Let the oxidation number of P be x.
We know that,
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = –2
$\stackrel{+1}{Na}\stackrel{+1}{H_2}\stackrel{x-2}{P{O}_4}$
Then, we have
$1(+1)+2(+1)+1\left(x\right)+4(-2)=0\Rightarrow 1+2+x-8=0\Rightarrow x=+5$
Hence, the oxidation number of P is +5.

$NaH\underset{–}{S}{O}_4$
Then, we have
$1(+1)+1(+1)+1\left(x\right)+4(-2)=0\Rightarrow 1+1+x-8=0\Rightarrow x=+6$
Hence, the oxidation number of S is + 6.

$H_4{\underset{–}{P}}_2{O}_7$
$\stackrel{+1}{H_4}\stackrel{x-2}{P_2{O}_7}$
Then, we have
$4(+1)+2\left(x\right)+7(-2)=0\Rightarrow 4+2x-14=0\Rightarrow 2x=+10\Rightarrow x=+5$
Hence, the oxidation number of P is + 5.

$K_2\underset{–}{Mn}{O}_4$
$\stackrel{+1}{K_2}\stackrel{x-2}{Mn{O}_4}$
Then, we have
$2(+1)+x+4(-2)=0\Rightarrow 2+x-8=0\Rightarrow x=+6$
Hence, the oxidation number of Mn is + 6.

$Ca{\underset{–}{O}}_2$
Then, we have
$\stackrel{+2}{C_a}\stackrel{x}{O_2}(+2)+2\left(x\right)=0\Rightarrow 2+2x=0\Rightarrow x=-1$
Hence, the oxidation number of O is -1.

$Na\underset{–}{B}{H}_4$
$\stackrel{+1}{Na}\stackrel{x-1}{B{H}_4}$
Then, we have
$1(+1)+1\left(x\right)+4(-1)=0\Rightarrow 1+x-4=0\Rightarrow x=+3$
Hence, the oxidation number of B is + 3.

$H_2{\underset{–}{S}}_2{O}_7$
$\stackrel{+1}{H_2}\stackrel{x-2}{S_2{O}_7}$
Then, we have
$2(+1)+2\left(x\right)+7(-2)=0\Rightarrow 2+2x-14=0\Rightarrow 2x=12\Rightarrow x=+6$
Hence, the oxidation number of S is + 6.

$KA1(\underset{–}{S}{O}_4{)}_2.12H_{2}O\stackrel{+13+}{KA1}{\left(\stackrel{x2-}{S{O}_4}\right)}_2.12\stackrel{+1-2}{H_{2}O}$
Then, we have
$1(+1)+1(+3)+2\left(x\right)+8(-2)+24(+1)+12(-2)=0\Rightarrow 1+3+2x-16+24-24=0\Rightarrow 2x=12\Rightarrow x=+6$
Or,
We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the water molecule, we have
$1(+1)+1(+3)+2\left(x\right)+8(-2)=0\Rightarrow 1+3+2x-16=0\Rightarrow 2x=12\Rightarrow x=+6$
Hence, the oxidation number of S is + 6.