8.1 Assign oxidation number to the underlined elements in each of the following species

(a) NaH_{2}\bar{P}O_{4}   

 (b)NaH\bar{S}O_{4} 

 (c) H_{4}\bar{P_{2}}O_{7}

  (d) K_{2}\bar{Mn}O_{4} 

 (e) Ca\bar{O_{2}} 

 (f) Na\bar{B}H_{4} 

  (g) H_{2}\bar{S_{2}}O_{7}

 (h) KAl(\bar{S}O_{4})_{2}.12H_{2}O

Answers (1)

solution- 

O.N is the oxidation number

O.N of Oxygen(O) = -2 ( In case of peroxide and superoxide it wil be different ON)

O.N of hydrogen(H)= +1 (In case of metalic hydride, -1)

O.N of sodium (Na) = +1

O.N of aluminium (Al) = +3

O.N of potassium (K)= +1

O.N of calcium (Ca) = +2

   In neutral compounds the sum of O.N of all the atoms is zero.

(a) Let the O.N of P be x

\therefore\:\:\:1\ast 1+2\ast 1+x+4\ast (-2) = 0 \Rightarrow x = +5

(b) Let the O.N of S be x

\therefore \:\:1\ast 1 + 1\ast 1+x +4\ast (-2) = 0\Rightarrow x = +6

(c)  Let the O.N of P be x

\therefore \:\:4*1 +2*x +7*(-2) = 0 \Rightarrow x = +5

(d) Let the O.N of Mn be x

\therefore \:\:2*1 + x + 4*(-2) = 0\Rightarrow x = +6

(e) Let the O.N of O be x

Ca is an alkaline earth metal so its O.N. is +2

\therefore \:\:1\ast 2 + 2\ast x = 0\Rightarrow x = -1

(f) Let the O.N of B be x

Note that in this H exists as hydride ion H^{-}  so its O.N. is -1

\therefore \:\:1*1+x+4*(-1) = 0 \Rightarrow x = +3

(g) Let the O.N of S be x

\therefore \:\: 2*1 +2*x+7*(-2) =0 \Rightarrow x = +6

(h) Let the O.N of S be x

\therefore \:\:1*1+1*3+[x+(-2)*4]*2 +12*[1*2+(-2)] = 0\Rightarrow x = +6

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