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

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

$(a) NaH_{2}\bar{P}O_{4}$

$(b)NaH\bar{S}O_{4}$

$(c) H_{4}\bar{P_{2}}O_{7}$

$(d) K_{2}\bar{Mn}O_{4}$

$(e) Ca\bar{O_{2}}$

$(f) Na\bar{B}H_{4}$

$(g) H_{2}\bar{S_{2}}O_{7}$

$(h) KAl(\bar{S}O_{4})_{2}.12H_{2}O$

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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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