Answer
a) In $Cu(NO_{3})_{2}$ , oxidation state of $Cu = +2, N = +5, O= -2$
b) In $Sr(OH)_{2}$ , oxidation state of $Sr = +2, O = -2, H= +1$
c) In $K_{2}Cr_{2}O_{7}$ , oxidation state of $K = +1, Cr= +6, O= -2$
d) In $NaHCO_ {3}$ , oxidation state of $Na = +1, H = +1, C= +4, O = -2$
Work Step by Step
Algebraic sum of oxidation state of all elements in a compound is zero. Also, algebraic sum of oxidation state for an ion is equal to the charge of the ion.
a) In $Cu(NO_{3})_{2}$, splits the compound in ionic forms. Here, Cationic part is $Cu^{2+} $ So, oxidation state of $Cu = +2$.
and anionic part is $NO_{3}^{-}$. oxidation state of $O = -2$. Let's say oxidation state of N = $x$.
Hence, $x + (-2) \times3 = -1$, by solving we get $x = +5$
Therefore, oxidation state of $N = +5$
b) In $Sr(OH)_{2}$, oxidation state of $O = -2, H = +1.$ Let's say oxidation state of Sr = $x$
Hence, $x + [(-2)+(+1)]\times2 = 0$ by solving we get $x = +2$
So, oxidation state of $Sr = +2$
c) In $K_{2}Cr_{2}O_{7}$, oxidation state of $O = -2, K = +1$, let's say oxidation state of Cr = $x$
Hence, $(+1)\times2 + x\times2 + (-2)\times7 = 0$ by solving we get $x = +6$
So, oxidation state of $Cr = +6$
d) In $NaHCO_ {3}$ oxidation state of $O = -2, H = +1, Na = +1$.let's say oxidation state of C = $x$
Hence, $(+1) + (+1) + x + (-2)\times3 = 0$ by solving we get $x = +4$
So, oxidation state of
$C = +4$
