Question

Find the oxidation numbers to the underlined species in the following compounds or ions.
A) $Na\underset{–}{B}{H}_4$
B) $H_2\underset{–}{Pt}{Cl}_6$

Answer 1

1) Oxidation no of Na is +1 (since oxidation no of alkali metal is +1) oxidation no of H is -1 (hydrogen in hydrides will have -1) sum of oxidation numbers in a compound is 0, so let oxidation no of boron be $x\times (+1)+x+4\times (-1)=0$

On solving it we get $x=+3$

So, oxidation no of B is +3

2)

It is 4+. because $H^{+}$ has two molecules and $C{l}^{-}$ has six, and the charge of the compound must be zero(0), so the charge of hydrogen $(+1\times 2=2)$ plus charge of chloride $(-1\times 6=6)$ and adding the Platinum charge must be positive four :

$H^{+}\times 2=2$

$C{l}^{-}6=-6$

$P{t}^{+}\times 4=4$

total charge of compound= 0

So charge on $H_{2}PtC{l}_6$

$6C{l}^{-}$ is -6

$2H^{+}$ 2

$Pt$ is 4+,becausethere is only one atom of it