Question
Oxidation number of nitrogen in $NaN{O_2}$ is
✓
Answer
Let the oxidation number of $N$ in $NaN{O_2}$ be $x$
$ + 1 + x + ( - 2) \times 2 = 0$
$1 + x - 4 = 0$; $x = + 3$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The observed magnetic moment of the complex $\left[ Mn (\underline{ NCS })_6\right]^{ x- }$ is $6.06\,BM$. The numerical value of $x$ is $.......$.
→$92\ gms$ and $32\ gms$ of $NO_2(g)$ and $O_2(g)$ respectivley are taken in a rigid container and at a constant temperature they react to produce $N_2O_5(g)$ . The pressure in the container after the reaction was observed to be $\frac{7}{8}P$ where $P$ is initial pressure of the gasesous mixture. The percentage yield of the reaction is ....$\%$
→How many geometrical isomers are possible for the above compound ?
What is entropy of vaporisation of water at $100\,^oC$ , If molar heat of vaporisation is $9710\,cal/mol$ .........$cal/K/mol$
→The solubility product of $PbI _{2}$ is $8.0 \times 10^{-9} .$ The solubility of lead iodide in $0.1$ molar solution of lead nitrate is $x \times 10^{-6} \,mol / L$. The value of $x$ is ....... .(Rounded off to the nearest integer)
→$[$ Given $: \sqrt{2}=1.41]$
The equivalent weight of $H_3PO_4$ in following reaction is
$H_3PO_4 + Ca(OH)_2 \to CaHPO_4 + 2H_2O$
→$H_3PO_4 + Ca(OH)_2 \to CaHPO_4 + 2H_2O$
The number of unpaired electrons in ferrous ion is
Van't Hoff factor of $Ca{(N{O_3})_2}$ is
→The number of moles of sodium oxide in $620\,g$ of it is ................ $\mathrm{moles}$
→To measure the quantity of $MnCl _2$ dissolved in an aqueous solution, it was completely converted to $KMnO _4$ using the reaction,
→$MnCl _2+ K _2 S _2 O _8+ H _2 O \longrightarrow KMnO _4+ H _2 SO _4+ HCl$ (equation not balanced).
Few drops of concentrated $HCl$ were added to this solution and gently warmed. Further, oxalic acid ( $225 mg$ ) was added in portions till the colour of the permanganate ion disappeared. The quantity of $MnCl _2$ (in $mg$ ) present in the initial solution is. . . . . . . . . (Atomic weights in $g mol ^{-1}: Mn =55, Cl =35.5$ )