Question
Oxidation number of $C$ in $HNC$ is
$+1 \; \stackrel{-1}{-2}\,+2$
direction of co-ordinate bond is from more $EN$ atom to less $EN$ atom. So there is no development of charge due to this co-ordinate bond.
So in $\mathrm{HNC} \Rightarrow$
$\mathrm{O} \cdot \mathrm{N} \text { of } \mathrm{H}=+1$
$\mathrm{O} \cdot \mathrm{N}$ of $\mathrm{N}=-3$
$\mathrm{O} \cdot \mathrm{N}$ of $\mathrm{C}=+2$
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Given: $\Delta H ^0=-54.07\,kJ\,mol ^{-1}$
$\Delta S ^{\circ}=10\,JK ^{-1}\,mol ^{-1}$
(Take $2.303 \times 8.314 \times 298=5705$ )
$CoCl _{3} . xNH _{3}+ AgNO _{3}( aq ) \rightarrow$
If two equivalents of $AgCl$ precipitate out, then the value of $x$ will be $....$
If all products formed in the above reaction undergo reaction with $Br_2(CCl_4)$ individually then number of products formed will be :
Given : $\Delta H _{ f }^{\ominus}\left( Al _2 O _3\right)=-1700\,kJ\,mol ^{-1}$
$\Delta H _{ f }^\theta\left( Fe _2 O _3\right)=-840\,kJ\,mol ^{-1}$
Molar mass of $Fe , Al$ and $O$ are $56,27$ and $16\,g\,mol ^{-1}$ respectively.