Oxidation state of C,N and H in HCN and HN

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The heats of neutralisation of $C H_{3} C O O H$ , HCOOH, HCN and $H_{2} S$ are -12.8, -13.2, -3.2 and -3.9 kcal per equivalent. Arrange these acids in increasing order of strength.

The oxidation state of the $C$ atom in $H C N$ molecule is

Q. If

H_{1}, H_{2}, . . ., H_{n}

are the

n

harmonic means between

a

and

b

, then

\frac{H_{1} + a}{H_{1} - a} + \frac{H_{n} + b}{H_{n} - b} =

Q. If

H_{1}, H_{2}, . . . ., H_{n}

are n harmonic means between a and b(

\neq

a), then the value of

\frac{H_{1} + a}{H_{1} - a} + \frac{H_{n} + b}{H_{n} - b} =

Q. Assertion :Oxidation number of carbon in HCN is

+ 2

. Reason: Carbon always shows an oxidation state of

+ 4

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