Q. Gold dissolves in $NaCN$ solution forming

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A

$Na _{3}\left[ Au ( CN )_{6}\right]$

B

$Na _{2}\left[ Au ( CN )_{4}\right]$

C

$Na _{2}\left[ Au ( CN )_{2}\right]$

D

$Na \left[ Au ( CN )_{2}\right]$

Solution:

Correct answer is (d) $Na \left[ Au ( CN )_{2}\right]$

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3. At standard conditions, if the change in the enthalpy for the following reaction is $-109 \,kJ\, mol^{-1}$ $H_{2(g)}+Br_{2(g)} \to 2HBr_{(g)}$ Given that bond energy of $H_{2}$ and $Br_{2}$ is $435\,KJ\,mol^{-1}$ and $192\,KJ\,mol^{-1}$ , respectively, what is the bond energy (in $KJ \,mol^{-1})$ of $HBr$ ?

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$\frac{1}{2} O _{2}+2 H ^{+}+2 e ^{-} \rightarrow H _{2} O ,$
$ E _{ O _{2} / H _{2} O }^{\circ}=+1.223 \,V$
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$\lambda^{o}_{m}\left(H_{2}SO_{4}\right) \,= \,x S\, cm^{2} mol^{-1}$
$\lambda^{o}_{m}\left(K_{2}SO_{4}\right) = y S\,cm^{2} mol^{-1}$
$\lambda^{o}_{m}\left(CH_{3}COOK\right) =z S\, cm^{2} mol^{-1} $
$\lambda^{o}_{m}\left(in \,S \,cm^{2} mol^{-1}\right)$ for $CH_3COOH$ will be-

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$H_3C-C\equiv CH\xrightarrow [{873\, K}] {\text{red hot iron tube}}$A,
the number of sigma(s) bonds present in the product A is :-

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