The electron gain enthalpy order for halogens is $$Cl > F > Br > I$$
Due to small size of fluorine the extra electron to be added feels more electron-electron repulsion. Therefore fluorine has less value for electron affinity than chlorine.
292.
Lead is not affected by $$dil.\,HCl$$ in cold because
A
$$Pb$$ is less electronegative than $$H$$
B
$$PbO$$ film is formed which resists chemical attack by acid
C
$$PbC{l_2}$$ protective coating gets formed on $$Pb$$ surface
D
$$Pb{O_2}$$ film is always present on $$Pb$$ surface, which resist chemical attack
Answer :
$$PbC{l_2}$$ protective coating gets formed on $$Pb$$ surface
The species called as pseudohalide ions are these are monovalent and made by electronegative atoms. They possess properties similar to halide ion. The
corresponding dimers of these pseudohalide ions are called pseudohalogens. $$RCO{O^ - }$$ is not is pseudohalide.
294.
Aluminium exhibits + 3 oxidation state. As we move down the group, + 1 oxidation state gets more stable. This is a consequence of
All the members form volatile halides of the type $$A{X_3}$$ . All halides are pyramidal in shape. The bond angle decreases on moving down the group due to decrease in bond pair-bond pair repulsion.
$$\mathop {NC{l_3}}\limits_{{{107}^0}} \,\,\,\,\,\,\,\,\mathop {PC{l_3}}\limits_{{{94}^0}} \,\,\,\,\,\,\,\,\mathop {AsC{l_3}}\limits_{{{92}^0}} $$
298.
One mole of calcium phosphide on reaction with excess of water gives
299.
The most stable form of carbon at high temperature is $$X.$$ The $$C-C$$ bond length in diamond is $$Y$$ while $$C-C$$ bond length in graphite is $$Z.$$ What are $$X, Y$$ and $$Z$$ respectively?
A
$${\text{Graphite,}}\,{\text{1}}{\text{.42}}\,\mathop {\text{A}}\limits^{\text{o}} ,1.54\mathop {\,{\text{A}}}\limits^{\text{o}} $$
B
$${\text{Coke,}}\,1.54\mathop {\,{\text{A}}}\limits^{\text{o}} ,1.84\,\mathop {\text{A}}\limits^{\text{o}} $$
C
$${\text{Diamond,}}\,1.54\,\mathop {\text{A}}\limits^{\text{o}} ,1.42\,\mathop {\text{A}}\limits^{\text{o}} $$
D
$${\text{Fullerene,}}\,1.54\mathop {\,{\text{A}}}\limits^{\text{o}} ,1.54\,\mathop {\text{A}}\limits^{\text{o}} $$
$${B_2}{H_6}$$ is electron deficient molecule because boron atom has three half-filled orbitals in excited state. The structure of $${B_2}{H_6}$$ is represented as follows:
In it two electrons of a $$B— H$$ bond are involved in formation of three centre bond, these bonds are represented as dotted lines.