Half filled orbitals are more stable because of symmetrical distribution of electrons and exchange energy is maximum.
$=50\times\frac{0.01}{100}=50\times10^{-4}$
$=5\times10^{-3}\text{ms}^{-1}$
$\Delta\text{x}.\Delta\text{V}=\frac{\text{h}}{4\pi\text{m}}$
$\Delta\text{x}\times5\times10^{-3}\text{ms}^{-1}=\frac{6.6\times10^{-34}\text{Js}}{4\times3.142\times9\times10^{-31}\text{kg}}$
$\Rightarrow\Delta\text{x}=\frac{6.6\times10^{-34}\text{Js}}{4\times3.142\times9.1\times10^{-31}\text{kg}\times5\times10^{-3}\text{ms}^{-1}}$
$=1.154\times10^{-2}\text{m}$
Scandium (Atomic number 21) is the element in which filling of 3d orbitals starts.
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$\ \ \ \ \ \ \ \ \text{H}\\ \ \ \ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{H(g)}+2\text{O}=\text{O(g)}\overrightarrow{\ \ \ \ \ }\ \text{O}=\text{C}=\text{O(g)}\\ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \text{H}\\+2\text{H}-\text{O}-\text{H(g)}$
Bond enthalpy $(\Delta\text{H}^\circ)/\text{kJ mol}^{-1}$ of C=O = 741; C-H = 414; H-O = 464; O=O = 498.
State what is oxidised to what and what is reduced to what in the reactions expressed by the equations?