The energy of an electron in the nth orbit is given by:
\[ E_n = - \frac{13.6}{n^2} \, \text{eV}. \] The energy in the first orbit (\(n = 1\)) is: \[ E_1 = - \frac{13.6}{1^2} = -13.6 \, \text{eV}. \] The energy in the fourth orbit (\(n = 4\)) is: \[ E_4 = - \frac{13.6}{4^2} = - \frac{13.6}{16} = -0.85 \, \text{eV}. \] The energy required to transfer the electron from the first orbit to the fourth orbit is: \[ \Delta E = E_4 - E_1 = -0.85 - (-13.6) = 12.75 \, \text{eV}. \]
The figures below show:
Which of the following points in Figure 2 most accurately represents the nodal surface shown in Figure 1?
But-2-yne and hydrogen (one mole each) are separately treated with (i) Pd/C and (ii) Na/liq.NH₃ to give the products X and Y respectively.
Identify the incorrect statements.
A. X and Y are stereoisomers.
B. Dipole moment of X is zero.
C. Boiling point of X is higher than Y.
D. X and Y react with O₃/Zn + H₂O to give different products.
Choose the correct answer from the options given below :
At 15 atm pressure, $ \text{NH}_3(g) $ is being heated in a closed container from 27°C to 347°C and as a result, it partially dissociates following the equation: $ 2\text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\text{H}_2(g) $ If the volume of the container remains constant and pressure increases to 50 atm, then calculate the percentage dissociation of $ \text{NH}_3(g) $
If equilibrium constant for the equation $ A_2 + B_2 \rightleftharpoons 2AB \quad \text{is} \, K_p, $ then find the equilibrium constant for the equation $ AB \rightleftharpoons \frac{1}{2} A_2 + \frac{1}{2} B_2. $
Consider the following reaction: $ \text{CO}(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CO}_2(g) $ At 27°C, the standard entropy change of the process becomes -0.094 kJ/mol·K. Moreover, standard free energies for the formation of $ \text{CO}_2(g) $ and $ \text{CO}(g) $ are -394.4 and -137.2 kJ/mol, respectively. Predict the nature of the above chemical reaction.