Interpretation:
The ground state electronic configuration of
Concept Introduction:
Electronic configuration: The electron configuration is the distribution of electrons of an atom or molecule in atomic or molecular orbitals.
Electrons occupy the lowest energy orbitals. The increasing order of orbital energy is
The energy order of the orbital for the first three periods is as follows,
The orbital which is closer to the nucleus has lower energy; therefore the
In general, the orbitals can hold maximum of two electrons, the two electrons must have opposite spin.
The subshell ordering by Aufbau principle is given below,
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Chapter 1 Solutions
CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM
- true or false The equilibrium constant for this reaction is 0.20. N2O4(g) ⇔ 2NO2(g) Based on the above, the equilibrium constant for the following reaction is 5. 4NO2(g) ⇔ 2N2O4(g)arrow_forwardtrue or false The equilibrium constant for this reaction is 0.20. N2O4(g) ⇔ 2NO2(g) Based on the above, the equilibrium constant for the following reaction is 0.4. 2N2O4(g) ⇔ 4NO2(g)arrow_forwardtrue or false Using the following equilibrium, if heat is added the equilibrium will shift toward the reactants. N2(g) + 3H2(g) ⇔ 2NH3(g) + heatarrow_forward
- True or False Using the following equilibrium, if heat is added the equilibrium will shift toward the products. N2O4(g) + heat ⇔ 2NO2(g)arrow_forwardtrue or false Using the following equilibrium, if solid carbon is added the equilibrium will shift toward the products. C(s) + CO2(g) ⇔ 2CO(g)arrow_forwardProvide the complete mechanism for the reaction below. You must include appropriate arrows,intermediates, and formal charges. Please also provide a reason to explain why the 1,4-adduct is preferred over the 1,3-adduct.arrow_forward
- Which of the following pairs are resonance structures of one another? I. III. || III IV + II. :0: n P !༠ IV. EN: Narrow_forwardPredict the major organic product(s) and byproducts (either organic or inorganic) for thefollowing reactions.arrow_forwardA 8.25 g sample of aluminum at 55°C released 2500 J of heat. The specific heat of aluminum is 0.900 J/g°C. The density of aluminum is 2.70 g/mL. Calculate the final temperature of the aluminum sample in °C.arrow_forward
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