c. A full s subshell is able to shield a newly filled p subshell from the nucleus, making the first electron in a p subshell easy to remove.
Explanation:
From the given options, a full s-sublevel is able to shield a newly filled p-subshell from the nucleus thereby making the first electron in a p-subshell easy to remove is correct.
What is ionization energy?
Ionization energy is a measure of the readiness of an atom to lose an electron.
First ionization energy is the energy required to remove the most loosely held electron in the gas phase.
The size of an atom/element depends on the number of electrons it contains. The more the electrons, the larger its size.
- The larger an atom becomes the lesser the ionization energy needed to remove the first electron from its outermost shell.
Electron - electron repulsion occurs when two electrons in the same sub-level repels one another.
Shielding effect is the ability of the inner electrons to protect the outer electrons from the pull of the nuclear charge.
In option C, a s-subshell has a greater shielding effect than the p,d and f sub-shell in that order.
A newly introduced electron in the p-sublevel will be loosely held and easier to remove.
Learn more:
First ionization energy brainly.com/question/2153804
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Answer:
The partial pressure of NO2 = 0.152 atm
Explanation:
Step 1: Data given
Pressure NO2 = 0.500 atm
Total pressure at equilibrium = 0.674 atm
Step 2: The balanced equation
2NO2(g) → 2NO(g) + O2(g)
Step 3: The initial pressure
pNO2 = 0.500 atm
pNO = 0 atm
p O2 = 0 atm
Step 4: Calculate pressure at the equilibrium
For 2 moles NO2 we'll have 2 moles NO and 1 mol O2
pNO2 = 0.500 - 2x atm
pNO =2x atm
pO2 = xatm
The total pressure = p(total) = p(NO2) + p(NO) + p(O2)
p(total) = (0.500 - 2x) + 2x + x= 0.674 atm
0.500 + x = 0.674 atm
x = 0.174 atm
This means the partial pressure of NO2 = 0.500 - 2*0.174 = 0.152 atm
