Answer:
Americium is: [Rn]5f^7s^2
Bismuth is: [Xe]6s^24f^145d^106p^3
Tin is: [Kr]4d^105s^25p^2
Vanadium is: [Ar]3d^34s^2
Aluminum is [Ne]3s^23p^1
It is harder to remove an electron from fluorine than from carbon because the size of the nuclear charge in fluorine is larger than that of carbon.
The energy required to remove an electron from an atom is called ionization energy.
The ionization energy largely depends on the size of the nuclear charge. The larger the size of the nuclear charge, the higher the ionization energy because it will be more difficult to remove an electron from the atom owing to increased electrostatic attraction between the nucleus and orbital electrons.
Since fluorine has a higher size of the nuclear charge than carbon. More energy is required to remove an electron from fluorine than from carbon leading to the observation that; it is harder to remove an electron from fluorine than from carbon.
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Answer:
Explanation:
M(s) → M (g ) + 20.1 kJ --- ( 1 )
X₂ ( g ) → 2X (g ) + 327.3 kJ ---- ( 2 )
M( s) + 2 X₂(g) → M X₄ (g ) - 98.7 kJ ----- ( 3 )
( 3 ) - 2 x ( 2 ) - ( 1 )
M( s) + 2 X₂(g) - 2 X₂ ( g ) - M(s) → M X₄ (g ) - 98.7 kJ - 2 [ 2X (g ) + 327.3 kJ ] - M (g ) - 20.1 kJ
0 = M X₄ (g ) - 4 X (g ) - M (g ) - 773.4 kJ
4 X (g ) + M (g ) = M X₄ (g ) - 773.4kJ
heat of formation of M X₄ (g ) is - 773.4 kJ
Bond energy of one M - X bond = 773.4 / 4 = 193.4 kJ / mole
<span>The radioactive uranium decays into its daughter product, lead. It would do this in magma as well, as nuclear decay depends on forces within the atom, not on the phase of the material in which the atom is a part.</span>
