Chemistry, asked by wahishaikh137, 4 months ago

Q3 which atom has the smaller
ionization energy ?
(a) B or N
(b) Be or mg
(c) C or Si​

Answers

Answered by samewib619
1

Answer:

a

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Answered by Nishantsinha90
0

Answer:

Ionization energies reported in unites of kilojoules per mole (kJ/mol).

Data taken from John Emsley, The Elements, 3rd edition. Oxford: Clarendon Press, 1998.

The ionization energy of an atom is the amount of energy that is required to remove an electron from a mole of atoms in the gas phase:

M(g) ® M+(g) + e-

It is possible to remove more electrons from most elements, so this quantity is more precisely known as the first ionization energy, the energy to go from neutral atoms to cations with a 1+ charge. The second ionization energy is the energy that is required to remove a second electron, to form 2+ cations from 1+ cations:

M+(g) ® M2+(g) + e-

The third ionization energy is the energy required to form 3+ cations:

M2+(g) ® M3+(g) + e-

and so on. Ionization energies are always positive numbers, because energy must be supplied (an endothermic energy change) to separate electrons from atoms. The second ionization energy is always larger than the first ionization energy, because it requires even more energy to remove an electron from a cation than it is from a neutral atom.

The first ionization energy varies in a predictable way across the periodic table. The ionization energy decreases from top to bottom in groups, and increases from left to right across a period. Thus, helium has the largest first ionization energy, while francium has one of the lowest.

From top to bottom in a group, orbitals corresponding to higher values of the principal quantum number (n) are being added, which are on average further away from the nucleus. Since the outermost electrons are further away, they are less strongly attracted by the nucleus, and are easier to remove, corresponding to a lower value for the first ionization energy.

From left to right across a period, more protons are being added to the nucleus, but the number of electrons in the inner, lower-energy shells remains the same. The valence electrons feel a higher effective nuclear charge — the sum of the charges on the protons in the nucleus and the charges on the inner, core electrons. The valence electrons are therefore held more tightly, the atom decreases in size (see atomic radius), and it becomes increasingly difficult to remove them, corresponding to a higher value for the first ionization energy.

The following charts illustrate the general trends in the first ionization energy:

There are some "fluctuations" in these general trends. For instance, the first ionization decreases from beryllium to boron, and from magnesium to aluminum, as electrons from the p-block start to come into play. In the case of boron, which has an electron configuration of 1s2 2s2 2p1, the 2s electrons shield the higher-energy 2p electron from the nucleus, making it slightly easier to remove. A similar effect occurs in aluminum, which has an electron configuration of 1s2 2s2 2p6 3s2 3p1.

Even though oxygen is to the right of nitrogen in period 2, its first ionization energy is slightly lower than that of nitrogen. Nitrogen has an electron configuration of 1s2 2s2 2p3, which puts one electron in each p orbital, making it a half-filled set of orbitals:

Half-filled sets of p orbitals are slightly more stable than those with 2 or 4 electrons, which makes it slightly harder to ionize a nitrogen atom. Oxygen has an electron configuration of 1s2 2s2 2p4, which puts another electron in one p orbital; since this is one electron away from being half-filled, it is slightly easier to remove this additional electron:

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