Periodic properties - JEE MCQ

On moving along the period, ionization enthalpy increases. In second period, the order of ionization enthalpy should be as follows:
F > O > N
But N has half-filled structure, therefore, it is more stable than O. That is why its ionization enthalpy is higher than O. Thus, the correct order of IE is
F > N > O

For isoelectronic species, size of anion increases as negative charge increases whereas size of cation decreases with increase in positive charge. Further ionic radii of anions is more than that of cations. Thus the correct order is 

for atoms
atoms of have been ionized, by energy
Thus, ionization energy per atom

for atoms
Thus, electron affinity of is per atom.

Helium (He) → Highest ionisation energy due to noble gas in nature.
Fluorine (F) High electronegativity in nature due to small size and -1 oxidation state.
Rubidium (Rb) → Most electropositive element due to large atomic size.
Lithium (Li) → Strongest reducing agent due to small size and positive oxidation state (+1)

He (Z = 2) is a noble gas and has highest I.E. The I.E. of Be(Z = 4) is more than that of Li(Z = 3)

Penetration effect order is s > p > d > f.

In all the above given cases, unpaired -electron has to be removed. In the case of , a electron is to be removed which is closer to the nucleus than the electron of .
So, of of Ag.
However, in case of Au, due to imperfect screening effect of of orbitals, the nuclear charge increases and therefore 5s of is more tightly held. Thus, the order of IE₁ is Cu > Ag < Au.

Electron affinity increases with the decrease in the size and it also depends on the electronic configuration

The electronic configuration of is and that of is . Their second ionisation enthalpies are higher than expected because after removal of one electron chromium gets a half-filled electronic configuration and get a fully-filled electronic configuration. These configuration are more stable than the partial-filled electronic configuration. So, it is very difficult to remove the second electron in both cases.
Thus, option (d) is correct.

The order of ionization enthalpy of elements of 2nd period in long form of periodic table is Li < B < Be < C < O < N < F < Ne.
Generally, on going left to right in a period, ionization energy increases, but there are some exceptions to this. For example, Be has more ionization energy than B, due to fulfilled s-orbital. And N has more ionization energy than O due to half filled p-subshell.
So,In second period of the long form of the periodic table, the element B has second lowest first ionization enthalpy and element F has second highest first ionization enthalpy values.

Among the given elements Cl has maximum tendency of dπ − pπ bonding On moving across period, nuclear charge increases and more s, p-electrons are added. Due to incomplete shielding of nuclear charge, the atomic size and so the size of 3 d-orbitals
decreaxes from Si < P < S < Cl
Decrease in size of 3 d-orbitals leads to stronger dπ−pπ overlap.