JEE Advanced (Single Correct MCQs): Classification of Elements & Periodicity in Properties - JEE MCQ

NOTE : Ionization potential is amount of ener gy required to take out most loosely bonded electron from neutral atom. Its value depends on stability of atom

(electronic configuration)
C – 1s² 2s² p²         C⁺ – 1s² 2s² p¹
N – 1s² 2s² p³          N⁺ – 1s² 2s² p²
O – 1s² 2s² p⁴          O⁺ – 1s² 2s² p³
F – 1s² 2s² p⁵           F⁺ – 1s² 2s² p⁴

(for second ionisation potential, IE₂) As for IE₂ the electron in all the cases is to be removed from 2p orbital so it must follow the order C < N < O < F (i.e. increase across a period) But in case of O⁺, the 2p orbital is half-filled and is more stable as compared to others. So the order becomes : C < N < F < O or O > F > N > C

Ionization potential is the energy needed to remove and electron

The elements with highest value of first ionization potential are the noble gases because they want to stick to their electrons as much as possible and retain their stability

Now, in the noble gases Helium has the highest IP (Ionization Potential) because its very small and the nucleus which is positive can attract its electrons a lot easier and retain them. Obviously it would be hard for the nucleus to retain an electron which is far away because the attractive forces would be less.

Ionisation potential of nitrogen is more than that of oxygen. This is because nitrogen has more stable halffilled p-orbitals. (N = 1s², 2s², 2p³,  O = 1s², 2s², 2p⁴)

TIPS/Formulae : (i) Noble gases do not have covalent radii. They have only vander waal’s radii.

(ii) Covalent radii is always larger than corresponding van-der Waal’s radii Atomic radius of neon being van der Waal’s radius is larger than that of fluorine which is in fact is its covalent radius.

NOTE : Electronegativity increases on moving from left to right in a period and decreases on moving from top to bottom in a group.
Si and P are placed in the 3^rd period while C and N are placed in the 2^nd period. Elements in 2^nd period have higher electronegativities than those in the 3^rd period.
Since N has smaller size and higher nuclear charge than C, its electronegativity is higher than that of C. Similarly, the electronegativity of P is higher than that of Si. Thus, the overall order is : Si, P, C, N.

NOTE : First ionisation potential increases from left to right in a period. But Mg is more stable than Al due to fully filled-3s orbitals.
IE₁ of Mg is higher than that of Na because of increased nuclear charge and also that of Al because in Mg a 3 s-electron has to be removed while in Al it is the 3 p-electron. The IE₁ of Si is, however, higher than those of Mg and Al because of its increased nuclear charge. Thus, the overall order is Na < Mg > Al < Si.

Nitrogen, being smallest in size, can give up its lone pair of electrons most easily.

TIPS/Formulae : For isoelectronic ions, ionic size 

∴ Na⁺ is largest in size.

NOTE :  Ionisation energy increases with increasing atomic number in a period, while it decreases on moving down a groups. IE of element with electronic configuration (d) is lowest because of its biggest size.
Among the remaining three elements of the same period (3rd). IE of element with electronic configuration (b) is the highest due to greater stability of the exactly halffilled 3 p-subshell.

The electrons are not filled in d-subshell monotonically with increase in atomic number, among transition elements.

TIPS/Formulae : (i) Ion having half filled or full filled orbital have extra stability. (ii) Larger the size of cation more will be its stability Pb²⁺ (5d¹⁰ 6s²), has the most stable +2 oxidation state because here the d-orbital is completely filled and is more stable than Fe²⁺ (3d⁶). Again Ag⁺ (4d¹⁰) is more stable as here again the d-orbital is completely filled and Ag²⁺ is not easily obtained. Pb²⁺ is more stable compared to Sn²⁺ (4d¹⁰ 5s²) because of its large size.

The electronic configuration of the given ions are as follows. ₁₂Mg²⁺ = 1s², 2s²2p⁶ (No unpaired electron)
₂₂Ti³⁺ = 1s², 2s²2p⁶, 3s²3p⁶3d¹ (One unpaired electron)
₂₃V³⁺ = 1s², 2s²2p⁶, 3s²3p⁶3d² (Two unpaired electrons)
₂₆Fe²⁺ = 1s², 2s²2p⁶,3s²3p⁶3d⁶ (Four unpaired electrons)

Effective nuclear charge (i,e. Z/e ratio) decreases from F^– to N^3– , hence the radii follows the order: F^– < O^2– < N^3–. Z/e for F^– = 9/10=0.9, for O^2– = 8/10 = .8, for N^{3 –} = 7/10= 0.7

Non-metallic oxides are acidic and acidic character decreases with increase in metallic character.

TIPS/Formulae : (i) Hydrogen bonding increases the boiling point. (ii) Hydrogen bonds are formed in compounds having F or O or N with hydrogen S, Se, Te cannot undergo hydrogen bond formation because of their larger size and lower electronegativity values.

Non-metallic oxides are acidic and metallic oxides are basic. Thus the acidic order is  CaO < CuO < H₂O < CO₂.