JEE Chemistry Classification of Elements and Periodicity in Properties MCQs Set 09

Answer: (d) \(\ce{Np}\)

Explanation:
1. Transuranic elements have atomic number greater than 92 (uranium)
2. Th = 90, Pa = 91, U = 92, Np = 93
3. So Np is transuranic

Teacher's Note:
1. All elements after uranium are man-made. Neptunium and plutonium were the first to be synthesised.

Answer: (c) F < O < Cl < S

Explanation:
1. Radii: F 64 pm, O 66 pm, Cl 99 pm, S 104 pm
2. Radius decreases across a period (F < O, and Cl < S) and increases down a group
3. So the order is F < O < Cl < S

Teacher's Note:
1. Across a period atoms shrink, but a period 3 atom is always bigger than a period 2 atom. That is why Cl and S come after F and O.

Answer: (b) Y, La

Explanation:
1. Lanthanoid contraction affects elements that come after the lanthanoids
2. Hf, W and Re (5d series) are almost the same size as Zr, Mo and Tc
3. Y and La both come before the lanthanoids, so La is simply larger than Y with no contraction effect

Teacher's Note:
1. Poor shielding by 4f electrons causes the contraction. It explains why Zr and Hf have nearly identical radii.

Answer: (d) None of the above

Explanation:
1. Removing the first electron from Na (3s) is easy, so IP1 is low (496 kJ/mol)
2. The second electron must come from the stable neon core, so IP2 is very high (4562 kJ/mol)
3. The correct reason is low IP1 and very high IP2, which none of options 1 to 3 states

Teacher's Note:
1. A large jump between successive ionisation energies shows that a noble gas core is being broken. This is why group 1 elements are always +1.

Answer: (c) \(\ce{N^3-}\), \(\ce{Mg^2+}\)

Explanation:
1. All four species are isoelectronic (10 electrons)
2. Ionisation potential rises with nuclear charge for the same number of electrons
3. \(\ce{N^3-}\) (Z = 7) has the lowest and \(\ce{Mg^2+}\) (Z = 12) the highest

Teacher's Note:
1. In an isoelectronic series the species with more protons holds its electrons more tightly. The anion with the fewest protons loses an electron most easily.

Answer: (b) IIA

Explanation:
1. There is a very large jump between IP2 and IP3
2. So two electrons are in the valence shell and the third is from an inner noble gas core
3. The element belongs to group IIA

Teacher's Note:
1. Look for the largest jump in the successive ionisation energies; the number of electrons removed before the jump equals the group number for s-block elements.

Answer: (a) The equation of the second ionisation energy is \(\ce{X(g) -> X^2+(g) + 2e^-}\)

Explanation:
1. Second ionisation energy refers to \(\ce{X^+(g) -> X^2+(g) + e^-}\), the removal of one electron from the uni-positive ion
2. The equation in option 1 removes two electrons from the neutral atom, so it describes IP1 + IP2
3. Statements 2, 3 and 4 are correct

Teacher's Note:
1. Successive ionisation energies always increase: IP1 < IP2 < IP3. Each step removes one electron from a species of higher positive charge.

Answer: (b) \(2s^22p^2\)

Explanation:
1. The configurations are Be, C, N and B
2. C can attain the stable half-filled \(2p^3\) configuration on gaining an electron, so it releases a lot of energy
3. N has a half-filled p subshell and Be a filled s subshell, so both resist gaining an electron; B gains less energy than C

Teacher's Note:
1. Electron affinity is high when the added electron leads to a half-filled or filled subshell. Elements with stable configurations, such as Be and N, have low or even positive values.

Answer: (a) \(\ce{Cl2O7}\) is the most acidic oxide of the third period

Explanation:
1. Across the third period the oxides change from basic to acidic; the right-most element (Cl) forms the most acidic oxide, \(\ce{Cl2O7}\) (gives \(\ce{HClO4}\))
2. Non-metallic character increases from left to right in a period, so statement 2 (right to left) is wrong
3. \(\ce{BeO}\) is amphoteric (statement 3 is wrong)

Teacher's Note:
1. Remember: oxides of metals are basic, of non-metals acidic, and of metalloids or of Be, Al, Zn amphoteric.

Answer: (c) Li < Na = K = Rb (screening constant)

Explanation:
1. Penetration power follows s > p > d > f, and the order of metallic character Li < Na < K > Ca is also correct
2. The screening constant increases steadily down a group as inner shells are added, so it is Li < Na < K < Rb
3. The statement claiming equal values for Na, K and Rb is incorrect

Teacher's Note:
1. Each new shell adds electrons that shield the valence electrons, so the screening constant keeps growing down the group.

Answer: (a) \(\ce{RbOH}\)

Explanation:
1. Basic strength of hydroxides increases with metallic character
2. Rb (group 1, bottom of the s-block) is the most electropositive of these
3. So RbOH is the strongest base

Teacher's Note:
1. Group 1 hydroxides are stronger bases than group 2 hydroxides of the same period. Al(OH)3 is amphoteric and P(OH)3 (H3PO3) is acidic.