JEE Questions for Chemistry Classification Of Elements And Periodicity In Properties Quiz 6 - MCQExams.com

Within a period from left to right, atomic volume first decreases and then increases

The four elements (Li, Be, B and C) belong to the 2^nd period. As we go form Li →Be, IE₁ increases due to smaller size and increased nuclear charge.The IE₁ of B is, however, lower than that of Be because in B a 2p-electron is to be removed while in Be a 2s-electron is to be removed.The IE₁ of C is however, higher than those of B and Be due to increased nuclear charge.Thus, the overall order is : C> Be > B > Li.

Abnormally high difference between 2^nd and 3^rd  ionization energy means that the element has two valence electrons,i.e, configuration (d).

The atomic radii of C, O, F which lie in the second period increase in the order: F, O, C. further, the atomic radii of F Cl Br. which lie in the same group (group17) increase in the order : F, Cl, Br. Thus, the overall increasing order is F, O, C, Cl, Br.

Argon has the highest ionization potential

Cerium is the most common lanthanide

H- (1s²) ion is isoelctronic with He (1s²).

Since atomic radii increase down a group, therefore, the atomic radii of Na, K and Rb follow the sequence : Na < K < Rb.

Further, the atomic radii decrease along a period. Since the atomic number of Mg is 12 and that of Na is 11, therefore, the atomic radius of Mg is lower than that of Na because of increased nuclear charge. Thus, the over all sequence is : Mg < Na < K < Rb.

Due to high ionization energy and weak metallic bonding, mercury is a liquid at room temperature

The atomic number of the element with electronic configuration 1s² 2 s² 2p⁶ 3 s² 3p³ is 15. It belongs to the 3^rd period. Therefore, the atomic number of the element which lies just below it in the 4^th period is 15 + 18 = 33

Cl has the highest E.A. value.

K⁺, Ca²⁺ and Ti⁴⁺ have all argon gas configuration i.e. they have three shells. On the other hand, Al³⁺ has neon gas configuration, i.e. it has only two shells. Since the increase in size due to the addition of an extra shell is not compensated by the contractive effect of an additional positive charge, therefore, the ionic radius of Al³⁺ (0.51 Å) is smaller than that of Ti⁴⁺ (0.68 Å).

If the configuration of a transition element X is [Ar] d⁴ in its +3 oxidation state, then the configuration of the neutral metal X is [Ar] 3d⁵ 4s² and hence its atomic number must be 18 + 5 + 2 = 25

Cl^‒ and S^2‒ are isoelectronic with Ar while Na⁺ and F^‒ are isoelectronic with Ne. Therefore, S^2‒ has the largest size. 

Transition elements have 1-10 electrons in the penultimate d-orbitals. Therefore, (a) represents the general electronic configuration of the transition elements

Na⁺, Mg²⁺, F^- are isoelectronic ions. Amongst isoelectronic ions,the ion with the highest positive charge has the smallest size, i.e. Mg²⁺

The atomic sizes of Li Na, Ca, S and Xe are 1.34 Å, 1.54 Å, 1.74 Å, 1.02 Å and 2.09 Å respectively Thus, the atomic size of S (1.02 Å) is approx. half than that of Xe(2.09 Å).

In accordance with aufbau principle, gadolinium with Z = 64 must have configuration (c).

IE₂ of Mg is lower than that of Na because in case of Mg³⁺ 3 s-electron has to be removed while in case of Na⁺, an electron from the stable inert gas configuration (Neon) has to be removed.

(d) Size of cation is always smaller and size of onion is always larger than the parent atom.

Second I.P. of an element will be the energy required to remove one mole of electrons from one mole of monovalent gaseous cations of the element.

Both Na and Mg belong to 3^rd period and group 1 and 2 respectively. Therefore atomic radius of Mg < Na Also Na, K, Rb belong to group 1 and their atomic radii would increase from top to bottom. Therefore, the overall sequence of increasing atomic radius is Mg, Na, K, Rb

Ionic radius of Fe³⁺(0.64 Å) is < that of Mn³⁺(0.66 Å) .Other properties vary as indicated. 

S^2- (16 + 2 = 18 electrons) has 2, 8, 8 electronic configuration where all the electrons are paired up and shows diamagnetism

IE increases from left to right in a period but Mg due to completely filled 3s² orbital has higher IE than that of Al ( 3s² 3p¹).

H belongs to first period while Li belongs to second period. Therefore, size of Li⁺ > H⁺. Both Li⁺ and H^‒ are isoelectronic. Thus the sequence of decreasing order of the size is H^‒ > Li⁺ > H⁺.

Cl^‒ ( 2, 8, 8) does not have unpaired electrons

P⁵⁺ has more effective nuclear charge than p³⁺ and is smaller in size than p³⁺

Atomic radius decreases on going from left to right in a period.Thus, size of O > F. As O^2‒ and F^‒ are isoelectronic, therefore, size of O² > F^‒