NCERT Based Test: Periodic Trends in Properties of Elements - Free MCQ

'a' Represents covalent radius which is the distance between the nuclei of two bonded atoms 'b' represents van der waals'radius which is the distance between nuclei of two closest molecules.

Atomic radius decreases in a period from left to right increases in a group from top to bottom.

Size of atoms decreases with increases in atomic number in a period.

Ionic radius in a group increases from top to bottom since one energy shell is added with each period

Option C is not correct.

Both ions have 18 electrons and the electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁶.

For isoelectronic species, ionic radius decreases with increasing nuclear charge because a larger positive nuclear charge pulls the electron cloud more strongly.

K⁺ has atomic number 19 whereas Cl^- has atomic number 17; therefore the effective nuclear attraction on the same electron cloud is greater for K⁺, making K⁺ smaller than Cl^-.

Hence the statement that K⁺ is bigger than Cl^- is false, so the correct answer (the statement that is not correct) is Option C.

M²⁺ → M³⁺ + e^- ; .IE₃ > IE₂ > IE₁

Third ionisation energy is higher than second ionisation energy which is higher than the first ionisation energy.

After losing one electron, the atom will get a stable configuration and its second ionisation energy will be very high. Hence, option (b) is correct.

Removal of electron from p-orbital is easier than from s-orbital because of less penetrating power of p-orbitals.

Beryllium halides are covalent in nature. This is due to small size and high charge of Be²⁺ ion i.e., it has high polarising power. However, the halides of the other alkaline earth metals (fluorides, chlorides, bromides and iodides) are ionic solids.

A sudden large jump between the values of 2^nd and 3^rd ionisation energies of an element indicates, that the element has two electrons in its valence shell. Then, the possible electronic configuration may be 1s², 2s², 2p⁶, 3s².

X has highest IE₁ and IE₂ hence, it is a noble gas. Y has low lE₁, but very high IE₂ hence, it is an alkali metal.
Z has low lE₁ than IE₂ and IE₂ is even lower than IE₂ of alkali metal hence, it is an alkaline earth metal.

(ii) is a reactive metal from second group, and (iii) is a reactive non-metal from 17^th group.
For second group elements, the diiFerence in  successive IE is less. The compound formed will be MX₂.

Given X = 1s² 2s² 2p⁶ 3s¹
Y = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
Less energy is required to remove an electron from X than Y, because the distance between the nucleus and valence shell in Y is greater as compared to X.

Half-filled orbitals have extra stability hence they have high ionisation energy.

Negative value of electron gain enthalpy of S is more than that of oxygen and for Cl it is more than that of fluorine.

I.E. in a group decreases and reactivity increases.

On moving down the group, the stability of-3 oxidation state decreases. This is due to the following reasons (i) On descending a group the size of the atom or ion increases. As a result, attraction of the nucleus per newly added electron decreases (ii) A large anion cannot fit easily into lattice ofa small cation, (iii) As the negative charge on the ion increases, it becomes more and more susceptible to polarisation.

Electron affinity of noble gases is zero.

Electron gain enthalpy becomes more negative across a period while it becomes less negative in a group.
However, electron gain enthalpy of O or F is less than that of the succeeding element. When electron is added to n = 2, the repulsion is more than when it is added to n = 3 in case of Cl or S.

There is more repulsion for the incoming electron when the size of atom is smaller.

Option C is correct.

Effective nuclear charge felt by an electron is given by Z_eff = Z - S, where Z is the nuclear charge and S is the screening (shielding) constant.

When an atom gains an electron to form an anion, the value of S increases because there are more electrons to screen the nucleus; hence the net Z_eff experienced by the electrons decreases, not increases.

Also, added electrons increase electron-electron repulsion, causing the outer electron cloud to expand and the ionic radius to increase. Therefore statements A, B and D are true, and statement C is the one that is not true.

Isotopes differ in physical properties.

Iodine can gain only one electron.
I + e^- → I^-

Due to stable configuration they do not accept an electron and hence they have positive electron gain enthalpy.

It is difficult to remove an electron from a highly electronegative element.

Ionisation enthalpy, electron gain enthalpy and electronegativity increases in a period from left to right while atomic radius decreases.

Electronegativity is measured on Pauling scale. Fluorine, the most electronegative element is given the value 4.0. Electronegativity increases from left to right across a period while decreases down a group.

Electronegativity decreases down the group.