Question 1

Match List-I with List-II. List-I Electronic configuration of neutral atom (where n=2) List-II 1^ st Ionization Energy (kJ mol^ -1 ) A. ns^2 I. 2080 B. ns^2np^1 II. 899 C. ns^2np^3 III. 800 D. ns^2np^

Accepted Answer

Given n=2, the electronic configurations correspond to elements of the second period. A. ns^2 2s^2 is Beryllium (Be). B. ns^2np^1 2s^2 2p^1 is Boron (B). C. ns^2np^3 2s^2 2p^3 is Nitrogen (N). D. ns^2np^6 2s^2 2p^6 is Neon (Ne). The first ionization energy generally increases across a period from left to right. However, Beryllium has a higher ionization energy than Boron due to the stable fully-filled 2s subshell. Nitrogen has a high ionization energy due to its stable half-filled 2p subshell. N

Question 2

In a period, the first ionisation enthalpy of the element at extreme left and the negative electron gain enthalpy of the extreme right element, except noble gases, are respectively.

Accepted Answer

In a period, as we move from left to right, the atomic size decreases and the effective nuclear charge increases. Due to the large atomic size and low effective nuclear charge, the element at the extreme left (alkali metal) has the lowest first ionisation enthalpy in the period. As we move to the extreme right (excluding noble gases), the elements (halogens) have the smallest atomic size and highest effective nuclear charge in the period. They require only one electron to complete their octet. T

Question 3

Given below are two statements: Statement I: Aluminium is more electropositive than thallium as the standard electrode potential value of E^ _ Al ^ 3+ / Al is negative and E^ _ Tl ^ 3+ / Tl is positiv

Accepted Answer

Statement I: The standard reduction potential E^ _ Al ^ 3+ / Al is -1.66 V, which is highly negative, indicating that aluminium has a strong tendency to lose electrons and is highly electropositive. The standard reduction potential E^ _ Tl ^ 3+ / Tl is +1.26 V, which is positive, indicating that thallium is less electropositive than aluminium. Thus, Statement I is true. Statement II: Boron has a very small atomic size, resulting in a very high sum of its first three ionization enthalpies. This h

Question 4

Given below are two statements : Statement I: The number of pairs among [Al_2O_3, Cr_2O_3], [Cl_2O_7, Mn_2O_7], [Na_2O, V_2O_3] and [CO, N_2O] that contain oxides of same nature (acidic, basic, neutra

Accepted Answer

Statement I: Al_2O_3 and Cr_2O_3 are both amphoteric oxides. Cl_2O_7 and Mn_2O_7 are both acidic oxides. Na_2O and V_2O_3 are both basic oxides. CO and N_2O are both neutral oxides. All 4 pairs contain oxides of the same nature. Hence, Statement I is true. Statement II: Na_2O is a strongly basic oxide. Al_2O_3 is an amphoteric oxide. CO is a neutral oxide. Cl_2O_7 is a strongly acidic oxide. Thus, the most basic oxide is Na_2O and the most acidic oxide is Cl_2O_7. Hence, Statement II is true. Bo

Question 5

Given below are two statements: Statement I: The correct order of electronegativity of fluorine, oxygen and nitrogen is F > O > N. Statement II: The oxidation state of oxygen in OF_2 is +2 and in Na_2

Accepted Answer

Electronegativity increases across a period from left to right. Nitrogen, oxygen, and fluorine belong to group 15, 16, and 17 respectively in the second period. Thus, the correct order of electronegativity is F > O > N. Statement I is true. In OF_2, fluorine is more electronegative than oxygen and always exhibits an oxidation state of -1. Let the oxidation state of oxygen be x. Then x + 2(-1) = 0 x = +2. In Na_2O, sodium is an alkali metal and exhibits an oxidation state of +1. Let the oxidation

Question 6

The 1^ st ionization enthalpy for Mg is +737 kJ/mol. The most probable estimated value of the 2^ nd ionization enthalpy of Mg is ______.

Accepted Answer

Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom or ion. It is always an endothermic process, so its value is always positive. The second ionization enthalpy is the energy required to remove an electron from a unipositive ion. Since the effective nuclear charge increases after the removal of the first electron, the remaining electrons are held more tightly by the nucleus. Thus, the second ionization enthalpy is always greater than the first ionizatio

Question 7

The electronegativity of a group 13 element 'E' is same as that of Ge (on Pauling scale and upto one decimal point). The CORRECT statements about E^ 3+ are A. It can act as a reducing agent. B. It can

Accepted Answer

The electronegativity of Ge on the Pauling scale is 1.8. In Group 13, the electronegativity values are B (2.0), Al (1.5), Ga (1.6), In (1.7), and Tl (1.8). Thus, the element 'E' is Thallium (Tl). Due to the inert pair effect, the +1 oxidation state is much more stable than the +3 oxidation state for the heavier elements of Group 13, especially Thallium. Therefore, Tl^+ is more stable than Tl^ 3+ . This makes statement C incorrect. Since Tl^ 3+ is relatively unstable, it readily accepts two elect

Question 8

Given below are two statements: Statement I: The second ionization enthalpy of B, Al and Ga is in the order of B > Al > Ga. Statement II: The correct order in terms of first ionization enthalpy is Si

Accepted Answer

Statement I: The second ionization enthalpy of Ga is higher than that of Al. This is due to the poor shielding effect of the ten 3d electrons in Ga, which increases the effective nuclear charge and holds the valence electrons more tightly. The correct order of second ionization enthalpy is B > Ga > Al. Thus, Statement I is false. Statement II: The first ionization enthalpy generally decreases down the group. However, Pb has a slightly higher ionization enthalpy than Sn due to the poor shielding

Question 9

Given below are two statements: Statement I: Among Zn, Mn, Sc and Cu, the energy required to remove the third valence electron is highest for Zn and lowest for Sc. Statement II: The correct order of t

Accepted Answer

Statement I: The third ionization energy (IE_3) corresponds to the removal of the third valence electron. Let us look at the electronic configurations of the divalent cations for the given metals: Sc^ 2+ : [Ar] 3d^ 1 Mn^ 2+ : [Ar] 3d^ 5 Cu^ 2+ : [Ar] 3d^ 9 Zn^ 2+ : [Ar] 3d^ 10 For Sc, removing the third electron involves taking it from the 3d^ 1 state, which results in a highly stable noble gas core ([Ar]). Hence, Sc has the lowest IE_3. For Zn, the third electron must be removed from a complete

Question 10

Given below are two statements : Statement (I) : The first ionisation enthalpy of the elements Na, Mg, Cl and Ar follows the order Na > Mg > Cl > Ar. Statement (II) : Among Ca, Al, Fe and B, the third

Accepted Answer

The first ionisation enthalpy generally increases across a period from left to right due to an increase in effective nuclear charge. For the third period elements Na , Mg , Cl , and Ar , the correct order of first ionisation enthalpy is Ar > Cl > Mg > Na . Thus, Statement I is false. The electronic configuration of Ca is [ Ar ] 4s^2. After the removal of two electrons, Ca ^ 2+ attains the highly stable noble gas configuration of Argon [ Ar ]. Removing a third electron requires breaking this stab

Question 11

Consider the elements N , P , O , S , Cl and F. The number of valence electrons present in the elements with most and least metallic character from the above list is respectively.

Accepted Answer

Valence electrons: N(5), P(5), O(6), S(6), Cl(7), F(7). Metallic character order in the list: P > S > N > Cl > O > F. Most metallic character: P with 5 valence electrons. Least metallic character (most nonmetallic): F with 7 valence electrons. F is the most electronegative and has the smallest atomic size, making it most nonmetallic.

Question 12

In period 4 of the periodic table, the elements with highest and lowest atomic radii are respectively.

Accepted Answer

In period 4, atomic radii decrease from left to right as nuclear charge increases while electron shell remains constant. K (potassium) is at the extreme left of period 4 metals and has the highest atomic radius. Br (bromine) is near the right end as a nonmetal and has the lowest atomic radius in this period. Therefore, the elements with highest and lowest atomic radii in period 4 are K and Br respectively.

Question 13

The correct order of C , N , O and F in terms of second ionisation potential is

Accepted Answer

Second ionization potential (I₂) is the energy to remove the second electron. After removing the first electron: C⁺ has 2p¹, N⁺ has 2p², O⁺ has 2p³ (half-filled, stable), F⁺ has 2p⁴. C has lowest I₂ as 2p electrons are least tightly bound with lowest nuclear charge. N has slightly higher I₂ than C due to increased nuclear charge. F has highest I₂ due to highest effective nuclear charge. O⁺ with half-filled 2p is relatively stable, making I₂ of O intermediate. Order: C < N < F < O

Question 14

Choose the INCORRECT statement

Accepted Answer

Analyzing each statement: 1 CO₂ is most acidic among group 14 dioxides - correct, it's a non-metal acidic oxide. 2 Carbon shows negative oxidation states (-4 in methane, -3 in ethane, etc.) along with +2 and +4 - correct. 3 Carbon covalency cannot exceed 4 - correct, as it's period 2 element unable to expand octet. 4 ¹³C is radioactive - incorrect. ¹³C is a stable, non-radioactive isotope. ¹⁴C is the radioactive isotope used in radiocarbon dating with half-life of 5730 years. Therefore, statemen

Question 15

Given below are two statements: Statement I : K > Mg > Al > B is the correct order in terms of metallic character. Statement II : Atomic radius is always greater than the ionic radius for any element.

Accepted Answer

Statement I: Metallic character decreases from left to right across a period and increases down a group. K (Group 1, Period 4) is the most metallic element among these. Mg (Group 2, Period 3) is less metallic than K. Al (Group 13, Period 3) is less metallic than Mg. B (Group 13, Period 2) is a metalloid with the least metallic character. The order K > Mg > Al > B correctly represents decreasing metallic character. TRUE. Statement II: Atomic radius compared to ionic radius depends on whether the

Question 16

Elements X and Y belong to Group 15. The difference between the electronegativity values of ' X ' and phosphorus is higher than that of the difference between phosphorus and ' Y '. ' X ' \& ' Y ' are

Accepted Answer

Group 15 elements have electronegativity values (Pauling scale): N = 3.04, P = 2.19, As = 2.18, Sb = 2.05, Bi = 2.02. The condition requires: |EN(X) - EN(P)| > |EN(P) - EN(Y)|. (1) N and As: |3.04 - 2.19| = 0.85 and |2.19 - 2.18| = 0.01. Since 0.85 > 0.01, the condition is satisfied. (2) As and Sb: |2.18 - 2.19| = 0.01 and |2.19 - 2.05| = 0.14. Since 0.01 (3) Bi and N: |2.02 - 2.19| = 0.17 and |2.19 - 3.04| = 0.85. Since 0.17 (4) As and Bi: |2.18 - 2.19| = 0.01 and |2.19 - 2.02| = 0.17. Since 0.

Question 17

Given below are two statements: Statement I : The second ionisation enthalpy of Na is larger than the corresponding ionisation enthalpy of Mg. Statement II : The ionic radius of O ^ 2- is larger than

Accepted Answer

Statement I examines second ionization enthalpy: Na (1s^2 2s^2 2p^6 3s^1) loses its first electron to achieve a stable noble gas configuration. The second ionization would break into the 2p core, requiring enormous energy. Mg (1s^2 2s^2 2p^6 3s^2) loses its second electron from the 3s orbital, which is less stable. Thus, IE_2 of Na is much larger than IE_2 of Mg. Statement I is TRUE. Statement II compares O^ 2- and F^-: both are isoelectronic with 10 electrons. O^ 2- has 8 protons while F^- has

Question 18

The correct trend in the first ionization enthalpies of the elements in the 3^ rd period of periodic table is :

Accepted Answer

First ionization enthalpy generally increases across a period due to increasing nuclear charge and decreasing atomic radius. However, exceptions occur based on electron configuration stability. For 3rd period elements, the approximate ionization enthalpy values are: Al (578), Si (787), P (1012), S (1000), Cl (1251) kJ/mol. (1) Al has lower IE than Si because removing 3p¹ is easier than removing one electron from 3p². (2) S has slightly lower IE than P despite being to the right because P has a h

Question 19

Given below are two statements : Statement I : C Statement II : S > Se > Te > Po > O is the correct order in terms of the magnitude of electron gain enthalpy values. In the light of the above statemen

Accepted Answer

Statement I: Ionization enthalpy generally increases across a period from left to right. However, Nitrogen (2s^2 2p^3) has a stable half-filled electronic configuration, making it harder to remove an electron compared to Oxygen (2s^2 2p^4). Thus, the order is C Statement I is true. Statement II: For Group 16 elements, Oxygen has an exceptionally low magnitude of electron gain enthalpy due to its small size and high inter-electronic repulsions in the 2p subshell. The magnitude of electron gain en

Question 20

Given below are two statements: Statement I : The halogen that makes longest bond with hydrogen in HX, has the smallest covalent radius in its group. Statement II : A group 15 element's hydride EH _ 3

Accepted Answer

HF F PH _3 Maximum possible covalency of phosphorous is 6

Classification of Elements and Periodicity in Properties — Previous Year Questions & Solutions