Q: Energy and radius of first Bohr orbit of He⁺ and Li²⁺ are [Given R_H = 2.18 \times 10^{-18} J, a_0 = 52.9 pm]

E_1(Li^{2+}) = -19.62 \times 10^{-18}\ J; r_1(Li^{2+}) = 17.6 pm; E_1(He^{+}) = -8.72 \times 10^{-18} J; r_1(He^{+}) = 26.4 pm. E_1 = -2.18 \times 10^{-18} Z^{2} J and r_1 = \dfrac{52.9}{Z} pm. For He⁺ (Z = 2): E_1 = -8.72 \times 10^{-18} J, r_1 = 26.4 pm. For Li²⁺ (Z = 3): E_1 = -19.62 \times 10^{-18} J, r_1 = 17.6 pm.

Question 1

The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes n = 2 	o n = 3 and n = 4 	o n = 6 transitions, respectively, is

Accepted Answer

\dfrac{1}{4}. \dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}}ight). For 2 	o 3: \dfrac{1}{\lambda_1} = R\left(\dfrac{1}{4} - \dfrac{1}{9}ight) = \dfrac{5R}{36}; for 4 	o 6: \dfrac{1}{\lambda_2} = R\left(\dfrac{1}{16} - \dfrac{1}{36}ight) = \dfrac{5R}{144}. So \dfrac{\lambda_1}{\lambda_2} = \dfrac{144}{36 	imes 4} \cdot \dfrac{1}{1} = \dfrac{1}{4}.

Question 2

Which of the following statements are true?
(A) Unlike Ga that has a very high melting point, Cs has a very low melting point.
(B) On Pauling scale, the electronegativity values of N and Cl are not the same.
(C) Ar, K⁺, Cl⁻, Ca²⁺, and S²⁻ are all isoelectronic species.
(D) The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is Si > Al > Mg > Na.
(E) The atomic radius of Cs is greater than that of Li and Rb.

Accepted Answer

C and E only. Ar, K⁺, Cl⁻, Ca²⁺ and S²⁻ each have 18 electrons, so (C) is true, and down the group Cs > Rb > K > Na > Li in size makes (E) true. Ga actually melts near room temperature (its melting point is low), N and Cl have the same Pauling electronegativity (3.0), and the ionization order is Si > Mg > Al > Na — so A, B and D are false.

Question 3

Match List I with List II.
List I (Ion): (A) Co²⁺  (B) Mg²⁺  (C) Pb²⁺  (D) Al³⁺
List II (Group Number in Cation Analysis): (I) Group-I  (II) Group-III  (III) Group-IV  (IV) Group-VI
Choose the correct answer from the option given below:

Accepted Answer

A-III, B-IV, C-I, D-II. In qualitative cation analysis: Co²⁺ precipitates as sulphide in Group-IV, Mg²⁺ falls in Group-VI, Pb²⁺ precipitates as chloride in Group-I, and Al³⁺ as hydroxide in Group-III. Hence A-III, B-IV, C-I, D-II.

Question 4

Predict the major product 'P' in the following sequence of reactions: 1-methylcyclopentene (i) HBr, benzoyl peroxide (ii) KCN (iii) Na(Hg)/C₂H₅OH → P (Major)

Accepted Answer

. HBr with peroxide adds anti-Markovnikov, putting Br on the ring carbon next to the methyl-bearing carbon. KCN substitutes Br by CN, and Na(Hg)/C₂H₅OH reduces the nitrile to –CH₂NH₂. The product is (2-methylcyclopentyl)methanamine — option (a).

Question 5

Energy and radius of first Bohr orbit of He⁺ and Li²⁺ are [Given R_H = 2.18 	imes 10^{-18} J, a_0 = 52.9 pm]

Accepted Answer

E_1(Li^{2+}) = -19.62 	imes 10^{-18}\ J; r_1(Li^{2+}) = 17.6 pm; E_1(He^{+}) = -8.72 	imes 10^{-18} J; r_1(He^{+}) = 26.4 pm. E_1 = -2.18 	imes 10^{-18} Z^{2} J and r_1 = \dfrac{52.9}{Z} pm. For He⁺ (Z = 2): E_1 = -8.72 	imes 10^{-18} J, r_1 = 26.4 pm. For Li²⁺ (Z = 3): E_1 = -19.62 	imes 10^{-18} J, r_1 = 17.6 pm.

Question 6

Which of the following are paramagnetic?
(A) [NiCl₄]²⁻  (B) Ni(CO)₄  (C) [Ni(CN)₄]²⁻  (D) [Ni(H₂O)₆]²⁺  (E) Ni(PPh₃)₄
Choose the correct answer from the options given below:

Accepted Answer

A and D only. In [NiCl₄]²⁻ and [Ni(H₂O)₆]²⁺ nickel is +2 (3d⁸) with weak-field ligands, leaving two unpaired electrons — both paramagnetic. Ni(CO)₄, [Ni(CN)₄]²⁻ and Ni(PPh₃)₄ have all electrons paired (strong-field or zero-oxidation-state cases), so they are diamagnetic.

Question 7

Given below are two statements:
Statement I: Like nitrogen that can form ammonia, arsenic can form arsine.
Statement II: Antimony cannot form antimony pentoxide.
In the light of the above statements, choose the most appropriate answer from the options given below:

Accepted Answer

Statement I is correct but Statement II is incorrect. All group-15 elements form EH₃ hydrides, so arsenic does form arsine (AsH₃) — Statement I is correct. They also form both E₂O₃ and E₂O₅ oxides, so antimony CAN form Sb₂O₅ — Statement II is incorrect.

Question 8

Which among the following electronic configurations belong to main group elements?
(A) [Ne]3s¹  (B) [Ar]3d³4s²  (C) [Kr]4d¹⁰5s²5p⁵  (D) [Ar]3d¹⁰4s¹  (E) [Rn]5f⁰6d²7s²
Choose the correct answer from the option given below:

Accepted Answer

A and C only. Main group means s- or p-block. [Ne]3s¹ (sodium) is s-block and [Kr]4d¹⁰5s²5p⁵ (iodine) is p-block. The others have partly filled d (or f) valence shells — transition or inner-transition elements.

Question 9

Dalton's Atomic theory could not explain which of the following?

Accepted Answer

Law of gaseous volume. Dalton's theory accounts for the mass laws but not Gay-Lussac's law of combining gaseous volumes — explaining that needed Avogadro's idea of molecules.

Question 10

Consider the following compounds: \underline{K}O_2, H_2\underline{O}_2, and H_2\underline{S}O_4. The oxidation states of the underlined elements in them are, respectively,

Accepted Answer

+1, -1, and +6. K is an alkali metal, always +1 (KO₂ is a superoxide). In H₂O₂ each oxygen is −1 (peroxide). In H₂SO₄, 2(+1) + x + 4(-2) = 0 gives S = +6.

Question 11

If the half-life (t_{1/2}) for a first order reaction is 1 minutes, then the time required for 99.9% completion of the reaction is closest to:

Accepted Answer

10 minutes. For a first-order reaction t_{99.9\%} \approx 10 	imes t_{1/2} (since 0.1\% = (1/2)^{10} of the start). With t_{1/2} = 1 min, that is 10 minutes.

Question 12

The correct order of the wavelength of light absorbed by the following complexes is,
(A) [Co(NH₃)₆]³⁺  (B) [Co(CN)₆]³⁻  (C) [Cu(H₂O)₄]²⁺  (D) [Ti(H₂O)₆]³⁺
Choose the correct answer from the options given below:

Accepted Answer

B < A < D < C. A stronger ligand field gives a larger \Delta_0, higher absorbed energy and hence shorter wavelength. Field strength runs CN⁻ > NH₃ > H₂O (and the Ti/Cu aqua complexes absorb at progressively longer wavelengths), so the wavelength order is B < A < D < C.

Question 13

Which one of the following compounds can exist as cis-trans isomers?

Accepted Answer

1, 2-Dimethylcyclohexane. Geometrical isomerism in a ring needs two ring carbons each carrying two different groups — true for 1,2-dimethylcyclohexane (the methyls can lie same side or opposite sides). Pent-1-ene and 2-methylhex-2-ene have a doubly-substituted terminal/identical-group carbon, and 1,1-substitution puts both methyls on one carbon.

Question 14

Phosphoric acid ionizes in three steps with their ionization constant values K_{a_1}, K_{a_2} and K_{a_3}, respectively, while K is the overall ionization constant. Which of the following statements are true?
(A) \log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}
(B) H₃PO₄ is a stronger acid than H₂PO₄⁻ and HPO₄²⁻
(C) K_{a_1} > K_{a_2} > K_{a_3}
(D) K_{a_1} = \dfrac{K_{a_3} + K_{a_2}}{2}
Choose the correct answer from the options given below:

Accepted Answer

A, B and C only. The overall constant is the product K = K_{a_1}K_{a_2}K_{a_3}, so its log is the sum (A). Each successive proton leaves a more negative ion, so ionization gets harder: K_{a_1} > K_{a_2} > K_{a_3} (C), which also makes H₃PO₄ the strongest acid of the three (B). (D) has no basis.

Question 15

Which one of the following reactions does NOT give benzene as the product?

Accepted Answer

Benzenediazonium chloride warmed with H₂O. Decarboxylation of sodium benzoate (a), aromatization of n-hexane (b) and cyclic polymerization of ethyne (c) all give benzene. Warming benzenediazonium chloride with water gives phenol, not benzene.

Question 16

If the molar conductivity (\Lambda_m) of a 0.050 mol L⁻¹ solution of a monobasic weak acid is 90 S cm² mol⁻¹, its extent (degree) of dissociation will be. (Assume \Lambda^{\circ}_+ = 349.6 S cm² mol⁻¹ and \Lambda^{0}_- = 50.4 S cm² mol⁻¹.)

Accepted Answer

0.225. \Lambda_m^{0} = 349.6 + 50.4 = 400 S cm² mol⁻¹, so \alpha = \dfrac{\Lambda_m}{\Lambda_m^{0}} = \dfrac{90}{400} = 0.225.

Question 17

Given below are two statements:
Statement I: A hypothetical diatomic molecule with bond order zero is quite stable.
Statement II: As bond order increases, the bond length increase.
In the light of the above statements, choose the most appropriate answer from the options given below:

Accepted Answer

Both Statement I and Statement II are false. Bond order zero (as in He₂, Be₂, Ne₂) means no bond — the molecule does not exist, so Statement I is false. Bond length is inversely related to bond order, so Statement II is also false.

Question 18

Out of the following complex compounds, which of the compound will be having the minimum conductance in solution?

Accepted Answer

[Co(NH₃)₃Cl₃]. Conductance depends on the number of ions furnished in solution. [Co(NH₃)₃Cl₃] and [Co(NH₃)₄Cl₂] (as written) are non-electrolytes with all chlorides inside the coordination sphere, so both show the minimum conductance — the official key accepts (a) and (b).

Question 19

Match List-I with List-II.
List-I: (A) XeO₃  (B) XeF₂  (C) XeOF₄  (D) XeF₆
List-II: (I) sp³d, linear  (II) sp³, pyramidal  (III) sp³d³, distorted octahedral  (IV) sp³d², square pyramidal
Choose the correct answer from the options given below:

Accepted Answer

A-II, B-I, C-IV, D-III. XeO₃ is sp³ with one lone pair — pyramidal; XeF₂ is sp³d with three lone pairs — linear; XeOF₄ is sp³d² with one lone pair — square pyramidal; XeF₆ is sp³d³ with one lone pair — distorted octahedral. Hence A-II, B-I, C-IV, D-III.

Question 20

C(s) + 2H_2(g) 	o CH_4(g); \Delta H = -74.8 kJ mol⁻¹. Which of the following diagrams gives an accurate representation of the above reaction? [R → reactants; P → products]

Accepted Answer

. The reaction is exothermic (\Delta H negative), so the products must sit 74.8 kJ mol⁻¹ BELOW the reactants on the energy axis, with an activation hump in between — the first diagram.

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