Question 1

Identify pair of orbitals having similar (n + l) values from the following.

Accepted Answer

The (n + l) value for an orbital is calculated by adding the principal quantum number n and the azimuthal quantum number l. For s, p, d, and f subshells, the values of l are 0, 1, 2, and 3 respectively. Calculating (n + l) for the given pairs: For 3s and 3p: (3 + 0) = 3 and (3 + 1) = 4 For 2p and 4s: (2 + 1) = 3 and (4 + 0) = 4 For 3d and 4p: (3 + 2) = 5 and (4 + 1) = 5 For 2p and 3p: (2 + 1) = 3 and (3 + 1) = 4 The pair 3d and 4p has the same (n + l) value of 5. Answer: 3d and 4p

Question 2

Calculate the shortest wavelength in hydrogen spectrum emission of Lymen series (R_H = 109677\, cm ^ -1 )

Accepted Answer

For the Lyman series, the transition is from n_2 to n_1 = 1. The shortest wavelength corresponds to the maximum energy transition, which occurs when n_2 = . Using the Rydberg formula: 1 = R_H ( 1 n_1^2 - 1 n_2^2 ) Substituting n_1 = 1 and n_2 = : 1 = R_H ( 1 1^2 - 1 ^2 ) = R_H = 1 R_H Given R_H = 109677 cm ^ -1 : = 1 109677 cm 9.117 10^ -6 cm Answer: 9.117 10^ -6 cm

Question 3

What is the wave number of lowest transition associated with Paschen series?

Accepted Answer

The wave number v for a transition in a hydrogen atom is given by the Rydberg formula: v = R_H ( 1 n_1^2 - 1 n_2^2 ) For the Paschen series, the lower energy level is n_1 = 3. The lowest transition corresponds to the minimum energy difference, which occurs for the transition from the immediately higher energy level, n_2 = 4. Substituting these values into the formula: v = R_H ( 1 3^2 - 1 4^2 ) v = R_H ( 1 9 - 1 16 ) v = R_H ( 16 - 9 144 ) v = R_H ( 7 144 ) cm ^ -1 Answer: R_H ( 7 144 )\ cm ^ -1

Question 4

Which from following colours has highest energy if wave lengths of violet, blue, yellow and red light are 410 nm, 470 nm, 580 nm and 750 nm respectively ?

Accepted Answer

The energy of a photon is given by the Planck-Einstein relation: E = hc where h is Planck's constant, c is the speed of light, and is the wavelength. From the relation, energy is inversely proportional to the wavelength (E 1 ). The given wavelengths are: Violet: 410 nm Blue: 470 nm Yellow: 580 nm Red: 750 nm Since violet light has the shortest wavelength among the given colours, it has the highest energy. Answer: Violet

Question 5

Which of the following colours of visible light has lowest energy?

Accepted Answer

The energy of a photon is given by the relation E = hc . This shows that energy is inversely proportional to the wavelength of light. In the visible spectrum (VIBGYOR), red light has the maximum wavelength. Therefore, red light has the lowest energy. Answer: Red

Question 6

Find the energy of third stationary orbit of Bohr's model of hydrogen atom if the energy of ground state is -E J

Accepted Answer

The energy of the n-th Bohr orbit of a hydrogen atom is given by E_n = E_1 n^2 Given that the energy of the ground state (n=1) is E_1 = -E For the third stationary orbit, n=3 Substituting the values, we get E_3 = -E 3^2 = - E 9 J Answer: - E 9 J

Question 7

What is uncertainty in velocity of an electron if uncertainty in measurement of position is 50 pm ? (m_e = 9.1 10^ -31 kg ,\ h = 6.63 10^ -34 Js ,\ = 3.142)

Accepted Answer

According to Heisenberg's uncertainty principle: x p h 4 x m v = h 4 v = h 4 m x Substituting the given values: v = 6.63 10^ -34 4 3.142 9.1 10^ -31 50 10^ -12 v = 6.63 10^ -34 5718.44 10^ -43 v = 1.1594 10^6 ms ^ -1 1.16 10^6 ms ^ -1 Answer: 1.16 10^6\ ms ^ -1

Question 8

What is the number of unpaired electrons present in ground state of Silicon and Chromium, respectively ?

Accepted Answer

The atomic number of Silicon (Si) is 14. The electronic configuration of Si is 1s^ 2 2s^ 2 2p^ 6 3s^ 2 3p^ 2 . In the 3p subshell, there are 2 electrons which occupy separate orbitals according to Hund's rule. Thus, Si has 2 unpaired electrons. The atomic number of Chromium (Cr) is 24. The electronic configuration of Cr is [Ar] 3d^ 5 4s^ 1 . In the 3d subshell, there are 5 unpaired electrons and in the 4s subshell, there is 1 unpaired electron. Thus, Cr has 6 unpaired electrons. The number of un

Question 9

What is the energy of an electron in a hydrogen atom in a stationary state corresponding to n = 2 ?

Accepted Answer

The energy of an electron in the nth stationary state of a hydrogen atom is given by the expression: E_n = -R_H ( 1 n^2 ) where R_H is the Rydberg constant for energy, which is equal to 2.18 10^ -18 J. For n = 2: E_2 = -2.18 10^ -18 ( 1 2^2 ) J E_2 = - 2.18 10^ -18 4 J E_2 = -0.545 10^ -18 J E_2 = -5.45 10^ -19 J Answer: -5.45 10^ -19 J

Question 10

What is the observed electronic configuration of Cu ?

Accepted Answer

The atomic number of copper (Cu) is 29. According to the Aufbau principle, the expected electronic configuration is [ Ar ]\ 3d^9\ 4s^2. However, a completely filled d-subshell (3d^ 10 ) provides extra stability due to symmetrical electron distribution and higher exchange energy. Therefore, one electron from the 4s orbital is transferred to the 3d orbital, resulting in the observed electronic configuration of [ Ar ]\ 3d^ 10 \ 4s^1. Answer: [ Ar ]\ 3d^ 10 \ 4s^1

Question 11

If the velocity of the electron in Bohr's first orbit is 2.19 10^6\ Ms ^ -1 , Calculate the de Broglie wavelength associated with it. [h = 6.626 10^ -34 \ J s & Mass of electrons = 9.10938 10^ -31

Accepted Answer

The de Broglie wavelength is given by the relation: = h mv Substituting the given values: = 6.626 10^ -34 9.10938 10^ -31 2.19 10^6 = 6.626 10^ -34 19.949 10^ -25 = 0.3321 10^ -9 \ m = 332.1 10^ -12 \ m 332\ pm Answer: 332 pm

Question 12

Calculate the de Broglie wavelength of an electron in the first Bohr orbit of a hydrogen atom if the velocity of an electron in the first orbit is 2.2 10^6 ms ^ -1 . [mass of electron = 9.1 10^ -31 kg

Accepted Answer

The de Broglie wavelength is given by the relation: = h mv Substituting the given values: = 6.626 10^ -34 9.1 10^ -31 2.2 10^6 = 6.626 10^ -34 20.02 10^ -25 = 0.331 10^ -9 m = 3.31 10^ -10 m Answer: 3.31 10^ -10 m

Question 13

If the energy of an electron in the first Bohr orbit of the H-atom is -2.18 10^ -18 J then the energy of the electron in the second orbit will be

Accepted Answer

The energy of an electron in the n-th Bohr orbit of a hydrogen atom is given by E_n = E_1 n^2 . Given E_1 = -2.18 10^ -18 J and n = 2. Substituting the values, we get: E_2 = -2.18 10^ -18 2^2 E_2 = -2.18 10^ -18 4 E_2 = -0.545 10^ -18 J. Answer: -0.545 10^ -18 J

Question 14

Which of the following pairs represents isobars?

Accepted Answer

Isobars are atoms of different elements having the same mass number but different atomic numbers. In the pair ^ 40 K _ 19 and ^ 40 Ca _ 20 , both have the same mass number 40 but different atomic numbers 19 and 20 respectively. Thus, they represent a pair of isobars. Answer: ^ 40 K _ 19 and ^ 40 Ca _ 20

Question 15

The electronic configurations of elements are as- A = 1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^2 B = 1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^5 Which of the following is formula of ionic compound that could be for

Accepted Answer

Element A has the electronic configuration 1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^2. It has 2 valence electrons in its outermost shell, so it loses 2 electrons to achieve a stable noble gas configuration, forming the A ^ 2+ ion. Element B has the electronic configuration 1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^5. It has 7 valence electrons in its outermost shell, so it gains 1 electron to achieve a stable noble gas configuration, forming the B ^ - ion. To form a neutral ionic compound, one A ^ 2+ ion comb

Question 16

What is the atomic number of an element having ns^1 electronic configuration and belonging to 3d transition series?

Accepted Answer

The 3d transition series belongs to the 4th period, so n=4. The general electronic configuration of the 3d transition series is [Ar] 3d^ 1-10 4s^ 1-2 . Elements with an ns^1 (i.e., 4s^1) configuration in this series are Chromium (Cr) and Copper (Cu) due to the extra stability of half-filled and fully-filled d-subshells. The electronic configuration of Cr (Z=24) is [Ar] 3d^5 4s^1. The electronic configuration of Cu (Z=29) is [Ar] 3d^ 10 4s^1. Thus, the atomic numbers are 24 and 29. Answer: 24 and

Question 17

What are the total number of electrons present in s and p orbitals respectively in the argon?

Accepted Answer

The atomic number of argon (Ar) is 18. The electronic configuration of argon is 1s^ 2 2s^ 2 2p^ 6 3s^ 2 3p^ 6 . The total number of electrons present in s-orbitals is 2 + 2 + 2 = 6. The total number of electrons present in p-orbitals is 6 + 6 = 12. Therefore, the number of electrons in s and p orbitals are 6 and 12 respectively. Answer: 6 and 12

Question 18

Match List I with List II List I Element List II Number of Neutron (A). ^ 24 _ 12 Mg (I). 5 (B). ^ 27 _ 13 Al (II). 12 (C). ^ 9 _ 4 Be (III). 6 (D). ^ 12 _ 6 C (IV). 14 Choose the correct answer from

Accepted Answer

The number of neutrons in an atom is calculated as the difference between its mass number (A) and atomic number (Z), i.e., n = A - Z. For ^ 24 _ 12 Mg, n = 24 - 12 = 12. This matches (II). For ^ 27 _ 13 Al, n = 27 - 13 = 14. This matches (IV). For ^ 9 _ 4 Be, n = 9 - 4 = 5. This matches (I). For ^ 12 _ 6 C, n = 12 - 6 = 6. This matches (III). The correct matching is (A) - (II), (B) - (IV), (C) - (I), (D) - (III). Answer: (A) - (II), (B) - (IV), (C) - (I), (D) - (III)

Structure of Atom — Previous Year Questions & Solutions