JEE Chemistry Structure Of Atom MCQs Set 05

Answer: (b) \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^9 4s^2\)

Explanation:
1. (a) is Ni, (c) is Cu and (d) is Cr, all correct ground states.
2. (b) would be Cu with a 3d9 4s2 arrangement, but a completely filled d subshell is more stable, so one 4s electron shifts to give \(3d^{10}4s^1\).
3. So (b) is not a ground state configuration.

Teacher's Note:
1. Cr and Cu are the two common exceptions in the 3d series because half-filled and fully filled d subshells are extra stable.
2. Always check the atomic number by adding up the electrons before judging a configuration.

Answer: (c) the sum of protons and neutrons is the same but the number of protons is different

Explanation:
1. Isobars are atoms of different elements with the same mass number.
2. Mass number is protons plus neutrons, so the sum is equal while the atomic numbers differ.
3. (a) describes isotopes and (c) describes isotones.

Teacher's Note:
1. Isotopes share Z, isobars share A, isotones share the number of neutrons.
2. Remember them as same protons, same mass number, same neutrons respectively.

Answer: (c) 1

Explanation:
1. Radial nodes = n - l - 1.
2. For 3p, n = 3 and l = 1, so nodes = 3 - 1 - 1 = 1.

Teacher's Note:
1. Total nodes are n - 1, of which l are angular nodes.
2. Radial nodes are spherical shells of zero probability.

Answer: (b) 2

Explanation:
1. The number of angular nodes equals the azimuthal quantum number l.
2. For a d orbital l = 2, so there are 2 angular nodes.

Teacher's Note:
1. Angular nodes are planes or cones through the nucleus.
2. Radial nodes for 4d are n - l - 1 = 1.

Answer: (d) \(\ce{Fe^{3+}}\), \(\ce{Mn^{2+}}\)

Explanation:
1. \(\ce{Fe^{3+}}\) has 26 - 3 = 23 electrons, i.e. \([Ar]3d^5\).
2. \(\ce{Mn^{2+}}\) has 25 - 2 = 23 electrons, also \([Ar]3d^5\).
3. Cr3+ is 3d3, Co3+ is 3d6 and Sc3+ is 3d0, so the others do not match.

Teacher's Note:
1. Remove 4s electrons first and then 3d when forming transition metal cations.
2. Same electron count does not guarantee the same configuration, so write each one out.

Answer: (b) The two electrons in the 2s orbital have opposite spin quantum numbers

Explanation:
1. By Pauli's principle two electrons in one orbital must have opposite spins.
2. In a many-electron atom 2s and 2p differ in energy and in effective nuclear charge because of penetration and shielding.
3. The 1s electron is much less shielded than 2s, so their \(Z_{eff}\) differ.

Teacher's Note:
1. In many-electron atoms energy depends on both n and l.
2. Penetration order is s greater than p greater than d.

Answer: (a) Both A and R are true and R is the correct explanation of A

Explanation:
1. Isotopes have the same number of electrons and hence the same electronic configuration.
2. Chemical behaviour depends on the electrons, so the reason explains the assertion.

Teacher's Note:
1. Isotopes differ in mass, not in chemistry.
2. Only rates of some reactions change slightly (isotope effect).

Answer: (b) \(1.214\times10^{-7}\) m

Explanation:
1. \(\Delta E = 2.178\times10^{-18}(1 - 1/4) = 1.6335\times10^{-18}\) J.
2. \(\lambda = hc/\Delta E = (6.62\times10^{-34})(3\times10^8)/(1.6335\times10^{-18})\).
3. This gives \(\lambda \approx 1.216\times10^{-7}\) m, closest to \(1.214\times10^{-7}\) m.

Teacher's Note:
1. This is the first line (Lyman alpha) of the hydrogen spectrum in the ultraviolet.
2. Use \(\lambda = hc/\Delta E\) with ΔE in joules per atom.

Answer: (b) 743 nm

Explanation:
1. Energy is conserved: \(1/\lambda_{abs} = 1/\lambda_1 + 1/\lambda_2\).
2. \(1/\lambda_2 = 1/355 - 1/680 = 0.0028169 - 0.0014706 = 0.0013463\) per nm.
3. So \(\lambda_2 \approx 743\) nm.

Teacher's Note:
1. The absorbed photon energy equals the sum of the emitted photon energies when the atom returns by two steps.
2. Work with reciprocals of wavelength, not wavelengths themselves.

Answer: (a) 494 nm

Explanation:
1. Energy per bond = 242000 / 6.022e23 = 4.02e-19 J.
2. \(\lambda = hc/E = (6.626\times10^{-34})(3\times10^8)/(4.02\times10^{-19}) = 4.95\times10^{-7}\) m.
3. So \(\lambda \approx 494\) nm.

Teacher's Note:
1. The longest wavelength corresponds to the minimum energy that is just enough to break the bond.
2. Always convert molar energy to energy per molecule first.

Answer: (c) \(1.92\times10^{-3}\) m

Explanation:
1. \(\Delta v = 600 \times 0.005/100 = 0.03\) m/s.
2. \(\Delta x = h/(4\pi m \Delta v) = 6.63\times10^{-34}/(4\pi \times 9.1\times10^{-31}\times 0.03)\).
3. This equals \(1.93\times10^{-3}\) m, i.e. about \(1.92\times10^{-3}\) m.

Teacher's Note:
1. Convert the percentage accuracy into an absolute uncertainty in velocity first.
2. Use \(\Delta x\,\Delta p = h/4\pi\) for the minimum uncertainty.

Answer: (d) \(1.92\times10^{-2}\) m

Explanation:
1. \(\Delta v = 300\times10^{-5} = 3\times10^{-3}\) m/s.
2. \(\Delta x = h/(4\pi m\Delta v) = 6.63\times10^{-34}/(4\pi\times9.1\times10^{-31}\times3\times10^{-3})\).
3. \(\Delta x \approx 1.93\times10^{-2}\) m, about \(1.92\times10^{-2}\) m.

Teacher's Note:
1. 0.001% means a fraction of \(10^{-5}\).
2. Mass of the electron is 9.1 x 10^-31 kg, so keep SI units throughout.

Answer: (d) 3s, 3p and 3d orbitals all have the same energy

Explanation:
1. In a one-electron species the energy depends only on n.
2. So all orbitals of the same shell are degenerate.
3. Statements (b), (c) and (d) wrongly separate orbitals of the same shell.

Teacher's Note:
1. Degeneracy within a shell is lifted only in many-electron atoms.
2. Hydrogen, He+ and Li2+ behave in this one-electron way.

Answer: (c) 12 and 5

Explanation:
1. Cr is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1\).
2. p electrons (l = 1): 6 in 2p + 6 in 3p = 12.
3. d electrons (l = 2): 5.

Teacher's Note:
1. Cr is an exception: \(4s^1 3d^5\) instead of \(4s^2 3d^4\).
2. Count electrons subshell by subshell.

Answer: (b) -1.51 eV

Explanation:
1. The second excited state is n = 3.
2. \(E_3 = -13.6/9 = -1.51\) eV.

Teacher's Note:
1. Ground state is n = 1, first excited n = 2 and second excited n = 3.
2. Energy in hydrogen is \(-13.6/n^2\) eV.

Answer: (b) energy change and frequency

Explanation:
1. Energies add: \(\Delta E = \Delta E_1 + \Delta E_2\).
2. Frequency is \(\Delta E/h\), so it adds too.
3. Wavelength is inversely related to energy, so wavelengths do not add.

Teacher's Note:
1. Wavelength adds only in reciprocal form: \(1/\lambda = 1/\lambda_1 + 1/\lambda_2\).
2. Remember \(E = h\nu = hc/\lambda\).

Answer: (c) 4

Explanation:
1. Bohr's condition gives \(2\pi r = n\lambda\).
2. So the number of waves in one revolution equals n.
3. For the fourth orbit n = 4.

Teacher's Note:
1. This is the de Broglie standing wave condition.
2. The number of waves equals the principal quantum number.

Answer: (a) -108.8 eV

Explanation:
1. Total energy of the first orbit of He+ is \(-13.6\times 2^2 = -54.4\) eV.
2. Potential energy is twice the total energy, so PE = -108.8 eV.
3. Kinetic energy is +54.4 eV.

Teacher's Note:
1. For Bohr orbits: KE = -E, PE = 2E.
2. Total energy scales with \(Z^2\).

Answer: (c) 3s < 3p < 4s < 3d

Explanation:
1. Use (n + l): 3s = 3, 3p = 4, 4s = 4, 3d = 5.
2. 3p and 4s tie, and the one with lower n (3p) is lower.
3. The order is 3s < 3p < 4s < 3d.

Teacher's Note:
1. This is the Madelung rule used for neutral atoms.
2. Within the same (n + l) the lower n is lower in energy.

Answer: (c) p electrons in Cl (Z = 17)

Explanation:
1. \(\ce{Fe^{2+}}\) has 6 d electrons.
2. Mg has 1s2 2s2 3s2, i.e. 6 s electrons. Ne has 6 p electrons. Fe has 6 d electrons.
3. Cl has 2p6 3p5, i.e. 11 p electrons, which is not 6.

Teacher's Note:
1. Write each configuration and count by subshell type.
2. Do not mix up total electrons with electrons of one subshell type.

Answer: (b) 1

Explanation:
1. Three quantum numbers n, l, m completely define an orbital.
2. So only one orbital, 3p with m = 0, fits.

Teacher's Note:
1. n sets the shell, l the subshell and m the orbital.
2. The spin quantum number s only decides which electron in the orbital.

Answer: (c) n = 3, l = 2, m = -3, s = -1/2

Explanation:
1. For l = 2, m can only be -2, -1, 0, +1, +2.
2. So m = -3 in (b) is not allowed.
3. The other three sets obey all the rules.

Teacher's Note:
1. m takes 2l + 1 values from -l to +l.
2. Check l less than n first, then m, then s.

Answer: (a) (ii), (iv) and (v)

Explanation:
1. (ii) is impossible because l must be less than n.
2. (iv) is impossible because l = 0 allows only m = 0.
3. (v) is impossible because l = 2 allows m only from -2 to +2.
4. (i) and (iii) are allowed.

Teacher's Note:
1. Test each set for l less than n, then m between -l and +l.
2. Spin is always +1/2 or -1/2.

Answer: (c) 1

Explanation:
1. n, l and m together specify a single orbital.
2. Here it is one particular 3d orbital.

Teacher's Note:
1. Spin is not needed to identify an orbital.
2. Two electrons of opposite spin can occupy that one orbital.

Answer: (d) 20

Explanation:
1. Power = (photons per second) x hc/λ.
2. Equal photon rates mean power is inversely proportional to wavelength.
3. Yellow power = 100 x 8000/4000 = 200 W = 20 x 10, so x = 20.

Teacher's Note:
1. Shorter wavelength photons carry more energy.
2. Power ratio equals the ratio of photon energies when photon rates are equal.

Answer: (a) 1

Explanation:
1. Let hc/λ = E. Then E - φ = 1 and 3E - φ = 5.
2. Subtracting gives 2E = 4, so E = 2 eV.
3. Hence φ = 2 - 1 = 1 eV.

Teacher's Note:
1. A third of the wavelength means three times the photon energy.
2. Subtracting the two equations eliminates the work function.

Answer: (a) 4

Explanation:
1. Photon energy = 1240/300 = 4.13 eV.
2. The effect occurs when the work function is less than 4.13 eV.
3. Li, Na, K and Mg qualify, so the number is 4.

Teacher's Note:
1. Use E (eV) = 1240 / λ (nm).
2. Equal work function and photon energy would give zero kinetic energy.