JEE Chemistry Structure Of Atom MCQs Set 14

Answer: (c) \( 8.72\times10^{-18} \) J

Explanation:
1. For a one-electron species, \( E_n=-2.18\times10^{-18}\,\frac{Z^2}{n^2} J \).
2. For \( He ^+ \), \( Z = 2 \) and \( n = 1 \), so the ionisation energy is \( 2.18\times10^{-18}\times4 \).
3. This equals \( 8.72\times10^{-18} J \).

Teacher's Note:
1. Ionisation energy of a hydrogen-like ion = (ionisation energy of H) x Z squared.
2. Remember that He+ has only one electron, so the Bohr formula applies.

Answer: (a) \( -\frac{6Ze^2}{r} \)

Explanation:
1. Electrostatic potential energy = \( -\frac{(\text{nuclear charge})(\text{electron charge})}{r} \) in Gaussian units.
2. Here the nuclear charge is Ze and the electron charge is \( 6e \).
3. So \( U=-\frac{Ze\times6e}{r}=-\frac{6Ze^2}{r} \).

Teacher's Note:
1. Potential energy depends only on the two charges and the distance, not on the mass or the speed.
2. The sign is negative because the charges attract.

Answer: (d) \( 1:3\sqrt3 \)

Explanation:
1. The kinetic energy gained is qV, so \( \lambda=\frac{h}{\sqrt{2mqV}} \).
2. For the same \( V \), \( \lambda\propto\frac{1}{\sqrt{mq}} \).
3. \( Be ^{3+} \) has mass 9 times and charge 3 times that of the proton: \( \frac{\lambda_{Be}}{\lambda_p}=\sqrt{\frac{1}{9\times3}}=\frac{1}{3\sqrt3} \).

Teacher's Note:
1. Use the accelerating-potential form of the de Broglie relation.
2. Both the mass number (9) and charge (3) of Be3+ enter under the root.

Answer: (b) \( 6.4\times10^{24} \)

Explanation:
1. Energy of one photon = \( \frac{hc}{\lambda}=\frac{6.62\times10^{-34}\times3\times10^8}{5893\times10^{-10}}=3.37\times10^{-19} J \).
2. Energy emitted in \( 10 h = 60\times10\times3600=2.16\times10^{6} J \).
3. Number of photons = \( \frac{2.16\times10^{6}}{3.37\times10^{-19}}=6.4\times10^{24} \).

Teacher's Note:
1. Total energy divided by energy per photon gives the number of photons.
2. Convert hours to seconds before multiplying by power.

Answer: (b) \( 2.646\times10^{-8} \) cm

Explanation:
1. Bohr radius \( r_n=0.529\,n^2 \) Å for hydrogen.
2. \( r_3-r_2=0.529(9-4)=5\times0.529=2.645 \) Å.
3. In cm this is \( 2.646\times10^{-8} \) cm.

Teacher's Note:
1. The radius grows as \( n \) squared, so gaps between orbits widen outward.
2. 1 Å = \( 10^{-8} \) cm.

Answer: (d) \( \sqrt{\frac{h}{m}} \)

Explanation:
1. Distance travelled in \( 1 s \) equals the speed \( v \) (numerically), so \( \lambda=v \).
2. The de Broglie relation gives \( \lambda=\frac{h}{mv} \).
3. Hence \( v=\frac{h}{mv} \), so \( v^2=\frac{h}{m} \) and \( v=\sqrt{h/m} \).

Teacher's Note:
1. Set the numerical value of the speed equal to the wavelength and solve.
2. Check units: \( h/m \) has units of \( m \) squared per second.

Answer: (a) n = 1, \( \ell \) = 0

Explanation:
1. The number of radial (spherical) nodes is \( n-\ell-1 \).
2. For \( 1s \): \( 1-0-1=0 \), so there is no spherical node.
3. \( 2s \) has \( 1, 3p \) has 1 and \( 3s \) has 2 radial nodes.

Teacher's Note:
1. Radial nodes = \( n - l - 1 \); angular nodes = \( l \).
2. Only the \( 1s \) orbital among the \( s \) orbitals has no radial node.

Answer: (c) \( \frac{h}{2\pi} \)

Explanation:
1. Angular momentum in the nth orbit is \( \frac{nh}{2\pi} \).
2. For orbits \( n+1 \) and \( n \) the difference is \( \frac{(n+1)h}{2\pi}-\frac{nh}{2\pi}=\frac{h}{2\pi} \).

Teacher's Note:
1. Angular momentum is quantised in whole multiples of \( h/2pi \).
2. The difference is independent of \( n \).

Answer: (c) \( \frac{1}{2m}\sqrt{\frac{h}{\pi}} \)

Explanation:
1. Heisenberg: \( \Delta x\,\Delta p\ge\frac{h}{4\pi} \); with equality and \( \Delta x=\Delta p \), \( (\Delta p)^2=\frac{h}{4\pi} \).
2. So \( \Delta p=\frac12\sqrt{\frac{h}{\pi}} \).
3. Then \( \Delta v=\frac{\Delta p}{m}=\frac{1}{2m}\sqrt{\frac{h}{\pi}} \).

Teacher's Note:
1. Use the minimum uncertainty product \( h/4pi \).
2. Divide the momentum uncertainty by the mass to get velocity uncertainty.

Answer: (a) 4

Explanation:
1. Bohr's quantisation gives \( 2\pi r_n=n\lambda \).
2. For the fourth orbit, the circumference contains \( n=4 \) de Broglie wavelengths.
3. So the electron makes 4 waves per revolution.

Teacher's Note:
1. The number of waves in an orbit equals the principal quantum number \( n \).
2. This is the standing-wave interpretation of Bohr's condition.

Answer: (d) \( 5.25\times10^{-30} \)

Explanation:
1. Here \( \Delta v=1\ \text{cm s}^{-1}=10^{-2}\ \text{m s}^{-1} \) and \( m=10^{-3} \) kg.
2. \( \Delta x=\frac{h}{4\pi m\Delta v}=\frac{6.626\times10^{-34}}{4\times3.14\times10^{-3}\times10^{-2}} \).
3. This gives \( \Delta x\approx5.25\times10^{-30} m \).

Teacher's Note:
1. Convert grams to kilograms and cm/s to \( m/s \) first.
2. For a macroscopic body the uncertainty is negligibly small.

Answer: (b) 1 : 4

Explanation:
1. \( E_1(H)=-13.6 \) eV.
2. For \( Be ^{3+} (Z = 4) \), first excited state \( n = 2 \): \( E=-13.6\times\frac{16}{4}=-54.4 \) eV.
3. Ratio = 13.6 : \( 54.4 = 1 \) : 4.

Teacher's Note:
1. Energy scales as \( Z \) squared over \( n \) squared.
2. The first excited state means \( n = 2 \), not \( n = 1 \).

Answer: (b) \( 4\to3 \)

Explanation:
1. \( r_n=0.529\,n^2 \) Å, so \( n_i^2-n_f^2=7 \).
2. The only pair of integers with squares differing by 7 is 16 and 9.
3. Hence the transition is from \( n = 4 \) to \( n = 3 \).

Teacher's Note:
1. Look for perfect squares that differ by 7.
2. Check the other options: \( 16-1 = 15, 16-4 = 12, 9-1 = 8 \).

Answer: (d) \( 10^{-30}:1 \)

Explanation:
1. \( \lambda=\frac{h}{mv} \), so for equal velocities \( \lambda\propto\frac1m \).
2. \( \frac{\lambda_{ball}}{\lambda_e}=\frac{m_e}{m_{ball}}=\frac{9.1\times10^{-31}}{0.91}=10^{-30} \).
3. The ratio is \( 10^{-30}:1 \).

Teacher's Note:
1. The heavier particle has the smaller wavelength.
2. This is why wave nature is not noticeable for everyday objects.

Answer: (a) 1.33 nm

Explanation:
1. The circumference holds a whole number of wavelengths: \( 2\pi r=n\lambda \).
2. For \( n = 4 \): \( \lambda=\frac{5.32}{4} \) nm = 1.33 nm.

Teacher's Note:
1. Circumference = \( n x \) wavelength.
2. The wavelength comes out comparable to atomic dimensions.

Answer: (c) The (n - 1)d subshell has higher energy than the ns subshell

Explanation:
1. Compare \( (n + l) \) values: for \( (n - 1)d \) it is \( (n-1)+2=n+1 \), for ns it is \( n \), so \( (n - 1)d \) is higher (for example \( 5d \) is above \( 6s \)).
2. Third statement: \( 6d (n + l = 8) \) lies above \( 4f (n + l = 7) \), so \( (n + 1)d \) is not lower than nf.
3. nf \( (n + 3) \) is higher than \( (n + 2)s (n + 2) \), for example \( 4f \) is above \( 6s \), so the fourth statement is false.

Teacher's Note:
1. Use the \( (n + l) \) rule to order subshell energies.
2. Examples: \( 5d \) above \( 6s, 6d \) above \( 4f, 4f \) above \( 6s \).

Answer: (c) An orbital is characterised by the three quantum numbers n, \( \ell \) and m

Explanation:
1. An orbital is a one-electron wave function, specified by \( n, l \) and \( m \).
2. It is not a trajectory, not always spherical (\( p \) and \( d \) orbitals are not), and refers to the region of high probability for one electron, not all electrons.

Teacher's Note:
1. Orbitals describe probability, not paths.
2. Spin is the fourth quantum number and is not needed to define an orbital.

Answer: (d) \( 1\times10^{-7} \) m

Explanation:
1. \( \lambda=\frac hp \), so \( \lambda\propto\frac1p \).
2. Halving the momentum doubles the wavelength: \( \lambda_B=2\times5\times10^{-8}=10^{-7} m \).

Teacher's Note:
1. Wavelength is inversely proportional to momentum.
2. Half the momentum means twice the wavelength.

Answer: (b) \( -4.41\times10^{-17} \) J per atom

Explanation:
1. Using the given value, \( E_1(\text{He}^+)=-19.6\times10^{-18} J \), so the value for hydrogen would be \( -19.6\times10^{-18}/4=-4.9\times10^{-18} J \).
2. For \( Li ^{2+} (Z = 3) \), \( E_1=-4.9\times10^{-18}\times9=-4.41\times10^{-17} J \).

Teacher's Note:
1. Ground-state energy scales as \( Z \) squared.
2. Use the data given in the question rather than the standard value.

Answer: (b) \( (n-1)d^5ns^1 \)

Explanation:
1. The maximum oxidation state equals the number of electrons that can be removed from the ns and \( (n - 1)d \) subshells.
2. \( d^5s^1 \) offers 6 electrons (\( +6 \), as in Cr); \( d^2s^2 \) offers only 4, while \( d^{10}s^2 \) and \( d^8s^2 \) are limited to about \( +2 \) or \( +3 \).
3. So \( d^5s^1 \) gives the highest state.

Teacher's Note:
1. \( d5s1 \) and \( d5s2 \) configurations give the highest oxidation states in the \( 3d \) series.
2. A completely filled \( d10 \) subshell is very stable and not involved in bonding.

Answer: (d) 2.84

Explanation:
1. \( Ni ^{2+} \) has the configuration [Ar] \( 3d ^8 \), with 2 unpaired electrons.
2. \( \mu=\sqrt{n(n+2)}=\sqrt{8}=2.83 \approx 2.84 \) BM.

Teacher's Note:
1. Count unpaired electrons, then use the spin-only formula.
2. Eight \( d \) electrons fill five orbitals leaving two unpaired.

Answer: (a) \( 1.8\times10^{-4} \) cm

Explanation:
1. \( \frac{nh}{2\pi}=4.2178\times10^{-34} \) gives \( n=4 \).
2. The transition is 4 to 3: \( \frac1\lambda=109678\left(\frac19-\frac1{16}\right)=5.33\times10^3 cm ^{-1} \).
3. So \( \lambda=1.88\times10^{-4} \) cm, closest to \( 1.8\times10^{-4} \) cm.

Teacher's Note:
1. First find \( n \) from the angular momentum.
2. Then use the Rydberg formula for the \( n \) to \( n - 1 \) line.