JEE Chemistry Structure Of Atom MCQs Set 01

Answer: (b) First line of the Lyman series

Explanation:
1. Energy of the H-atom transition \(=13.6\left(\frac1{n_1^2}-\frac1{n_2^2}\right)\) eV
2. First Lyman line (2 to 1): 10.2 eV
3. Last Balmer line (infinity to 2): 3.4 eV; first Balmer line (3 to 2): 1.89 eV; first Paschen line (4 to 3): 0.66 eV
4. The first Lyman line is the highest

Teacher's Note:
1. Transitions to lower n-levels release more energy. Even the first line of the Lyman series beats the series limit of the Balmer series.

Answer: (d) In the hydrogen atom the energy of the 3s orbital is lower than that of the 3p orbital

Explanation:
1. The first two are forms of Heisenberg's uncertainty principle and the third is the photon energy relation, all correct
2. In a one-electron species like hydrogen, orbital energy depends only on n, so 3s and 3p have equal energy
3. The fourth statement is incorrect

Teacher's Note:
1. The order 3s < 3p < 3d holds only in multi-electron atoms because of penetration and shielding. In H the subshells of one shell are degenerate.

Answer: (a) \(3Y\)

Explanation:
1. Bohr's condition gives \(2\pi r_n=n\lambda_n\), so \(\lambda_n=\frac{2\pi r_n}{n}\)
2. \(r_n\propto n^2\), so \(\lambda_n\propto n\)
3. \(\lambda_3=3\lambda_1=3Y\)

Teacher's Note:
1. The circumference of the n-th orbit holds n de Broglie wavelengths. Since the radius grows as n squared, wavelength grows directly as n.

Answer: (b) 3

Explanation:
1. Total nodes \(=n-1=3\)
2. Radial nodes \(=n-l-1=4-0-1=3\) and angular nodes \(=l=0\)
3. Total = 3

Teacher's Note:
1. Radial nodes = n - l - 1 and angular nodes = l. An s orbital has no angular nodes.

Answer: (c) 10

Explanation:
1. The 3d subshell has \(l=2\) and so \(2l+1=5\) orbitals
2. Each orbital holds two electrons
3. Maximum \(=5\times2=10\) electrons

Teacher's Note:
1. A subshell with azimuthal quantum number l holds at most \(2(2l+1)\) electrons. A single d orbital holds only two.

Answer: (d) 16

Explanation:
1. The shell \(n=4\) has \(n^2=16\) orbitals
2. Each orbital holds exactly one electron with \(m_s=-\tfrac12\)
3. So at most 16 electrons

Teacher's Note:
1. A shell holds \(2n^2\) electrons, half with each spin. So the number with one given spin is \(n^2\).

Answer: (a) 3.70

Explanation:
1. \(\ce{Be^2+}\) has 2 electrons, \(1s^2\)
2. For a 1s electron the other 1s electron screens 0.30
3. \(Z_{eff}=Z-\sigma=4-0.30=3.70\)

Teacher's Note:
1. Slater's rule: an electron in the same 1s shell screens 0.30, not 1. After removing both outer electrons the 1s shell is the outermost shell.