JEE Chemistry Structure of the Atom MCQs Set 07

Answer: (c) 16

Explanation:
1. Orbitals in a shell = \(n^2\).
2. For n = 4 that is 16 (1 + 3 + 5 + 7).

Teacher's Note:
1. Each (\(\ell\), m) pair is one orbital.
2. The shell holds 32 electrons.

Answer: (d) 1

Explanation:
1. The four quantum numbers together identify a unique electron.
2. By the Pauli exclusion principle no two electrons share all four.
3. So only 1 electron.

Teacher's Note:
1. A complete set of four quantum numbers describes only one electron.
2. This is the Pauli exclusion principle.

Answer: (b) \(\ell\) = 1

Explanation:
1. 4d has n = 4 and \(\ell\) = 2.
2. m can be -2, -1, 0, +1, +2, so m = 1 and m = 2 are possible.
3. \(\ell\) = 1 corresponds to a p orbital, so it is not possible for 4d.

Teacher's Note:
1. d orbitals have \(\ell\) = 2.
2. The magnetic quantum number runs from -\(\ell\) to +\(\ell\).

Answer: (b) 8

Explanation:
1. The configuration is 1s2 2s2 2p6 3s2 3p6 3d3 4s2.
2. s electrons: 2 (1s) + 2 (2s) + 2 (3s) + 2 (4s) = 8.

Teacher's Note:
1. Write the full configuration from the shell populations.
2. The atom has 23 electrons (vanadium).

Answer: (c) 5f, 7

Explanation:
1. \(\ell\) = 3 is an f subshell with \(2\ell + 1 = 7\) orbitals.
2. Option (b) gives 14, which is the electron capacity of the subshell and not the number of orbitals.
3. Options (a) and (d) have the wrong \(\ell\) (p has \(\ell\) = 1 and d has \(\ell\) = 2).
4. So the correct choice is 5f with 7 orbitals.

Teacher's Note:
1. Orbitals = \(2\ell + 1\), electrons = \(2(2\ell + 1)\).
2. 14 is the electron capacity, not the orbital count.

Answer: (d) two electrons with opposite spin

Explanation:
1. One f orbital is a single orbital, and any orbital holds at most 2 electrons.
2. By the Pauli principle they must have opposite spins.
3. The whole f subshell (7 orbitals) holds 14 electrons.

Teacher's Note:
1. Distinguish between an orbital and a subshell.
2. An orbital: 2 electrons. The f subshell: 14 electrons.

Answer: (c) both (a) and (b) are correct

Explanation:
1. 3p means n = 3 and \(\ell\) = 1.
2. The orbitals p\(_x\) and p\(_y\) are linear combinations of the m = +1 and m = -1 wave functions, so either value can be assigned by convention.
3. Hence both (a) and (b) can be correct.

Teacher's Note:
1. Only p\(_z\) corresponds to a pure m = 0 orbital.
2. m = +1 and -1 are used for p\(_x\) and p\(_y\) by convention.

Answer: (c) d

Explanation:
1. The number of angular nodes equals \(\ell\).
2. A d orbital has \(\ell\) = 2, so two angular nodal surfaces.
3. s has none, p has one and f has three.

Teacher's Note:
1. Angular nodes = \(\ell\).
2. In d\(_{xy}\) the two nodal planes are the xz and yz planes.

Answer: (d) \(\sum_{\ell=0}^{\ell=n-1} 2(2\ell + 1)\)

Explanation:
1. Each subshell has \(2\ell + 1\) orbitals and each orbital holds 2 electrons.
2. \(\ell\) runs from 0 to n - 1.
3. Total = \(\sum_{\ell=0}^{n-1} 2(2\ell + 1) = 2n^2\).

Teacher's Note:
1. The sum equals \(2n^2\).
2. Check with n = 2: 2 + 6 = 8.

Answer: (b) [up-down] [up-up] [up]

Explanation:
1. Pauli's principle forbids two electrons with all four quantum numbers equal.
2. In arrangement (b) the second box contains two electrons with the same spin, which is not allowed.
3. The other arrangements have at most two electrons per box with opposite spins.

Teacher's Note:
1. Two electrons in one orbital must have opposite spins.
2. Check each box in turn.

Answer: (a) 2s [up-down]; 2p [up] [up] [empty]

Explanation:
1. Hund's rule says orbitals of equal energy are singly occupied with parallel spins before pairing begins.
2. In (a) the two 2p electrons are in different orbitals with parallel spins, so the rule is obeyed.
3. In (b) and (d) the electrons in separate orbitals have opposite spins, and in (c) an electron pairs up while a 2p orbital is still empty, so all three violate the rule.

Teacher's Note:
1. Fill degenerate orbitals one electron at a time, all with the same spin.
2. Pair up only when every orbital in the subshell has one electron.

Answer: (d) 6d

Explanation:
1. (n + \(\ell\)) values: 3d = 5, 5p = 6, 4s = 4, 6d = 8.
2. The largest value belongs to 6d.

Teacher's Note:
1. Higher (n + \(\ell\)) means higher energy.
2. If equal, the larger n has higher energy.

Answer: (a) B is more than A

Explanation:
1. A is 3d with (n + \(\ell\)) = 5; B is 5s with (n + \(\ell\)) = 5.
2. When (n + \(\ell\)) is equal, the orbital with the lower n has the lower energy.
3. So A (n = 3) is lower and B (n = 5) is higher in energy.

Teacher's Note:
1. Use the (n + \(\ell\)) rule first.
2. For equal (n + \(\ell\)), the orbital with the lower n has the lower energy.

Answer: (a) 4

Explanation:
1. Possible pairs with n + \(\ell\) = 7 and \(\ell\) < n: (n = 7, \(\ell\) = 0) 7s, (n = 6, \(\ell\) = 1) 6p, (n = 5, \(\ell\) = 2) 5d and (n = 4, \(\ell\) = 3) 4f.
2. (n = 3, \(\ell\) = 4) is not allowed since \(\ell\) must be less than n.
3. So there are 4 subshells.

Teacher's Note:
1. List the allowed (n, \(\ell\)) pairs one by one.
2. The restriction \(\ell\) < n limits the count.

Answer: (a) 1/2

Explanation:
1. d9 has four paired orbitals and one orbital with a single electron.
2. Paired electrons cancel their spins, leaving one unpaired electron with spin +1/2.
3. The total spin is 1/2.

Teacher's Note:
1. Total spin = (number of unpaired electrons) x 1/2.
2. d9 has 1 unpaired electron.

Answer: (d) all of the above

Explanation:
1. In the hydrogen atom the energy depends only on n.
2. All the orbitals 3s, 3p and 3d have the same energy.
3. So none of these transitions changes the energy, and there is no absorption or emission.

Teacher's Note:
1. Orbitals with the same energy are called degenerate.
2. Hydrogen is a one-electron system with no subshell splitting.

Answer: (b) n = 3, \(\ell\) = 0, m = 0, s = +1/2

Explanation:
1. Sodium is 1s2 2s2 2p6 3s1.
2. The last electron is in the 3s orbital: n = 3, \(\ell\) = 0, m = 0.
3. By convention its spin is +1/2.

Teacher's Note:
1. For \(\ell\) = 0, m can only be 0.
2. The first electron in an orbital is conventionally given s = +1/2.

Answer: (a) 4

Explanation:
1. Cr is [Ar] 3d5 4s1.
2. The 4s electron and one 3d electron are lost, giving [Ar] 3d4.
3. d4 has 4 unpaired electrons.

Teacher's Note:
1. For transition metals, the 4s electron is removed first, but Cr has only one 4s electron.
2. The second electron comes from 3d.

Answer: (b) N

Explanation:
1. Three electrons in 2p give 1s2 2s2 2p3, so Z = 7 (nitrogen).
2. Atomic weight = 2 x 7 = 14, which matches nitrogen.
3. The element is N.

Teacher's Note:
1. Fill the electrons from the configuration to get Z.
2. Check with the mass condition.

Answer: (c) 3

Explanation:
1. The element is chlorine, 1s2 2s2 2p6 3s2 3p5.
2. The valence shell n = 3 has 3s2 (one paired orbital) and 3p5 (two paired orbitals and one half-filled).
3. Total orbitals with pairs = 1 + 2 = 3.

Teacher's Note:
1. Chlorine has 7 valence electrons.
2. 3p5 gives two filled p orbitals and one singly occupied.

Answer: (b) 26

Explanation:
1. \(\ce{X^{3+}}\) has [Ar] 3d5, i.e. 18 + 5 = 23 electrons.
2. The neutral atom has 3 more, so Z = 26.
3. The element is iron.

Teacher's Note:
1. Add back the electrons lost to find the atomic number.
2. Fe is [Ar] 3d6 4s2 in the neutral state.

Answer: (c) 30

Explanation:
1. Zn (Z = 30) is [Ar] 3d10 4s2, so its outermost shell is 4s2.
2. Cu (29) and Cr (24) have 4s1 and K (19) has 4s1.
3. So the answer is Z = 30.

Teacher's Note:
1. The outer shell is the shell with the largest n.
2. Cu and Cr are exceptions with one 4s electron.

Answer: (b) 12

Explanation:
1. \(\ce{Fe^{2+}}\) is d6, which has 1 pair, i.e. 2 paired electrons. \(\ce{Co^{2+}}\) is d7, with 2 pairs = 4 paired electrons. \(\ce{Ni^{2+}}\) is d8, with 3 pairs = 6 paired electrons.
2. Total = 2 + 4 + 6 = 12.

Teacher's Note:
1. Fill d orbitals singly first, then pair.
2. Paired electrons = 2 x number of pairs.

Answer: (a) 15

Explanation:
1. A p subshell has at most 3 unpaired electrons, when it is half filled (p3).
2. The 3p3 configuration belongs to phosphorus, Z = 15.

Teacher's Note:
1. Half-filled p subshells have the maximum number of unpaired electrons.
2. 3p3 comes after 3s2: 1s2 2s2 2p6 3s2 3p3.

Answer: (a) 6

Explanation:
1. The atom is 1s2 2s2 2p6 3s2 3p6 4s2 (calcium).
2. m = 0 orbitals: 1s, 2s, one 2p, 3s, one 3p, 4s, which is 6 orbitals.
3. Each such orbital has one electron with s = +1/2, so 6 electrons.

Teacher's Note:
1. In each orbital with m = 0, one electron has +1/2 spin and one has -1/2.
2. Count the orbitals with m = 0.

Answer: (d) 13

Explanation:
1. Cr is 1s2 2s2 2p6 3s2 3p6 3d5 4s1.
2. The M shell (n = 3) has 3s2 + 3p6 + 3d5 = 13 electrons.

Teacher's Note:
1. Cr has the exceptional configuration 3d5 4s1.
2. Count the electrons with n = 3.