CBSE Class 11 Chemistry Structure of Atom Worksheet Set 10

Structure of Atom Worksheet

Teacher's Note:
1. Orbits come from the Bohr model; orbitals come from the quantum mechanical model.
2. Learn any three points of difference.

Teacher's Note:
1. Number of orbitals in a subshell = 2l + 1.
2. Number of subshells in a shell = n.

Question 3. Explain, with reasons, which of the following sets of quantum numbers (n, l, m_l, m_s) are not possible: (a) 0, 0, 0, +1/2; (b) 1, 1, 0, +1/2; (c) 1, 0, 0, −1/2; (d) 2, 1, 0, +1/2; (e) 3, 2, −3, +1/2.
Answer:
(a) Not possible: n cannot be zero (n = 1, 2, 3 ...).
(b) Not possible: for n = 1 only l = 0 is allowed, because l runs from 0 to n − 1.
(c) Possible: it describes a 1s electron with spin −1/2.
(d) Possible: it is a 2p electron (l = 1, m_l = 0).
(e) Not possible: for l = 2, m_l can be only −2 to +2, so −3 is not allowed.
In simple words: Check n ≥ 1, then l < n, then |m_l| ≤ l, and m_s = ±1/2.

Teacher's Note:
1. Test the four rules in order for every set.
2. Sets (c) and (d) are valid; the other three break a rule.

Question 4. Using s, p, d, f notation, name the orbital with (a) n = 1, l = 0, (b) n = 3, l = 1, (c) n = 4, l = 2 and (d) n = 4, l = 3. How many orbitals are there in each of the shells n = 1 to 4?
Answer:
(a) 1s (b) 3p (c) 4d (d) 4f.
Number of orbitals in a shell = n²: n = 1: 1; n = 2: 4; n = 3: 9; n = 4: 16.
In simple words: l = 0, 1, 2, 3 are s, p, d, f, written after the value of n.

Teacher's Note:
1. A shell with principal quantum number n has n subshells.
2. The lowest n with a g (l = 4) orbital is n = 5.

Question 5. How many electrons in an atom can have (a) n = 4 and m_s = −1/2, (b) n = 3 and l = 0, (c) n = 3, l = 1 and m_l = −1?
Answer:
(a) The n = 4 shell holds 2n² = 32 electrons, half with m_s = −1/2: 16.
(b) n = 3, l = 0 is the 3s orbital, which holds 2 electrons.
(c) n = 3, l = 1, m_l = −1 is one particular 3p orbital: 2 electrons (m_s = +1/2 and −1/2).
In simple words: Count the orbitals that fit the given numbers, then allow two electrons in each.

Teacher's Note:
1. Each orbital holds a maximum of two electrons with opposite spins.
2. Half of the electrons in any full shell have each spin.

Teacher's Note:
1. Total nodes always equal n − 1.
2. A p orbital has one nodal plane and a d orbital has two.

Question 7. Describe the shapes of s, p and d orbitals. How many of each are there in a shell, and which are non-directional?
Answer:
s orbital. Spherical and non-directional, because the probability depends only on the distance from the nucleus. The size increases with n. 1s has no node; 2s has one radial node.
p orbitals. Dumb-bell shaped, with two lobes on either side of the nucleus and a nodal plane through the nucleus. There are three (p_x, p_y, p_z), at right angles along the x, y and z axes. They first appear at n = 2.
d orbitals. Five orbitals (d_xy, d_yz, d_zx, d_x²−y² and d_z²), first appearing at n = 3. Four have a cloverleaf of four lobes; d_z² has two lobes along the z axis and a ring in the xy plane.
Only the s orbitals are non-directional.
In simple words: s is a ball, p is a dumb-bell and most d orbitals look like four-leaf clovers.

Teacher's Note:
1. The three p orbitals of a shell have equal energy (degenerate).
2. d_x²−y² has lobes along the axes, d_xy between them.

Teacher's Note:
1. 4d, 5p and 6s tie at 6, so compare their n values.
2. This is why 4f comes after 6s.

Question 9. Write the electronic configurations of Fe, Fe²⁺ and Fe³⁺ (Z = 26). Find the number of unpaired electrons in each and say which of Fe²⁺ and Fe³⁺ is more paramagnetic.
Answer:
Fe: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² → 4 unpaired electrons in 3d.
Fe²⁺: remove two electrons from the outermost 4s: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ → 4 unpaired electrons.
Fe³⁺: remove one more from 3d: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ → 5 unpaired electrons.
Fe³⁺ has more unpaired electrons, so it is more paramagnetic (and more stable because of its half-filled 3d subshell).
In simple words: Electrons leave the highest-n shell (4s) first when a cation forms.

Teacher's Note:
1. Write the neutral atom first, then remove electrons.
2. Spin-only moment of Fe³⁺ is √35 = 5.92 B.M.

Question 10. Why are the configurations of chromium and copper written as [Ar] 3d⁵ 4s¹ and [Ar] 3d¹⁰ 4s¹ instead of 3d⁴ 4s² and 3d⁹ 4s²?
Answer:
Half-filled (d5) and completely filled (d10) subshells are extra stable for two reasons.
Symmetrical distribution. Symmetry gives stability, and electrons in such subshells shield each other less, so they are held more strongly.
Exchange energy. Electrons of parallel spin in degenerate orbitals can exchange positions, releasing exchange energy. For Cr (3d⁵ 4s¹) six electrons have parallel spin, giving 15 possible exchanges, against 10 for 3d⁴ 4s².
So one 4s electron shifts to 3d in Cr and Cu.
In simple words: A half-filled or full d subshell is so stable that it is worth moving an electron from 4s.

Teacher's Note:
1. The 3d and 4s levels are very close in energy.
2. Other examples are Mo and Ag in later periods.

Question 11. Write the electronic configurations and the number of unpaired electrons for (a) P (Z = 15), (b) Si (Z = 14), (c) Cr (Z = 24) and (d) Kr (Z = 36).
Answer:
(a) P: 1s² 2s² 2p⁶ 3s² 3p³ → 3 unpaired electrons (one in each 3p orbital).
(b) Si: 1s² 2s² 2p⁶ 3s² 3p² → 2 unpaired electrons.
(c) Cr: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹ → 6 unpaired electrons.
(d) Kr: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ → 0 unpaired electrons (noble gas).
In simple words: Fill the orbitals in order, then spread electrons singly (Hund's rule) before pairing.

Teacher's Note:
1. Count the singly occupied orbitals in the last subshell.
2. Noble gases have no unpaired electrons.

Question 12. A compound of vanadium has a magnetic moment of 1.73 B.M. Find the number of unpaired electrons and the configuration of the vanadium ion. Also calculate the spin-only magnetic moments of Fe³⁺ and Ti³⁺.
Answer:
μ = √[n(n + 2)] B.M. With μ = 1.73: n(n + 2) = 3, so n = 1.
V (Z = 23) is [Ar] 3d³ 4s². One unpaired electron means the ion is V⁴⁺: [Ar] 3d¹.
Fe³⁺ (3d⁵, n = 5): μ = √35 = 5.92 B.M.; Ti³⁺ (3d¹, n = 1): μ = √3 = 1.73 B.M.
In simple words: Magnetic moment tells you how many unpaired electrons there are.

Teacher's Note:
1. More unpaired electrons means a larger magnetic moment.
2. Species with unpaired electrons are paramagnetic; with none, diamagnetic.

Question 13. Give the number of electrons in H₂⁺, H₂, O₂, O₂⁺ and O₂⁻.
Answer:
H₂⁺: 1 + 1 − 1 = 1.
H₂: 1 + 1 = 2.
O₂: 8 + 8 = 16.
O₂⁺: 16 − 1 = 15.
O₂⁻: 16 + 1 = 17.
In simple words: Add the atomic numbers, then subtract one electron for each positive charge or add one for each negative charge.

Teacher's Note:
1. Electrons are lost to make a positive ion.
2. The atomic number of O is 8 and of H is 1.