NJC 3 Atomic Structure Checkpoint

National Junior College SH1 H2 Chemistry 1 Checkpoint 1 1 State the number of protons, electrons and neutrons for the following particles: Particles Nucleon number No. of protons No. of neutrons No. of electrons P31 15 31 15 16 15 +240 20Ca 40 20 20 18 −232 16 S 32 16 16 18 Generally, except for 𝑯𝟏 𝟏 , number of neutrons  number of protons for other elements. 2 Which of the ions has more electrons than protons and more protons than neutrons? [Given: H]DO,H,He, 2 1 16 8 1 1 4 2 = A He+ B OH− C D3O+ D OD− Nucleon number No. of protons No. of neutrons No. of electrons He+ 4 2 2 1 OH− 17 9 8 10 D3O+ 22 11 11 10 OD− 18 9 9 10 3 Oxygen has two isotopes, 16O and 18O. An oxide ion, −yxO8 has the same number of neutrons and electrons. What could be the values of x and y? x y A 16 1 B 16 2 C 18 1 D 18 2 x y No. of neutrons No. of electrons A 16 1 8 9 B 16 2 8 10 C 18 1 10 9 D 18 2 10 10

National Junior College SH1 H2 Chemistry 2 4 In 1999 Russian chemists claimed to be the first to identify atoms of a new element of proton number 114. This was produced by bombarding atoms of plutonium, Pu, with the atoms of an isotope of a group 2 element, X. The reaction taking place is as shown below. 𝑃𝑢 + 𝑿 → [𝑛𝑒𝑤 𝑒𝑙𝑒𝑚𝑒𝑛𝑡] + 3 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 ( 𝑛0 1 )𝟏𝟏𝟒 𝟐𝟖𝟗 94 244 What is X? A Mg B Ca C Sr D Ba [N03/1/4] Nucleon number and charge are always conserved. Proton number is also conserved since electrons are not involved in this nuclear reaction. To identify X, determine its proton number. proton number of 114 – 94 = 20.  X is Ca.

National Junior College SH1 H2 Chemistry 3 Checkpoint 2 The angle of deflection, θ, is proportional to the charge mass of the particles. The charge mass ratio of the O2– ion is the highest amongst the four. Hence O2– will deflect the most 2 When passed through an electric field, the 1H nucleus is deflected as shown below. [Note: 1H nucleus refers to 1H+ (all electrons removed)] Which of the above beams represents the deflection for an ion 2X2–? θ = k × charge mass for 1H+, +4 = k × +1 1 , k = 4 for 2X2–, angle of deflection for 2X2– = 4 × ( −2 2 ) = – 4o 3 A plasma is a gaseous mixture in which atoms have been completely stripped of their electrons, leaving bare nuclei. When passed through an electric field, the 1H nucleus is deflected at an angle of +4 . What will be the angle of deflection for the 3H nucleus in the same plasma? A +0.75 B +1.3o C +4o D +12o Angle of deflection = k × charge mass For the 1H nucleus, +4 = k ×( +1 1 ). Hence k = 4. For the 3H nucleus, angle = 4 × ( +1 3 )= +1.3 1 Use of the Data Booklet is relevant to this question. Which of the ions will be deflected the most in an electric field? A S2– B Br– C F– D O2– source B C D A _ + ⬚1𝐻 nucleus +𝟒𝒐

National Junior College SH1 H2 Chemistry 4 4 The five isotopes of krypton occur in the following abundances: Isotopic mass % abundance 80 2 82 12 83 12 84 57 86 17 Use these data to calculate a value, to one decimal place, for the Ar of the atmospheric krypton. Ar = 2 100 × 80 + 12 100 × 82 + 12 100 × 83 + 57 100 × 84 + 17 100 × 86 = 83.9 5 The relative atomic mass of boron, which consists of the isotopes 𝐵5 10 and 𝐵5 11 is 10.8. What is the percentage of 𝐵5 11 atoms in the isotopic mixture? Let A be the fraction of 𝐵5 11 , thus the fraction of 𝐵5 10 is (1-A). 11A + (1−A) 10 = 10.8 A = 0.8 % abundance = 0.8 x 100 = 80%

National Junior College SH1 H2 Chemistry 5 Checkpoint 3 (Think-pair-share) 1 Discuss if each statement is true (T) or false (F). Provide an explanation for statement that is false. (a) There are two 2s orbitals. T / F Explanation (if any): There is only one 2s orbital. All s subshells contain one s orbital. (b) 1s subshell and 2s subshell are in the same principal quantum shell. T / F Explanation (if any): They are in different principal quantum shell. 1s subshell is in the 1st principal quantum shell, 2s subshell is in the 2nd principal quantum shell. (c) 3px orbital is at a higher energy level than 2px orbital. T / F Explanation (if any): The larger the principal quantum number, the higher the energy level, hence 3px is of a higher energy level than 2px. (d) 3px orbital is at a higher energy level than 3py orbital. T / F Explanation (if any): All the three 3p orbitals are at same energy level (degenerate). (e) 2py and 3py orbitals have the same shape but different in size. T / F Explanation (if any): All p orbitals have dumbbell shapes and 3py orbital is bigger in size (electrons are more diffused). (f) 3s, 3p and 3d orbitals are in the same subshell. T / F Explanation (if any): They are in the same principal quantum shell but different subshells. (g) Electrons orbit around the nucleus. T / F Explanation (if any): Electrons do not revolve around the nucleus at particular distance from the nucleus in one plane, instead they are found in atomic orbitals (region of 3- dimensional space where there is 95% chance of finding an electron).

National Junior College SH1 H2 Chemistry 6 Checkpoint 4(a) 1 Complete the table below with reference to the periodic table. In terms of s, p, d, f notations electron-in-boxes representation Full electronic configuration 1s 2s 2p 3s 3p 3d 4s 1H 1s1 2He 1s2 3Li 1s2 2s1 4Be 1s2 2s2 5B 1s2 2s2 2p1 6C 1s2 2s2 2p2 7N 1s2 2s2 2p3 8O 1s2 2s2 2p4 9F 1s2 2s2 2p5 10Ne 1s2 2s2 2p6 19K 1s2 2s2 2p6 3s2 3p6 4s1 20Ca 1s2 2s2 2p6 3s2 3p6 4s2 22Ti 1s2 2s2 2p6 3s2 3p6 3d2 4s2 *24Cr 1s2 2s2 2p6 3s2 3p6 3d5 4s1 *29Cu 1s2 2s2 2p6 3s2 3p6 3d10 4s1

National Junior College SH1 H2 Chemistry 7 Checkpoint 4(b) 1 Which atom has 2 unpaired electrons? 2 Which atom has a half-filled set of 4p electrons? A Al …3p1 B P ….3p3 C Ga ….4p1 D As ….4p3 ( ) 3 The table refers to the electron distribution in the second shell of an atom with eight protons. Which row is correct for this atom? Orbital shape Orbital shape Orbital type Number of electrons Orbital type Number of electrons A p 2 s 4 B p 4 s 2 C s 2 p 4 D s 4 p 2 ( ) An atom with eight protons: oxygen → electronic configuration: 1s22s22p4 Second shell : 2s22p4 5 Using the ground state valence electronic configuration of atom given below, identify the elements with reference to the periodic table. (i) 4s24p4 (ii) 5s25p1 Se and In A Ba [Xe] 6s2 (0 unpaired e−) B Ge [Ar] 3d10 4s2 4p2 (2 unpaired e− in 4p orbitals) C Fe [Ar] 3d6 4s2 (4 unpaired e− in 3d orbitals) D Sc [Ar] 3d1 4s2 (1 unpaired e− in 3d orbitals) ( )

National Junior College SH1 H2 Chemistry 8 Checkpoint 5 1. Write the electronic configuration and draw the electron-in-boxes representation to represent the electronic configurations of each of the following species. (a) 15P3− 1s22s22p63s