Engineering Chemistry (BBS00010) Study Material - 2024-2025 PDF

B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 STUDY MATERIAL ENGINEERING CHEMISTRY (BBS00010) Table of Contents MODULE III: PERIODIC PROPERTIES...

B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 STUDY MATERIAL ENGINEERING CHEMISTRY (BBS00010) Table of Contents MODULE III: PERIODIC PROPERTIES Effective nuclear charge, penetration of orbitals, variations of s, p, d and f orbital energies of atoms in the periodic table, electronic configurations. Atomic and ionic radii, ionization energies, electron affinity and electro negativity. Polarizability, oxidation states. Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 1 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 MODERN PERIODIC TABLE Modern Periodic Law: The physical and chemical properties of the elements are the periodic function of their atomic numbers. Groups When the elements are arranged in increasing order of atomic number then the elements with same chemical properties come under the same vertical columns. These vertical columns are known as groups. Periods In the modern periodic table, there are successive horizontal rows from left to right. These rows are known as periods. Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 2 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 Features of Modern Periodic Table Consists of 7 periods and 18 groups. Each period is characterized by principal quantum number (n). First period contains 2 elements: H (1) & He (2) (1s orbital is filled up). Second period contains 8 elements: From Li (3) to Ne (10) (2s and 2p orbitals are filled up). Third period contains 8 elements: From Na (11) to Ar (18) (3s and 3p orbitals are filled up). Fourth period contains 18 elements: From K (19) to Kr (36) (4s, 4p and 3d orbitals are filled up). It contains 10 transition elements (Sc (21) to Zn (30)). Fifth period contains 18 elements: From Rb (37) to Xe (54) (5s, 5p and 4d are filled up). It contains tentransition elements (Y (39) to Cd (48)). Sixth period contains 32 elements: From Cs (55) to Rn (86) (6s, 5d, 6p and 4f orbitals are filled up). Among them 14 elements are Lanthanides (from Ce (58) to Lu (71)) and 10 elements are transition elements (La (57) and Hf (72) to Hg (80)). Seventh period contains 32 elements: From Fr (87) to Og (118) (7s, 6d, 7p and 5f orbitals are filled up). Among them 14 elements are Actinides (from Th (90) to Lw (103)) and 10 elements are transition elements (Ac (89) and Rf (104) to Cn (112)). They are radioactive. Groups Group 1: Alkali Metals: Li, Na, K, Rb, Cs Group 2: Alkaline Earth Metals: Be, Mg, Ca, Sr, Ba Group 16: Chalcogens: O, S, Se, Te, Po Group 17: Halogens: F, Cl, Br, I, At Group 18/Group 0: Noble/Inert gases: He, Ne, Ar, Kr, Xe, Rn Group 11: Coinage metals: Cu, Ag, Au Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 3 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 Transition elements (d-block elements) The elements in which atoms in ground state or in stable oxidation state contain incompletely filled d orbital are known as transitional elements. General electronic configuration of transition elements: (n-1)d1-10 ns1-2 First transition series: Scandium (Sc) (21) to Zinc (Zn) (30) (Zn is a non-transitional element). Second transition series: Yttrium (Y) (39) to Cadmium (Cd) (48) (Cd is a non-transitional element). Third transition series: Lanthanum (La) (57) and Hafnium (Hf) (72) to Mercury (Hg) (80) (Hg is a non-transitional element). Fourth transition series: Actinium (Ac) (89) and Rutherfordium (Rf) (104) to Copernicium (Cn) (112) (Cn is a non-transitional element). Inner transition elements (f-block elements) In the sixth period of the periodic table, 14 elements starting from Cerium (Ce) (58) to Lutecium (Lu) (71) are placed together in the same position of the periodic table along with Lanthanum (57). They are called Lanthanides in which 4f and 5d orbitals are incompletely filled. Similarly, in the seventh period, 14 elements starting from Thorium (Th) (90) to Lawrencium (Lw) (103) have the properties similar to that of Actinium and placed along with Actinium (89). They are called Actinides in which 5f and 6d orbitals are incompletely filled. Lanthanides and Actinides are also known as rare earth metals. General electronic configuration of inner transition elements is (n-2)f1-14 (n-1)d0-1 ns2 Variation of s, p, d and f orbital energies of atom in the periodic table Orbitals can be ranked in the increasing order of orbital energy as follows: 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 4 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 Electronic Configuration Aufbau’s Principle Electrons will be filled up in orbitals having lower energy fast and when the lower energy orbitals are filled up, then they will occur orbitals of higher energy. Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 5 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 The general order of filling orbitals follows the sequence: The energy order of the subshells is generally as follows: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p Hund’s rule When electrons occupy orbitals of equal energy they don’t pair up until they have a single electron in each subshell. Pauli's Exclusion Principle It states that no two electrons in the same atom can have identical values for all four of their quantum numbers. Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 6 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 Effective nuclear charge The attractive positive charge of nuclear protons acting on valence electrons of any atom. The effective nuclear charge is always less than the total number of protons present in a nucleus due to shielding effect of inner electrons. Effective nuclear charge is behind all other periodic table tendencies. The effective nuclear charge may be approximated by the equation: Zeff = Z – S Where Z is the total nuclear charge and S is the shielding constant. Trends of effective nuclear charge Effective nuclear charge increases across the period (due to increasing nuclear charge with no significant increase in shielding effect). Effective nuclear charge decreases down a group (although nuclear charge increases down a group, shielding effect is more to counter its effect). Example: A 2s lithium electron can have 2 1s electrons between itself and the lithium nucleus. Measurements indicate the effective nuclear charge experienced by a 2s lithium electron is 0.43 times the charge of the lithium nucleus. Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 7 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 The effective nuclear charge is calculated using SLATER’S RULE: Step 1: Write the electron configuration of the atom in the following form: (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p)... Step 2: Identify the electron of interest, and ignore all electrons in higher groups (to the right in the list from Step 1). These do not shield electrons in lower groups. Step 3: Slater's Rules is now broken into two cases: If the shielding is experienced by an ns or np valence electron, i.e., placed in (ns, np) group, a. Electrons within same group shield 0.35 nuclear charge, except the 1s which shield 0.30 nuclear charge. b. Electrons within the (n-1) group(s) shield 0.85 nuclear charge. c. Electrons within the (n-2) or lower groups shield 1.00 nuclear charge. If the shielding is experienced by an (nd) or (nf) valence electron, i.e., placed in (nd) or (nf) group, a. Electrons within the same group (same n and l value) shield 0.35 nuclear charge. b. Electrons within the lower groups (same n but less l/ less n and l values) shield 1.00 nuclear charge. Summary Table Other electrons are present in Electronic Same group groups (n – 1) group (n – 2) group Same n and l Same n but less l value value (ns, np) 0.35 Not applicable 0.85 1.0 (nd) 0.35 1.0 1.0 1.0 (nf) 0.35 1.0 1.0 1.0 Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 8 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 Example 1: Calculate the effective nuclear charge for the last electron in an atom with electronic configuration 1s22s22p63s23p5. Answer: Here Z = 17 and Z* = Z – S The electronic configuration according to Slater’s rule is (1s)2 (2s, 2p)8 (3s, 3p)7, where the desired electron is present in (ns, np) group. Thus, Shielding constant (S) = (0.35 × number of other electrons in same group) + (0.85 × number of electrons in (n – 1) group) + (1.00 × total number of electrons in (n – 2) or lower groups) = (0.35 × 6) + (0.85 × 8) + (1.0 × 2) = 10.9 Hence, effective nuclear charge (Z*) = 17 − 10.9 = 6.1 Example 2: Calculate the effective nuclear charge for the 3d electron of Fe. Answer: Here Z = 26 and Z* = Z – S The electronic configuration of Fe according to Slater’s rule is (1s)2 (2s, 2p)8 (3s, 3p)8 (3d)6 (4s)2, where the desired electron is present in (nd) group. So, the two 4s electrons will not contribute in shielding as they present on the right side of 3d. Thus, Shielding constant (S) = (0.35 × 5 + 1.0 × 8 + 1.0 × 10) = 19.75 Hence, effective nuclear charge (Z*) = 26 – 19.75 = 6.25 Example 3: Calculate the effective nuclear charge for the 4p electron of Br. Answer: Here Z = 35 and Z* = Z – S The electronic configuration of Br according to Slater’s rule is (1s)2 (2s, 2p)8 (3s, 3p)8 (3d)10 (4s, 4p)7, where the desired electron is present in (ns, np) group. Thus, Shielding constant (S) = (0.35 × 6 + 0.85 × 18 + 1.0 × 10) = 27.4 Hence, effective nuclear charge (Z*) = 35 – 27.4 = 7.6 Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 9 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 Example 3: Calculate the effective nuclear charge for the 3d electron of Cr2+. Answer: Here Z = 24 and Z* = Z – S The electronic configuration of Cr2+ according to Slater’s rule is (1s)2 (2s, 2p)8 (3s, 3p)8 (3d)4, where the desired electron is present in (nd) group. Thus, Shielding constant (S) = (0.35 × 3 + 1.0 × 18) = 19.05 Hence, effective nuclear charge (Z*) = 24 – 19.05 = 4.95 PERIODIC PROPERTIES I. Atomic Radii It is the distance between the centre of the nucleus and outermost shell containing electrons of any atom. Types of atomic radii a. Covalent radius: It is equal to the half of the distance between the nuclei of two covalently bonded atoms of the same element. For homonuclear diatomic molecule, Covalent radius, (rAA) = ½ (internuclear distance). For heteronuclear diatomic molecule, rAB = rA + rB. Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 10 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 b. Van Der Waal’s radius: It is equal to the half of the distance between the nuclei of two non- bonded atoms. Trends of atomic radii along a period As we move along a period, the atomic number increases and as a result increase of nuclear charge takes place. Thus the increased nuclear charge attracts the electrons more strongly towards the nucleus and thus the atomic radii decreases. In case of inert gases, the radii is not covalent radii but Van Der Waals radii. Van Der Waals radii are larger than covalent radii. Trends of atomic radii along a group As we move downward in a group, the electrons are added to the new quantum shell i.e., principal quantum number increases by one. So the distance between the outermost electron and nucleus increases. Thus the atomic radii increases. Again the inner electron of an atom acts as a kind of screen protecting the outer electron from the influence of nuclear charge. This is the screening effect. The screening effect increases with the addition of a new shell. So the attraction between the outermost electron and nucleus decreases. This increases atomic radii. II. Ionic Radii: It is the distance between the nucleus of the ion and its outermost shell. The ionic radii of a cation is always smaller than the parent atom while the ionic radii of an anion is always larger than the parent atom. Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 11 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 Trend of ionic radii A cation is formed by the loss of electrons. So the same number of protons attract less electrons more tightly and radius of cation decreases. Thus Na+ < Na. An anion is formed by the gaining of electrons. So the same number of protons attract greater electrons less tightly and radius of anion increases. Thus Cl– > Cl. Greater the +ve charge, lesser will be the ionic radii: Fe3+ < Fe2+. So the trend of ionic radii of a given isoelectronic series is: cations < atoms < anions. Arrange Mg2+, O2–, Na+, F– in increasing order of ionic radii. Ans. Mg – 2e = Mg2+ O + 2e = O2– Na – e = Na+ F + e = F– Here, Mg2+, O2–, Na+ and F– are isoelectronic species containing 10 electrons each. Now the number of protons present in Mg2+, O2–, Na+ and F– are 12, 8, 11 and 9 respectively. 12 protons can attract 10 electrons more strongly than 8 protons. So ionic radii of above species increases in the order: Mg2+ < Na+ < F– < O2–. III. Ionization Potential (I.P.) First I.P. (IP-1) is defined as the minimum amount of energy required to remove valence electrons from isolated neutral gaseous atom in its ground state to produce a monovalent cation in its lowest energy state. Factors affecting I.P. Nuclear charge: Greater the nuclear charge, greater will be the force of attraction between valence electrons and nucleus. So more energy is required to remove the electrons. Thus I.P. increases. Principal quantum no.: Greater the principal quantum no, greater will be the distance between electron and nucleus. Thus, lower will be the force of attraction between electron and nucleus. Thus, lower will be the I.P. Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 12 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 Penetration of orbitals: Penetration of orbitals follows the order s > p > d > f. Greater the penetration of orbitals, greater will be the force of attraction between electron and nucleus. Thus, greater will be the I.P. Screening effect: Greater the no. of inner shell, greater will be the screening effect, thus lower will be the force of attraction between electron and nucleus, and thus lower will be the I.P. Stable electronic configuration: Half-filled or full filled orbitals have higher stability. Thus, greater will be the I.P. General trend of I.P. (IP-1) As we move from left to right along a period, the force of attraction between electron and nucleus increases. Thus, greater will be the I.P. Order: N > O > C > Be > B > Li Here, N has higher IP-1 than O due to the extra stability of half-filled 2p orbital while Be has higher IP-1 than B due to the extra stability of full-filled 1s orbital. As we move from top to bottom along a group, the force of attraction between electron and nucleus decreases, thus I.P. decreases. Order: Li > Na > K > Rb > Cs Why is IP-3 > IP-2 > IP-1 for a given atom? Ans. After the removal of the 1st electron from a neutral gaseous atom, a monovalent cation is formed. Thus, no of electrons decreases but total nuclear charge remains the same. This leads to an increase in effective nuclear charge. So the removal of second electron, i.e., the formation of a divalent cation form a monovalent cation is very difficult. Thus IP-2 > IP-1. The same reason is applicable for the formation of a trivalent cation form a divalent cation. Thus IP-3 > IP-2. So, IP-3 > IP-2 > IP-1 for a given atom. Which one has higher I.P. Be or B and why? Ans. Be = 1s22s2 and B=1s22s22p1 Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 13 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 In B, the last electron is in 2p whereas in Be, the last electron is in 2s. Now penetration power follows the order s > p. Greater the penetration power, greater will be the force of attraction between electron and nucleus, greater will be the I.P. For Be, 2s orbital is completely filled and it is a full filled configuration. We know full filled configuration is extremely stable. Thus I.P of Be > B. Which one has higher I.P. Cu or K and why? Ans. Cu = [Ar] 3d104s1 and K = [Ar] 4s1 In Cu, 10 3d electrons are present which screen the nucleus to a small extent. Lower the screening effect, greater will be the force of attraction between electron and nucleus. Thus I.P of Cu > K. Arrange the followings in increasing order of I.P. C, N, O. Ans. C = 1s22s22p2, N = 1s22s22p3 and O: 1s22s22p4 N has a half-filled 2p orbital. So a huge amount of energy is required to remove the electron. Thus I.P of N > C and O. Also, C has a 2p2 and O has 2p4 configuration. Removing one electron from each of the above atoms, O achieves a half-filled 2p configuration. Thus, C > O. Hence the increasing order of I.P. C, N, O is O < C < N. IV. Electron affinity (EA) The amount of energy released when an electron is added to the valence shell of a neutral gaseous atom to form an anion. Factors on which electron affinity depends: i. Atomic size: Larger size atoms have less effective nuclear charge. Hence the incoming electron experiences less forces of attraction towards the nucleus of an atom. Therefore, the electron affinity value will be smaller (more negative). Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 14 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 ii. Nuclear charge: If the nuclear charge increases, the attraction for the incoming electron increases. Hence the electron affinity value increases (becomes less negative). iii. Electronic configuration: Atoms with stable electronic configurations (half-filled and full- filled) have a less tendency to accept an electron. Hence, they possess smaller electron affinity values. Trends in electron affinity along a period As we travel from left to right on the periodic table, electron affinities become more negative meaning that the electron attachment process is more exothermic, i.e., EA increases. As elements trend to the right, the added electrons sit closer to their nuclei. These electrons exhibit a stronger attraction to the nuclei as a result of this proximity, explaining the exothermic nature of their electron affinities. Trend in electron affinity along a group As we travel down groups, electron affinities become less negative, meaning the process is more endothermic. Electron proximity to these respective nuclei also influences this phenomenon, but contrary to the previous trend, electrons are placed in higher energy levels. As we travel down a group, elements contain electrons further from their nuclei, and these electrons are bound less tightly, i.e., EA decreases. Why EA2 is +ve? Ans. Conversion of A– to A2– does not occur easily due to increased interelectronic repulsion between the –ve charge of A– and additional electrons. So work is to be done to insert the 2nd electron. So instead of releasing energy, a second electron insertion requires a huge amount of energy. So EA2 is +ve. Be, Mg and N have very less electron affinity? Ans. Be and Mg have stable full filled 2s and 3s configuration whereas N has half-filled 2p configuration. Thus they have a very less tendency to gain electrons. So E.A. is very less. Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 15 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 Which one has higher electron affinity among F and Cl and why? Ans. F has smaller size than Cl. When an electron is added to F, there exists a strong repulsion between the added electron and the electrons present in F. Cl has larger size, so there will be no repulsion, i.e., Cl can easily accommodate the incoming electron in its larger 3p valence orbital. Thus, electron affinity of Cl > F. Why do inert gases have zero electron affinity? Ans. Noble gas elements have completely filled outer-shells. Such electronic configurations are highly stable and as such noble gases find it very difficult to accept electrons. Thus, electron affinity of noble gas elements is zero. V. Electronegativity (E.N.) Electronegativity is defined as the attractive force by which atom in a molecule attracts the electrons forming a covalent bond towards itself. Trends along a period As we move from left to right along a period, electronegativity increases. Trends along a group As we move from top to bottom along a group, electronegativity decreases. Explanation Let us consider the HCl molecule. In this case the bonded pair of electrons between H and Cl will be dragged more towards Cl atom. As a result, Cl atom acquires a partial –ve charge and H atom acquires a partial +ve charge. This proves that Cl is more electronegative than H. Electronegativity is a unit less quantity. It is generally measured in Pauling’s scale (0 to 5). Fluorine is the highest electronegative element with E.N. value 4.0 and Cesium is the lowest electronegative element with E.N. value 0.8 in the Pauling’s scale. Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 16 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 VI. Polarization Due to difference in E.N. in between anions and cations, the electron density of cation is dragged towards the more electronegative anion. Thus, there occurs a repulsion in between the cation and the nucleus of anion as well as an attraction between the cations and the electron cloud of anion. This results in bulging of anion towards cation and the anion gets deformed. This phenomenon is known as polarization. The tendency of anion to get deformed is known as polarizability. The extent to which a cation can deform the anion is known as polarizability power. Fajans’ Rule Fajans formulated a rule to predict whether a chemical bond is predominantly covalent or ionic. He assumed that all compounds are ionic in nature and will develop the following covalent character: a. The size of the cation is smaller and has higher +ve charge. b. The size of anion is larger and it has higher –ve charge. c. Cations with pseudo inert gas configuration will polarize the anions to a greater extent to those having inert shell configuration. Examples: Increasing ionic character: SiCl4 < AlCl3 < MgCl2 < NaCl Increasing ionic character: NaI < NaBr < NaCl < NaF Increasing ionic character: NaCl < CuCl VII. Oxidation States (Oxidation number/O.N.) It can be defined as the total number of electrons that an atom either gains or losses in order to form a chemical bond with another atom. It could be positive, negative, or zero, or even fraction. Zero O.N.: Neutral compounds that contains atoms of only one element. Example: H2, O2, Cl2, P4, S8, etc. Positive O.N.: Atom loses electron(s) to form cations. Example: Ca to Ca2+, Na to Na+, etc. Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 17 B.Tech. CSE, B.Tech. CSE (DS/AIML), B.Tech. RA – 2024 Engineering Chemistry (BBS00010) Academic Session: 2024-2025 Negative O.N.: Atom gains electron(s) to form anions. Example: O to O2–, Cl to Cl–, CO32–, etc. Partial O.N.: KO2, Fe3O4, etc. Model Questions 1. Classify Cu and Zn as transition and non-transition elements. Justify your answer. 2. Differentiate electron affinity and electronegativity. Compare the electronegativity of nitrogen and phosphorus, and also compare their electron affinity. 3. F is higher electronegative than Cl but the electron affinity of F is less than Cl. Explain. 4. Calculate the effective nuclear charge for sulphur atom. 5. What is the oxidation number of Mn in MnO4– ion? Name of the faculty Assistant Professor, Department of Chemistry Brainware University, Kolkata 18