Crystal field stabilization energy (CFSE) - Advanced
Inorganic Chemistry Problems
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
1. Write the Lewis structure of .
2. Identify the steric number of the central atom (S).
3. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
4. Hybridization of S = sp^3d^2.
5. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
1. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
2. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
1. Identify the oxidation states and possible products.
2. Fe in is in the +2 oxidation state.
3. acts as an oxidizing agent, converting Fe from +2 to +3.
4. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
1. Fe in is in the +2 oxidation state.
2. The electron configuration of Fe2+ is [Ar] 3𝑑6.
3. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
1. The complex ion is .
2. Coordination number: 6 (number of ligands).
3. Overall charge of the complex ion = +3.
4. NH3 is a neutral ligand.
5. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
1. Identify the oxidation states of Mn in and .
2. Mn in : +7, in : +2.
3. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
4. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
5. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
1. is blue in color.
2. When treated with concentrated HCl, chloride ions () replace water ligands.
3. New complex formed: , which is yellow-green.
4. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
1. Ni in is in the +2 oxidation state.
2. The electron configuration of Ni2+ is [Ar] 3𝑑8.
3. CN− is a strong field ligand, causing pairing of electrons.
4. Configuration: (t2𝑔)6(e𝑔)2.
5. Number of unpaired electrons: 0.
6. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
7. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
1. Draw the molecular orbital diagram for .
2. Count the number of bonding and antibonding electrons.
3. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
4. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
1. NH3 is a stronger field ligand than H2O.
2. Stronger field ligands cause greater splitting of d-orbitals.
3. Greater splitting leads to more stabilization.
4. Hence, is more stable than .
5. Coordination number: 6 (number of ligands).
6. Overall charge of the complex ion = +3.
7. NH3 is a neutral ligand.
8. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
9. Write the Lewis structure of .
10. Identify the steric number of the central atom (S).
11. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
12. Hybridization of S = sp^3d^2.
13. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
14. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
15. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
16. Identify the oxidation states and possible products.
17. Fe in is in the +2 oxidation state.
18. acts as an oxidizing agent, converting Fe from +2 to +3.
19. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
20. Fe in is in the +2 oxidation state.
21. The electron configuration of Fe2+ is [Ar] 3𝑑6.
22. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
23. The complex ion is .
24. Coordination number: 6 (number of ligands).
25. Overall charge of the complex ion = +3.
26. NH3 is a neutral ligand.
27. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
28. Identify the oxidation states of Mn in and .
29. Mn in : +7, in : +2.
30. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
31. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
32. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
33. is blue in color.
34. When treated with concentrated HCl, chloride ions () replace water ligands.
35. New complex formed: , which is yellow-green.
36. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
37. Ni in is in the +2 oxidation state.
38. The electron configuration of Ni2+ is [Ar] 3𝑑8.
39. CN− is a strong field ligand, causing pairing of electrons.
40. Configuration: (t2𝑔)6(e𝑔)2.
41. Number of unpaired electrons: 0.
42. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
43. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
44. Draw the molecular orbital diagram for .
45. Count the number of bonding and antibonding electrons.
46. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
47. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
48. NH3 is a stronger field ligand than H2O.
49. Stronger field ligands cause greater splitting of d-orbitals.
50. Greater splitting leads to more stabilization.
51. Hence, is more stable than .
52. Coordination number: 6 (number of ligands).
53. Overall charge of the complex ion = +3.
54. NH3 is a neutral ligand.
55. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
56. Write the Lewis structure of .
57. Identify the steric number of the central atom (S).
58. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
59. Hybridization of S = sp^3d^2.
60. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
61. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
62. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
63. Identify the oxidation states and possible products.
64. Fe in is in the +2 oxidation state.
65. acts as an oxidizing agent, converting Fe from +2 to +3.
66. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
67. Fe in is in the +2 oxidation state.
68. The electron configuration of Fe2+ is [Ar] 3𝑑6.
69. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
70. The complex ion is .
71. Coordination number: 6 (number of ligands).
72. Overall charge of the complex ion = +3.
73. NH3 is a neutral ligand.
74. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
75. Identify the oxidation states of Mn in and .
76. Mn in : +7, in : +2.
77. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
78. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
79. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
80. is blue in color.
81. When treated with concentrated HCl, chloride ions () replace water ligands.
82. New complex formed: , which is yellow-green.
83. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
84. Ni in is in the +2 oxidation state.
85. The electron configuration of Ni2+ is [Ar] 3𝑑8.
86. CN− is a strong field ligand, causing pairing of electrons.
87. Configuration: (t2𝑔)6(e𝑔)2.
88. Number of unpaired electrons: 0.
89. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
90. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
91. Draw the molecular orbital diagram for .
92. Count the number of bonding and antibonding electrons.
93. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
94. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
95. NH3 is a stronger field ligand than H2O.
96. Stronger field ligands cause greater splitting of d-orbitals.
97. Greater splitting leads to more stabilization.
98. Hence, is more stable than .
99. Coordination number: 6 (number of ligands).
100. Overall charge of the complex ion = +3.
101. NH3 is a neutral ligand.
102. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
103. Write the Lewis structure of .
104. Identify the steric number of the central atom (S).
105. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
106. Hybridization of S = sp^3d^2.
107. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
108. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
109. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
110. Identify the oxidation states and possible products.
111. Fe in is in the +2 oxidation state.
112. acts as an oxidizing agent, converting Fe from +2 to +3.
113. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
114. Fe in is in the +2 oxidation state.
115. The electron configuration of Fe2+ is [Ar] 3𝑑6.
116. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
117. The complex ion is .
118. Coordination number: 6 (number of ligands).
119. Overall charge of the complex ion = +3.
120. NH3 is a neutral ligand.
121. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
122. Identify the oxidation states of Mn in and .
123. Mn in : +7, in : +2.
124. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
125. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
126. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
127. is blue in color.
128. When treated with concentrated HCl, chloride ions () replace water ligands.
129. New complex formed: , which is yellow-green.
130. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
131. Ni in is in the +2 oxidation state.
132. The electron configuration of Ni2+ is [Ar] 3𝑑8.
133. CN− is a strong field ligand, causing pairing of electrons.
134. Configuration: (t2𝑔)6(e𝑔)2.
135. Number of unpaired electrons: 0.
136. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
137. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
138. Draw the molecular orbital diagram for .
139. Count the number of bonding and antibonding electrons.
140. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
141. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
142. NH3 is a stronger field ligand than H2O.
143. Stronger field ligands cause greater splitting of d-orbitals.
144. Greater splitting leads to more stabilization.
145. Hence, is more stable than .
146. Coordination number: 6 (number of ligands).
147. Overall charge of the complex ion = +3.
148. NH3 is a neutral ligand.
149. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
150. Write the Lewis structure of .
151. Identify the steric number of the central atom (S).
152. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
153. Hybridization of S = sp^3d^2.
154. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
155. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
156. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
157. Identify the oxidation states and possible products.
158. Fe in is in the +2 oxidation state.
159. acts as an oxidizing agent, converting Fe from +2 to +3.
160. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
161. Fe in is in the +2 oxidation state.
162. The electron configuration of Fe2+ is [Ar] 3𝑑6.
163. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
164. The complex ion is .
165. Coordination number: 6 (number of ligands).
166. Overall charge of the complex ion = +3.
167. NH3 is a neutral ligand.
168. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
169. Identify the oxidation states of Mn in and .
170. Mn in : +7, in : +2.
171. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
172. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
173. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
174. is blue in color.
175. When treated with concentrated HCl, chloride ions () replace water ligands.
176. New complex formed: , which is yellow-green.
177. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
178. Ni in is in the +2 oxidation state.
179. The electron configuration of Ni2+ is [Ar] 3𝑑8.
180. CN− is a strong field ligand, causing pairing of electrons.
181. Configuration: (t2𝑔)6(e𝑔)2.
182. Number of unpaired electrons: 0.
183. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
184. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
185. Draw the molecular orbital diagram for .
186. Count the number of bonding and antibonding electrons.
187. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
188. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
189. NH3 is a stronger field ligand than H2O.
190. Stronger field ligands cause greater splitting of d-orbitals.
191. Greater splitting leads to more stabilization.
192. Hence, is more stable than .
193. Coordination number: 6 (number of ligands).
194. Overall charge of the complex ion = +3.
195. NH3 is a neutral ligand.
196. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
197. Write the Lewis structure of .
198. Identify the steric number of the central atom (S).
199. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
200. Hybridization of S = sp^3d^2.
201. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
202. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
203. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
204. Identify the oxidation states and possible products.
205. Fe in is in the +2 oxidation state.
206. acts as an oxidizing agent, converting Fe from +2 to +3.
207. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
208. Fe in is in the +2 oxidation state.
209. The electron configuration of Fe2+ is [Ar] 3𝑑6.
210. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
211. The complex ion is .
212. Coordination number: 6 (number of ligands).
213. Overall charge of the complex ion = +3.
214. NH3 is a neutral ligand.
215. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
216. Identify the oxidation states of Mn in and .
217. Mn in : +7, in : +2.
218. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
219. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
220. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
221. is blue in color.
222. When treated with concentrated HCl, chloride ions () replace water ligands.
223. New complex formed: , which is yellow-green.
224. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
225. Ni in is in the +2 oxidation state.
226. The electron configuration of Ni2+ is [Ar] 3𝑑8.
227. CN− is a strong field ligand, causing pairing of electrons.
228. Configuration: (t2𝑔)6(e𝑔)2.
229. Number of unpaired electrons: 0.
230. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
231. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
232. Draw the molecular orbital diagram for .
233. Count the number of bonding and antibonding electrons.
234. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
235. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
236. NH3 is a stronger field ligand than H2O.
237. Stronger field ligands cause greater splitting of d-orbitals.
238. Greater splitting leads to more stabilization.
239. Hence, is more stable than .
240. Coordination number: 6 (number of ligands).
241. Overall charge of the complex ion = +3.
242. NH3 is a neutral ligand.
243. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
244. Write the Lewis structure of .
245. Identify the steric number of the central atom (S).
246. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
247. Hybridization of S = sp^3d^2.
248. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
249. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
250. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
251. Identify the oxidation states and possible products.
252. Fe in is in the +2 oxidation state.
253. acts as an oxidizing agent, converting Fe from +2 to +3.
254. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
255. Fe in is in the +2 oxidation state.
256. The electron configuration of Fe2+ is [Ar] 3𝑑6.
257. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
258. The complex ion is .
259. Coordination number: 6 (number of ligands).
260. Overall charge of the complex ion = +3.
261. NH3 is a neutral ligand.
262. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
263. Identify the oxidation states of Mn in and .
264. Mn in : +7, in : +2.
265. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
266. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
267. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
268. is blue in color.
269. When treated with concentrated HCl, chloride ions () replace water ligands.
270. New complex formed: , which is yellow-green.
271. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
272. Ni in is in the +2 oxidation state.
273. The electron configuration of Ni2+ is [Ar] 3𝑑8.
274. CN− is a strong field ligand, causing pairing of electrons.
275. Configuration: (t2𝑔)6(e𝑔)2.
276. Number of unpaired electrons: 0.
277. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
278. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
279. Draw the molecular orbital diagram for .
280. Count the number of bonding and antibonding electrons.
281. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
282. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
283. NH3 is a stronger field ligand than H2O.
284. Stronger field ligands cause greater splitting of d-orbitals.
285. Greater splitting leads to more stabilization.
286. Hence, is more stable than .
287. Coordination number: 6 (number of ligands).
288. Overall charge of the complex ion = +3.
289. NH3 is a neutral ligand.
290. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
291. Write the Lewis structure of .
292. Identify the steric number of the central atom (S).
293. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
294. Hybridization of S = sp^3d^2.
295. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
296. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
297. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
298. Identify the oxidation states and possible products.
299. Fe in is in the +2 oxidation state.
300. acts as an oxidizing agent, converting Fe from +2 to +3.
301. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
302. Fe in is in the +2 oxidation state.
303. The electron configuration of Fe2+ is [Ar] 3𝑑6.
304. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
305. The complex ion is .
306. Coordination number: 6 (number of ligands).
307. Overall charge of the complex ion = +3.
308. NH3 is a neutral ligand.
309. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
310. Identify the oxidation states of Mn in and .
311. Mn in : +7, in : +2.
312. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
313. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
314. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
315. is blue in color.
316. When treated with concentrated HCl, chloride ions () replace water ligands.
317. New complex formed: , which is yellow-green.
318. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
319. Ni in is in the +2 oxidation state.
320. The electron configuration of Ni2+ is [Ar] 3𝑑8.
321. CN− is a strong field ligand, causing pairing of electrons.
322. Configuration: (t2𝑔)6(e𝑔)2.
323. Number of unpaired electrons: 0.
324. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
325. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
326. Draw the molecular orbital diagram for .
327. Count the number of bonding and antibonding electrons.
328. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
329. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
330. NH3 is a stronger field ligand than H2O.
331. Stronger field ligands cause greater splitting of d-orbitals.
332. Greater splitting leads to more stabilization.
333. Hence, is more stable than .
334. Coordination number: 6 (number of ligands).
335. Overall charge of the complex ion = +3.
336. NH3 is a neutral ligand.
337. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
338. Write the Lewis structure of .
339. Identify the steric number of the central atom (S).
340. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
341. Hybridization of S = sp^3d^2.
342. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
343. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
344. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
345. Identify the oxidation states and possible products.
346. Fe in is in the +2 oxidation state.
347. acts as an oxidizing agent, converting Fe from +2 to +3.
348. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
349. Fe in is in the +2 oxidation state.
350. The electron configuration of Fe2+ is [Ar] 3𝑑6.
351. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
352. The complex ion is .
353. Coordination number: 6 (number of ligands).
354. Overall charge of the complex ion = +3.
355. NH3 is a neutral ligand.
356. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
357. Identify the oxidation states of Mn in and .
358. Mn in : +7, in : +2.
359. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
360. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
361. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
362. is blue in color.
363. When treated with concentrated HCl, chloride ions () replace water ligands.
364. New complex formed: , which is yellow-green.
365. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
366. Ni in is in the +2 oxidation state.
367. The electron configuration of Ni2+ is [Ar] 3𝑑8.
368. CN− is a strong field ligand, causing pairing of electrons.
369. Configuration: (t2𝑔)6(e𝑔)2.
370. Number of unpaired electrons: 0.
371. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
372. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
373. Draw the molecular orbital diagram for .
374. Count the number of bonding and antibonding electrons.
375. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
376. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
377. NH3 is a stronger field ligand than H2O.
378. Stronger field ligands cause greater splitting of d-orbitals.
379. Greater splitting leads to more stabilization.
380. Hence, is more stable than .
381. Coordination number: 6 (number of ligands).
382. Overall charge of the complex ion = +3.
383. NH3 is a neutral ligand.
384. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
385. Write the Lewis structure of .
386. Identify the steric number of the central atom (S).
387. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
388. Hybridization of S = sp^3d^2.
389. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
390. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
391. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
392. Identify the oxidation states and possible products.
393. Fe in is in the +2 oxidation state.
394. acts as an oxidizing agent, converting Fe from +2 to +3.
395. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
396. Fe in is in the +2 oxidation state.
397. The electron configuration of Fe2+ is [Ar] 3𝑑6.
398. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
399. The complex ion is .
400. Coordination number: 6 (number of ligands).
401. Overall charge of the complex ion = +3.
402. NH3 is a neutral ligand.
403. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
404. Identify the oxidation states of Mn in and .
405. Mn in : +7, in : +2.
406. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
407. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
408. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
409. is blue in color.
410. When treated with concentrated HCl, chloride ions () replace water ligands.
411. New complex formed: , which is yellow-green.
412. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
413. Ni in is in the +2 oxidation state.
414. The electron configuration of Ni2+ is [Ar] 3𝑑8.
415. CN− is a strong field ligand, causing pairing of electrons.
416. Configuration: (t2𝑔)6(e𝑔)2.
417. Number of unpaired electrons: 0.
418. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
419. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
420. Draw the molecular orbital diagram for .
421. Count the number of bonding and antibonding electrons.
422. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
423. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
424. NH3 is a stronger field ligand than H2O.
425. Stronger field ligands cause greater splitting of d-orbitals.
426. Greater splitting leads to more stabilization.
427. Hence, is more stable than .
428. Coordination number: 6 (number of ligands).
429. Overall charge of the complex ion = +3.
430. NH3 is a neutral ligand.
431. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
432. Write the Lewis structure of .
433. Identify the steric number of the central atom (S).
434. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
435. Hybridization of S = sp^3d^2.
436. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
437. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
438. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
439. Identify the oxidation states and possible products.
440. Fe in is in the +2 oxidation state.
441. acts as an oxidizing agent, converting Fe from +2 to +3.
442. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
443. Fe in is in the +2 oxidation state.
444. The electron configuration of Fe2+ is [Ar] 3𝑑6.
445. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
446. The complex ion is .
447. Coordination number: 6 (number of ligands).
448. Overall charge of the complex ion = +3.
449. NH3 is a neutral ligand.
450. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
451. Identify the oxidation states of Mn in and .
452. Mn in : +7, in : +2.
453. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
454. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
455. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
456. is blue in color.
457. When treated with concentrated HCl, chloride ions () replace water ligands.
458. New complex formed: , which is yellow-green.
459. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
460. Ni in is in the +2 oxidation state.
461. The electron configuration of Ni2+ is [Ar] 3𝑑8.
462. CN− is a strong field ligand, causing pairing of electrons.
463. Configuration: (t2𝑔)6(e𝑔)2.
464. Number of unpaired electrons: 0.
465. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
466. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
467. Draw the molecular orbital diagram for .
468. Count the number of bonding and antibonding electrons.
469. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
470. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
471. NH3 is a stronger field ligand than H2O.
472. Stronger field ligands cause greater splitting of d-orbitals.
473. Greater splitting leads to more stabilization.
474. Hence, is more stable than .
475. Coordination number: 6 (number of ligands).
476. Overall charge of the complex ion = +3.
477. NH3 is a neutral ligand.
478. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
479. Write the Lewis structure of .
480. Identify the steric number of the central atom (S).
481. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
482. Hybridization of S = sp^3d^2.
483. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
484. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
485. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
486. Identify the oxidation states and possible products.
487. Fe in is in the +2 oxidation state.
488. acts as an oxidizing agent, converting Fe from +2 to +3.
489. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
490. Fe in is in the +2 oxidation state.
491. The electron configuration of Fe2+ is [Ar] 3𝑑6.
492. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
493. The complex ion is .
494. Coordination number: 6 (number of ligands).
495. Overall charge of the complex ion = +3.
496. NH3 is a neutral ligand.
497. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
498. Identify the oxidation states of Mn in and .
499. Mn in : +7, in : +2.
500. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
501. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
502. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
503. is blue in color.
504. When treated with concentrated HCl, chloride ions () replace water ligands.
505. New complex formed: , which is yellow-green.
506. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
507. Ni in is in the +2 oxidation state.
508. The electron configuration of Ni2+ is [Ar] 3𝑑8.
509. CN− is a strong field ligand, causing pairing of electrons.
510. Configuration: (t2𝑔)6(e𝑔)2.
511. Number of unpaired electrons: 0.
512. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
513. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
514. Draw the molecular orbital diagram for .
515. Count the number of bonding and antibonding electrons.
516. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
517. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
518. NH3 is a stronger field ligand than H2O.
519. Stronger field ligands cause greater splitting of d-orbitals.
520. Greater splitting leads to more stabilization.
521. Hence, is more stable than .
522. Coordination number: 6 (number of ligands).
523. Overall charge of the complex ion = +3.
524. NH3 is a neutral ligand.
525. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
526. Write the Lewis structure of .
527. Identify the steric number of the central atom (S).
528. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
529. Hybridization of S = sp^3d^2.
530. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
531. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
532. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
533. Identify the oxidation states and possible products.
534. Fe in is in the +2 oxidation state.
535. acts as an oxidizing agent, converting Fe from +2 to +3.
536. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
537. Fe in is in the +2 oxidation state.
538. The electron configuration of Fe2+ is [Ar] 3𝑑6.
539. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
540. The complex ion is .
541. Coordination number: 6 (number of ligands).
542. Overall charge of the complex ion = +3.
543. NH3 is a neutral ligand.
544. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
545. Identify the oxidation states of Mn in and .
546. Mn in : +7, in : +2.
547. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
548. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
549. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
550. is blue in color.
551. When treated with concentrated HCl, chloride ions () replace water ligands.
552. New complex formed: , which is yellow-green.
553. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
554. Ni in is in the +2 oxidation state.
555. The electron configuration of Ni2+ is [Ar] 3𝑑8.
556. CN− is a strong field ligand, causing pairing of electrons.
557. Configuration: (t2𝑔)6(e𝑔)2.
558. Number of unpaired electrons: 0.
559. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
560. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
561. Draw the molecular orbital diagram for .
562. Count the number of bonding and antibonding electrons.
563. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
564. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
565. NH3 is a stronger field ligand than H2O.
566. Stronger field ligands cause greater splitting of d-orbitals.
567. Greater splitting leads to more stabilization.
568. Hence, is more stable than .
569. Coordination number: 6 (number of ligands).
570. Overall charge of the complex ion = +3.
571. NH3 is a neutral ligand.
572. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
573. Write the Lewis structure of .
574. Identify the steric number of the central atom (S).
575. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
576. Hybridization of S = sp^3d^2.
577. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
578. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
579. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
580. Identify the oxidation states and possible products.
581. Fe in is in the +2 oxidation state.
582. acts as an oxidizing agent, converting Fe from +2 to +3.
583. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
584. Fe in is in the +2 oxidation state.
585. The electron configuration of Fe2+ is [Ar] 3𝑑6.
586. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
587. The complex ion is .
588. Coordination number: 6 (number of ligands).
589. Overall charge of the complex ion = +3.
590. NH3 is a neutral ligand.
591. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
592. Identify the oxidation states of Mn in and .
593. Mn in : +7, in : +2.
594. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
595. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
596. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
597. is blue in color.
598. When treated with concentrated HCl, chloride ions () replace water ligands.
599. New complex formed: , which is yellow-green.
600. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
601. Ni in is in the +2 oxidation state.
602. The electron configuration of Ni2+ is [Ar] 3𝑑8.
603. CN− is a strong field ligand, causing pairing of electrons.
604. Configuration: (t2𝑔)6(e𝑔)2.
605. Number of unpaired electrons: 0.
606. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
607. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
608. Draw the molecular orbital diagram for .
609. Count the number of bonding and antibonding electrons.
610. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
611. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
612. NH3 is a stronger field ligand than H2O.
613. Stronger field ligands cause greater splitting of d-orbitals.
614. Greater splitting leads to more stabilization.
615. Hence, is more stable than .
616. Coordination number: 6 (number of ligands).
617. Overall charge of the complex ion = +3.
618. NH3 is a neutral ligand.
619. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
620. Write the Lewis structure of .
621. Identify the steric number of the central atom (S).
622. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
623. Hybridization of S = sp^3d^2.
624. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
625. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
626. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
627. Identify the oxidation states and possible products.
628. Fe in is in the +2 oxidation state.
629. acts as an oxidizing agent, converting Fe from +2 to +3.
630. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
631. Fe in is in the +2 oxidation state.
632. The electron configuration of Fe2+ is [Ar] 3𝑑6.
633. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
634. The complex ion is .
635. Coordination number: 6 (number of ligands).
636. Overall charge of the complex ion = +3.
637. NH3 is a neutral ligand.
638. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
639. Identify the oxidation states of Mn in and .
640. Mn in : +7, in : +2.
641. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
642. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
643. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
644. is blue in color.
645. When treated with concentrated HCl, chloride ions () replace water ligands.
646. New complex formed: , which is yellow-green.
647. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
648. Ni in is in the +2 oxidation state.
649. The electron configuration of Ni2+ is [Ar] 3𝑑8.
650. CN− is a strong field ligand, causing pairing of electrons.
651. Configuration: (t2𝑔)6(e𝑔)2.
652. Number of unpaired electrons: 0.
653. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
654. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
655. Draw the molecular orbital diagram for .
656. Count the number of bonding and antibonding electrons.
657. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
658. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
659. NH3 is a stronger field ligand than H2O.
660. Stronger field ligands cause greater splitting of d-orbitals.
661. Greater splitting leads to more stabilization.
662. Hence, is more stable than .
663. Coordination number: 6 (number of ligands).
664. Overall charge of the complex ion = +3.
665. NH3 is a neutral ligand.
666. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
667. Write the Lewis structure of .
668. Identify the steric number of the central atom (S).
669. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
670. Hybridization of S = sp^3d^2.
671. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
672. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
673. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
674. Identify the oxidation states and possible products.
675. Fe in is in the +2 oxidation state.
676. acts as an oxidizing agent, converting Fe from +2 to +3.
677. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
678. Fe in is in the +2 oxidation state.
679. The electron configuration of Fe2+ is [Ar] 3𝑑6.
680. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
681. The complex ion is .
682. Coordination number: 6 (number of ligands).
683. Overall charge of the complex ion = +3.
684. NH3 is a neutral ligand.
685. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
686. Identify the oxidation states of Mn in and .
687. Mn in : +7, in : +2.
688. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
689. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
690. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
691. is blue in color.
692. When treated with concentrated HCl, chloride ions () replace water ligands.
693. New complex formed: , which is yellow-green.
694. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
695. Ni in is in the +2 oxidation state.
696. The electron configuration of Ni2+ is [Ar] 3𝑑8.
697. CN− is a strong field ligand, causing pairing of electrons.
698. Configuration: (t2𝑔)6(e𝑔)2.
699. Number of unpaired electrons: 0.
700. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
701. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
702. Draw the molecular orbital diagram for .
703. Count the number of bonding and antibonding electrons.
704. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
705. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
706. NH3 is a stronger field ligand than H2O.
707. Stronger field ligands cause greater splitting of d-orbitals.
708. Greater splitting leads to more stabilization.
709. Hence, is more stable than .
710. Coordination number: 6 (number of ligands).
711. Overall charge of the complex ion = +3.
712. NH3 is a neutral ligand.
713. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
714. Write the Lewis structure of .
715. Identify the steric number of the central atom (S).
716. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
717. Hybridization of S = sp^3d^2.
718. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
719. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
720. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
721. Identify the oxidation states and possible products.
722. Fe in is in the +2 oxidation state.
723. acts as an oxidizing agent, converting Fe from +2 to +3.
724. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
725. Fe in is in the +2 oxidation state.
726. The electron configuration of Fe2+ is [Ar] 3𝑑6.
727. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
728. The complex ion is .
729. Coordination number: 6 (number of ligands).
730. Overall charge of the complex ion = +3.
731. NH3 is a neutral ligand.
732. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
733. Identify the oxidation states of Mn in and .
734. Mn in : +7, in : +2.
735. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
736. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
737. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
738. is blue in color.
739. When treated with concentrated HCl, chloride ions () replace water ligands.
740. New complex formed: , which is yellow-green.
741. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
742. Ni in is in the +2 oxidation state.
743. The electron configuration of Ni2+ is [Ar] 3𝑑8.
744. CN− is a strong field ligand, causing pairing of electrons.
745. Configuration: (t2𝑔)6(e𝑔)2.
746. Number of unpaired electrons: 0.
747. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
748. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
749. Draw the molecular orbital diagram for .
750. Count the number of bonding and antibonding electrons.
751. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
752. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
753. NH3 is a stronger field ligand than H2O.
754. Stronger field ligands cause greater splitting of d-orbitals.
755. Greater splitting leads to more stabilization.
756. Hence, is more stable than .
757. Coordination number: 6 (number of ligands).
758. Overall charge of the complex ion = +3.
759. NH3 is a neutral ligand.
760. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
761. Write the Lewis structure of .
762. Identify the steric number of the central atom (S).
763. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
764. Hybridization of S = sp^3d^2.
765. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
766. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
767. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
768. Identify the oxidation states and possible products.
769. Fe in is in the +2 oxidation state.
770. acts as an oxidizing agent, converting Fe from +2 to +3.
771. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
772. Fe in is in the +2 oxidation state.
773. The electron configuration of Fe2+ is [Ar] 3𝑑6.
774. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
775. The complex ion is .
776. Coordination number: 6 (number of ligands).
777. Overall charge of the complex ion = +3.
778. NH3 is a neutral ligand.
779. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
780. Identify the oxidation states of Mn in and .
781. Mn in : +7, in : +2.
782. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
783. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
784. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
785. is blue in color.
786. When treated with concentrated HCl, chloride ions () replace water ligands.
787. New complex formed: , which is yellow-green.
788. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
789. Ni in is in the +2 oxidation state.
790. The electron configuration of Ni2+ is [Ar] 3𝑑8.
791. CN− is a strong field ligand, causing pairing of electrons.
792. Configuration: (t2𝑔)6(e𝑔)2.
793. Number of unpaired electrons: 0.
794. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
795. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
796. Draw the molecular orbital diagram for .
797. Count the number of bonding and antibonding electrons.
798. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
799. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
800. NH3 is a stronger field ligand than H2O.
801. Stronger field ligands cause greater splitting of d-orbitals.
802. Greater splitting leads to more stabilization.
803. Hence, is more stable than .
804. Coordination number: 6 (number of ligands).
805. Overall charge of the complex ion = +3.
806. NH3 is a neutral ligand.
807. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
808. Write the Lewis structure of .
809. Identify the steric number of the central atom (S).
810. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
811. Hybridization of S = sp^3d^2.
812. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
813. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
814. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
815. Identify the oxidation states and possible products.
816. Fe in is in the +2 oxidation state.
817. acts as an oxidizing agent, converting Fe from +2 to +3.
818. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
819. Fe in is in the +2 oxidation state.
820. The electron configuration of Fe2+ is [Ar] 3𝑑6.
821. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
822. The complex ion is .
823. Coordination number: 6 (number of ligands).
824. Overall charge of the complex ion = +3.
825. NH3 is a neutral ligand.
826. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
827. Identify the oxidation states of Mn in and .
828. Mn in : +7, in : +2.
829. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
830. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
831. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
832. is blue in color.
833. When treated with concentrated HCl, chloride ions () replace water ligands.
834. New complex formed: , which is yellow-green.
835. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
836. Ni in is in the +2 oxidation state.
837. The electron configuration of Ni2+ is [Ar] 3𝑑8.
838. CN− is a strong field ligand, causing pairing of electrons.
839. Configuration: (t2𝑔)6(e𝑔)2.
840. Number of unpaired electrons: 0.
841. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
842. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
843. Draw the molecular orbital diagram for .
844. Count the number of bonding and antibonding electrons.
845. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
846. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
847. NH3 is a stronger field ligand than H2O.
848. Stronger field ligands cause greater splitting of d-orbitals.
849. Greater splitting leads to more stabilization.
850. Hence, is more stable than .
851. Coordination number: 6 (number of ligands).
852. Overall charge of the complex ion = +3.
853. NH3 is a neutral ligand.
854. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
855. Write the Lewis structure of .
856. Identify the steric number of the central atom (S).
857. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
858. Hybridization of S = sp^3d^2.
859. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
860. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
861. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
862. Identify the oxidation states and possible products.
863. Fe in is in the +2 oxidation state.
864. acts as an oxidizing agent, converting Fe from +2 to +3.
865. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
866. Fe in is in the +2 oxidation state.
867. The electron configuration of Fe2+ is [Ar] 3𝑑6.
868. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
869. The complex ion is .
870. Coordination number: 6 (number of ligands).
871. Overall charge of the complex ion = +3.
872. NH3 is a neutral ligand.
873. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
874. Identify the oxidation states of Mn in and .
875. Mn in : +7, in : +2.
876. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
877. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
878. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
879. is blue in color.
880. When treated with concentrated HCl, chloride ions () replace water ligands.
881. New complex formed: , which is yellow-green.
882. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
883. Ni in is in the +2 oxidation state.
884. The electron configuration of Ni2+ is [Ar] 3𝑑8.
885. CN− is a strong field ligand, causing pairing of electrons.
886. Configuration: (t2𝑔)6(e𝑔)2.
887. Number of unpaired electrons: 0.
888. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
889. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
890. Draw the molecular orbital diagram for .
891. Count the number of bonding and antibonding electrons.
892. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
893. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
894. NH3 is a stronger field ligand than H2O.
895. Stronger field ligands cause greater splitting of d-orbitals.
896. Greater splitting leads to more stabilization.
897. Hence, is more stable than .
898. Coordination number: 6 (number of ligands).
899. Overall charge of the complex ion = +3.
900. NH3 is a neutral ligand.
901. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
902. Write the Lewis structure of .
903. Identify the steric number of the central atom (S).
904. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
905. Hybridization of S = sp^3d^2.
906. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
907. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
908. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
909. Identify the oxidation states and possible products.
910. Fe in is in the +2 oxidation state.
911. acts as an oxidizing agent, converting Fe from +2 to +3.
912. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
913. Fe in is in the +2 oxidation state.
914. The electron configuration of Fe2+ is [Ar] 3𝑑6.
915. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
916. The complex ion is .
917. Coordination number: 6 (number of ligands).
918. Overall charge of the complex ion = +3.
919. NH3 is a neutral ligand.
920. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
921. Identify the oxidation states of Mn in and .
922. Mn in : +7, in : +2.
923. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
924. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
925. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
926. is blue in color.
927. When treated with concentrated HCl, chloride ions () replace water ligands.
928. New complex formed: , which is yellow-green.
929. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
930. Ni in is in the +2 oxidation state.
931. The electron configuration of Ni2+ is [Ar] 3𝑑8.
932. CN− is a strong field ligand, causing pairing of electrons.
933. Configuration: (t2𝑔)6(e𝑔)2.
934. Number of unpaired electrons: 0.
935. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
936. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
937. Draw the molecular orbital diagram for .
938. Count the number of bonding and antibonding electrons.
939. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
940. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
941. NH3 is a stronger field ligand than H2O.
942. Stronger field ligands cause greater splitting of d-orbitals.
943. Greater splitting leads to more stabilization.
944. Hence, is more stable than .
945. Coordination number: 6 (number of ligands).
946. Overall charge of the complex ion = +3.
947. NH3 is a neutral ligand.
948. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
949. Write the Lewis structure of .
950. Identify the steric number of the central atom (S).
951. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
952. Hybridization of S = sp^3d^2.
953. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
954. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
955. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
956. Identify the oxidation states and possible products.
957. Fe in is in the +2 oxidation state.
958. acts as an oxidizing agent, converting Fe from +2 to +3.
959. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
960. Fe in is in the +2 oxidation state.
961. The electron configuration of Fe2+ is [Ar] 3𝑑6.
962. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
963. The complex ion is .
964. Coordination number: 6 (number of ligands).
965. Overall charge of the complex ion = +3.
966. NH3 is a neutral ligand.
967. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
968. Identify the oxidation states of Mn in and .
969. Mn in : +7, in : +2.
970. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
971. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
972. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
973. is blue in color.
974. When treated with concentrated HCl, chloride ions () replace water ligands.
975. New complex formed: , which is yellow-green.
976. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
977. Ni in is in the +2 oxidation state.
978. The electron configuration of Ni2+ is [Ar] 3𝑑8.
979. CN− is a strong field ligand, causing pairing of electrons.
980. Configuration: (t2𝑔)6(e𝑔)2.
981. Number of unpaired electrons: 0.
982. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
983. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
984. Draw the molecular orbital diagram for .
985. Count the number of bonding and antibonding electrons.
986. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
987. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
988. NH3 is a stronger field ligand than H2O.
989. Stronger field ligands cause greater splitting of d-orbitals.
990. Greater splitting leads to more stabilization.
991. Hence, is more stable than .
992. Coordination number: 6 (number of ligands).
993. Overall charge of the complex ion = +3.
994. NH3 is a neutral ligand.
995. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
996. Write the Lewis structure of .
997. Identify the steric number of the central atom (S).
998. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 = 6.
999. Hybridization of S = sp^3d^2.
1000. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
1001. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
1002. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
1003. Identify the oxidation states and possible products.
1004. Fe in is in the +2 oxidation state.
1005. acts as an oxidizing agent, converting Fe from +2 to +3.
1006. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
1007. Fe in is in the +2 oxidation state.
1008. The electron configuration of Fe2+ is [Ar] 3𝑑6.
1009. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
1010. The complex ion is .
1011. Coordination number: 6 (number of ligands).
1012. Overall charge of the complex ion = +3.
1013. NH3 is a neutral ligand.
1014. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
1015. Identify the oxidation states of Mn in and .
1016. Mn in : +7, in : +2.
1017. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
1018. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
1019. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
1020. is blue in color.
1021. When treated with concentrated HCl, chloride ions () replace water ligands.
1022. New complex formed: , which is yellow-green.
1023. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
1024. Ni in is in the +2 oxidation state.
1025. The electron configuration of Ni2+ is [Ar] 3𝑑8.
1026. CN− is a strong field ligand, causing pairing of electrons.
1027. Configuration: (t2𝑔)6(e𝑔)2.
1028. Number of unpaired electrons: 0.
1029. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
1030. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
1031. Draw the molecular orbital diagram for .
1032. Count the number of bonding and antibonding electrons.
1033. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
1034. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
1035. NH3 is a stronger field ligand than H2O.
1036. Stronger field ligands cause greater splitting of d-orbitals.
1037. Greater splitting leads to more stabilization.
1038. Hence, is more stable than .
1039. Coordination number: 6 (number of ligands).
1040. Overall charge of the complex ion = +3.
1041. NH3 is a neutral ligand.
1042. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
1043. Write the Lewis structure of .
1044. Identify the steric number of the central atom (S).
1045. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 =
6.
1046. Hybridization of S = sp^3d^2.
1047. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
1048. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
1049. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
1050. Identify the oxidation states and possible products.
1051. Fe in is in the +2 oxidation state.
1052. acts as an oxidizing agent, converting Fe from +2 to +3.
1053. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
1054. Fe in is in the +2 oxidation state.
1055. The electron configuration of Fe2+ is [Ar] 3𝑑6.
1056. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
1057. The complex ion is .
1058. Coordination number: 6 (number of ligands).
1059. Overall charge of the complex ion = +3.
1060. NH3 is a neutral ligand.
1061. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
1062. Identify the oxidation states of Mn in and .
1063. Mn in : +7, in : +2.
1064. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
1065. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
1066. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
1067. is blue in color.
1068. When treated with concentrated HCl, chloride ions () replace water ligands.
1069. New complex formed: , which is yellow-green.
1070. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
1071. Ni in is in the +2 oxidation state.
1072. The electron configuration of Ni2+ is [Ar] 3𝑑8.
1073. CN− is a strong field ligand, causing pairing of electrons.
1074. Configuration: (t2𝑔)6(e𝑔)2.
1075. Number of unpaired electrons: 0.
1076. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
1077. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
1078. Draw the molecular orbital diagram for .
1079. Count the number of bonding and antibonding electrons.
1080. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
1081. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
1082. NH3 is a stronger field ligand than H2O.
1083. Stronger field ligands cause greater splitting of d-orbitals.
1084. Greater splitting leads to more stabilization.
1085. Hence, is more stable than .
1086. Coordination number: 6 (number of ligands).
1087. Overall charge of the complex ion = +3.
1088. NH3 is a neutral ligand.
1089. Oxidation state of Co: +3.
Problem 1
Determine the hybridization and geometry of the central atom in .
Solution
1090. Write the Lewis structure of .
1091. Identify the steric number of the central atom (S).
1092. Steric number of S = number of sigma bonds + number of lone pairs = 6 + 0 =
6.
1093. Hybridization of S = sp^3d^2.
1094. Geometry: Octahedral.
Problem 2
Calculate the crystal field stabilization energy (CFSE) for a d^6 ion in an octahedral field.
Solution
1095. In an octahedral field, the d-orbitals split into t_2g and e_g orbitals.
1096. For a d^6 configuration:
– In a low-spin octahedral field:
(t2𝑔)6(e𝑔)0
– CFSE = (number of electrons in t_2g)(-0.4_0) + (number of electrons in
e_g)(0.6_0)
– CFSE = 6(-0.4_0) + 0(0.6_0) = -2.4_0.
Problem 3
Predict the products of the following reaction and balance it:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> ?}$$
Solution
1097. Identify the oxidation states and possible products.
1098. Fe in is in the +2 oxidation state.
1099. acts as an oxidizing agent, converting Fe from +2 to +3.
1100. The balanced reaction is:
$$\ce{K4[Fe(CN)6] + H2SO4 + H2O2 -> K3[Fe(CN)6] + KHSO4 + 2H2O}$$
Problem 4
Determine the number of unpaired electrons in .
Solution
1101. Fe in is in the +2 oxidation state.
1102. The electron configuration of Fe2+ is [Ar] 3𝑑6.
1103. In an octahedral field:
– High-spin configuration: (t2𝑔)4(e𝑔)2.
– Number of unpaired electrons: 4.
Problem 5
What is the coordination number and oxidation state of the central metal in ?
Solution
1104. The complex ion is .
1105. Coordination number: 6 (number of ligands).
1106. Overall charge of the complex ion = +3.
1107. NH3 is a neutral ligand.
1108. Oxidation state of Co: +3.
Problem 6
Write the balanced equation for the reaction between and in acidic medium.
Solution
1109. Identify the oxidation states of Mn in and .
1110. Mn in : +7, in : +2.
1111. Oxidation half-reaction:
$$\ce{C2O4^{2-} -> CO2}$$
1112. Reduction half-reaction:
$$\ce{MnO4^- -> Mn^{2+}}$$
1113. Balance the half-reactions and combine:
$$\ce{2MnO4^- + 5H2C2O4 + 6H^+ -> 2Mn^{2+} + 10CO2 + 8H2O}$$
Problem 7
Explain the color change observed when is treated with concentrated HCl.
Solution
1114. is blue in color.
1115. When treated with concentrated HCl, chloride ions () replace water ligands.
1116. New complex formed: , which is yellow-green.
1117. The color change is due to the different ligand field strengths of and .
Problem 8
Calculate the magnetic moment (𝜇) of .
Solution
1118. Ni in is in the +2 oxidation state.
1119. The electron configuration of Ni2+ is [Ar] 3𝑑8.
1120. CN− is a strong field ligand, causing pairing of electrons.
1121. Configuration: (t2𝑔)6(e𝑔)2.
1122. Number of unpaired electrons: 0.
1123. Magnetic moment (𝜇) = √𝑛(𝑛 + 2) where 𝑛 = number of unpaired electrons.
1124. 𝜇 = √0(0 + 2)= 0 BM.
Problem 9
Determine the bond order in using molecular orbital theory.
Solution
1125. Draw the molecular orbital diagram for .
1126. Count the number of bonding and antibonding electrons.
1127. Bond order = Number of bonding electrons−Number of antibonding electrons
2.
1128. For , bond order = 8−2
2= 3.
Problem 10
Explain why is more stable than .
Solution
1129. NH3 is a stronger field ligand than H2O.
1130. Stronger field ligands cause greater splitting of d-orbitals.
1131. Greater splitting leads to more stabilization.
1132. Hence, is more stable than .
1133. Coordination number: 6 (number of ligands).
1134. Overall charge of the complex ion = +3.
1135. NH3 is a neutral ligand.
1136. Oxidation state of Co: +3.
