(a) A coordinate covalent bond occurs when one atom donates a pair of electrons to another atom or ion. The atom or ion that donates the electrons is called the donor or Lewis base, while the atom or ion that accepts the electrons is called the acceptor or Lewis acid. The shared electrons in the bond are typically shown as a coordinate arrow pointing from the donor to the acceptor.
(b) (i) FeCl3 dissolves in water to form Fe(H2O)6^3+ complex ions. In this process, the water molecules act as Lewis bases, donating electron pairs to the Fe3+ ion to form coordinate covalent bonds. The Fe3+ ion acts as a Lewis acid, accepting the electron pairs from the water molecules.
(ii) When HCl gas dissolves in water, it forms H3O+ and Cl- ions. In this process, the water molecule donates a pair of electrons to the HCl molecule, forming a coordinate covalent bond. The HCl molecule acts as the Lewis acid by accepting the electron pair from the water molecule, which acts as the Lewis base.
Q2. According to valence bond theory, for a period 4 metal ion:
(a) A square planar complex is formed using dsp2 orbitals. The d orbitals hybridize with one s and two p orbitals to form four equivalent hybrid orbitals, which are then used for bonding in a square planar arrangement.
(b) A tetrahedral complex is formed using sp3 hybrid orbitals. The d orbitals hybridize with one s and three p orbitals to form four equivalent hybrid orbitals, which are then used for bonding in a tetrahedral arrangement.
Q3. If a metal ion uses d2sp3 orbitals when forming a complex, it has a coordination number of 6 and the geometry of the complex is octahedral. The d orbitals hybridize with one s, three p, and two d orbitals to form six equivalent hybrid orbitals, which are used for bonding in an octahedral arrangement.
(a) What is a coordinate covalent bond? (b) Does
coordinate bond formed when (i) FeCl3 dissolve in water?
Explain (ii) HCl gas dissolves in water? Explain
•
• Q2. According to valence bond theory, what set of orbitals is
used by period 4 metal ion in forming (a) a square planar
complex; (b) a tetrahedral complex?
• Q3. A metal ion uses d2
sp3 orbitals when forming a complex.
What is its coordination number and the geometry of the
complex?
3
3 answers
. Give the number of d electrons for the central metal ion
in (a) [TiCl6]2-
, (b) K[AuCl4
], (c) [RhCl6]3-,
(d) [Mn(CN)6]2-
, (e)
[HgI4]2-
(f) [Co(EDTA)]2-
• Q5. Why [CoF6]3-
is paramagnetic and Co(NH3)
3+
is
diamagnetic? Explain
• Q6. Draw bonding diagram using valence bond theory of the
following complexes. Predict the hybridization, geometry and
magnetic properties of each.
• (a) [NiBr4]2-
(b) Fe[(CN6
)]4-
(c) Ni(CO)4
(d) [Ni(CN)4]2-
35
in (a) [TiCl6]2-
, (b) K[AuCl4
], (c) [RhCl6]3-,
(d) [Mn(CN)6]2-
, (e)
[HgI4]2-
(f) [Co(EDTA)]2-
• Q5. Why [CoF6]3-
is paramagnetic and Co(NH3)
3+
is
diamagnetic? Explain
• Q6. Draw bonding diagram using valence bond theory of the
following complexes. Predict the hybridization, geometry and
magnetic properties of each.
• (a) [NiBr4]2-
(b) Fe[(CN6
)]4-
(c) Ni(CO)4
(d) [Ni(CN)4]2-
35
(a) [TiCl6]2-: Titanium (Ti) is a transition metal and it has a configuration of 4s^2 3d^2. Therefore, it has 2 d electrons.
(b) K[AuCl4]: Gold (Au) is also a transition metal and it has a configuration of 5s^2 4d^10. Therefore, it has 10 d electrons.
(c) [RhCl6]3-: Rhodium (Rh) is a transition metal and it has a configuration of 5s^2 4d^8. Therefore, it has 8 d electrons.
(d) [Mn(CN)6]2-: Manganese (Mn) is a transition metal and it has a configuration of 4s^2 3d^5. Therefore, it has 5 d electrons.
(e) [HgI4]2-: Mercury (Hg) is a transition metal and it has a configuration of 6s^2 5d^10 4f^14. Therefore, it has 10 d electrons.
(f) [Co(EDTA)]2-: Cobalt (Co) is a transition metal and it has a configuration of 4s^2 3d^7. Therefore, it has 7 d electrons.
Q5. [CoF6]3- is paramagnetic because it has unpaired electrons. Cobalt (Co) in [CoF6]3- has a configuration of 4s^2 3d^7. When it forms the complex with six fluoride (F) ions, the d orbitals split into two sets, with three orbitals in the higher energy level and two orbitals in the lower energy level. The three remaining d orbitals are now available to accept electrons from the fluorine ions, resulting in the formation of three coordinate covalent bonds. Since there are three unpaired electrons in the d orbitals, [CoF6]3- is paramagnetic.
Co(NH3)3+ is diamagnetic because it has all the electrons paired. Cobalt (Co) in Co(NH3)3+ has a configuration of 4s^2 3d^7. When it forms the complex with three ammonia (NH3) molecules, the d orbitals do not split significantly. Therefore, there are no unpaired electrons in the d orbitals, making Co(NH3)3+ diamagnetic.
Q6.
(a) [NiBr4]2-: The bonding diagram of [NiBr4]2- can be represented as follows:
Ni: [Ar] 3d^8 4s^2
Br: [Kr] 4d^10 5s^2 5p^5
In this complex, nickel (Ni) donates electrons from the 3d orbitals to form coordinate covalent bonds with four bromine (Br) ions. The hybridization of Ni in this case is dsp2, and the complex has a square planar geometry. Since there are unpaired electrons in the d orbitals, [NiBr4]2- is paramagnetic.
(b) Fe[(CN6)]4-: The bonding diagram of Fe[(CN6)]4- can be represented as follows:
Fe: [Ar] 3d^6 4s^2
CN: [He] 2s^2 2p^6
In this complex, iron (Fe) donates electrons from the 3d orbitals to form coordinate covalent bonds with six cyanide (CN) ions. The hybridization of Fe in this case is d^2sp3, and the complex has an octahedral geometry. Since all the electrons in the d orbitals are paired, Fe[(CN6)]4- is diamagnetic.
(c) Ni(CO)4: The bonding diagram of Ni(CO)4 can be represented as follows:
Ni: [Ar] 3d^8 4s^2
CO: [He] 2s^2 2p^4
In this complex, nickel (Ni) donates electrons from the 3d orbitals to form coordinate covalent bonds with four carbon monoxide (CO) molecules. The hybridization of Ni in this case is dsp3, and the complex has a tetrahedral geometry. Since there are unpaired electrons in the d orbitals, Ni(CO)4 is paramagnetic.
(d) [Ni(CN)4]2-: The bonding diagram of [Ni(CN)4]2- can be represented as follows:
Ni: [Ar] 3d^8 4s^2
CN: [He] 2s^2 2p^6
In this complex, nickel (Ni) donates electrons from the 3d orbitals to form coordinate covalent bonds with four cyanide (CN) ions. The hybridization of Ni in this case is dsp2, and the complex has a square planar geometry. Since there are unpaired electrons in the d orbitals, [Ni(CN)4]2- is paramagnetic.
(b) K[AuCl4]: Gold (Au) is also a transition metal and it has a configuration of 5s^2 4d^10. Therefore, it has 10 d electrons.
(c) [RhCl6]3-: Rhodium (Rh) is a transition metal and it has a configuration of 5s^2 4d^8. Therefore, it has 8 d electrons.
(d) [Mn(CN)6]2-: Manganese (Mn) is a transition metal and it has a configuration of 4s^2 3d^5. Therefore, it has 5 d electrons.
(e) [HgI4]2-: Mercury (Hg) is a transition metal and it has a configuration of 6s^2 5d^10 4f^14. Therefore, it has 10 d electrons.
(f) [Co(EDTA)]2-: Cobalt (Co) is a transition metal and it has a configuration of 4s^2 3d^7. Therefore, it has 7 d electrons.
Q5. [CoF6]3- is paramagnetic because it has unpaired electrons. Cobalt (Co) in [CoF6]3- has a configuration of 4s^2 3d^7. When it forms the complex with six fluoride (F) ions, the d orbitals split into two sets, with three orbitals in the higher energy level and two orbitals in the lower energy level. The three remaining d orbitals are now available to accept electrons from the fluorine ions, resulting in the formation of three coordinate covalent bonds. Since there are three unpaired electrons in the d orbitals, [CoF6]3- is paramagnetic.
Co(NH3)3+ is diamagnetic because it has all the electrons paired. Cobalt (Co) in Co(NH3)3+ has a configuration of 4s^2 3d^7. When it forms the complex with three ammonia (NH3) molecules, the d orbitals do not split significantly. Therefore, there are no unpaired electrons in the d orbitals, making Co(NH3)3+ diamagnetic.
Q6.
(a) [NiBr4]2-: The bonding diagram of [NiBr4]2- can be represented as follows:
Ni: [Ar] 3d^8 4s^2
Br: [Kr] 4d^10 5s^2 5p^5
In this complex, nickel (Ni) donates electrons from the 3d orbitals to form coordinate covalent bonds with four bromine (Br) ions. The hybridization of Ni in this case is dsp2, and the complex has a square planar geometry. Since there are unpaired electrons in the d orbitals, [NiBr4]2- is paramagnetic.
(b) Fe[(CN6)]4-: The bonding diagram of Fe[(CN6)]4- can be represented as follows:
Fe: [Ar] 3d^6 4s^2
CN: [He] 2s^2 2p^6
In this complex, iron (Fe) donates electrons from the 3d orbitals to form coordinate covalent bonds with six cyanide (CN) ions. The hybridization of Fe in this case is d^2sp3, and the complex has an octahedral geometry. Since all the electrons in the d orbitals are paired, Fe[(CN6)]4- is diamagnetic.
(c) Ni(CO)4: The bonding diagram of Ni(CO)4 can be represented as follows:
Ni: [Ar] 3d^8 4s^2
CO: [He] 2s^2 2p^4
In this complex, nickel (Ni) donates electrons from the 3d orbitals to form coordinate covalent bonds with four carbon monoxide (CO) molecules. The hybridization of Ni in this case is dsp3, and the complex has a tetrahedral geometry. Since there are unpaired electrons in the d orbitals, Ni(CO)4 is paramagnetic.
(d) [Ni(CN)4]2-: The bonding diagram of [Ni(CN)4]2- can be represented as follows:
Ni: [Ar] 3d^8 4s^2
CN: [He] 2s^2 2p^6
In this complex, nickel (Ni) donates electrons from the 3d orbitals to form coordinate covalent bonds with four cyanide (CN) ions. The hybridization of Ni in this case is dsp2, and the complex has a square planar geometry. Since there are unpaired electrons in the d orbitals, [Ni(CN)4]2- is paramagnetic.
1 answer