JEE Advance - Chemistry (2021 - Paper 2 Online - No. 5)
The pair(s) of complexes wherein both exhibit tetrahedral geometry is(are)
(Note : py = pyridine)
Given : Atomic numbers of Fe, Co, Ni and Cu are 26, 27, 28 and 29, respectively)
(Note : py = pyridine)
Given : Atomic numbers of Fe, Co, Ni and Cu are 26, 27, 28 and 29, respectively)
[FeCl4]$$-$$ and [Fe(CO)4]2$$-$$
[Co(CO)4]$$-$$ and [CoCl4]2$$-$$
[Ni(CO)4] and [Ni(CN)4]2$$-$$
[Cu(py)4]+ and [Cu(CN)4]3$$-$$
Explanation
(a) $${[FeC{l_4}]^ - }$$
Atomic number of Fe = 26
Oxidation number of Fe = x + 4 ($$-$$1) = $$-$$1
$$\Rightarrow$$ x = 3
Electronic configuration of Fe = 1s2 2s2 2p6 3s2 3p6 3d6 4s2
Electronic configuration of Fe3+ = 1s2 2s2 2p6 3s2 3p6 3d5 4s0
Cl- is a weak field ligand, so no pairing of electrons occurs.
Hybridisation is sp3
Geometry is tetrahedral.
$${[Fe{(CO)_4}]^{2 - }}$$ Oxidation number of Fe = x + 4(0) = - 2 $$\Rightarrow$$ x = - 2
Electronic configuration of Fe2- = 1s2 2s2 2p6 3s2 3p6 3d8 4s2
CO is a strong field ligand and cause electrons of 3d orbitals to pair up and shift electrons of 4s to 3d orbital.
One 4s and three 4p orbitals undergoes hybridisation to form four sp3 hybrid orbitals which arrange into a tetrahedral geometry.
Each of the four sp3 hybrid orbitals accept a pair of electrons from carbonyl ligand.
Geometry is tetrahedral.
(b) $${[Co{(CO)_4}]^ - }$$ Atomic number of Co = 27
Oxidation number of Co = x + 4(0) = $$-$$1 $$\Rightarrow$$ x = $$-$$1
Electronic configuration of Co = 1s22s22p63s23p63d74s2
Electronic configuration of Co- = 1s22s22p63s23p63d84s2
CO is a strong field ligand, so pairing of electrons occurs.
Hybridisation is sp3
Geometry is tetrahedral.
$${[CoC{l_4}]^{2 - }}$$
Oxidation number of Co = x + 4($$-$$1) = $$-$$2
$$\Rightarrow$$ x = + 2
Electronic configuration of Co2+ = 1s22s22p63s23p63d74s0
Cl- is a weak field ligand, so pairing of electrons do not occur.
Hybridisation is sp3
Geometry is tetrahedral
(c) $$[Ni{(CO)_4}]$$ Oxidation number of Ni = x + 4(0) = 0 $$\Rightarrow$$ x = 0
Atomic number of Ni = 28
Electronic configuration of Ni = 1s22s22p63s23p63d84s2
CO is a strong field ligand, so pairing of electrons occur.
Hybridisation is sp3
Geometry is tetrahedral.
$${[Ni{(CN)_4}]^{2 - }}$$
Oxidation number of Ni = x + 4($$-$$1) = $$-$$2 $$\Rightarrow$$ x = + 2
Electronic configuration of Ni2+ = 1s22s22p63s23p63d84s0
CN is a strong field ligand, so pairing of electrons occur.
Hybridisation is dsp2.
Geometry is square planar.
(d) $${[Cu{(py)_4}]^ + }$$
Oxidation number of Cu = x + 4(0) = + 1 $$\Rightarrow$$ x = + 1
Atomic number of Cu = 29
Electronic configuration of Cu = 1s22s22p63s23p63d104s1
Electronic configuration of Cu+ = 1s22s22p63s23p63d104s0
Hybridisation is sp3
Geometry is tetrahedral.
$${[Cu{(CN)_4}]^{3 - }}$$
Oxidation number of Cu = x + 4($$-$$1) = $$-$$ 3 $$\Rightarrow$$ x = + 1
Hybridisation is sp3
Geometry is tetrahedral.
Atomic number of Fe = 26
Oxidation number of Fe = x + 4 ($$-$$1) = $$-$$1
$$\Rightarrow$$ x = 3
Electronic configuration of Fe = 1s2 2s2 2p6 3s2 3p6 3d6 4s2
Electronic configuration of Fe3+ = 1s2 2s2 2p6 3s2 3p6 3d5 4s0
Cl- is a weak field ligand, so no pairing of electrons occurs.
Hybridisation is sp3
Geometry is tetrahedral.
$${[Fe{(CO)_4}]^{2 - }}$$ Oxidation number of Fe = x + 4(0) = - 2 $$\Rightarrow$$ x = - 2
Electronic configuration of Fe2- = 1s2 2s2 2p6 3s2 3p6 3d8 4s2
CO is a strong field ligand and cause electrons of 3d orbitals to pair up and shift electrons of 4s to 3d orbital.
One 4s and three 4p orbitals undergoes hybridisation to form four sp3 hybrid orbitals which arrange into a tetrahedral geometry.
Each of the four sp3 hybrid orbitals accept a pair of electrons from carbonyl ligand.
Geometry is tetrahedral.
(b) $${[Co{(CO)_4}]^ - }$$ Atomic number of Co = 27
Oxidation number of Co = x + 4(0) = $$-$$1 $$\Rightarrow$$ x = $$-$$1
Electronic configuration of Co = 1s22s22p63s23p63d74s2
Electronic configuration of Co- = 1s22s22p63s23p63d84s2
CO is a strong field ligand, so pairing of electrons occurs.
Hybridisation is sp3
Geometry is tetrahedral.
$${[CoC{l_4}]^{2 - }}$$
Oxidation number of Co = x + 4($$-$$1) = $$-$$2
$$\Rightarrow$$ x = + 2
Electronic configuration of Co2+ = 1s22s22p63s23p63d74s0
Cl- is a weak field ligand, so pairing of electrons do not occur.
Hybridisation is sp3
Geometry is tetrahedral
(c) $$[Ni{(CO)_4}]$$ Oxidation number of Ni = x + 4(0) = 0 $$\Rightarrow$$ x = 0
Atomic number of Ni = 28
Electronic configuration of Ni = 1s22s22p63s23p63d84s2
CO is a strong field ligand, so pairing of electrons occur.
Hybridisation is sp3
Geometry is tetrahedral.
$${[Ni{(CN)_4}]^{2 - }}$$
Oxidation number of Ni = x + 4($$-$$1) = $$-$$2 $$\Rightarrow$$ x = + 2
Electronic configuration of Ni2+ = 1s22s22p63s23p63d84s0
CN is a strong field ligand, so pairing of electrons occur.
Hybridisation is dsp2.
Geometry is square planar.
(d) $${[Cu{(py)_4}]^ + }$$
Oxidation number of Cu = x + 4(0) = + 1 $$\Rightarrow$$ x = + 1
Atomic number of Cu = 29
Electronic configuration of Cu = 1s22s22p63s23p63d104s1
Electronic configuration of Cu+ = 1s22s22p63s23p63d104s0
Hybridisation is sp3
Geometry is tetrahedral.
$${[Cu{(CN)_4}]^{3 - }}$$
Oxidation number of Cu = x + 4($$-$$1) = $$-$$ 3 $$\Rightarrow$$ x = + 1
Hybridisation is sp3
Geometry is tetrahedral.
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