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Number of d-orbitals used in the hybridisation of ICl3 is = x and number of lone pair at central atom = y find x + y = ?
- d)
Correct answer is '3'. Can you explain this answer?
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Number of d-orbitals used in the hybridisation of ICl3is = x and numbe...
sp3d [x = 1; y = 2]
lp = 2
x + y = 1 + 2 = 3
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Number of d-orbitals used in the hybridisation of ICl3is = x and numbe...
Explanation:
To determine the number of d-orbitals used in the hybridization of ICl3, we need to first determine the hybridization of the central atom (Iodine, I).
Hybridization of Iodine (I):
1. The electronic configuration of iodine (I) is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p5.
2. Since iodine has 5 valence electrons in its outermost shell (5s2 5p5), it can form 3 covalent bonds with chlorine (Cl) atoms, as each chlorine atom contributes 1 electron to form a bond.
3. In order to form 3 covalent bonds, iodine must promote one of its electrons from the 5s orbital to the empty 5d orbital. This results in the formation of 3 hybrid orbitals.
Therefore, the number of d-orbitals used in the hybridization of ICl3 is 3.
Lone Pair at Central Atom:
1. In ICl3, there are three chlorine atoms bonded to the central iodine atom.
2. Each chlorine atom contributes one electron to form a covalent bond with iodine, resulting in three covalent bonds.
3. The remaining two valence electrons on iodine are a lone pair.
4. Therefore, the number of lone pairs at the central iodine atom is 2.
Therefore, x (the number of d-orbitals used in the hybridization of ICl3) is 3, and y (the number of lone pairs at the central iodine atom) is 2.
To determine the number of d-orbitals used in the hybridization of ICl3, we need to first determine the hybridization of the central atom (Iodine, I).
Hybridization of Iodine (I):
1. The electronic configuration of iodine (I) is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p5.
2. Since iodine has 5 valence electrons in its outermost shell (5s2 5p5), it can form 3 covalent bonds with chlorine (Cl) atoms, as each chlorine atom contributes 1 electron to form a bond.
3. In order to form 3 covalent bonds, iodine must promote one of its electrons from the 5s orbital to the empty 5d orbital. This results in the formation of 3 hybrid orbitals.
Therefore, the number of d-orbitals used in the hybridization of ICl3 is 3.
Lone Pair at Central Atom:
1. In ICl3, there are three chlorine atoms bonded to the central iodine atom.
2. Each chlorine atom contributes one electron to form a covalent bond with iodine, resulting in three covalent bonds.
3. The remaining two valence electrons on iodine are a lone pair.
4. Therefore, the number of lone pairs at the central iodine atom is 2.
Therefore, x (the number of d-orbitals used in the hybridization of ICl3) is 3, and y (the number of lone pairs at the central iodine atom) is 2.
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Number of d-orbitals used in the hybridisation of ICl3is = x and number of lone pair at central atom = y find x + y = ?d)Correct answer is '3'. Can you explain this answer?
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